## TN Board 12th Chemistry Important Questions Chapter 5 Coordination Chemistry

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## TN State Board 12th Chemistry Important Questions Chapter 5 Coordination Chemistry

Question 1.
What is the coordination number of the central metal ions in the following complexes?
(i) [Cu(NH3)4]+2
(ii) [Fe(C2O4)3]-3
(iii) [Pt (en)2Cl2]
(iv) [Mo(CN)8]-4
(v) Fe[EDTA]]
(vi) [Pd(H2O)2(ONO)2I2]
(i) NH3 is a monodentate ligand.
Point of attachment with Cu+2 = 4 x 1 = 4
C.N of Cu+2 = 4
(ii) C2O4-2 is bidentate ligand.
Point of attachment with Fe+3 = 3 x 2 = 6
C.N of Fe+3 = 6
(iii) ‘en’ is a bidentate ligand and Cl is a monodentate ligand.
Point of attachment with
Pt+2 = 2 x 2 + 2 x 1 = 6
C.N of Pt+2 = 6
(iv) CN is a monodentate ligand.
Point of attachment with Mo+4 = 8 x 1 = 8
C.N of Mo+4 = 8
(v) EDTA is a hexadentate ligand.
Point of attachment with Fe+2 = 6 x 1 = 6
C.N of Fe+2 = 6
(vi) Point of attachment with
Pd+4 = 2 x 1 + 2 x 1 + 2 x 1 = 6
C.N of Pd+2 = 6

Question 2.
Calculate the oxidation state, of the central metal atom, in the following:
(i) [CO(NH3)5Cl]+2
(ii) K4[Fe(CN)6]
(iii) [Co(NO2)2(Py)2(NH3)2] NO3
(iv) Ni[CO]4/
(v) [Fe(EDTA)]
(i) x + 5 × (0) – 1 = + 2
x = 2 + 1 = 3
oxidation state of cobalt = +3
(ii) 4 × (+1) + x + 6 (-1) = 0
x = +6 – 4 = +2
oxidation state of Iron = + 2.
(iii) x + 2 (-1) + 2(1) + 2(0) – 1 = 0
x = 2 + 1 = 3
oxidation state of cobalt = +3
(iv) x + 4 (0) = 0 or x = 0:
oxidation State of nickel = 0
(v) x + 1 × x (-4) = -1
x = 4 – 1 = +3
oxidation state of Iron = +3

Question 3.
Give the IUPAC names of the following compounds:
(i) K3[Al(C2O4)3]
(ii) [Pt(NH3)4(NO2)Cl] SO4
(iii) K3[Cr(CN)6]
(iv) [CO(NH3)5ONO]Cl2
(v) [Cr(NH3)5CO3]Cl
(vi) [Cr(NH3)5 (NCS)][ZnCl4]
(vii) K[Au(CN)2]
(i) Pottasium trioxalatoalurhinate (III)
(ii) Tetraamminechloridonitroplatinum (IV) sulphate.
(iii) Potassiumhexacyanochromate (III)
(iv) Pentaammine nitritocobalt (III) chloride.
(v) Pentaammine carbonate chromium (III) chloride.
(vi) Pentaammine isothiocyanato chromium (III) chloride.
(vii) Potassium dicyanoaurate (I).

Question 4.
Write the formula of the following co-ordination compounds.
(i) Tetraammine diaqua cobalt (III) chloride.
(ii) Potassium tetracyano nickelate (II).
(iii) Tris (ethane 1,2, diamine) chromium (III) chloride.
(iv) Amminebromidochloridonitrito-N- platinate (III).
(v) Dichlorido bis (ethane 1, 2, diamine) platinum (IV) nitrate.
(vi) Iron (III) hexacyanoferrate (III).
(i) [Co(NH3)4(H2O)2]Cl3
(ii) K2[Ni(CN)4]
(iii) (Cr (en)3]Cl3
(iv) [Pt(NH3)Br Cl (NO2)]‘-
(v) [Pt Cl2 (en)2] (NO2)2
(vi) Fe4(Fe(CN)2]3

Question 5.
Write the IUPAC names of the following
co-ordination compounds:
(i) [Co(NH3)6]Cl3
(ii) [Co(NH3)5Cl]Cl2
(iii) K3[Fe(CN)6]
(iv) K3[Fe (C2O4)3]
(v) K2[Pd Cl4]
(vi) Pt [(NH3)2 Cl (NH2CH3)]Cl.
(i) Hexaammine cobalt (III) chloride.
(ii) Pentaammine chloridocobalt (III) chloride.
(iii) Potassium hexacyano ferrate (III).
(iv) Potassium trioxalato ferrate (III).
(vi) Diammine chlorido (methylamine) platinum (II) chloride.

Question 6.
Using IUPAC names, write the formula of the following:
(i) Tetra hydrazozincate (II)
(ii) Hexaammine cobalt (III) sulphate.
(iii) Potassium trioxalato chromate (III)
(iv) Diammine dichlorido platinum (II)
(v) Tetra bromido cuprate (II)
(vi) Pentaamine nitrito-O-cobalt (III)
(vii) Pentaamine nitrito-N-cobalt (III)
(i) [Zn(OH)4]-2
(ii) [CO(NH3)6]2SO4
(iii) K3[Cr (C2O4)3]
(iv) [Pt(NH3)2Cl2]
(v) [Cu Br4]-2
(vi) [CO(NH3)5 (ONO)]+2
(vii) [CO(NH3)5NO2]+2

Question 7.
Specify the oxidation numbers of the metal ions in the following coordination entities.
(i) [CO(H2O) (CN) (en)2]+2
(ii) [Pt Cl4]-2
(iii) [Cr(NH3)3 Cl3]
(iv) [Co Br2 (en)2]+
(v) K3[Fe(CN)6]
(i) x + 0 + (-1) + 0 = + 2; x = +3
(ii) x – 4 = – 2 ; or x = +2
(iii) x + 0 + 3(-1) = 0; x = +3
(iv) x – 4 = -2 or x = +2
(v) 3(1) + x + 6(-1) = 0; or x = +3

Question 8.
Aqueous copper sulphate (blue in colour) gives (i) a green precipitate with aqueous potassium fluoride and (ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
Aqueous copper sulphate exists as [Cu(H2O)4] SO4. It is a labile complex. The blue colour is due to [Cu(H2O)4]+2 ions.
(i) When KF is added, weak H2O ligands are replaced by F ligands forming [CuF4]-2 ions which is a green precipitate.

(ii) When KCl is added, Cl ligands replaced by weak H2O ligands forming [CuCl4]-2 ion which has a bright green colour.

Question 9.
Give two examples for each of the following:
(i) Cationic complex
(ii) Anionic complex
(iii) Neutral complex
(i) [Ag(NH3)2]+, [Co(NH3)6]+3
(ii) [Co (CN)6]-3, [Fe(CN)6]-4
(iii) [Ni(CO)4], [CO(NH3)3 Cl3]

Question 10.
What are homoleptic and heteroleptic complexes? Give one example for each.
A coordination compound in which the central metal atom / ion is co-ordinated to only one kind of ligand is called homoleptic complex. Examples (Co(NH3)6]+3, [Fe(H2O)6]+3.
A coordination compound in which the central metal atom / ion is coordinated to more than 6ne kind of ligand is called heteroleptic complex. Example [CO(NH3)5Cl]+2, [Pt(NH3)2Cl2].

Question 11.
Identify the following complexes as homoleptic or heteroleptic complexes.
(i) [Fe(CN)6]-4
(ii) [CO(NH3)4Cl2]+
(iii) [Cr(en)3]+3
(iv) [Ag(NH2)2]+
(v) [Fe(CN)5NO]-3
(i) Homoleptic complex ion.
(ii) Heteroleptic complexion.
(iii) Homoleptic complexion.
(iv) Homoleptic complexion.
(v) Heteroleptic complex ion.

Question 12.
Explain the following giving an example in each case: (i) Linkage isomerism, (ii) Coordination isomerism, (iii) Ionisation isomerism, (iv) Solvate or hydrate isomerism.
(i) Linkage isomerism This type of isomers arises when an ambidentate ligand is bonded to the central metal atom/ion through either of its two different donor atoms, eg: In the nitrite ion is bound to the central metal ion Co3+ through a nitrogen atom in one complex,and through oxygen atom in other complex.
[CO(NH3)5(NO2)]2-

(ii) Coordination isomerism: This type of isomerism arises in coordination compounds having both complex cations and complex anions. The interchange of one or more ligands between the cationic and the anionic coordination entities gives different isomers, eg:
[CO(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [CO(CN)6]
[Cr(NH3)6] [Co(C2O4)3] and [Co(NH3)6]
[Cr(C2O4)3]
[Co(en)3] [Cr(CN)6] and [Cr(en)3] [CO(CN)6]
(iii) Ionisation isomerism: This type of isomerism arises when the coordination compounds give different ions in solution. For example, the complex having the formula [Co(NH3)5 BrSO4] exists in two isomeric forms.
[CO(NH3)3 Br]SO4 and [Co(NH3)5SO4] Br. They give different ions in an aqueous solution.

(iv) Solvate or hydrate isomerism: This isomerism arises when free solvent molecules like H20, ammonia or alcohol present in the coordination entity are exchanged with the ligands outside the coordination entity.
eg: [Cr(H2O)6]Cl3, [Cr(H2O)5Cl]Cl2 H2O [Cr(H2O)4Cl2] Cl2H2O.

Question 13.
Give the ionisation isomer of the following and write their IUPAC names.
(i) [Pt(NH3)4 Cl2] Br2
(ii) [CO(NH3)4Cl2] NO2
(i) The ionisation isomer of [Pt(NH3)4 Cl2] Br2 is [Pt(NH3)4 Br2] Cl2. Its IUPAC name is tetraammine dibromido platinum (IV) chloride.
(ii) The ionisation isomer of [CO(NH3)4 Cl2] NO2 is [CO(NH3)4Cl.NO2]Cl. Its IUPAC name is tetraammine chlorido nitro cobalt (III) chloride.

Question 14.
Briefly outline the geometrical isomerism exhibited by square planar and with an example.
Square planar complexes: In square planar complexes of the form [MA2B2] and [MA2BC] (where A, B and C are mono dentate ligands and M is the central metal ion/ atom), Similar groups (A or B) present either on same side or on the opposite side of the central metal atom (M) give rise to two different geometrical isomers and they are called, cis and trans isomers respectively. The square planar complex of the type [M(xy)2] where xy is a bidentate ligand with two different coordinating atoms also shows cis-trans isomerism. Square planar complex of the form [MABCD] also shows geometrical isomerism. In this case, by considering any one of the ligands (A, B, C or D) as a reference, the rest of the ligands can be arranged in three different ways leading to three geometrical isomers.
eg:

Question 15.
Write the structure of geometrical isomers in octahedral complexes of the type Ma4b2, Ma2b4, Ma4bc, Ma3b3 where a + b are monodentate ligands. Give example for each type.
(i) Ma4b2 type:

(ii) Ma3b3 type:

(iii) M(aa3)2b2 or M(aa)2 bc type:
where aa is a symmetrical bidentate ligand, b and c are monodentate ligands, eg: [Co(en)2Cl2]+

Question 16.
Mention the types of stereoisomerism exhibited by coordination compounds.
Coordination compounds exhibit geometrical and is optical isomerism.

Question 17.
Explain the geometrical isomerism exhibited by the [MA3B3]±n where A and B are monodentate ligands. [OR]
Explain the terms facial and meritorial isomers.
Octahedral complex of the type [MA3B3] also shows geometrical isomerism. If the three similar ligands (A) are present in the comers of one triangular face of the octahedron and the other three ligands (B) are present in the opposing triangular face, then the isomer is referred as a facial isomer (fac isomer). If the three similar ligands are present around the meridian which is an imaginary semicircle from one apex of the octahedral to the opposite apex as shown in the figure, the isomer is called as a meridional isomer (mer isomer). This is called meridional because each set of ligands can be regarded as lying on a meridian of an octahedron.

Question 18.
Write the formula of coordination isomers of the following. Write their IUPAC names.
(i) [Pt(NH3)4] [CuCl4]
(ii) [Cr(NH3)6] [Co(C2O4)3]
(iii) [Cr(NH3)6] [Cr(SCN)6]
(iv) Co(en)3[Cr(CN)6]
(i) The coordination isomer of [Pt(NH3)4] [CuCl4] is [Cu(NH3)4] [PtCl]4. Its IUPAC name is Tetraammine copper (II) tetrachlorido platinate (II).
(ii) The coordination isomer of [Cr(NH3)6] [Co(C2O4)3] is [CO(NH3)6] [Cr(C2O4)3]. Its IUPAC name is Hexaammine cobalt (III) trioxalato chromate (III).
(iii) The coordination isomer of [Cr(NH3)6] [Cr(SCN)6] is [Cr(NH3)4(SCN)2] [Cr(NH3)2(SCN)4] Tetraammine dithiocyanatochromium (III) diammine tetrathiocyanatochromate (III).
(iv) The coordination isomer of [Co(en)3] [Cr(CN)6] is [Cr(en)3]4 [Co(CN)6], Its ionisation isomer is Tri (ethane 1, 2, diamine) chromium (III) hexacyano cobaltate (III).

Question 19.
Discuss the bonding in metal carbonyls.
The metal-carbon bond in metal carbonyls have both σ and π bond character. The metal – carbon sigma bond is formed by the donation of lone pair of electrons from the carbonyl carbon into a vacant orbital of the metal. The metal- carbon pi bond is formed by the donation of a pair of electrons from a filled d orbital of the metal to the vacant antibonding pi molecular orbital of carbon monoxide. The metal ligand bonding causes a synergic effect which strengthens the bond between Co and the metal.

Question 20.
Indicate the type of isomerism exhibited by the following complexes and draw structures for these isomers.
(i) KCr (H2O)2(C2O4)3
(ii) [Co(en)3]Cl3
(iii) [ CO(NH3)5(NO2)(NO3)2]
(iv) [Pt (NH3) (H2O)Cl2]
(i) Both show geometrical (cis-trans) isomerism. Optical isomerism is shown by cis form.

(ii) Two optical isomers can exist. [Co(en)3]+

(iii) There are 10 possible (geometrically ionisation and linkage) isomers.
Ionisation isomers:

(iv) Geometrical (cis-trans) isomerism can exist.

Question 21.
Give evidence to show that [CO(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl are ionisation isomers.
The ionisation isomers dissolve in water to yield different ions and thus react differently with different reagents.

A white precipitate is than obtained.

Question 22.
Draw the structure of optical isomers of (i) PtCl2 (en)2, (ii) [Cr(NH3)2 Cl2 (en)2]+

Question 23.
Write all the geometrical isomers of [Pt (NH3) BrCl Py] and how many of these exhibit optical isomerism?
Three isomers are possible.

Isomers of this type do not show optical isomerism. Optical isomerism rarely occurs in square planar or tetrahedral complexes and that too when they contain asymmetrical chelating ligand.

Question 24.
Draw the structure of all the isomers (geometrical and optical) of (i) [CO(NH3)Cl (en)2]+2 (ii) [Co(NH3)2Cl2(en)]+
(i)

I and II are optical isomers.
II and III are geometrical isomers.
(ii)

I and II are geometrical isomers.
II and III are optical isomers.

Question 25.
Draw the structure of geometrical isomers of [Fe(NH3)2(CN)4].

Question 26.
Out of the following two coordination entities which is optically active? (i) cis [CrCl3(oX)2]-3(ii) trans [Cr Cl2 (oX)2]-3 The two entities are represented as follows:

Of these cis form is optically active.

Question 27.
Write the structures of isomers, if any and write the names of the following complexes. (i) [Cr (NH3)4 Cl2]+, (ii) [Co(en)3]+
(i) Geometrical isomers of [Cr(NH3)4Cl2]+

(ii) Optical isomers of OX [Co(en)3]+3

Question 28.
Platinum (II) forms’square planar complexes and platinum (IV) gives octahedral complexes.. How many geometrical isomers are possible for each of the following. Draw their structures. (a) [Pt(NH3)3Cl]+, (b) [Pt(NH3)Cl5], (c) [Pt(NH3)2ClNO2], (d) [Pt(NH3)4Cl Br]+2.
(a) No isomer is possible for a square planar complex MA3B.

(b) No isomer is possible for an octahedral complex of the type MAB5.

(c) cis-trans isomers are possible for a square planar complex MA2BC.

(d) cis-trans isomers are possible for an octahedral complex of the type MA2BC.

Question 29.
Explain the term (i) inner orbital complex and (ii) outer orbital complex with an example for each.
(i) Bonding in complexes are explained in terms of hybridisation. If the hybridisation of central metal atom/ion involves the use of (n-1) d, ns and np orbitals, then d2sp3 hybridisation is possible for octahedral complexes. Complexes formed by d2sp3 hybridisation of the central metal atom or ion is known as inner orbital complex. [Fe(CN)6]-4, [Fe(CN)6]-3, [Cr(NH3)6]+3 are examples of inner orbital complexes.
(ii) If the hybridisation of the metal ion involved the ns, np and nd orbitals and for octahedral complexes sp3d2 hybridisation involved, the complex formed is known as outer orbital complex. [CoF6]-3 is an example for outer orbital complex.

Question 30.
[Co(CN)6]-3 and [CoF6]-3 are both octahedral complexes. Then what is the difference between the two?
In both complexes, Co is in +3 oxidation state. It has 6 electrons in the ‘d’ subshell (d6).

In the formation of [Co(CN)6], the ‘d’ electrons in 3d level gets paired leaving behind two 3d orbital vacant.

The Co+3 ion makes use of two 3d orbitals, one 4s and two 4p orbitals for hybridisation. d2 sp3 hybridisation and two 4p orbitals for hybridisation.

The CN ion donates the electron pairs to the d2spi hybrid orbitals and form six covalent bond. Thus [(CO(CN)6]-3 is an inner orbital or low spin complex.
On the other hand, in the formation of [CoF6]-3 the Co+3 ion makes use of outer 4s, 4p and 4d orbitals for hybridisation.

In the presence of F ion, electron pairing does not occur in 3d orbitals, and Co+3 ion makes use of one 4s, three 4p and two 4d orbitals for hybridisation.

The six fluoride ion donates a pair of electrons to the six sp3d2 hybrid orbitals and form six-coordinate covalent bonds.

Since outer orbitals are used in the formation it is an outer orbital complex. Since more unpaired electrons are present it is also known as high spin complex.

Question 31.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory.
(i) [Fe(CN)6]-4
(ii) [FeF6]-3
(iii) [Co(C2O4)3]-3
(iv) [NiCl4]-2
(i) [Fe(CN)6]-4: d2sp3 hybridisation octahedral: diamagnetic.
In this complex, the oxidation state of iron is +2.
Electronic configuration of Fe = [Ar]3d46s2
Electronic configuration of Fe+2 = [Ar]3d6
To accommodate six pairs of electrons from six cyanide ions, the iron (II) must make available six empty orbitals. This can be achieved by the d2sp3 hybridisation.

In the presence of CN ion, a strong ligand, the electrons in 3d subshell gets paired and two of the 3d one 4s and three 4p orbitals hybridise to produce six d2sp3 hybrid orbitals.
Fe+3 (d6) excited state.

Thus, the six pairs of electrons from six cyanide ion occupy the six hybridised orbitals of iron (II) ion. [Fe(CN)6]-4

Since, there are no unpaired electron present, it is diamagnetic and its geometry is octahedral.

(ii) [FeF6]-3: This complex is high spin or octahedral, since the central atom Fe(II) utilises nd orbitals for hybridisation. It is an octahedral complex involving sp3d2 hybridisation. Each orbital accommodates a lone pair of electrons from six fluoride ions as shown below. [Fe+3(d5) ground state]

In the presence of F ions (weak ligand) the electrons in 3d orbitals do not get paired and the iron (II) utilises ns, np and nd orbitals for hybridisation. Thus six sp3d2 hybrid orbitals are formed.
[Fe+3 excited state]

The six pairs of electrons from F ions accommodate the six vacant spid1 hybrid orbitals
[FeF6]-3

Thus, the complex is outer orbital complex, octahedral and paramagnetic.

(iii) [Co(C2O4)3]-3: d2sp3 hybridisation diamagnetic.
The cobalt ion is in +3 oxidation state. It has six electrons in d subshell (d6).
[Co+3(d6)]

Oxalate ion being a relatively strong ligand pairs of the electrons in ‘d’ subshell, learning two of the ‘d’ orbital for hybridisation. Thus Co+3 ion under goes d2sp3 hybridisation.
Co+3(ground state)

The three oxalate anions (bidentate) donate two pairs of electrons to two of the hybrid orbitals.

Thus [Co(C2O4)3]-3 is an inner orbital, octahedral, complex and diamagnetic.

(iv) [NiCl4]-2: In this complex Ni is in +2 oxidation state and has d8 configuration.

The chloride ions donate pair of electrons to each of the four sp3 hybrid orbitals.

Thus, the complex is tetrahedral and paramagnetic.

Question 32.
[NiCl4]-2 is paramagnetic while Ni(CO)4 is diamagnetic though both are tetrahedral why?
In Ni(Co)4, Ni is in the zero oxidation state whereas in [NiCl4]-2, it is in +2 oxidation state. In the presence of a CO being a strong ligand, the unpaired ‘d’ electrons of Ni pair up but Cl being a weak ligand it is unable to pair up the unpaired electrons. Hence [NiCl4]-2 is paramagnetic where as Ni(CO)4 is diamagnetic.

Question 33.
Explain [Co(NH3)6]+3 is an inner orbital complex whereas [Ni(NH3)6]+2 is an outer orbital complex.
In the presence of NH3, the 3d electrons pair up learning two ‘d’ orbitals empty to be involved in d2sp3 hybridisation forming inner orbital complex.
In [Ni(NH3)6]+2, Ni is in +2 oxidation state and has d8 configuration. The hybridisation involved in sp3d2 and hence, it forms an outer orbital complex.

Question 34.
[Cr(NH3)6]+3 is paramagnetic while [Ni(CN)4]-2 is diamagnetic explain why?
(i) Formation of [Cr(NH3)6]+3:
The oxidation state of chromium in [Cr(NH3)6]+3 ion is +3. The electronic configuration of chromium is [Ar] 3d5 4s1. The hybridisation scheme is shown below:

Ammonia being a weak ligand is unable to pair of up the electrons of Cr+3 and hence Cr+3 undergoes d2sp3 hybridisation.

Thus the complex is inner orbital, octahedral complex and paramagnetic because of the presence of three unpaired electrons.

(ii) Formation of [Ni(CN)4]-2: In the square planar complexes, the hybridisation involved is dsp2. In [Ni(CN)4]-2, nickel is in +2 oxidation state and has the electronic configuration of d8.

CN ion being a strong ligand, pairs up the 3d electrons leaving one 3d orbital vacant that is used for dsp2 hybridisation.

Thus the complex is diamagnetic as there are no impaired electron.

Question 35.
A solution of [Ni(H2O)6]+2 is green but a solution of [Ni(CN)4]-2 is colourless explain.
In [Ni(H2O)6]+3, Ni is in +2 oxidation state with configuration d8. i.e., it has 2 unpaired electrons which do not pair in the presence of weak H2O ligand. Hence d-d transition is possible and so it is coloured. The d-d transition absorbs red light and the complimentary light emitted is green.
In case of [Ni(CN)4]+2, Ni is in +2 oxidation state with d8 configuration. But in the presence of strong CN ion as ligand, the two unpaired electrons in 3d orbitals are paired up. Hence there are no unpaired electrdhs present and so it is colourless.

Question 36.
[Fe(CN)6]-4 and [Fe(H2O)6]+2 are of different colours in dilute solution why?
In both complexes, Fe is in +2 oxidation state with d6 configuration. In the presence of a weak ligand, they do not pair up. In the presence of strong CN ion, they paired up learning no unpaired electrons. Due to the difference in the number of unpaired electrons they have different colours.

Question 37.
Write down the IUPAC name of each of the following complexes and indicate the oxidation state, electronic configuration and coordination number of the central metal ion. Also, give the stereochemistry and magnetic moment of each complex.
(i) K4[Cr(H2O)2(C2O4)2]
(ii) [CrCl3(Py3)]
(iii) K2[Mn(CN)4]
(iv) [CO(NH3)5Cl]Cl2
(v) CS[FeCl4]
(i) Potassiumdiaquaoxalatochromate (III) hydrate.
Coordination number = 6
Shape: Octahedral
Oxidation state of Cr = x + 0 + 2 (- 2) = -1 or x = +3

(ii) Trischloridopyridinechromium (III)
Coordination number of Cr = 6
Shape: Octahedral
Oxidation state of Cr = x – 3 + 0 + 0 = 0 or x = +3

(iii) Potassiumhexacyanomanganate (II)
Coordination number of Mn = 6.
Shape: Octahedral
Oxidation state of Mn = x – 6 = – 4 (or) x = +2

(iv) Pentaammine chloridocobalt (III) chloride.
Coordination number of Co = 6.
Shape: Octahedral
Oxidation state of Cr = x + 0 – 1 = +2 or x = +3

Magnetic moment = 0

(v) Caesiumtetrachloridoferrate (III).
Coordination number of Fe = 4.
Shape: Tetrahedral
Oxidation number of Fe = x – 4 = -1 or x = +3

Question 38.
Calculate the magnetic moment of the following coordination entities.
(i) [Cr(H2O)6]+3,
(ii) [Fe(H2O)6]+3,
(iii) [Zn(H2O)6]+2
The oxidation states are (i) Cr(III); (ii) Fe(II) and (iii) Zn(II)
(i) E.C of Cr+3 = 3d3:
no ofunpaired electrons = 3
Magnetic moment = $$\sqrt{3(3+2)}$$ = √15
(ii) E.C of Fe+2 = 3d6:
no of unpaired electrons = 4
Magnetic moment = $$\sqrt{4(4+2)}$$ = √24
(iii) E. C of Zn+2 = 3d10:
no of unpaired electrons = 0
Magnetic moment = 0

Question 39.
What will be the correct order for the wavelength of absorption in the visible region of the following:
[Ni(NO2)6]-4, [Ni(NH3)6]+2, [Ni(H2O)6]+2
As the metal ion is fixed the increasing field strength (CFSE values) of the ligands from the spectrochemical series are in the order.
H2O<NH3<NO2
Thus the energies absorbed for excitation will be in the same order
[Ni(H2O)6]+2 < [Ni(NH3)6]+2 < [Ni(H2O)6]+2
As, E = $$\frac{h c}{\lambda}$$, the wavelength absorbed will be in the opposite order.

Question 40.
Give the number of unpaired electrons in the following complex ions: (i) [FeF6]-4, (ii) [Fe(CN)6]-4
(i) In [FeF6]-4, Fe is in +2 oxidation state. It has d6 configuration. As F is a weak ligand, the electrons in 3d orbital are not paired.
i-e.,
Hence, it has 4 unpaired electrons.

(ii) In [Fe(CN)6]-4 also, Fe is in +2 oxidation state and it has d6 configuration. Because CN ion is a strong ligand, the electrons in 3d orbital of Fe+2 are paired and so there are no unpaired electrons.

Question 41.
Explain as how two complexes [Ni(CN)4]-2 and [Ni(CO)4] have different structures but do not differ in their magnetic behaviour. (Ni = 28)
(i) In Ni(CO)4, nickel is in zero oxidation state. (Ni+2 = 3d8 4s2)

E.C of Ni in hybridised state

It is diamagnetic and has a tetrahedral geometry.

(ii) In [Ni(CN)4]-2, Ni is in +2 oxidation state. Electronic configuration of Ni+2 ion is 3d8.

CN ion being a strong liquid and in its presence, the electrons in 3d orbital gets paired leaving are 3d orbital for hybridisation. Thus, it undergoes dsp2 hybridisation.

The CN ions denote a pair of electrons to the valent dsp2 hybrid orbitals. Thus, four Ni CN bonds are formed.

Thus the geometry is square planar and the complex is diamagnetic.

Question 42.
Using valence bond theory explain [Co(NH3)6]+3 in relation to the complex given below:
(i) type of hybridisation [Co(NH3)6]+3 : Co+3 = d6 configuration
(ii) inner or outer orbital complex
(iii) magnetic behaviour
(iv) spin only magnetic moment value.
(i) In the presence of ammonia ligand, the electrons in 3d level gets paired up leaving two of the 3d orbitals for hybridisation. Hence, the type of hybridisation is d 2sp3

(ii) Since inner ‘d’ orbitals are used for hybridisation, it is an inner orbital complex.
(iii) The complex is diamagnetic as there are no unpaired electrons.
(iv) Since, there are no unpaired electrons the magnetic moment is zero.

Question 43.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory.
(i) [FeCN6]-4
(ii) [FeF6]2-
(iii) [Co(C2O4)3]-3
(iv) [COF6]-3
(i) Oxidation state of Fe in the complex is +2. Its electronic configuration is [Ar] 3d6. Iron undergoes d2sp3 hybridisation giving octahedral, inner orbital complex.

(ii) Oxidation state of iron in the complex is +3. Its electronic configuration is [Ar] 3d5. In the presence of F (weak ligand), the 3d electrons do not get paired. Hence it undergoes sp3d2 hybridisation and forms an outer orbital, octahedral complex.

(iii) The oxidation state of cobalt in the complex is +3. Its electronic configuration is [Ar] 3d6. Since oxalate ion is relatively a strong ligand, the 3d electrons get paired up leaving two of 3d orbital for hybridisation. Thus, cobalt undergoes d2sp3 hybridisation giving octahedral, inner orbital complex.
(iv) Oxidation state of cobalt in the complex is +3. Its electronic configuration is [Ar] 3d6. The F ion being a weak ligand is unable to pair up the electrons in 3d level. It undergoes sp3d2 hybridisation giving an octahedral, outer orbital complex.

Question 44.
What are crystal fields?
The ligands especially anionic (or polar neutral ligands) has around them negatively charged field because of which they are called crystal fields.

Question 45.
What do you understand by the term crystal field splitting?
When the ligands approach the central metal ion, the electrons in the ‘d’ orbital of central metal ion will be repelled by the lone pairs of the ligands. Because of these interactions, the degeneracy of ‘d’ orbitals of the metal ion is lost and they split into two sets of orbitals having different energies. This is known as crystal field splitting. The extent of crystal field splitting depends on the nature of the ligand.

Question 46.
Briefly discuss the crystal fieid splitting in octahedral complexes.
As a result of repulsive forces between the ‘d’ electrons of the metal ion and that of ligands (negative or polar neutral), the degeneracy of the ‘d’ orbitals is lost. Since the two lobes of two eg{dx2-y2, dz2) orbitals lie in the path of approaching ligands these orbitals experience greater forces of repulsion than those of t2g (dxy, dyz and dxz) orbitals is decreased.
Thus, an energy difference exists between two sets of orbitals. This energy difference is called crystal field splitting energy and is represented by Δo. (the subscript ‘o’ stands for octahedral). It measures the crystal field strength of the ligands. The crystal field splitting occurs in such a way that the average energy of the ‘d’ orbitals do not change.

(a) Five degenerate ‘d’ orbitals of the free metal ion.
(b) Hypothetical degenerate ‘GP orbitals at higher energy levels under spherically symmetrical ligand field.
(c) Splitting of ‘d’ orbitals under the influence of ligands.
Thus, three orbitals lie at an energy i.e., 2/5 Δo below the average ‘d’ orbital energy and two ‘d’ orbitals lie at an energy 3/5 Δo above the average energy. The energy gap between t2g and eg sets also denoted by 10Dq.
The energy of t2g orbitals is 4Dq less than that of hypothetical degenerate ‘d’ orbitals and that of eg orbitals is 6Dq above that of hypothetical degenerate ‘d’ orbitals: Thus, t2g set loses energy equal to 0.4 Δo or 4 Dq white eg sets gain energy equal to 0.6 Δo or 6 Dq.

Question 47.
For the complex
[Fe(en)2Cl2]Cl (en = ethylene diamine), identify The complex is [Fe(en)2Cl2]Cl.
(i) The oxidation number of Fe.
(ii) The hybrid orbital and shape of the complex.
(iii) The magnetic behaviour of the complex.
(iv) The number of geometrical isomers.
(v) Whether there is an optical isomer also.
(vi) The name of the complex (At. No of Fe 26)
(i) Oxidation number of Fe
= x + 2 x 0 + 2(-1)
= +1
x – 2 = +1
⇒ x = 3
(ii) Orbitals in Fe+3(d5)

In the presence of a strong ligand (en), the electrons in 3d orbital get paired.
Fe+3 excited state

Thus, Fe+3 ion undergoes d2sp3 hybridisation therefore the shape is octahedral.
(iii) The complex is paramagnetic due to the presence of one unpaired electron.
(iv) It exhibits cis-trans isomerism

(v) Cis isomer will show optical isomerism.

(vi) Dichlorido bis(ethane 1, 2, diamine) iron, (III) chloride.

Question 48.
Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved.
(i) [CoF4]-2, (ii) [Cr(H2O)2(C2O4)2], (iii) [Ni(Co)4], (At No: Co = 27, Cr = 24, Ni = 28).

Question 49.
Briefly discuss the crystal field splitting in tetrahedral complexes.
The three orbitals i.e., t2g orbitals are close to approaching ligands. As a result of this, the t2g electrons suffer more repulsions than eg electrons. The energy of t2g orbitals increases more than eg orbitals. The splitting is shown below:

The energy gap between two sets of orbitals is designated as Δt. It is observed that Δt, is considerably less than Δo. It has been found that.
Δt = $$\frac{4}{9}$$ Δo

Question 50.
Briefly outline crystal field splitting in square planar complexes.
As the lobes of dx2-y2 point towards the ligands, this orbital has the highest energy. The lobes of dxy orbital lie between the ligands but are coplanar with them this orbital is the next highest energy. The lobes of dz2 orbital point out of the plane of the complex but the belt around the centre of the orbital (which contain 1/3 of the electron density) lies in the plane. Therefore dz2 orbital is the next highest in energy. The lobes of dx2 and dy2 orbitals point out of the plane of the complex. Hence, they are least affected by the electrostatic field of the ligands. They are degenerate and lowest in energy.

Question 51.
What is crystal field splitting energy? How does the magnitude of Δo decide the actual configuration of ‘d’ orbitals in a coordination entity?
When ligands approach a transition metal ion, the ‘rf orbital splits into two sets one with lower energy and the other with higher energy. The difference in energy between two sets (t2g and eg) of orbitals is called crystal field splitting energy Δo for an octahedral field.
If Δo < p (pairing energy), the fourth electron enters one of the eg orbitals giving the configuration t2g3 eg1, thereby forming high spin complexes. Such ligands for which Δo < p are called weak field ligands.
If Δo > p, the fourth electron pair up in one of the t2g orbital giving the configuration t2g4 eg0, thus forming low spin complexes. Such ligands for which Δo > p are called strong field ligands.

Question 52.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
The arrangement of ligands in order of their increasing field strength i.e increasing crystal field splitting energy (CFSE) values is called spectrochemical series.
The ligand with a small value of Δo CFSE (Δo) is called weak field ligands whereas those with a large value of CFSE(Δo) are called strong field ligands.

Question 53.
How does the oxidation state of a metal ion influence or affect the crystal field splitting energy?
Generally higher the oxidation state of the metal, greater is the crystal field splitting. For example most of the cobalt (II) complexes have low value of Δ whereas all cobalt (III) complexes have high value of Δ.

Question 54.
Briefly explain the distribution of ‘d’ electrons in t2g and eg orbitals in octahedral complexes in a weak ligand field.
Under the influence of weak ligands the energy difference, Δo between t2g and eg sets is relatively small and hence all the five ‘d’ orbitals may be supposed to be degenerate. The distribution of electrons in t2g and eg sets occurs according to Hund’s rule.
When the ligands are weak, the first three electrons numbered as 1,2,3 go to t2g set those numbered 4, 5 go to eg set, those numbered 6, 7, 8 go to t2g set and the remaining two electrons numbered 9 and 10 will occupy eg set. i.e.,
In complexes of weak ligands, Δo is less than P(P is called pairing energy which is the energy required to pair two electrons in the same orbital). Δo is the octahedral crystal field splitting energy, tends to force as many electrons into t2g set while P tends to prevent the electrons to pair up in the t2g level.

Question 55.
The Hexa aqua manganese (II) ions contain five unpaired electrons while hexacyano ions contain only one unpaired electron. Explain using crystal field theory.
Mn is in +2 oxidation state and has 3d5 configuration. H2O is a weak ligand. In the presence of water molecules, the distribution of electrons is t2g3 eg2 i.e., all the electrons are unpaired.
In the other hand CN ion is a strong ligand. The distribution of electrons is t2g5 eg0.
i.e., and it has one unpaired electron.

Question 56.
Why do compounds having similar geometry have a different magnetic moment?
It is due to the presence of strong and weak ligands in complexes. If GFSE is high, the complex will show a low value of the magnetic moment and vice versa. eS: [CoF6]-3 and [CO(NH3)6]+3, the former is paramagnetic and the latter is diamagnetic.

Question 57.
Briefly explain how the electrons are distributed between t2g and eg sets in an octahedral complex in the presence of a strong field ligand.
Under the influence of strong ligands, the energy difference between t2g and eg sets is relatively high and thus the distribution of ‘d’ electrons in t2g and eg sets does not obey Hund’s rule. The first electrons numbers 1, 2, 3, 4, 5, 6 will go to t2g set remaining four electrons numbered 7, 8, 9, 10 will go to eg set.

For these complexes, Δo is greater than p.

Question 58.
Using crystal field theory, draw an energy level diagram of the central metal atom/ion and determine the magnetic moment value of the following.
(i) [COF6]-3,
(ii) [Co(H2O)6]+2,
(iii) [Cu(CN)6]-3
(i) F is a weak field ligand and cobalt is in a +3 oxidation state. It has d6 configuration. The electron distribution is t2g4 eg2.

i.e., Number of unpaired electron = 4
Magnetic moment = $$\sqrt{4(4+2)}$$ = √24
= 4.9 BM
(ii) In this complex cobalt is in +2 oxidation state. Its electronic configuration is d1. Since water is a weak ligand, electron filling follows Hund’s rule, i.e., t2g5 eg2.

i.e., Number of unpaired electron = 3 Magnetic moment = $$\sqrt{3(3+2)}$$ = √15
= 3.87BM
(iii) In this complex cobalt is in +3(d6) oxidation state and CN ion a strong ligand. Electron filling is not obeyed by Hund’s rule. All the six electrons get paired in t2g set.

Since, there is no unpaired electron the complex is diamagnetic.

Question 59.
In the basis of crystal field theory explain why cobalt (III) forms a paramagnetic octahedral complex with weak field ligands whereas it forms a diamagnetic octahedral complex with strong field ligands.
With weak field ligand, Δo < P, the electronic configuration (of cobalt (III)) i.e., d6 will be t2g4 eg2 and it has 4 unpaired electrons and is paramagnetic. With strong field ligands, Δo > P, the electron distribution is t2g6 eg0. It has no unpaired electron and hence diamagnetic.

Question 60.
Differentiate between weak field and strong field coordination entity.

 Weak field coordination entity Strong field coordination entity They are formed when the crystal field stabilisation energy (CFSE) in octahedral complexes is less than the energy required for electron pairing in a single orbital (P). They are formed when crystal field stabilisation energy (CFSE) Δo is greater than pairing energy (P). They are also called high spin complexes. They are also called low spin complexes. They are mostly paramagnetic. They are mostly diamagnetic or less paramagnetic than weak field. Never formed by CN– ions. Formed by CN– like ligands.

Question 61.
State for a d6 configuration, how the actual configuration of the split ‘d’ orbitals in an octahedral field is decided by relative values of Δo and P.
For d6 configuration, three electrons will first enter t2g. Now, if P < Δo electrons will pair up in t2g orbitals giving the configuration t2g6 eg0. If Δo< P, two electrons will first enter into eg orbitals and one will pair up in t2g, giving the configuration t2g4 eg2.

Question 62.
Arrange the following ligands in the increasing order of crystal field splitting power.
H2O, OH, Cl, F, CN
Cl < F < OH < H2O < CN

Question 63.
Using crystal field theory, show energy level diagrams, write the electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following. (i) [FeF6]-3, (ii) [Fe(H2O)6]+2,
(iii) [Fe(CN)6]-4
(i) In this complex, Fe is +3 oxidation state and Fe+3 has 3d5 configuration. Since F is a weak field ligand, electron filling in t2g and eg sets follow Hund’s rule.

i.e., t2g4 eg2. The number of unpaired electron is 5. The magnetic moment value is $$\sqrt{5(5+2)}$$ = √35 = 5.92 BM.
(ii) In [Fe(H2O)6]+2, Fe is in +2 oxidation state and has d6 configuration. Since water is a weak ligand electron distribution in t2g and eg orbitals take place according to Hund’s rule.

i.e., t2g4 eg2. The number of unpaired electrons is 4. Its magnetic moment value is $$\sqrt{4(4+2)}$$ = √24 = 4.9 BM.
(iii) In [Fe(CN)6]-4, Fe is in +2 oxidation state and has 3d6 configuration. As CN is a strong ligand, electron distribution between t2g and eg sets take place not in accordance with Hund’s rule. All the t2g orbitals are filled with six electrons.

Since there is no unpaired electron, the complex is diamagnetic.

Question 64.
Give the oxidation state, ‘d’ orbital occupation and coordination number of the central metal ion in the following complexes.
(i) K3[Co(C2O4)3],
(ii) (NH4)[CoF4],
(iii) Cis – [Cr(en)2Cl2]Cl,
(iv) [Mn(H2O)6]SO4.

Question 65.
With the help of crystal field theory, product the number of unpaired electrons in [Fe(CN)6]-4 and [Fe(H6O)2]+2 complexes.
In both complexes, Fe is present in +2 oxidation state and has d6 configuration. As CN is a strong field ligand, these electrons will pair up giving the configuration t2g6 eg0. As there are no unpaired electrons, the complex is paramagnetic. But H2O is a weak ligand and electron filling follow Hund’s rule i.e., t2g4 eg2. It has 4 unpaired electron and hence paramagnetic.

Question 66.
CO is stronger ligand than NH3 for many metals. Explain giving appropriate reason.
Ligands such as CO, CN, NO+ have empty π orbitals which overlap with the filled d orbitals (t2g) of transition metals forming π bonds (back bonding). These π interactions increase the value of Δo. This account for the position
of these ligands as strong ligand. NH3 cannot form π bonds by back bonding.

Question 67.
What is meant by the stability of a coordination compound in solution?
The stability of a complex in a solution refers to the degree of association between two species involved in a state of equilibrium. The magnitude of the equilibrium constant for the association is a quantitative measure of stability.

For the above equilibrium, the larger the stability constant, the higher is the proportion of ML4 that exists in the solution.

Question 68.
The stability constant of some complexes are given below:

Among CN and NH3 which is a stronger ligand? Why?
The numerical value of the stability constant is a measure of the stability of the complex. Greater the magnitude of the stability constant more stable the complex. Thus CN ion is a stronger ligand than ammonia.

Question 69.
The dissociation constants of [Cu(NH3)4]+2 and [CO(NH3)6]+3 are 1.10 x 10-12 and 6.2 x 10-36 respectively. Which complex would be more stable and why?
Smaller the values of the dissociation constant more stable the complexion in solution Hence [CO(NH3)6]+3 is more stable than [Cu(NH3)4]+2 ion.

Question 70.
Calculate the overall complex dissociation equilibrium constant for the [Cu(NH3)4]+2 ion given that β4 for this complex is 2.14 x 1013.
Overall stability constant (B4)
= 2.1 x 1013. Overall dissociation constant is the reciprocal of overall stability constant. Hence, overall dissociation constant.

= 4.7 x 10-14.

Question 71.
Calculate the ratio of [Ag(S2O3)2]-3 and [Ag+] in 0.1 M S2O3 solution. Given the stability or formation constant of [Ag(S2O3)2]-3 is 1.0 x 1013.
The equilibrium is

Question 72.
Mention the use of coordination compounds in metallurgy.

1. Purification of nickel by Mond’s process involves the formation of nickel carbonyl Ni(CO)4 which yields 99.5% pure nickel and decomposition.
2. Coordination compounds are used in the extraction of silver and gold from their ores by forming a soluble cyano complex. These cyano complexes are reduced by zinc to yield pure metals.

Question 73.
Mention the use of complexes used as catalysts in organic reactions.

1. Wilkinson’s catalyst [(PPh3)3RhCl] is used for the hydrogenation of alkenes.
2. Ziegler-Natta catalyst – [TiCl4] + Al[C2H5]3 is used in the polymerization of ethene.
3. In order to get a fine and uniform deposit of superior metals (Ag, Au, Pt etc.,) over base metals, coordination complexes [Ag(CN)2] and [Au(CN)2] etc., are used in an electrolytic bath.

Question 74.
Mention the use of complexes in photography.
In photography, when the developed film is washed with sodium thio sulphatesolution (hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called sodiumdithiosulphatoargentate(I) which can be easily removed by washing the film with water.

Question 75.
Mention the use of metal complexes in biological systems.

1. A red blood corpuscle (RBC) is composed of the heme group, which is Fe2+ – Porphyrin complex. It plays an important role in carrying oxygen from the lungs to tissues and carbon dioxide from tissues to the lungs.
2. Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing Mg2+ as a central metal ion surrounded by a modified Porphyrin ligand called a corrin ring. It plays an important role in photosynthesis, by which plants converts CO2 and water into carbohydrates and oxygen.
3. Vitamin B12 (cyanocobalamine) is the only vitamin consist of the metal ion. It is a coordination complex in which the central metal ion is Co+ surrounded by a Porphyrin ligand.
4. Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion contains a zinc ion coordinated to the protein.

Question 76.
What is Cisplatin? Mention its use.
Cisplatin is a square planar coordination complex (cis- [Pt (NH3)2C12]), in which two similar ligands are in adjacent positions. It is a platinum-based anticancer drug. This drug undergoes hydrolysis and reacts with DNA to produce various crosslinks. These crosslinks hinder DNA replication and transcription, which results in cell growth inhibition and ultimately cell death. It also crosslinks with cellular proteins and inhibits mitosis.

Question 77.
What are metal carbonyls? Mention their uses.
Metal carbonyls are transition metal complexes of carbon monoxide, containing metal-carbonyl (M ← CO) bonds. They are used as catalysts.

Question 78.
What are mononuclear and polynuclear carbonyls? Give examples.
Mono nuclear carbonyls contain only one metal atom.eg: Ni(CO)4, ie(CO)5, Cr(CO)6. Polynuclear carbonyls contain two or more metal atoms. They may be homonuclear or heteronuclear.

Question 79.
What are non-bridged metal carbonyls? Give examples.

1. Non bridged metal carbonyls contain only terminal carbonyls, eg: [Ni (Co)4] [Fe(Co)5] and [Cr(Co)6]
2. Non bridged metal carbonyl which contains terminal carbonyls as well as metal- metal bonds, eg: Mn2(Co)10. It contains Mn—Mn bonds and also Mn ← 10 bonds.

Question 80.
What are bridged carbonyls? Give example.
These metal carbonyls contain one or more bridging carbonyl ligands along with terminal carbonyl ligands and one or more Metal-Metal bonds. For example,

The structure of Fe2(CO)9, di-ironennea carbonyl molecule consists of three bridging CO ligands, six terminal CO groups and a single Fe-Fe bond formed by the weak coupling of the unpaired electrons present in two 3d orbitals of two Fe atoms. This bond is represented by a dotted line and is called a fractional single bond.

1. Which of the following complexes formed by Cu+2 ions is the most stable?

(b)
Hint: Greater the log k value, more stable the complex.

2. When 1 mol of CrCl3 6H2O is treated with excess of AgNO3 3 mol of AgCl are obtained. The formula of the complex is:

(d)

3. The correct IUPAC name of [Pt (NH3)2Cl2] is:
(a) Diammine dichloridoplatinum (II)
(b) Diammine dichloridoplatinum (IV)
(c) Diammine dichlorido platinum (0)
(d) Dichlorodiammine platinum (IV)
(a)

4. The stabilisation of coordination compounds due to chelation is called chelate effect. Which of the following is the most stable complex species?

(c)
Hint: C2O4-2 is a bidentate ligand forming chelate.

5. The compounds [Co(SO4)(NH3)5] Br and [Co(SO4)(NH3)5]Cl represent:
(b) ionisation isomerism
(c) coordination isomerism
(d) no isomerism
(d)

6. Which of the following species is not expected to be a ligand?
(a) NO
(b) NH4+
(c) NH2CH2CH2NH2
(d) Co
(b)
Hint: It has no lone pair of electrons.

7. The oxidation state of Fe in the brown ring complex [Fe(H2O)5NO]SO4 is:
(a) +1
(b) +2
(c) +3
(d) +4
(a)
Hint: [Fe(H2O)5NO]+2 : x + 0 + 1 = 2
x = 1.

8. [CO(NH3)4 (NO2)2]Cl exhibits:
(a) ionisation isomerism, geometrical isomerism and optical isomerism.
(b) linkage isomerism, geometrical isomerism and optical isomerism.
(c) linkage isomerism, ionisation isomerism and optical isomerism.
(d) linkage isomerism, ionisation isomerism and geometrical isomerism.
(d)
Hint:

9. Which of the following complex ions has geometrical isomers?

(c)
Hint: Geometrical isomerism is shown by disubstituted octahedral complexes such as [Co(NH3)2 (en)2]+3

10. The correct statement with respect to the complexes Ni(CO)4 and [Ni(CN)4]-2 is:
(a) nickel is in the same oxidation state in both.
(b) both have tetrahedral geometry
(c) both have square planar geometry
(d) have tetrahedral and square planar geometry respectively.
(d)
Hint: Ni(CO)4: sp3hybridisation: Tetrahedral
[Ni(CN)4]-2: dsp2 hybridisation: square planar.

11. When 0.01 mole of a cobalt complex is treated with excess AgNO3 solution, 4.305g of silver chloride is precipitated. The formula of the complex is:

(c)
Hint: 4.305g AgCl = $$\frac{4.305}{143.5}$$ mol = 0.03 mol As 0.01 mol of the complex gives 0.03 mol of AgCl this shows that there are 3 ionisable chlorine atoms i.e., c.

12. Amongst Ni(CO)4, [Ni(CN)4]-2 and [NiCl4]-2

(c)
Hint: Ni(CO)4 and [Ni(CN)4]-2 do not contain any unpaired electrons while [NiCl4]-2 contain two unpaired electrons.

13. Which of the following complexes are not correctly matched with the hybridisation of their central metal ion?

(a) (i) and (ii)
(b) (ii), (iii) and (iv)
(c) (i), (ii) and (iii)
(d) (ii) and (iv)
(b)
Hint:

is a weak ligand and cannot force ‘d’ electrons to pair up forming outer orbital complex.

14. Which of the following complexes has a minimum magnitude of ΔO?

(c)
Hint:

Hence, the minimum CFSE is for [CoCl6]-3.

15. Hybridisation, shape and magnetic moment of K3 [Co(CO3)3] is:
(a) d2sp3, octahedral, 4.9 BM
(b) sp3d2, octahedral, 4.9 BM
(c) dsp2, square planar, 4.9 BM
(d) sp3, tetrahedral, 4.9 BM
(b)
Hint: In each case, the magnetic moment is 4.9 BM, which shows that each of them has 4 unpaired electrons. Further, in the given complex, oxidation state of cobalt is +3. Hence, it will have d6 configuration.

Thus the possible hybridisation in sp3d2. i.e., outer orbital (high spin) octahedral complex.

16. The spin only magnetic moment value (in Bohr magneton) of Cr(CO)6 is:
(a) 0
(b) 2.84
(c) 4.90
(d) 5.92
(d)
Hint.In Cr(CO)6, Cr is in zero oxidation state, [Ar]3d5 4s1. As CO is a strong ligand, all the six electrons will pair up i.e, there will be no unpaired electron. Hence, u = 0.

17. Among the following complexes (K – P)

the diamagnetic complexes are:
(a) K, L, M, N
(b) K, M, O, P
(c) L, M, O, P
(d) L, M, N, O
Hint:
K = K3[Fe(CN)6] : Fe+3 (3d5) d2sp3 hybridisation.
No. of unpaired electron = 1; Paramagnetic
L = [CO(NH3)6Cl3]; CO+3 (3d6): d2sp3 hybridisation.
No. of unpaired electrons = 0; Diamagnetic
M: Na2 [CO(oX)3]: CO+3 (3d6): d2sp3 hybridisation.
No. of unpaired electron = 0; Diamagnetic
N: [Ni(H2O)6] Cl2 ; Ni+3 (3d8): sp3d2 hybridisation.
No. of unpaired electron = 2; paramagnetic
O: K2 [Pt (CN)1]; Pt+2 (3d8): dsp2 hybridisation.
No. of unpaired electron = 0; Diamagnetic.
P: [Zn(H2O)6] (NO3)2; Zn+2 (3d10); sp3d2 hybridisation;
No. of unpaired electron = 0; Diamagnetic.
Hence, diamagnetic complexes are L, M, O and P.

18. Which of the following facts about the complex [Cr(NH3)6] Cl3 is wrong?
(a) The complex is paramagnetic
(b) The complex is an outer orbital complex
(c) The complex gives white precipitate with silver nitrate solution.
(d) The complex involves d2sp3 hybridisation and is octahedral in shape.
(b)
Hint: In [Cr(NH3)6]Cl3, Cr+3(3d3): d2sp3 hybridisation. No. of unpaired electron = 3 paramagnetic. The complex is inner orbital complex as it involves (n -1) d orbitals for hybridisation. [Cr(NH3)6]+3

19. The ‘d’ electron configuration of Cr+2, Mn+2, Fe+2 and Co+2 are d4, d5, d6 and d1 respectively. Which one of the following will exhibit minimum paramagnetic behaviour?

(d)
Hint:
In [Mn(H2O)6]+2, Mn+2 = 3d5
So, the number of unpaired electron = 5.
In [C(H2O)6]+2, Fe+2 = 3d6
So, the number of unpaired electron = 4.
In [Co(H2O)6]+2, Co+2 = 3d7
So, the number of unpaired electron = 3.
In [Cr(H2O)6]+2, Cr+2 = 3d4
So, the number of unpaired electron = 4.
All these show sp3d2 hybridisation forming outer orbital complexes. Hence, minimum paramagnetic behaviour is shown by [Co(H2O)6]+2

20. Crystal field stabilisation energy for high spin d4 octahedral complex is:
(a) -0.6 Δo
(b) -1.8 Δo
(c) -1.66 Δo + P
(d) -1.2 Δo
(a)
Hint: d4 octahedral complex has the configuration t2g3 eg1
CFSE = (-0.4x + 0.6y) Δo
x = no. of electron in t2g
y = no. of electron in eg
CFSE = [-0.4 x 3+ 0.6 x 1] Δo
= [-1.2 + 0.6] Δo
= – 0.6 Δo

21. In spectrochemical series, chlorine is above H2O i.e, Cl > H2O. This is due to:
(a) Good π acceptor properties of chlorine
(b) Strong σ donor and good π acceptor properties of chlorine.
(c) Good π donor properties of chlorine.
(d) Larger size of chlorine than H2O.
(c)

22. The magnitude of crystal field stabilisation energy (CFSE or Δt) in tetrahedral complexes is considerably less than in the octahedral field. Because
(a) There are only 4 ligands instead of six.
So the ligand field in only 2/3 the size, hence Δt, is only 2/3 the size.
(b) the direction of orbitals does not coincide with the direction of the ligands. This reduces the crystal field stabilisation energy (Δt) by further 2/3.
(c) both (a) and (b) are correct
(d) both (a) and (b) are wrong.
(c)

23. Which of the following complex ions is expected to absorb visible light?

(At. No: Zn = 30, Sc = 21, Ti = 22; Cr = 24) ,
(b)
Hint: d0 (Sc+3, Ti+4) and d10 (Zn+2) species will not absorb visible light and are colurless whereas Cr+3 (d3) can undergo d-d transition on absorption of visible light of particular wave length and will be coloured.

24. Which of the following is high spin complex?
(a) [CoCl6]-3
(b) [FeF6]-3
(c) [Co(NH3)6]+2
(d) All of these.
(d)
Hint: All the given ligands are weak field ligands. Hence they form high spin complexes.

25. Which one of the following ligand is capable of forming a low spin as well as high spin complex?
(a) CO
(b) F
(c) NH3
(d) CN
(c)
Hint:CO and CN are strong ligands and F is a weak ligand. NH3 is neither very weak nor very strong ligand.

26. Complexes with halide ions are generally:
(a) low spin complexes
(b) high spin complexes
(c) both (a) and (b)
(d) neither (a) and (b)
(b)
Hint: Halide ions have low CFSE (A) values. They cannot pair up the electrons of the metal atom / ion. Hence they form high spin complexes.

27. Which of the following shall form an octahedral complex?
(a) d4 (low spin)
(b) d8 (high spin)
(c) d6 (low spin)
(d) all of these
(c)
Hint: In d6 (low spin) electrons pair up thereby making two empty ‘d orbitals of the metal atom / ion and undergo d2sp3 hybridisation.

28. Match the entities of column I with appropriate entities of column II.

Code:

(a)

29. Assertion (A): The complex [CO(NH3)3Cl3] does not give precipitate with AgNO3.
Reason (R): The complex does not contain counter ions.
(a) If both assertion and reason are true and reason is the correej explanation of assertion.
(b) If both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If both assertion and reason are false.
(a)

30. Assertion (A): The [Ni(en)3]Cl2 (en = ethylene diamine) has lower stability than (Ni (NH3)6] Cl2.
Reason (R): In [Ni(en)3] Cl2, the geometry of nickel is trigonal bi-pyramidal.
(a) If both assertion and reason are true and the reason is the correct explanation of assertion.
(b) If both assertion and reason are true, but the reason is not the correct explanation of assertion.
(c) If the assertion is true but the reason is false.
(d) If both assertion and reason are false.
(d)
Hint:
Correct assertion: [Ni(en)3]Cl2 has greater Stability than [Ni (NH3)6]Cl2.
Correct Reason: [Ni (en)3]Cl2 has a bidentate ligand (en). It forms chelates.

## TN Board 12th Chemistry Important Questions Chapter 4 Transition and Inner Transition Elements

Students get through the TN Board 12th Chemistry Important Questions Chapter 4 Transition and Inner Transition Elements which is useful for their exam preparation.

## TN State Board 12th Chemistry Important Questions Chapter 4 Transition and Inner Transition Elements

Question 1.
Write the electronic configuration of (a) Cr+3, (b) Cu+, (c) CO+2, (d) Mn+2, (e) Pm+3, (f) Ce+4, (g) Lu+2, (h) Th+4.

Question 2.
To what extent do the electronic configuration decide the stability of oxidation states in the first series of transition elements. Illustrate your answer with an example.
In a transition, the series the oxidation state which leads to exactly half filled or completely filled ‘d’ orbitals are more stable. For example, the electronic configuration of Fe(z = 96) is [Ar] 3d6 4s2. It shows +2 and +3 oxidation states. The +3 oxidation state is more stable because it has the configuration [Ar] 3d5.

Question 3.
Explain why transition elements have many irregularities in their electronic configuration.
In the transition elements, there is a little difference in the energy of (n – 1 )d and ns orbitals. The incoming electron may occupy either of these orbitals. Hence, there is irregularities in their electronic configurations.

Question 4.
Name the oxometal anoins of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Cr2O7-2 (dichromate) and CrO4-2 (Chromate):
Group number of Cr=Oxidation state of Cr = 6
MnO4 (permanganate):
Group number of Mn=Oxidation state of Mn=7
Group number of V = Oxidation state of V = 5.

Question 5.
What are the stable oxidation states of the transition elements with the electronic configuration in their ground states of their atoms 3d3, 3d5, 3d8, and 3d4.

Question 6.
For M+2/M and M+3/M+2 systems, the E° values for some metals are as follows:
Cr+2/Cr = -0.9V; Cr+3/Cr+2 = -0.4V
Mn+2/Mn = -1.2V; Mn+3/Mn+2 = +1.5 V
Fe+2/Fe – -0.4V; Fe+3/Fe+2 = +0.80V
Use this data to comment upon:
(i) the stability of Fe+3 in acid solution as compared to Cr+3 or Mn+3 and
(ii) the case with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
(i) Higher the reduction potential of species, greater is the tendency for its reduction to take place. Therefore, Mn+3, with the highest reduction potential would be readily reduced to Mn+2 (Mn+3 + e → Mn+2) and hence is least stable. Thus, from the values of reductions potential, it is clear that the stability of Fe+3 in acidic solution is more than Mn+3, but less than that of Cr+3.
(ii) Lower the reduction potential or higher the oxidation potential of a species, greater the case with which its oxidation will take place. Thus, the order of tendency to undergo oxidation is Fe < Cr < Mn.
Note: Oxidation and reduction potentials have the same magnitude but opposite in sign.

Question 7.
How is the variability of oxidation states of transition metals different from that of non-transition metals? Illustrate with examples.
In transition elements, the successive oxidation states differ by unity. For example, manganese shows all oxidation states +2 to +7. On the other hand, non-transition metals exhibit variable oxidation states which differs by two units.eg: Pb(II) and Pb (IV); Sn(II) and Sn(IV).

Question 8.
How would you account for the following?
(i) Many of the transition elements and then compounds can act as good catalysts.
(ii) The metallic radii of the third (5d) series of transition elements are literally the same as those of the corresponding members of the second series.
(iii) There is a greater range of oxidation states among actinoids than those of lanthanoids.
(i) The catalytic activity of transition elements is attributed to the following reasons:

• Because of their variable oxidation states transition metals form unstable intermediates and provide a new path of lower activations energy for the reaction.
• In some cases, the transition metal provides a suitable large surface area with free vacancies, on which reactants are absorbed.

(ii) This is due to filling of 4/orbitals which have poor shielding effect or due to lanthanoid contraction.
(iii) This is due to comparable energies of 5f, 5d and 7s orbitals in actinoids.

Question 9.
Calculate the number of unpaired electrons in the following gaseous ions.
Mn+3, Cr+3, V+3 and Ti+3 Which one of these is the most stable in an aqueous solution?
Mn+3 = [Ar] 3d4 – 4 unpaired electrons
Cr+3 = [Ar] 3d3 – 3 unpaired electrons
V+3 = [Ar] 3d2 – 2 unpaired electrons
Ti+3 = [Ar] 3d1 – 1 unpaired electron
Cr+3 is most stable as it is half-filled t2g level.

Question 10.
The electronic configuration of chromium and copper are [Ar] 3d5 4s1 and [Ar] 3d10 4s1 respectively instead of [Ar] 3d4 4s2 and [Ar] 3d9 3s2. Explain.
The electronic configuration [Ar] 3d5 4s1 and [Ar] 3d10 4s1 is more stable than the other because of the extra stability of half filled and completely filled ‘d’ orbitals. This extra stability is due to the symmetrical distribution of electron density.

Question 11.
The melting and boiling points of Zn, Cd and Hg are low why? (or) Why Zn, Cd and Hg are soft and have low melting points?
In Zn, Cd and Hg, all the electrons in the ‘d’ orbitals are paired. Hence metallic bonding present in than are weak. Hence, the attraction between the constituent atoms are weak, i.e., they have low melting and boiling points.

Question 12.
Account for the trend in melting points of 3d series.
The melting point first increases as the number of unpaired ‘d’ electrons available for metallic bonding increases, reaches a maximum and then decreases, as the ‘d’ electrons pair up and becomes unavailable for metallic bonding. The maximum melting point is in the middle, where the transition metal has d5 configuration. This results in the formation of strong metallic bonding, which results in the maximum melting point in the middle of the series.

Question 13.
Explain why manganese has a lower melting point than chromium.
The electronic configuration of chromium is [Ar] 3d5 4s1 and that of manganese is [Ar] 3d5 4s2 i.e., in manganese-all the ‘d’ orbitals are singly filled and the 4s orbital is completely filled. As the ‘d’ electrons are more tightly held by the nucleus, the electrons are not available for bonding resulting in weaker metallic bonding in manganese compared to chromium.

Question 14.
The second and third members of each group of transition elements have similar atomic radii- Explain.
This is due to lanthanide contraction by members of 4f series, which occupy a position between lanthanum (at. no. 57) a first member of 3rd transition series and hafnium (Z = 72), the second member of third transition series.

The pairs of elements Zr – Hf, Nb – Ta, Mo – Wo, possess nearby atomic radii and almost have the same properties.

Question 15.
The atomic radii of the elements in a transition series do not vary much while they do vary in case of ‘s’ and ‘p’ block elements explain.
In transition elements two effects are operating, viz nuclear charge effect and screening effect which oppose each other. Due to an increase in nuclear charge from member to member in a transition series the atomic radii tend to decrease. At the same time, the addition of extra electrons in (n – 1 )d orbital provides the screening effect. As the number of ‘d’ electrons increases, the screening effect increases and this tends to increase the size. Due to these opposing tendencies, there is a small change in the atomic radii in a transition series.
In ‘s’ and ‘p’ block elements, the extra electron is added to the same ‘s’ or ‘p’ subshell which does not exert a screening effect and hence, the atomic radii considerably in a period due to an increase in nuclear charge.

Question 16.
How would you account for the following?
(i) The atomic radius of metals of the third (5d) series of elements are virtually the same as those of corresponding members of the second (4d) series.
(ii) Chromium is a typical hard metal while mercury is a liquid.
(i) Due to lanthanoid contraction.
(ii) Chromium has five unpaired electrons in the ‘d’ subshell (3d5 5s1). Hence metallic bonds are very strong. In mercury, all the ‘d’ orbitals are fully filled (3d10 4s2). Hence, metallic bonding is weak and thus it exists as a liquid.

Question 17.
How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of transition metals?
Reason for irregular variation of first IE: Generally moving from left to right in a period IE increase because of increase in nuclear charge. The irregular trend is observed, when an electron is removed, the relative energies of 3d and 4s orbitals are altered. Thus there is a reorganisation of energy accompanying ionisation. This results in the release of exchange energy which increases as the number of electrons increases in the ‘d’ configuration and also from the transfer of ‘s’ electrons into ‘d’ orbitals. Chromium has the first low IE because of the loss of one electron gives a stable configuration (d5). Zn has high IE because the electron has to be removed from 4s orbital of the stable configuration (3d10 4s2)
Reason for irregular variation of second IE: After the loss of one electron, the removal of the second electron is difficult. Hence, second IE is much higher and in a general increase from left to right. Cr and Cu show much higher values because the second electron has to be removed from the stable configuration. Cr+(3d5) and Cu+(3d10).

Question 18.
Compare the stabilities of Ni+4 and Pt+4 from their ionisation enthalpy values.

 IE Ni Pt I 737 864 II 1753 1791 III 3395 2800 IV 5297 4150

The smaller the magnitude of the ionisation energy, the more stable is the compound in a particular oxidation state. Nickel (II) compounds are more stable than Pt(II) compounds while Pt(IV) compounds are more stable than Ni(IV) compounds. The sum of the first two and first four IE of nickel and platinum are given below.

Since the first two ionisation energies is less for nickel than platinum, Ni(II) compounds are more stable than platinum.
On the other hand, Pt(IV) compounds are more stable than Ni(IV) compounds as the seem of the first four IE values are less than for Pt the Ni.

Question 19.
The silver atom has completely ‘d’ orbitals (4d10) in its ground state. How can you say that it is a transition element?
The electronic configuration of Ag (z = 47) to 4d10 5s1. In addition to +1, it shows an oxidation state of +2 (eg: Ag2O, and F2). i.e., d9 configuration i.e., the ‘d’ orbital is incompletely filled. Hence, it is a transition element.

Question 20.
Which of the 3d transition metals exhibit the largest number of oxidation states? and why?
Manganese (z = 25) shows a maximum number of oxidation states. This is because its electronic configuration is 3d5 4s2. As 3d and 4s are close in energy, it has a maximum number of electrons to lose or share (all the 3d electrons are unpaired) Hence, it shows oxidation state +2 to +7.

Question 21.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Oxygen and fluorine have small sizes and high electronegativity. Hence they can oxidise the metal to its highest oxidation state.

Question 22.
What may be the stable oxidation state of the transition element with the following ‘d’ electronic configuration in the ground state of their atoms? 3d3, 3d5, 3d8 and 3d4.

The maximum oxidation state corresponds to the sum of the ‘5’ and ‘cf electrons in Upto Mn. After that there is an abrupt decrease in the stability of the higher oxidation state.

Question 23.
Why are Mn+2 compounds are more stable than Fe+2 compounds towards oxidation to +3 state?
The electronic configuration of Mn is 3d5. Which is half-filled and hence stable. So, the third ionisation enthalpy is very high, i.e., 3rd electron cannot be removed easily. In the case of Fe+2, (3d6), it can lose one more electron to attain the stable (d5) configuration.

Question 24.
Explain briefly how +2 oxidation state becomes more and more stable in the first half of first-row transition elements with increasing atomic number.
The sum of IE1, and IE2 increases. Thus, the standard reduction potential becomes less and less negative. Therefore, the tendency to form M+2 ion decreases. The greater stability of +2 state, say for Mn, is due to half-filled (d5) configuration and that for Zn is due to completely filled ‘d’ orbitals. That for nickel is due to the highest negative enthalpy of hydration.

Question 25.
To what extent do the electronic configuration decide the stability of oxidation states in the first series of transition elements? Illustrate your answer with an example.
The stability of the oxidation state in the first transition elements are related to their electronic configuration.

The first five elements of the first transition series up to Mn in which 3d subshell is not more than half-filled, the minimum oxidation state is given by the number of electrons in the outer s – subshell and the maximum oxidation state is given by the sum of outer ‘s’ and ‘d’ electrons.

Question 26.
Explain why E°(Mn+3/Mn+2) couple is more positive than for E°(Fe+3/Fe+2) (At.no.of Mn = 25; Fe = 26).
The electronic configuration of Mn+2 is Mn+2 = [Ar] 3d5; Mn+3 is 3d4 and that of Fe+2 = [Ar] 3d6; Fe+3 is [Ar] 3d5.
Thus Mn+2 has a more stable configuration than Mn+3 while Fe+3 has a more stable configuration than Fe+2. Consequently, large third ionisation enthalpy is required to change Mn+2 to Mn+3. As E° values is the sum of the enthalpy of atomisation, ionisation enthalpy and hydration enthalpy, therefore E° for Mn+3/ Mn+2 couple is more positive than Fe+3/Fe+2.
Note: The large positive E° for Mn+3/ Mn+2 means that Mn+3 can be easily reduced to Mn+2 (Mn+3 + e → Mn+2) i.e., Mn+3 is less stable, i.e., Fe+3 can be reduced to Fe+2 (Fe+3 + e → Fe+2 but ion easily. Thus, Fe+3 is more stable than Mn+3.

Question 27.
The E°values in respect of the electrodes of chromium (z = 24), manganese (z = 25) and iron (z = 26) are: Cr+3/Cr+2 = – 0.4V, Mn+3/Mn+2 = 1.5V; Fe+3/Fe+2 – 0.80V. On the basis of the above information compare the feasibilities of further oxidation of their +2 oxidation states.
Cr+3 + e -> Cr+2: E° value is -0.4V. The negative value means Cr+3 is more stable or Cr+2 is less stable. Mn+3 + e → Mn+2: E° value is 1.5V. Fe+3 + e → Fe+2: E° value is 0.80V. Greater positive value for Mn+3/Mn+2 than that for Fe+3/Fe+2 shows that Mn+2 is more stable than Fe+2. Hence, the stability of +2 oxidation state is in the order Cr+2 < Fe+2 < Mn+2 or oxidation of their +2 state is in the order Cr+2 > Fe+2 > Mn+2.

Question 28.
The E°(M+2/M) value for copper is positive 0.34V. Explain why?
E°(M+2/M) for any metal is related to the sum of the enthalpy change taking place in the following steps.

Copper has a high enthalpy of atomisation and low enthalpy of hydration. Thus, high energy required to transform Cu(s) to Cu+2(aq) is not balanced by its hydration enthalpy Hence E°(Cu+2/Cu) is positive.

Question 29.
Which is a stronger reducing agent Cr+2 or Fe+2 and why?
E°(Cr+3/Cr+2) = -0.41V and E°(Fe+3/Fe+2) is 0.77V.
In general, for the reduction reaction Mn+ + ne → M,
Larger the value of standard reduction potential, greater the extent to which Mn+ is reduced to M. i.e., Mn+ acts as an oxidising agent. Compare
Cr+3 + e → Cr+2 E° = -0.41V
Fe+3 + e → Fe+2 E° = 0.77V.
The negative value indicates that Cr+2 is readily oxidised to Cr+3 i.e., it is a stronger reducing agent than Fe+2.

Question 30.
Explain the terms paramagnetism and diamagnetism with suitable examples.

1. A paramagnetic substance is one that is weakly attracted into a magnetic field and a diamagnetic substance is one that is repelled by a magnetic field.
2. The paramagnetic behaviour arises due to the presence of one or more unpaired electrons, while a diamagnetic substance is due to the presence of paired electrons.
3. For example Sc+3 has no unpaired electron and hence diamagnetic whereas Ti+3 has one unpaired electron and hence paramagnetic.

Question 31.
Calculate the magnetic moment of the following ions.

Question 32.
Explain how transition metals and their compounds act as catalysts.
Transition metal has energetically available ‘d’ orbitals that can accept electrons from reactant molecule or metal can form a bond with reactant molecule using its ‘d’ electrons. For example, in the catalytic hydrogenation of an alkene, the alkene bonds to an active site by using its π electrons with an empty ‘d’ orbital of the catalyst.
The σ bond in the hydrogen molecule breaks and each hydrogen atom forms a bond with a ‘d’ electron on an atom in the catalyst. The two hydrogen atoms then bond with the partially broken π -bond in the alkene to form an alkane.

Question 33.
Give examples wherein transition metal compounds act as catalysts in various industrial processes.

1. In the manufacture of sulphuric acid from sulphur trioxide, vanadium pentoxide is used as a catalyst.
2. Hydroformylation of olefins
3. Preparation of acetic acid from acetaldehyde.
4. Zeigler – Natta catalyst
A mixture of TiCl4 and trialkyl aluminium is used for polymerization.

Question 34.
What is an alloy? How are they formed?
An alloy is a mixture of two or more metals. They are formed when molten metals are mixed together and allowed to crystallise, eg: ferrous alloys, gold – copper alloy, chrome alloys etc.

Question 35.
Explain why transition metals have a tendency to form alloys.
For the formation of alloy, the following conditions have to be met.

1. Both the solute and the solvent must have the same crystal structure.
2. Their valence and their electronegativity difference must be close to zero. Transition metals satisfy these conditions. Their atomic sizes are similar and one metal atom can be easily replaced by another metal atom from its crystal lattice to form an alloy.

Question 36.
What are interstitial compounds? Give examples.
An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are usually non-stoichiometric compounds. Transition metals form a number of interstitial compounds such as TiC, ZrH1.92, Mn4N etc., The elements that occupy the metal lattice provide them new properties.

Question 37.
Mention the properties of interstitial compounds.

1. They are hard and show electrical and thermal conductivity.
2. They have high melting points higher than those of pure metals.
3. Transition metal hydrides are used as powerful reducing agents.
4. Metallic carbides are chemically inert.

Question 38.
What are complexes or coordination compounds? Give a brief account of complexes formed by transition metals.
Complexes or coordination compounds are these in which a transition metal ion accept a pair of electrons, from a fixed number of compounds in which the central atom can denote a pair of electrons and form coordinate covalent bonds.
eg: [Fe(CN)6]4-. In this complex, Fe+2 ion accepts lone pair of electrons from the cyanide ion and form 6 coordinate covalent bonds. Transition metal ions form complexes due to their small size and high charge and have vacant ‘d’ orbitals to accept electron pairs denoted by other groups.

Question 39.
Give a brief account of oxides formed by transition metals.

1. The metals of the first transition series form oxides with oxygen at high temperatures.
2. The oxidation state of the metal in the oxides vary from +1 to +7.
3. The highest oxidation state in the oxides of any transition metal is equal to its group number eg: +7 in Mn2O7. Beyond group 7. No higher oxides of iron above Fe2O3 are known.
4. All the metals except scandium form oxides of the formula MO which are ionic in nature. As the oxidation number of the metal increases, ionic character decreases, eg: Mn2O7 is covalent even CrO3, V2O5 are covalant and have low melting point.
5. In general, the oxides in the lower oxidation states of the metals are basic and in their higher oxidation states they are acidic. Whereas in the intermediate oxidation states they are amphoteric.

Question 40.
Classify the following oxides as acidic, basic and amphoteric
(a) Mn2O7, (b) CrO3, (c) Cr2O3, (d) CrO.
(a) Mn2O7 → acidic
(b) CrO3 → acidic
(c) Cr2O3 → amphoteric
(d) CrO → basic.

Question 41.
Give example and suggest the reason for the following:
(i) The lowest oxide of transition metal is basic, the highest is acidic.
(ii) A transition metal exhibits higher oxidation states in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxo anion of a metal.
(i) The lower oxide of a transition metal is basic because the metal atom has a low oxidation state whereas the highest oxide is acidic due to a higher oxidation state for example, MnO is basic while Mn2O7 is acidic.
(ii) A transition metal exhibits a higher oxidation state in oxides and fluorides because oxygen and fluorine are highly electro negative elements, small in size (and strongest oxidising agents). For example, osmium shows an oxidation state of +6 in O5F6 and vanadium shows an oxidation state of +5 in V2O5.
(iii) Oxo metal anion have the highest oxidations state, eg: Cr in Cr2O7-2, has an oxidation state of +6, whereas Mn in MnO4 ion has an oxidation state of +7. This again due to the combination of the metal with oxygen which is a highly electronegative and oxidising element.

Question 42.
Indicate the steps involved in the preparation of (i) K2Cr2O7 from chromite (ii) KMnO4 from pyrolusite
(i) K2Cr2O7 from chromite are

K2Cr2O7 is separated by fractional crystallisation.
(ii) KMnO4 from pyrolusite are

KMnO4 is prepared from pyrolusite. MnO2 with KOH in the presence of an oxidising agent like KNO3. This produces dark brown potassium manganate which disproportionates in a neutral or acidic solution to give purple permanganets.

Question 43.
Give chemical oxidation and electrolytic oxidation of MnO4-2 to MnO4.
Chemical oxidation: In this method potassium manganate is treated with ozone (O3) or chlorine to get potassium permanganate.

Electrolytic oxidation: In this method aqueous solution of potassium manganate is electrolyzed in the presence of little alkali.

Manganate ions are converted into permanganate ions at anode.

H2 is liberated at the cathode.
2H+ + 2e → H2 ↑
The purple coloured solution is concentrated by evaporation and forms crystals of potassium permanganate on cooling.

Question 44.
What happens when potassium dichromate is heated? Give equation.
It decomposes to give green Cr2O3 and oxygen.

Question 45.
Discuss the action of alkali on potassium dichromate.
When an alkali is added to an orange-red solution of potassium dichromate, a yellow solution results due to the formation of chromate.

Question 46.
What is the effect of pH on a solution of potassium dichromate?
In aqueous solution

In acidic medium (i.e., decreasing pH), the equilibrium shifts backwards and the colour is orange-red. In a basic medium, (i.e., increasing pH) the equilibrium will shift forward and the solution is yellow.

Question 47.
Draw the structure of the dichromate ion.

Question 48.
Write the ionic equation to show that K2Cr2O7 is an oxidising agent in an acid medium.

Question 49.
Complete and balance the following equation: (all in acid medium)

Question 50.
Describe the chromyl chloride test.
When potassium dichromate is heated with any chloride salt in the presence of Conc. H2SO4, orange-red vapours of chromyl chloride (CrO2Cl2) is evolved. This reaction is used to confirm the presence of chloride ions in inorganic qualitative analysis.

The chromyl chloride vapours are dissolved in sodium hydroxide solution and then acidified with acetic acid and treated with lead acetate. A yellow precipitate of lead chromate is obtained.

Question 51.
Mention the uses of potassium dichromate.

1. It is used as a strong oxidizing agent.
2. It is used in dyeing and printing.
3. It used in leather tanneries for chrome tanning.
4. It is used in quantitative analysis for the estimation of iron compounds and iodides.

Question 52.
Draw the structure of MnO4 ion.
Permanganate ion has tetrahedral geometry in which the central Mn7+ is sp3 hybridised.

Question 53.
What happens when:
(i) potassium permanganate is heated.
(ii) is heated with cold conc. H2SO4
(iii) is heated with hot conc. H2SO4?
(i) It decomposes to form potassium
manganate and manganese dioxide.
2KMnO4 → 2K2MnO4 + MnO2 + O2
(ii) It decomposes to form manganese heptoxide which subsequently decomposes explosively.

(iii) It gives manganous sulphate

Question 54.
How does potassium permanganate act as an oxidising agent in the neutral medium? Explain with examples.
In neutral medium, KMnO4 is reduced to MnO2.
MnO4 + 2H2O + 3e → MnO2 + 4OH
(i) It oxidises H2S to sulphur

(ii) It oxidises thiosulphate into sulphate

Question 55.
Write an ionic equation or the half-reaction for the oxidising action of KMnO4 in an alkaline medium.
MnO4 + 2H2O + 3e → MnO2 + 4OH

Question 56.
What is Bayer’s reagent? Mention its use.
Cold dilute alkaline KMnO4 is known as Bayer’s reagent. It is used to oxidise alkenes into diols. For example, ethylene can be converted into ethylene glycol and this reaction is used as a test for unsaturation.

Question 57.
What is the action of acidified KMnO4 on
(i) ferrous salts,
(ii) Potassium iodide,
(iii) Oxalic acid
(iv) Sulphide ion,
(v) nitrite ion,
(vi) Sulphite ion.
(i) It oxidises ferrous salts to ferric salts

(ii) It oxidises iodide ion to iodine

(iii) It oxidises oxalic acid to carbon dioxide.

(iv) It oxidises sulphide ion to sulphur.

(v) It oxides nitrite ion to nitrates

(vi) It oxidises sulphite ion to sulphates.

Question 58.
Mention the uses of potassium permanganate.

1. It is used as a strong oxidizing agent.
2. It is used for the treatment of various skin infections and fungal infections of the foot.
3. It is used in water treatment industries to remove iron and hydrogen sulphide from well water.
4. It is used as a Bayer’s reagent for detecting unsaturation in an organic compound.
5. It is used in quantitative analysis for the estimation of ferrous salts, oxalates, hydrogen peroxide and iodides.

Question 59.
Find the equivalent weight of KMnO4 is an acidic, basic and neutral medium.
The ionic equation is

In basic medium:

In neutral medium:

Question 60.
What are inner transition elements?
Decide which of the given atomic numbers are the numbers of inner transition elements: 29, 59, 74, 95, 102, 104.
The ‘f’ block elements i.e., in which the last electron enters the ‘f’ sub shell are called inner transition elements. These include lanthanides (58 – 71) and actinides (90 – 103). Thus, the elements with atomic numbers 59,95 and 102 are inner transition elements.

Question 61.
The chemistry of actinoid elements is not so smooth as that of lanthanoids justify then statement by giving some examples from the oxidation state of these elements.
Lanthanoids show a limited number of oxidation states viz +2, +3, +4 out of which +3 is most common. This is because of the energy gap between 4f and 5d sub shells. The dominant oxidation state of actinides is also +3 but they show a number of oxidation states also eg: uranium (Z = 92) and polonium (Z = 92) show +3, +4 and +5 and +6, neptunium (Z = 94) shows +3, +4, +5 and +7 etc., This is due to small energy diffidence between 5f, 6d and7s subshells.

Question 62.
What is the last element in the series f actinoids? Comment on the possible oxidation state of the element.
Last actinoid: Lawrencium (Z = 103).
Electronic configuration; [Rn]86 5f14 6d1 7s2. possible oxidation state = +3.

Question 63.
Name of the members of the lanthanoid series which exhibit +4 oxidation state. Try to correlate this type of behaviour with the electronic configuration of these elements.
+4: Ce(Z=58), Pr(Z=59), Nd(Z=60), Tb(Z=65), Dy(Z = 66)
+2: Nd(Z = 60); Sm(Z = 62); Eu(Z = 63), Tm(Z = 69); Yb(Z = 70)
+2 Oxidation state is exhibited when lanthanoid has the configuration 5d0 6s2. So that 2 electrons one easily lost. +4 state is exhibited when the electronic configuration left to close to 4f0 (eg: 4f0, 4f1, 4f2) or close to 4f7 (eg: 4f7 or 4f8)

Question 64.
Write the electronic configuration of the elements with atomic numbers 61, 91, 101 and 109.

Question 65.
Compare the chemistry of actinoids with that of lanthanoids with special reference to
(i) electronic configuration
(ii) Oxidation state
(iv) Chemical reactivity
(i) The general electronic configuration of lanthanoids is [Xe]54 4f1-14 5d0-1 6s2 where as that of actinoids in [Rn]86 5f1-14 6d0-1 7s2. Thus lanthanoids belong to 4f series whereas actinoids belong to 5f series.
(ii) Lanthanoids show limited oxidation states (+2, +3, +4) out of which +3 is most common. This is because of the large energy gap between 4f and 5d subshells. On the other hand, actinoids show a large number of oxidation states because of a small energy gap between 5f, 6d and 7s subshells.
(iii) Both show a decrease in the size of their atoms or ions in the +3 oxidation state. In lanthanoids, this decrease is called lanthanoid contraction, whereas in actinoids it is called actinoid contraction. However, the contraction is greater from element to element in actinoids due to poor shielding of 5f electrons than that by 4f electrons in lanthanoids.
(iv) Chemical reactivity of lanthanoids

The chemical behaviour of actinoids:

1. They react with boiling water to give a mixture of hydride and oxide.
2. They combine with most of the non-metals at high temperatures.
3. All these metals are attacked by hydro chloric acid, but the effect of nitric acid is very small due to the formation of a protective oxide layer on their surface.
4. Alkalies have no action on actinoids.

Question 66.
The +3 oxidation state of lanthanium (Z = 57), gadoliniun (Z = 64) and lutetium (Z = 71) are especially stable. Why?
This is because that is the +3 oxidation state, they have empty, half-filled and completely filled 4f sub shell respectively.

Question 67.
The outer electronic configuration of two members of the lanthanoid are as follows:
(i) 4f1 5d1 6s2 and (ii) 4f7 5d0 6s2
What are their atomic numbers predict the oxidation state exhibited by these elements in their compounds?
Complete Electronic configuration of the (i) lanthanoid:
[Xe]54 4f1 5d1 6s2: Atomic number = 58
The element is cerium.
Complete Electronic configuration of the (ii) lanthanoid:
[Xe]54 4f7 5d0 6s2: Atomic number = 68.
The element is Europium (Eu).
Oxidation state of 1st lanthanoid +2(4f2) +3(4f1) and +4(4f0)
Oxidation state of the 2nd lanthanoid:
+2(4f7) and +3(4f6)

Question 68.
Give an explanation for each of the following observations.
(i) The gradual decrease in size (actinoid contraction) from element to element is greater among actinides than that among lanthanides.
(ii) The actinoids exhibit a greater range of oxidation states than the lanthanoids.
(iii) Zv and Hf have identical sizes.
(iv) Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and Cl. Why?
(v) Ce+4 is used as an oxidising agent in volumetric analysis.
(i) This is due to poor shielding by 5f electrons from element to element in actinoids than by 4f electrons in lanthanoid series.
(ii)This is because there is less energy difference between 5f and 6d orbitals belonging to actinoids than the energy difference between 4f and 5d Orbitals in the case of lanthanoids.
(iii) This is due to lanthanoid contraction.
(iv) It is because at the beginning, where 5f orbitals begin to be occupied they will penetrate less into the inner core of electrons. The 5felectron will therefore be more effectively shielded from the nuclear charge than 4f electrons of the corresponding lanthanoids. Therefore the outer electrons are less firmly held and they are available for bonding in the actinoids.
(v) Ce+4 has a tendency to attain +3 oxidation state and so it is used as an oxidising agent in volumetric analysis.

Question 69.
Account for the following:
(i) Europium (II) is more stable than cerium(II)
(ii) Actinoid ions are generally coloured.
(i) Europium has stable configuration i.e., [Xe] 4f7 5d0 6s0.
(ii) Unpaired electrons are present in their ions which undergo f-f transition.

Question 70.
On the basis of lanthanoid contraction, explain the following:
(i) Nature of bonding in La2O3 and Lu2O3.
(ii) Trends in the stability of oxosaltes of lanthanoids from La to Lu.
(iii) Stability of complexes of lanthanides.
(iv) Radii of 4d and 5d block elements.
(v) Trends in the acidic character of lanthanoid oxides.
(i) As the size decreases, the covalent character increases. Therefore La2O3 is more ionic and Lu2O3 is more covalent.
(ii) As the size decreases from La to Lu, the stability of oxo salts also decreases.
(iii) Stability of the complexes increases as the size of the lanthanoid decreases.
(iv) Radii of 4d and 5d block elements are will be almost the same.
(v) Acidic character of the oxides increases from La to Lu.

Question 71.
What is meant by the term Lanthanoid contraction? What is it due to and what consequences does it have on the chemistry of lanthanides in the periodic table.
Lanthanide contraction: The steady decrease in atomic and ionic radii with an increase in atomic number is known as lanthanide contraction.
Causes of lanthanide contraction: As we move along the lanthanoid series, for every additional proton in the nucleus, the corresponding electron goes into 4f sub shell, there is poor shielding of one electron by another in the sub shell due to the shapes of these orbitals. This imperfect shielding is not able to counterbalance the effect on the increased nuclear charge. Thus, the net result is a decrease in size, with an increase in atomic number.
Consequences:

1. 5d series elements have nearly the same radii as that of the 4d series.
2. The basic strength of hydroxides decrease from La(OH)3 to Lu(OH)3

1. The correct order of decreasing second ionisation enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is:
(a) Cr > Mn > V > Ti
(b) V > Mn > Cr > Ti
(c) Mn > Cr > Ti > V
(d) Ti > V > Cr > Mn
(a) Cr > Mn > V > Ti
Hint: Electronic configuration of the given elements are
Ti = [Ar] 3d2 4s2; V = [Ar] 3d3 4s2 Cr = [Ar] 3d5 4s1; Mn = [Ar] 3d5 4s2
Their effective nuclear charge increases from Ti to Mn. Hence their first IE increases in the same order, i.e., Mn > Cr > V > Ti. However after the removal of first electron, chromium attains the stable configuration [3d5] and hence, it has very high second IE. For the remaining elements the trend remains the same. Thus, the second IE will be in the order.
Cr > Mn > V >Ti

2. The electronic configuration of Cu(II) is 3d9 whereas that of Cu(I) is 3d10. Which of the following is correct?
(a) Cu(II) is more stable.
(b) Cu(II) is less stable.
(c) Cu(I) and Cu(II) are equally stable.
(d) Stability of Cu(I) and Cu(II) salts depends on the nature of copper salts.
(a) Cu(II) is more stable.
Hint: Though Cu(I) has 3d10 configurations Cu(II) is more stable. This is due to greater effective nuclear charge of Cu(II). i.e., to hold 17 electrons instead of 18 electrons is Cu(I).

3. Although zirconium belongs to 4d transition series and Hafnium belongs to 5d transition series even their they have similar physical and chemical properties because:
(a) both belong to ‘cf block
(b) both have same number of electrons
(c) both have similar atomic radius
(d) both belong to the same group of the periodic table.
(c) both have similar atomic radius

4. Why HCl is not used to make the medium acidic in oxidation reactions of KMnO4 in acidic medium?
(a) Both HCl and KMnO4 act as oxidising agents.
(b) KMnO4 oxidises HCl, to Cl2, which is also an oxidising agent.
(c) KMnO4 is a weaker oxidising agent than HCl.
(d) KMnO4 acts as a reducing agent in the presence of HCl.
(b) KMnO4 oxidises HCl, to Cl2, which is also an oxidising agent.

5. KMnO4 acts as an oxidising agent in acidic medium. The number of moles of KMnO4 that will be needed to react one mole of sulphide ions in acidic solution is:
(a) $$\frac{2}{5}$$
(b) $$\frac{3}{5}$$
(c) $$\frac{4}{5}$$
(d) $$\frac{1}{5}$$
(a) $$\frac{2}{5}$$
Hint:

2 mol of KMnO4 ≡ 5 mol H2S
$$\frac{2}{5}$$ mol of KMnO4 ≡ 1 mol of H2S

6. Which one of the following ions is the most stable in aqueous solution:
Note: Atomic No: Ti = 22, V = 23, Cr – 24, Mn = 25.
(a) Mn+2
(b) Cr+3
(c) V+3
(d) Ti+3
(b) Cr+3
Hint: Out of the given species, Cr+3 has the highest negative reduction potential. Hence, it cannot be reduced to Cr+2 and therefore is the most stable in aqueous solution.
Alternatively, Mn+3 = [Ar] 3d4, Cr+3 = [Ar] 3d3, V+3 = [Ar] 3d2, Ti+2 = [Ar] 3d1. In Cr+3, all the three electrons enter the lowest energy 3d1(t2g ) orbitals. The lowering of the energy is maximum and hence the stability is maximum.

7. In acidic medium MnO4-2:
(a) disproportionates to MnO2 and MnO4
(b) is oxidised to MnO4
(c) reduced to MnO2
(d) is reduced to Mn+2
(a) disproportionates to MnO2 and MnO4
Hint:

8. Consider the following statements:
(i) La(OH)3 is the least basic among hydroxides of lanthanides
(ii) Zn+4 and Hf+4 possess almost same ionic radii
(iii) Ce+4 can act as oxidising agent. Which of the above is/are true?
(a) All
(b) (ii) and (iii)
(c) (ii) only
(d) (i) and (iii)
(b) (ii) and (iii)
Hint: La(OH)5, is most basic, i.e., (i) is wrong.
(ii) is correct due to lanthanide contraction.
(iii) is correct because Ce+4 tends to change to stable Ce+3.

9. Large number of oxidation states are exhibited by actinoids than these by the lanthanoids, the main reason being:
(a) more energy difference between 5f and 6d than between 4f and 5d orbitals.
(b) more reactive nature of the actinoids than the lanthanoids.
(c) 4f orbitals more diffused than 5f orbitals.
(d) less energy difference between 5f and 6d than between 4f and 5d orbitals.
(d) less energy difference between 5f and 6d than between 4f and 5d orbitals.

10. Irregular trend in the standard reduction potential value of the first row transition elements is due to:
(a) regular variation of first and second row enthalpies.
(b) irregular variation of sublimation enthalpies.
(c) regular variation of sublimation enthalpies.
(d) increase in number of unpaired electrons.
(b) irregular variation of sublimation enthalpies.

11. Match the property in column I with the metals in column II.

(a) (i) – (C); (ii) – (A); (iii) – (B)
(b) (i) – (C); (ii) – (D); (in) – (A)
(c) (i) – (B); (ii) – (C); (iii) – (D)
(d) (i) – (D); (ii) – (C); (iii) – (A)
(a) (i) – (C); (ii) – (A); (iii) – (B)

12. Match statements given in column I with the oxidation states given in column II.

(a) (i) – (C); (ii) – (A); (iii) – (E); (iv) – (B)
(b) (i) – (B); (ii) – (C); (iii) – (D); (iv) – (A)
(c) (i) – (D); (ii) – (B); (iii) – (C); (iv) – (A)
(d) (i) – (E); (ii) – (C); (iii) – (A); (iv) – (B)
(a) (i) – (C); (ii) – (A); (iii) – (E); (iv) – (B)

13. Knowing that the chemistry of lanthanoids
(Ln) is determined by its +3 oxidation state, which of the following statement is incorrect?
(a) The ionic size of Ln(III) decrease in general with increasing atomic number.
(b) Ln(III) compounds are generally colourless.
(c) Ln(III) hydroxides are mainly basic in character.
(d) Because of large size of Ln(III) ions the bonding in its compounds is predominantly ionic in character.
(b) Ln(III) compounds are generally colourless.
Hint: Ln(III) compounds are generally coloured due to partly filled f orbitals which permit f-f transition.

14. Which of the following statements regarding cerium (atomic no.58) is incorrect?
(a) The common oxidation states of cerium are +3 and +4.
(b) +3 oxidation state of Ce is more stable than +4.
(c) The +4 oxidation state of Ce is not known in solution.
(d) Cerium (IV) acts as an oxidising agent.
(c) The +4 oxidation state of Ce is not known in solution.

15. MnO4 reacts with Br in alkaline pH to give:
(a) BrO3, MnO2
(b) Br2, MnO4-2
(c) Br2, MnO2
(d) BrO, MnO4-2
(a) BrO3, MnO2
Hint:

16. Amount of oxalic acid present in a solution can be determined by titration with KMnO4 solution in the presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl because HCl:
(a) reduces permanganate to Mn+2.
(b) oxidises oxalic acid to carbondioxide and water.
(c) gets oxidised by oxalic acid to chlorine.
(d) furnishes H+ ions in additional to those from oxalic acid.
(a) reduces permanganate to Mn+2.
Hint: In the presence of HCl, KMnO4 not only oxidises oxalic acid but also oxidises HCl to Cl2 and itself reduced to Mn+2.

17. The spin only magnetic moment of Fe+2 (in Bμ) is approximately:
(a) 4
(b) 1
(c) 5
(d) 6
(c) 5
Hint: Fe+2 = 3d6 4s0. It has 4 unpaired electron.

18. Electronic configuration 6f a transition element X in +3 oxidation state is [Ar]3d5. What is its atomic number?
(a) 25
(b) 26
(c) 27
(d) 24
(b) 26
Hint: X+3 = [Ar]18 3d5 = 23 electrons As X+3 ion is formed by the loss of 3 electrons from X, X will have 26 electrons. Hence its at. no. = 26 V

19. Metallic radii of some elements are given below: Which of these elements will have highest density?

 Element Fe Co Ni Cu Metallic radii/pm 126 125 125 128

(a) Fe
(b) Ni
(c) Co
(d) Cu
(d) Cu
Hint: $$\text { Density }=\frac{\text { mass }}{\text { volume }}$$
As we move from Fe to Cu mass increases and volume decreases. Hence, density increases. Increase in mass of Cu dominates over small increase in volume.

20. The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element which shows highest magnetic moment:
(a) 3d7
(b) 3d5
(c) 3d8
(d) 3d2
(b) 3d5
Hint: 3d5 has maximum number of impaired electron.

21. There are 14 elements in actinoid series which of the following elements does not belong to this series?
(a) Cl
(b) Np
(c) Tm
(d) Fm
(c) Tm
Hint: Actinoid series is with atomic numbers 90 to 103. Thulium (Tm) has atomic number 69.

22. Gadolinium belongs to 4/series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?
(a) [Xe] 4f7 5d1 6s2
(b) [Xe] 4f6 5d2 6s2
(c) [Xe] 4f8 6s2
(d) [Xe] 4f9 5s1
(a) [Xe] 4f7 5d1 6s2
Hint: [Xe]54 4f7 5d1 6s2 due to extra stability of half filled 4f sub shell.

23. Highest oxidation state of manganese in fluoride is +4(MnF4) but the highest oxidation state in oxides is +7(Mn2O7) because:
(a) fluorine is more electronegative than oxygen.
(b) fluorine does not possess ‘d’ orbital.
(c) fluorine stabilises lower oxidation state.
(d) in covalent compounds, fluorine can form only single bonds white oxygen can form double bonds.
(d) in covalent compounds, fluorine can form only single bonds white oxygen can form double bonds.

24. Assertion: Mercury is not considered as a transition element.
Reason: Mercury is a liquid.
(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but the reason is not the correct explanation of assertion.
(c) If an assertion is true, but the reason is false.
(d) If both assertion and reason are false.
(b) If both assertion and reason are true but the reason is not the correct explanation of assertion.
Hint: Mercury has completely filled ‘d’ orbitals.

25. Assertion: in any transition series the magnetic moment of M+2 ions first increases and then decreases.
Reason: in a transition series, the number of unpaired electrons increases and then decreases.
(a) if both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) if both assertion and reason are true but the reason is not the correct explanation of assertion.
(c) if the assertion is true, but the reason is false.
(d) if both assertion and reason are false.
(a) if both assertion and reason are true and reason is the correct explanation of the assertion.

26. Match the ions in list I with their corresponding property in list II.

(a) (i) – (B) & (C); (ii) – (A) & (D); (iii) – (B) & (C); (iv) – (C)
(b) (i) – (C); (ii) – (A) & (B); (iii) – (D); (iv) – (B) & (C)
(c) (i) – (B) & (C); (ii) – (B); (iii) – (A) & (D); (iv) – (B)
(d) (i) – (D); (ii) – (C); (iii) – (B); (iv) – (A), (D)
(a) (i) – (B) & (C); (ii) – (A) & (D); (iii) – (B) & (C); (iv) – (C)

27. The complex forming tendency of a transition metal depends upon:
(a) the availability of number of vacant ‘d’ orbitals.
(b) high ionisation energy.
(c) large size of the cation or high charge density.
(d) variable oxidation states.
(a) the availability of a number of vacant ‘d’ orbitals.

28. Which lanthanide is most commonly used?
(a) La
(b) No
(c) Th
(d) Ce
(a) La

29. In which of the following pairs are both the ions coloured in an aqueous solution?
(a) Ni+2, Ti+3
(b) Ni+2, Cu+
(c) Sc+3, CO+2
(d) Sc+3, Ti+3
(a) Ni+2, Ti+3
Hint: Ni+2 = [Ar] 3d8 t2g6 eg2. It has 2 unpaired electron in e level. Hence it is coloured.
Ti+3 = [Ar]3d1 one unpaired electron and hence coloured.

30. When manganese dioxide is fused with KOH or K2CO3 in air, it gives:
(a) potassium permanganate
(b) manganese oxide
(c) manganese heptoxide
(d) potassium manganate
(d) potassium manganate

## TN Board 12th Chemistry Important Questions Chapter 3 p-Block Elements – II

Students get through the TN Board 12th Chemistry Important Questions Chapter 3 p-Block Elements – II which is useful for their exam preparation.

## TN State Board 12th Chemistry Important Questions Chapter 3 p-Block Elements – II

Question 1.
Briefly account for the trend in atomic radius of elements in the nitrogen family.
Atomic radii increase on going down the group. The increase is due to addition of new energy level in each succeeding element. There is a considerable increase in covalent radius from N to P but from As to Bi only a small increase is observed.
The increase in nitrogen to phosphorus is attributed to strong shielding effect of ‘s’ and ‘p’ electrons present in inner shells. Small increase in covalent radii from As to Bi is due to poor shielding effect of ‘d’ and ‘f’ electrons present in inner shells on valency electrons. The increased nuclear charge reduces the effect of the addition of new shell to same extent.

Question 2.
Briefly explain the trend in melting and boiling point in the nitrogen family elements.
The melting point first increase from nitrogen to arsenic and then decreases from antimony and bismuth. The boiling point gradually increases from nitrogen to antimony. Lower values of melting points for antimony and bismuth may be due to availability of only three electrons instead of five for metallic bonding due to inert pair effect.

Question 3.
What happens when (i) Sodium azide is heated? (ii) Ammonia is treated with bromine. Give equations.
In both cases, nitrogen is formed.

Question 4.
Explain why nitrogen is inert at room temperature.
This is due to high bond enthalpy of N ≡ N. The triple bond is short and hence large amount of heat is necessary to break the bond. Hence N2 is unreactive at room temperature.

Question 5.
What are nitrides? Give the preparation of (i) Lithium nitride, (ii) Calcium nitride, (iii) Boron nitride by means of chemical equation.
Nitrides are binary compounds of nitrogen which contains N-3 ion. They are formed by direct combination of metals with nitrogen.

Question 6.
Mention the conditions under which maximum amount of ammonia is formed from nitrogen to hydrogen.
The formation of NH3 is represented by

is an exothermic reaction. The reaction is favoured at high temperatures and at an optimum temperature in the presence of iron as catalyst.

Question 7.
Give equation for (i) hydrolysis of urea,
(ii) heating ammonium chloride with CaO,
(iii) Heating magnesium nitride with water.

Question 8.
How is ammonia manufactured?
Ammonia is manufactured by passing nitrogen and hydrogen over iron catalyst (a small amount of K2O and Al2O3 is also used to increase the rate of attainment of equilibrium) at 750 K at 200 atm pressure. In the actual process the hydrogen required is obtained from water gas and nitrogen from fractional distillation of liquid air.

Question 9.
Compare the properties of liquid ammonia and water.
(i) Both liquid ammonia and water are highly associated through strong hydrogen bonding.
(ii) Both ammonia and water are good ionising solvents.
(iii) The ionisation of ammonia and water is given as

Both OH and NH2 are strong bases,
eg: NaOH and NaNH2.

Question 10.
Give equations for the following reactions.
(i) Ammonia is heated over 500°C
(ii) Ammonia is burnt in oxygen.
(iii) Ammonia is burnt in oxygen in the presence of a metal catalyst (pt)
(iv) Ammonia is treated with excess of chlorine.
(v) Excess of ammonia is treated with chlorine.

Question 11.
Give an example for a reducing property of ammonia. [OR] Ammonia is a reducing agent. Give an example to prove this statement.
It reduces metal oxides to metals when passed over heated metallic oxide.

In this reaction, ammonia is oxidised to nitrogen.

Question 12.
What are amides and nitrides? Give an example for each. How are they formed?
Amides are salts formed by ammonia and a strong electropositive metal like Na, eg: NaNH2 (sodamide). It is formed by the action of ammonia on sodium.
2Na + 2NH3 → 2NaNH2 + H2
Nitrides are salts which contain N-3 ion.
eg: Mg3N2 (magnesium nitride). It is formed by the action of magnesium with ammonia.
3Mg + 2NH3 → Mg3 N2 + 3H2

Question 13.
What happens when an aqueous solution of ammonia is treated with (i) aqueous solution of ferric chloride, (ii) an aqueous solution of cupric chloride, (iii) an aqueous solution of aluminium chloride.
An aqueous solution of ammonia or ammonium hydroxide precipitates metal hydroxides from metallic salts solution.

This property is made use of in detecting group III metals in qualitative analysis.

Question 14.
Explain with examples, that ammonia acts as a Lewis base.
Due to the presence of lone pair of electrons on the nitrogen atom, ammonia acts as a Lewis base. It can donate its electron pair to form coordinate covalent bond with electron deficient molecules eg: BF3 or transition metal cation, having vacant ‘d’ orbitals) to form complexes. For example,

Question 15.
Explain why (i) insoluble silver chloride dissolves in aqueous ammonia?
This is due to the formation of a complex Ag (NH3)2 Cl (diammine silver (I) chloride)

Question 16.
When aqueous ammonia is treated with a copper sulphate solution, a blue precipitate is formed It dissolves an adding excess ammonia. Explain this observation.
The precipitate formed is cupric hydroxide, which dissolves in excess ammonia to form a blue solution due to the formation of soluble [Cu (NH3)4] SO4 complex, i.e., tetrammine copper(II) sulphate.

Question 17.
Explain the structure of ammonia.
The nitrogen in ammonia is sp3 hybridised. Each of the sp3 hybrid orbital containing one unpaired electron forms a covalent bond with the 1s orbital of hydrogen. The fourth sp3 hybrid orbital contain a lone pair of electrons. Because of the lone pair – bond pair repulsion, the tetrahedral bond angle is reduced to 107°. Hence, it assumes a pyramidal shape.

Question 18.
How is nitric acid prepared?
Nitric acid is prepared by heating equal amounts of potassium or sodium nitrate with concentrated sulphuric acid.
KNO3 + H2SO4 → KHSO4 + HNO3
The temperature is kept as low as possible to avoid decomposition of nitric acid.

Question 19.
Give reason for the following:
The nitric acid prepared by heating potassium nitrate and conc. H2SO4 is brown coloured.
The acid formed by heating KNO3 and H2SO4 dioxide formed by the decomposition of nitric acid.
4HNO3 → 4NO2 + 2H2O + O2

Question 20.
Explain the manufacture of Oswalld’s process of nitric acid.
The method is based as upon the catalystic oxidation of ammonia by atmospheric oxygen.

Nitric oxide thus formed readily combines with oxygen to form nitrogen dioxide (NO2).
2NO(g) + O2(g) → 2NO2(g)
Nitrogen dioxide, thus formed dissolves in water to form nitric acid.
3NO2 (g) + H2O (l) → 2HNO3 (aq) + NO (s)
The nitric oxide formed is recycled and the aqueous nitric acid is concentrated by distillation to give 68% nitric acid by mass. Further concentration to 98% can be achieved by dehydration with conc. H2SO4.

Question 21.
Explain with examples to show that nitric acid acts as an (i) acid, (ii) oxidising agent and (iii) as nitrating agent.
(i) It reacts with bases and basic oxides to form salts and water.

(ii) It oxidises non metals to their highest oxy acids.

(iii) As a nitrating agent, it forms nitronium ion (NO2+) in aromatic electrophillic substitution reactions.

Question 22.
Briefly explain the action of nitric acid on metals.
Nitric acid, both dilute and concentrated are powerful oxidising agents and metals are good reducing agents. The products of oxidation depending on the concentration of the acid and the nature of the metal.
(i) Gold, platinum, rhodium, iridium and tantalum do not react with nitric oxide.
(ii) Metals which are more electropositive than hydrogen (Na, K, Ca, Mg, Al, Mn, Zn, Cr, Cd, Fe, CO, Ni, Sn, Pb, etc.,) reduces HNO3 to a variety of products such as NO2, NO, NH3, NH4 NO3 and N2O, depending upon the conditions.

(iii) Magnesium and manganese are the only metals that produce H2 with very dilute (1-2%) HNO3.

(iv) More active metals like magnesium, zinc, tin and iron react with cold, dilute nitric acid to form ammonium nitrate.

(v) With hot, dilute nitric acid, ammonium nitrate thus formed undergoes decomposition to form nitrous oxide (N2O)

(vi) Metals like Zn, Mg, Bi, Pb, etc., form metallic nitrate and nitrogen dioxide when treated with conc. HNO3.

(vii) Similarly,

Exception: Tin forms metastannic acid H2SnO3 and NO2

(viii) Metals which are less electropositive than hydrogen, reduce nitric acid to NO2 or NO.

eg:

(ix) Metals like iron, chromium, nickel and aluminium because passive as treatment with cone. HNO3.
(x) Noble metals like gold, platinum do not react with conc. HNO3.

Question 23.
Give uses of nitric acid.

1. Nitric acid is used as a oxidising agent and in the preparation of aquaregia.
2. Salts of nitric acid are used in photography (AgNO3) and gunpowder for firearms. (NaNO3).

Question 24.
Complete and balance the following equations.

Question 25.
Give the oxidation states of nitrogen in the following oxides.
(i) N2O, (ii) NO, (iii) N2O3, (iv) NO2, (v) N2O4, (vi) N2O5.

Question 26.
Write the structure of the following oxides of nitrogen, (i) N2O, (ii) NO, (iii) N2O3, (iv) NO2, (v) N2O4, (vi) N2O5.

Question 27.
Give equation for the preparation of the following acids.
(i) Hyponitrous acid,
(ii) Nitrous acid,
(iii) Pemitrous acid,
(iv) Nitric acid,
(v) Pemitric acid.

Question 28.
Give reason for the following:
(i) Freshly prepared phosphorus becomes yellow on standing.
(ii) Yellow phosphorus glows in dark.
(iii) Nitrogen is a gas while phosphorus is a solid.
(i) This is due to the formation of a red phosphorus upon standing.
(ii) This is due to oxidation which is known as phosphorus.
(iii) The nitrogen has N ≡ N structure while in phosphorus four atoms in phosphorus have a polymeric structure with chains of P4 linked tetrahedrally. P ≡ P is less stable than P — P bonds. Hence nitrogen is a gas while phosphorus is a solid.

Question 29.
Write the structure of the following:
(i) Hyponitrous acid, (ii) Hydronitrous acid, (iii) Nitrous acid, (iv) Pemitrous acid, (v) Nitric acid, (vi) Pemitric acid.

Question 30.
Give the formula and the oxidation state of nitrogen in the following acids.
(i) Hyponitrous acid, (ii) Nitrous acid, (iii) Pemitrous acid, (iv) Nitric acid, (v) Pemitric acid.

Question 31.
Explain the structure of white phosphorus.
It exists as P4 units. The four sp3 hybridised phosphorus atoms lie at the corners of a regular tetrahedral with ∠PPP=60° as shown in figure.

Each phosphorus atom is linked to other three phosphorus atoms by covalent bonds so that each P-atom completes its octet.

Question 32.
Explain why white phosphorus is kept under water.
The P4 units of white phosphorus are held together by weak vanderwaals forces of attraction. As a result, its ignition temperature (303K) is very low and easily catches fire. Hence, it is kept under water.

Question 33.
The red phosphorus is less reactive than white phosphorus. Explain with examples.
(i) Yellow phosphorus readily catches fire in air forming phosphorus pentoxide whereas red phosphorus forms phosphorus trioxide or phosphorus pentoxide only on heating.

(ii) With chlorine, phosphorus forms phosphorus tri and penta chlorides. White phosphorus reacts violently at room temperature while red phosphorus react on heating.

(iii) Yellow phosphorus reacts with alkali an boiling in an inert atmosphere to form phosphine. Red phosphorus has no action an alkali.

Question 34.
Distinguish between white and red phosphorus interms of their properties.

Question 35.
Compare the chemical properties of white and red phosphorus.

Question 36.
How phosphine is formed from (i) White phosphorus, (ii) Calcium phosphide, (iii) Phosphorus acid, (iv) Phosphonium iodide. Give equations.
(i) By heating white phosphorus and sodium hydroxide in an inert atmosphere of carbon dioxide or hydrogen.

(ii) By the hydrolysis of calcium phosphide with water or dilute mineral acids.

(iii) By heating phosphorus acid (H3PO3)

(iv) By heating phosphorus iodide with sodium hydroxide.

Question 37.
Give equations for the following:
(i) Phosphine is heated at 317 K in the absence of air.
(ii) Phosphine is heated with oxygen.
(iii) Phosphine is treated with chlorine.

(iii) PH3 + 4Cl3 → PCl5 + 3HCl

Question 38.
Give examples for (i) basic nature and (ii) reducing property of phosphine.
(i) Basic nature of PH3: Because it contains a lone pair of electrons as phosphorus, it accepts a proton and form phosphorium ion.

It is a weaker base than NH3 because of larger size and lesser electronegativity of phosphorus than nitrogen.

(ii) Reducing property of PH3: Phosphine reduces aqueous solutions of copper and mercury salts to their corresponding phosphides.

Question 39.
Briefly explain the structure of phosphine.
The phosphorus in phosphine is sp3 hybridised. The three sp3 hybrid orbitals each containing an electron overlaps with 1s orbital of hydrogen, containing are electron form three P — H covalent bonds. The fourth sp3 hybrid orbital contains a lone pair of electron. The lone pair-bond pair interaction, due to the larger size and lesser electro negativity of phosphorus, compared to nitrogen, reduces the tetrahedral bond angle to 94°. Hence phosphine has a pyramidal shape, like ammonia.

Question 40.
What are Holmes signals? Mention its use.
Phosphine (PH3) is used in Holme’s signals in deep seas and oceans for signalling danger points to steamers. Containers containing a mixture of calcium phosphide and calcium carbide are pierced and thrown into the sea.
In contact with water, a mixture of acetylene and phosphine is produced. Phosphine contains traces of highly inflammable P2H4 which catches five spontaneously. This ignites acetylene with a luminous flame and thus serves as a signal to the approaching ship.

Question 41.
How is phosphorus trichloride prepared? [OR] What happens when white phosphorus is treated with a stream of chlorine gas.
P4 + 3Cl2 → P4Cl6

Question 42.
What happens when (i) White phosphorus is treated with thionyl chloride.
(i) White phosphorus is treated with thionyl chloride.
(ii) Phosphorus trichloride is hydrolysed by cold water.
(i) Phosphorus trichloride is formed.
4P4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2
(ii) Hydrolysis of phosphorus trichloride gives phosphorus acid. (H3PO3)
PCl3 + 3H2O → H3PO3 + 3HCl

Question 43.
How does phosphorus trichloride react with
(i) ethanol and (ii) propionic acid? Give equations.
The ‘OH’ group in alcohols and acids are replaced by ‘Cl’.

Question 44.
Explain the structure of phosphorus trichloride.
Its structure is similar to ammonia. P atoms undergoes sp3 hybridisation. In the tetrahedral configuration one of the positions is occupied by the lone pair. Thus, it is pyramidal in shape. The Cl — P — Cl bond angle is 100.4° and P — Cl bond length is 204 pm.

Question 45.
Give equations for (i) the formation of PCl5 from PCl3 (ii) H3PO4 from PCl5.

Question 46.
How does phosphorus pentachloride react with metals? Give examples.
Phosphorus penta chloride (PCl5) reacts with metals to give metal chlorides.

Question 47.
Briefly outline the structure of PCl5.
It has a trigonal bipyramidal shape in which P undergoes sp3d hybridisation.

Question 48.
Mention the uses of phosphorus pentachloride.
Phosphorus pentachloride is a chlorinating agent and is useful for replacing hydroxyl groups by chlorine atom.

Question 49.
Hypophosphorus acid is monobasic and a good reducing agent. Explain.
The structure of hypophosphorus acid is

It has One P — OH bond, two P — H and P → O or P = O bonds. Of the three hydrogen atoms, only one hydrogen atom attached to the oxygen atom (OH group) is replaceable. Hence it is a monobasic acid. Because of the presence of P — H bond, it acts as a reducing agent.

Question 50.
Briefly explain the structure of phosphorus trioxide and phosphorus pentoxide.
(i) Structure of phosphorus trioxide: The phosphorus atom lie at tetrahedral positions with respect to each other and six oxygen atoms are inserted between them. Each phosphorus atom is covalently bonded to three oxygen atoms and each oxygen is bonded to two phosphorus atom. The P — O bond length is 165.6 pm.

(ii) Structure of phosphorus pentoxide: In P4O10, each P atom form three bonds with oxygen atom an also an additional coordinate bond with an oxygen atom. Terminal coordinate P — O bond length is 143 pm.

Question 51.
What is the basicity of orthophosphorus acid? Explain with its structure. Will it act as a reducing agent or not?
The structure of orthophosphorus acid is

It has two P — P — OH bonds, one P → O or P = O bond, and one P — H bond. There are two ‘OH’ groups attached to the phosphorus atom, and the hydrogen atoms are replaceable.
Hence it is a dibasic acid. The acid and their salts are good reducing agent because of P —H bond.

Question 52.
Write the formula and structure of hypo phosphoric acid. What information do you get from its structure regarding its basicity and its reducing action?
The formula is H4P2O6 and its structure is

It contains ‘4’ OH group and 4 replaceable hydrogen atoms. Hence it is a tetrabasic acid or its basicity is four. It does not contain a P — H bond. Hence, it does not act as a reducing agent.

Question 53.
Orthophosphoric acid is tribasic and is not a reducing agent. Explain.
Orthophosphoric acid (H3PO4) is tribasic, because it contains three OH groups attached to the phosphorus atom where all the hydrogens are replaceable.
Hence, it is a tribasic acid.

It forms three series salt, by successive replacement of H atom of the ‘OH’ group,

These salts are known as dihydrogen phosphate, hydrogen phosphate and phosphate respectively. It is not a reducing agent because it does not contain a P — H bond.

Question 54.
Write the structure of pyrophosphoric acid and explain its basicity on the basis of the structure.
Phosphoric acid (H4P2O7) has the structure

It is a tetra basic acid.

Question 55.
Write the formula and the oxidation state of phosphorus in the following oxyacids.
(i) Hypophosphorus acid
(ii) Orthophosphorus acid
(iii) Hypophosphoric acid
(iv) Orthophosphoric acid
(v) Pyrophosphoric acid

Question 56.
Complete and balance the following equations:

Question 57.
Briefly explain the trend in physical properties of group 16 elements.
Group 16 elements consist of oxygen (O), sulphur (S), selenium (Se), tellurium (Te) and polonium (PO). They are also called oxygen family or chalcogens, because many metals occur as oxides and sulphides.

1. The electronic configuration of these elements are:
2. The atomic and ionic radii of group 16 elements are smaller than those of the corresponding elements of group 15. Further, the atomic radii increases down the group. The increase in atomic radii of group 16 elements in primarily due to increase in the number of electron shells.
3. The melting, boiling points and densities increase regularly as we go down the group upto tellunium. However, the melting and boiling points of polonium are lower than those of tellurium. This is because, the atomic size increases down the group. As a result, vanderwaals forces of attraction among their atoms also increase and hence melting and boiling points regularly increase from oxygen to tellurium. However due to presence of inner ‘d’ and ‘f’ electrons, the inert pair effect is maximum in polonium. Consequently, the ‘s’ valence electrons in polonium are less available as compared to those in tellurium for bonding. As a result, vanderwaals forces of attraction will be weaker in Po than in Te and therefore melting point and boiling point of Po will be lower than that of Te.
4. Oxygen exists as a diatomic gas while other elements exist as octa atomic solids.
5. All the elements of this group show allotropy. Oxygen exists in two non metallic forms viz dioxygen (O2), tri oxygen O3. Sulphur exists in crystalline as well as amorphous allotropic forms. The crystalline form includes rhombic sulphur (α sulphur) and mono clinic sulphur (β sulphur). Amorphous allotropic forms include plastic sulphur (γ sulphur), milk of sulphur, and colloidal sulphur.

Question 58.
Give two methods of preparation of oxygen in the laboratory.
In the laboratory, oxygen is prepared by one of the following methods.
The decomposition of hydrogen peroxide in the presence of a catalyst (MnO2) or by oxidation with potassium permanganate.

The thermal decomposition of certain metallic oxides or oxoanions gives oxygen.

Question 59.
Give reasons for
(i) Oxygen exists as a diatomic gas.
(ii) Oxygen is paramagnetic.
(iii) Oxygen forms strong hydrogen bonds.
(i) Due to small size and high electronegativity, oxygen forms pπ — pπ, double bond with another oxygen atom to form O = O molecule. The intermolecular forces of attraction between oxygen molecules are weak and hence oxygen exists as a diatomic gas.
(ii) The oxygen is paramagnetic because it contains two unpaired electrons in the antibonding molecular orbitals.
(iii) The high electronegativity of the oxygen atom is responsible to form hydrogen bonding eg: H2O.

Question 60.
How is ozone prepared in the laboratory?
Ozone is prepared in the laboratory by passing silent electrical discharge through oxygen. At a potential of 20,000 V about 10% of oxygen is converted into ozone it gives a mixture known as ozonised oxygen. Pure ozone is obtained as a pale blue gas by the fractional distillation of liquefied ozonised oxygen.

Question 61.
Write a note on the structure of ozone molecule.
The central oxygen atom in ozone is sp2 hybridised containing a lone pair of electrons. As a result, ozone has an angular structure.

It is actually a resonance hybrid of the following two resonating structural (I and II) as shown is figure.

Question 62.
How does oxygen react with metals and non metals? Give examples.
Oxygen reacts with metals and non metals and form oxides.
(i) Oxygen combines with many metals and non metals to form oxides.

(ii) Some less reactive metals react when powdered finely and made to react exothermically with oxygen at room temperature.

Question 63.
Explain why ozone is a more powerful
oxidising agent than oxygen.
This is due to the reason that ozone has higher energy content than oxygen and undergo decomposition.
2O3 (g) → O2 (g) + [O]
The atomic oxygen, thus liberated brings about oxidation.

Question 64.
Give an example for the oxidising action of ozone.
Ozone oxidises potassium iodide to iodine.

Question 65.
Explain how ozone can be estimated quantitatively.
Ozone gas is passed through an aqueous solution of potassium iodide. The liberated iodine is titrated with a standard solution of theo sulphate. From the volume of the standard solution of sodium theo sulphate, the amount of ozone can be calculated.

Question 66.
Mention the uses of oxygen.

1. Oxygen is one of the essential components for the survival of living organisms.
2. It is used in welding (oxy acetylene welding).
3. Liquid oxygen is used as fuel in rockets etc.

Question 67.
Name the stable allotrope of sulphur at ordinary temperature and pressure. What happens when it is heated to avoid 96°C?
Rhombic sulphur is the most stable allotrope of sulphur at ordinary temperature and pressure. When heated slowly at 96°C. It is converted to monoclinic sulphur.

Question 68.
Give the characteristics of rhombic and monoclinic sulphur.
Rhombic sulphur consist of yellow coloured crystals and made up of S8 molecules. Monoclinic sulphur also contains S8 molecules in addition to S6 molecules (small amount). It exists as a long needle like prism. It is stable between 96 – 119°C.

Question 69.
How is sulphur dioxide prepared from
(i) sulphur, (ii) galena, (iii) iron pyrites.
(i) Burning sulphur in air.
S + O2 → SO2

Question 70.
Give equation for the preparation of sulphur dioxide in the laboratory.

Question 71.
With an example, prove sulphur dioxide in an acidic oxide.
It dissolves in water producing sulphurous acid. (H2SO3)

It reacts with sodium hydroxide and sodium carbonate to form sodium hydrogen sulphate and sodium sulphate respectively.

Question 72.
Give two examples for the oxidising property of sulphur dioxide.
It oxidises hydrogen sulphide to sulphur and magnesium to magnesium oxide.

Question 73.
Give two examples to show that sulphur dioxide acts as a reducing agent.
(i) It reduces chlorine water to hydrochloric acid.

(ii) It reduces potassium permanganate and potassium dichromate to Mn+2 and Cr+3 respectively.

Question 74.
Give equation for the reaction in which sulphur dioxide is used for the manufacture of sulphuric acid by contact process.

Question 75.
Explain the use of sulphur dioxide as a bleaching agent.
In presence of water, sulphur dioxide bleaches coloured wool, silk, sponges and straw into colourless due to its reducing property.

However, the bleached product (colourless) is allowed to stand in air, it is reoxidised by atmospheric oxygen to its original colour. Hence bleaching action of sulphur dioxide is temporary.

Question 76.
Draw the structure of sulphur dioxide.
In sulphur dioxide, sulphur atom undergoes sp2 hybridisation. A double bond arises between S and O is due to pπ- dπ overlapping.

It is a resonance hybrid of the canonical form I and II.

Question 77.
Mention the uses of sulphur dioxide.

1. Sulphur dioxide is used in bleaching hair, silk, wool etc…
2. It can be used for disinfecting crops and plants in agriculture.

Question 78.
Discuss the various steps involved in the manufacture of sulphuric acid by contact process?
The contact process involves the following steps.

1. Initially sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen/air.
2. Sulphur dioxide formed is oxidised to sulphur trioxide by air in the presence of a catalyst such as V2O5 or platinised asbestos.
3. The sulphur trioxide is absorbed in concentrated sulphuric acid and produces oleum (H2S2O7). The oleum is converted into sulphuric acid by diluting it with water.

To maximise the yield the plant is operated at 2 bar pressure and 720 K. The sulphuric acid obtained in this process is over 96 % pure.

Question 79.
How do you prove that sulphuric acid is dibasic acid?
The structure of sulphuric acid is

It has two replaceable hydrogen atoms. Hence, it forms two types of salts viz bi sulphates and sulphates. Hence H2SO4 is a dibasic acid.

Question 80.
Give two examples to show that cone, sulphuric acid is an oxidising agent.
(i) It oxidises carbon to carbon dioxide and gets reduced to sulphur dioxide

(ii) It oxidises hydrogen sulphide to sulphur. It gets reduced to sulphur dioxide.

Question 81.
Complete and balance the following equations.

Question 82.
Give a brief account of the action of sulphuric acid on metals.
Sulphuric acid reacts with metals and gives different products depending on the reactants and reacting condition. Dilute sulphuric acid reacts with metals like tin, aluminium, zinc to give corresponding sulphates.

Hot concentrated sulphuric acid reacts with copper and lead to give the respective sulphates as shown below.

Sulphuric acid doesn’t react with noble metals like gold, silver and platinum.

Question 83.
What happens when
(i) Potassium chloride is heated with concentrated sulphuric acid.
(ii) Potassium nitrate is heated with concentrated sulphuric acid.
(iii) Sodium carbonate is heated with dilute sulphuric acid.
(iv) Sodium bromide is heated with cone, sulphuric acid?
(i) Hydrogen chloride is produced.
KCl + H2SO4 → KHSO4 + HCl
(ii) Sulphuric acid, being a strong acid displaces relatively weaker nitric acid from its salt.
KNO3 + H2SO4 → KHSO4 + HNO3
(iii) It decomposes sodium carbonate to carbon dioxide and water.

(iv) It oxidises Br to Br2.

Question 84.
How will you detect sulphate radical in qualitative analysis.
Dilute solution of sulphuric acid/aqueous solution of sulphates gives white precipitate (barium sulphate) with barium chloride solution. It can also be detected using lead acetate solution. Here a white precipitate of lead sulphate is obtained.

Question 85.
Write the formula and the oxidation state of sulphur in the following oxoacids.
(i) Sulphurous acid,
(ii) Sulphuric acid,
(iii) Thio sulphuric acid,
(iv) Peroxy mano sulphuric acid,
(v) Peroxydithionic acid,
(vi) Dithionic acid,
(vii) Pyrosulphuric acid.

Question 86.
Write the structure of the following:

1. Sulphurous acid,
2. Sulphuric acid,
3. Thiosulphuric acid,
4. Dithionous acid,
5. Pyro sulphuric acid,
6. peroxy mono sulphuric acid,
7. Peroxy di sulphuric acid,
8. Dithionic acid,
9. Polythionic acid.

Question 87.
Briefly outline the trend in the physical properties of halogens.

1. Fluorine, chlorine, bromine, iodine and astatine constitute group 17 elements or halogens.
2. The electronic configuration of these elements are
3. Halogens have the smallest atomic radii in their respective periods. Both atomic and ionic radii increase from fluorine to iodine as the number of shells increases.
4. All halogens exist as diatomic molecules in the elemental state. The different molecules of halogens are held together by vanderwaals forces of attraction. The strength of vanderwaals forces increases as. the size of the halogen atom increases from fluorine to iodine. As a result, F2 and Cl2 are gases, at room temperature, Br2 is a liquid, whereas I2 is a solid. Due to increase in vanderwaals force of attraction increases down the group, melting and boiling points increase with increase in atomic mass from fluorine to chlorine.
5. All halogens are highly electronegative and electronegativity decreases down the group.

Question 88.
How is chlorine prepared from (i) NaCl (ii) HCl (iii) bleaching powder?
(i) By the action of conc.H2SO4 in the presence of manganese dioxide an sodium chloride.

(ii) Oxidation of hydrochloric acid using oxidising agents such as MnO2, PbO2, KMnO4 or K2Cr2O7.

(iii) By treating to bleaching powder with a mineral acid.

Question 89.
Complete and balance the following equations.

Question 90.
Briefly outline the manufacture of chlorine by
(i) Electrolytic process, (ii) Deacon’s process.
(i) Electrolytic process: When a solution of brine (NaCl) is electrolysed, Na+ and Cl ions are formed. Na+ ion reacts with OH ions of water and forms sodium hydroxide. Hydrogen and chlorine are liberated as gases.

(ii) Deacon’s process: In this process, a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed. Hot gases at about 723 K are passed through a jacket that surrounds the chamber.

The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is given below,

Question 91.
Give example for the reaction of chlorine an (i) Aluminium, (ii) Sulphur, (iii) boron, (iv) arsenic, (v) Phosphorus.

Question 92.
What happens when (i) chlorine is treated with excess ammonia (ii) ammonia is treated with an excess of chlorine? Give equation.
(i) With excess ammonia, nitrogen gas is evolved.

(ii) With excess of chlorine, ammonium chloride is formed.

Question 93.
Give examples for the oxidising power of chlorine.
Chlorine oxidises hydrogen sulphide to sulphur and liberates bromine and iodine from iodides and bromides. However, it doesn’t oxidise fluorides

Question 94.
What happens when chloride is treated with (i) cold, dilute sodium hydroxide and (ii) hot, concentrated solution of sodium hydroxide? Give equations.
(i) When chlorine is treated with cold, dilute solution of sodium hydroxide, sodium chloride, sodium hypochlorite and water are formed.

(ii) When chlorine is treated with hot, concentrated solution of sodium hydroxide, sodium chloride, sodium chlorate and water are formed.

Question 95.
Explain the bleaching action of chlorine.
Chlorine bleaches by oxidation

Colouring matter + Nascent oxygen → Colourless oxidation product The bleaching of chlorine is permanent.

Question 96.
Complete and balance the equation:

Question 97.
How is bleaching powder prepared? Give equation.
Bleaching powder is produced by passing chlorine gas through dry slaked lime (calcium hydroxide).

Question 98.
Give a brief account of displacement reactions of halogens.
In general, the halogen of lower atomic number will.displace the halide ion of higher atomic number.
Fluorine displaces other halogens from their corresponding halides.

chlorine displaces Br and I ions from their solutions and bromine displaces I ion from their solutions.

Question 99.
Mention the uses of chlorine.
Uses of chlorine:

1. Purification of drinking water.
2. Bleaching of cotton textiles, paper and rayon.
3. It is used in the extraction of gold and platinum.

Question 100.
How is hydrochloric acid prepared?
Hydrochloric acid is prepared by heating sodium chloride with concentrated sulphuric acid and dissolving the hydrogen chloride formed in water.

Question 101.
How does hydrochloric acid react with (i) Zn, (ii) Na2CO3, (iii) Na2SO4? Give equation.
(i) Hydrogen gas is liberated.
Zn + 2HCl → ZnCl2 + H2
(ii) Sodium carbonate is decomposed.
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
(iii) Liberates sulphur dioxide from sodium sulphate.
Na2SO4 + 2HCl → 2NaCl + H2O + SO2.

Question 102.
What is aquaregia? Mention its uses.
A mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio 3:1 is known as aquaregia. This is used for dissolving gold, platinum,

Question 103.
Mention the uses of hydrochloric acid.
Uses of hydrochloric acid:

1. Hydrochloric acid is used for the manufacture of chlorine, ammonium chloride, glucose from com starch etc.,
2. Extraction of glue from bone and for purification of bone black.

Question 104.
Briefly outline the trend in physical and chemical properties of hydrogen halides.

1. At room temperature, hydrogen halides are gases but hydrogen fluoride can be readily liquefied. The gases are colourless but, with moist air gives white fumes due to the production of droplets of hydrohalic acid. In HF, due to the presence of strong hydrogen bond it has high melting and boiling points. This effect is absent in other hydrogen halides.
2. Stability: The bond strength H—X decreases from HF to HI. Thus, HF is the most stable while HI is the least stable. The decrease is stability is due
to decrease in electronegativity from fluorine to iodine. This is reflected in the values of dissociation energy of H—X bond.
3. Volatility of the hydrides: A more volatile liquid must have a lower boiling point. The volatility of the hydrides shows the order HF < HI < HBr < HCl. The boiling point of HF is the highest due to extensive hydrogen bonding. As we move from HCl to HI, then boiling points show a regular increase due to a corresponding increase in the magnitude of vanderwaals force of attraction as the size of the halogen increases.
4. Thermal stability: Thermal stability increases in the order HI< HBr < HCl < HF. Thermal stability is directly proportional to the bond dissociation energy. Since the bond dissociation energy of HF is the highest and that of HI is the least, therefore HF is the most stable halogen acid while the HI is the least stable halogen acid.
5. Acid Strength: Aci(i strength increases in the order HF < HCl < HBr < HI. The strength of an acid depends upon its degree of ionisation which, in turn, depends upon the bond strength.

Thus higher the bond dissociation energy, lower is its degree of ionisation and weaker in the acid. Since bond dissociation energies of halogen acids increases in the order HI < HBr < HCl < HF the strength of acids increases in the reverse direction, i.e., HF < HCl < HBr < HI.

Question 105.
How are hydrogen fluoride and hydrogen chloride prepared from CaF2 and NaCl respectively? Give equation.
HF and HCl are prepared by heating fluorides and chlorides, respectively with cone. H2SO4.

Question 106.
Hydrogen bromide and hydrogen iodide cannot be prepared by treating their bromides and iodides with cone. H2SO4. Why?
HBr and HI are reducing agent and H2SO4 is an oxidising agent. The HBr or HI formed is oxidised to bromine or iodine.

Question 107.
Give a method of preparation of hydrogen bromide and hydrogen iodide from sodium bromide and sodium iodide respectively.
Phosphoric acid is used instead of cone H2SO4, because, conc H2SO4 oxidise HBr and HI formed to Br2 and I2 respectively.

Question 108.
How is hydrogen bromide prepared from red phosphorus?
To a paste of red phosphorus and water, bromine is added drop wise.

Question 109.
Give a method of preparation of hydrogen iodide from red phosphorus.
By adding water drop wise to a mixture of red phosphorus and iodine.

Question 110.
Hydrogen fluoride has a high melting and boiling point compared to other hydrogen halides. Give reason.
In hydrogen fluoride, strong halogen bonding is present. This type of hydrogen bonding is not present in other halides. Hence, hydrogen fluoride has high melting or boiling point.

Question 111.
Fluorine forms hydrogen dihalides while other hydrogen halides do not form hydrogen dihalides. Give reason.
HF is a stronger acid at high concentration

At high concentrations, the equilibrium involves the removal of fluoride ions is important, since, it affects the dissociation of hydrogen fluoride and increases the concentration of hydrogen ions. Hence, salts like NaHF2, KHF2, NH4HF2 are known. The other hydrogen halides do not form hydrogen dihalides.

Question 112.
Mention a reaction which is given by hydrogen fluoride alone, but not other hydrogen halides.
Moist hydrofluoric acid (not dry) rapidly react with silica and glass.

Question 113.
What is etching? Explain with an example.
Glass being a mixture of sodium and calcium silicates react with hydro fluoric acid and forming sodium and calcium fluorosilicates respectively.

The etching of glass is based on these reactions. Thus, when hydrofluoric acid is added to glass, that portion is affected due to the formation of sodium and calcium fluorosilicates. This phenomenon is known as etching.

Question 114.
Compare to reducing power of the hydrogen halides.
The reducing nature increases from HF to HI as the stability decreases from HF to HI. HF does not show any reducing nature. It cannot be oxidised by even by strong oxidising agents. HI is the strongest reducing agent.
It aqueous solution gets oxidised even by atmospheric oxygen.
The reducing action can be explained on the basis of increasing sizfe of the halide ion from F to I. The bigger ion can lose electron easily. HCl can be oxidised by strong oxidising agents like MnO2, KMnO4, K2Cr2O7
PbO2, Pb3O4 etc.
HBr acts as a strong reducing agent than HCl.
It can be oxidised by H2SO4 and atmospheric oxygen also

HI is the strongest reducing agent. It reduces H2SO4 to SO2, S, H2S, nitric acid to NO2, nitrous acid to NO, etc.

Question 115.
Mention the conditions for the formation of inter halogen compounds.

1. The central atom will be the larger one
2. It can be formed only between two halogen and not more than two halogens.

Question 116.
Briefly outline the structure of interhalogens based on VSEPR theory,
(i) AB type,
(ii) AB3 type,
(iii) AB5 type,
(iv) AB7 type.
(i) AB type: (eg: ClF, Br F, etc). In these molecules, the bigger atom undergoes sp3 hybridisation.

These hybrid orbitals contain one lone pair of electrons each,while the fourth orbital contain one electron. This hybrid orbital overlaps axially with the atomic orbital of B atom forming sigma bond. Thus theshape of the molecule is linear.
(ii) AB3 type: The central atom A undergoes sp3d hybridisation. The result is trigonal bi-pyramid structure.

Out of the five hybrid orbitals, two (equatorial) contain lone pair of electrons while the remaining three contain one electron each. The singly occupied hybrid orbitals overlap with the singly filled ‘p’ orbitals of an three B atoms forming three sigma bonds. The shape of the molecule AB3 is slightly bent T shaped with bond angles equal to 87.5°. eg: ClF3
(iii) Structure of AB5 type: The central atom A undergoes sp3d2 hybridisation.

The result is an octahedral configuration one hybrid orbital contains a lone pair of electrons, while the remaining five have one electron each. Those singly occupied orbitals overlap with orbital of B atoms forming five sigma bonds. B atoms utilize their singly filled p orbital. The shape of the molecule is AB5 is square pyramidal. eg:BrF5
(iv) AB7 type: The central atom A undergoes sp3d3 hybridisation.

The structure is pentagonal bipyramid. Each of these seven hybrid orbitals is singly filled. Each of these hybrid orbitals overlaps with the singly filled ‘p’ orbitals of B atoms forming seven sigma bonds. Thus, the molecule, AB, has a pentagonal bi-pyramidal structure, eg: IF7

Question 117.
What happens when (i) BrF5 and (ii) ICl are treated with NaOH? Give equation.
When heated with alkalies, the larger halogen form oxohalogen and the smaller forms halide ion.

Question 118.
Write a short note on oxides of halogens?

1. Fluorine form two oxygen compounds F2O and F2O2. These compounds are not called oxides but oxygen fluorides as fluorine is more electronegative than oxygen. The compounds of the rest of the halogens are termed as oxides.
2. Halogens and oxygen do not combine readily with each other. The oxides of halogens are obtained indirectly.
3. The important compounds of the halogens are listed below.

In these compounds, except fluorine (-1, oxidation state) all the other halogens have positive oxidation states.
4. Most of the halogen oxides are unstable and tend to explode when kept an standing or exposure to light.
5. The iodine oxides are the most stable than chlorine oxides, but all the oxides of bromine decompose below the room temperature. The higher oxidation states are more to stable than the lower oxidation states.
6. Oxygen fluorides do not form oxo acids. The oxides of other halogens are acidic. Acidic nature increases as the percentage of oxygen increases, i.e, Cl2O is less acidic than Cl2O7.
7. All oxides are good oxidising agent. I2O5 is a very good oxidising agent.

Question 119.
Write a short note on oxy acids of halogens.
Fluorine does not form any oxy acid as it is more electronegative than oxygen. Other halogens forms oxyacids of the type, HXO (hypohalous acid, +1 oxidation state), HXO2 (halous acid, + 3 oxidation state), HXO3 (halic acid, + 5 oxidation state, and HXO4 (perhalic acid, + 7 oxidation state. Some of these acids are unstable and are known 6nly in solutions or their salts.

Question 120.
Write a brief account of the trends in properties of group 18 (noble gases) elements.

1. Helium, neon, argon, krypton, xenon and Radan are the elements belong this group.
2. Noble gases have the largest ionisation energy in a given period, as they have a completely filled orbital in their outer most shell. The first ionisation energy decreases from helium to radan.
3. The electronic configuration of these elements are
4. Their atomic radius is known as vanderwaals radii. It is the highest in any period. The vanderwaals radii increases down the group.
5. All have low values of melting and boiling points. These values increase gradually as atomic number increases. The low melting points is due to weak vanderwaals forces present between the atoms of the noble gases in liquid and solid states. These weak molecular forces, however, increase with increase in atomic size (mass) from He to Rn and therefore, the melting and boiling points increase gradually.

Question 121.
How are the following prepared? Give equations, (i) XeF2, (ii) XeF4, (iii) XeF6.
Xenon fluorides are prepare by direct reaction of xenon and fluorine under different conditions as shown below.

Question 122.
Explain what happens when
(i) When XeF6 is heated at 50°C in a sealed quartz vessel.
(ii) When vapours of XeF6 is treated with water vapour.
(iii) When XeF6 reacts with 2.5 M sodium hydroxide.

Question 123.
Give an example for the oxidising property of sodium perxenate.
It oxidises Mn+2 ion to MnO4 ion as shown by the equation.

Question 124.
Name the addition compounds formed by the Xenon difluoride.
Xenon difluoride forms addition compounds XeF2.2SbF5 and XeF2.2TaF5. Xenon hexa fluorides forms compound with boron and alkali metals. eg: XeF6.BF3, XeF6MF, M-alkali metals.

Question 125.
Explain the structure of Xenon fluoride (XeF2).

1. The Xenon in XeF undergoes sp3d hybridisation and the expected geometry is trigonal bi-pyramid. It has three lone pairs and two bond pairs.
2. The Xenon and two fluorine atoms lie in a straight line while the three equatorial positions are occupied by lone pairs. The actual shape is linear.

Question 126.
Explain the structure of Xenon tetrafluoride (XeF4).
In the compound, Xenon is sp3d2 hybridised, i.e., The molecule is octahedral.
The Xenon and four fluorine atoms are coplanar, while two axial positions are occupied by the two lone pairs. The actual shape is square planar.

Question 127.
Explain the structure of Xenon hexa fluoride (XeF6).
In the formation of XeF6, the XenOn atom undergoes sp3d3 hybridisation which gives the molecule is pentagonal bipyramid structure. Six positions are occupied by fluorine atoms and one position is occupied by lone pair of electrons. Due to the presence of lone pair of electron, the actual shape is distorted octahedral.

Question 128.
Explain the structure of Xenon oxy difluoride (XeOF2).
It has a T shaped structure. Xenon undergoes sp3d3 hybridisation giving trigonal bipyramid configuration in which two position are occupied by lone pairs.

Question 129.
Explain the structure of Xenon oxy tetra fluoride (XeOF4).
XeOF4 has a square pyramidal shape as shown below.

This shape results from sp2d3 hybridisation of Xenon atom. One position is occupied by a lone pair of electrons.

Question 130.
Explain the structure of Xenon trioxide (XeO3).
The Xenon undergoes sp2 hybridisation. The structure is tetrahedral. Three oxygen atoms occupy three comers of a tetrahedral forming a sigma and a pi bond. Each with a hybrid and ‘d’ orbital of the xenon atom. The actual shape of the molecule is pyramidal as one position of a tetrahedron is occupied by lone pair.

Question 131.
Mention the uses of Xenon.

1. Xenon is used in fluorescent bulbs, flash bulbs and lasers.
2. Xenon emits an intense light in discharge tubes instantly. Due to this it is used in high speed electronic flash bulbs used by photographers.

Question 132.

1. Radon is radioactive and used as a source of gamma rays.
2. Radon gas is sealed as small capsules and implanted in the body to destroy malignant i.e., cancer growth.

1. Which of the following is tribasic?
(a) H2PO2
(b) H3PO3
(c) H4P2O7
(d) H3PO4
(d)
Hint: The structure of H3PO4 is

It has three replaceable hydrogen and hence it is tribasic.

2. Which of the following oxides of nitrogen is thermally most stable?
(a) N2O5
(b) NO2
(c) NO
(d) N2O
(c)

3. P2O5 is extensively used as:
(a) reducing agent
(b) preservative
(c) oxidising agent
(d) dehydrating agent
(d)

4. The range of oxidation states shown by phosphorus is from:
(a) – 3 to + 5
(b) – 3 to 0
(c) 0 to +5
(d) – 4 to +2
(a)

5. The structural formula of hypo phosphorus acid is:

(a)

6. Which of the following oxides is the most acidic?
(a) P2O5
(b) N2O5
(c) Sb2O3
(d) AS2O3
(b)
Hint: Down the group acidic character decreases.

7. In white phosphorus molecule (P4) which one is not correct?
(a) Six P – P single bonds are present
(b) Four P – P single bonds are present
(c) Four lone pairs of electrons are present
(d) P – P – P bond angle is 60°
(b)

8. In the preparation of sulphuric acid, V2O5 is used as a catalyst in the reaction?
(a) S + O2 → SO2
(b) 2SO2 + O2 → 2SO3
(c) SO3 + H2O → H2SO4
(d) none of these
(b)

9. HCOOH reacts with conc. H2SO4 to produce:
(a) CO
(b) CO2
(c) SO2
(d) SO3
(a)

10. The geometry of H2S and its dipole moment are:
(a) angular and non zero
(b) angular and zero
(c) linear and non zero
(d) linear and zero
(a)

11. Among H2O, H2S, H2Te, H2Se, the one with maximum boiling point is:
(a) H20 because of hydrogen bonding
(b) H2Te because of higher molecular mass
(c) H2S because of hydrogen bonding
(d) H2Se because of lower molecular mass
(a)

12. Which of the following has the highest bond energy?
(a) O – O
(b) S – S
(c) Se – Se
(d) Te – Te
(a)

13. Which of the following reaction is not feasible?
(a) 2KI + Br2 → 2KBr + I2
(b) 2KBr + I2 → 2KI + Br2
(c) 2KBr + Cl2 → 2KCl + Br2
(d) 2H2O + F2 → 4HF + O2
(b)
Hint: F2 can displace Cl2, Br2 and I2 from HCl, KBr and KI.
Cl2 can displace Br2 and I2 from KBr and KI.
Br2 can displace only I2 from KI.
I2 can displace none.

14. [X] + H2SO4 → [Y]. a colourless gas with initiating smell.
[Y] + K2Cr2O7 + H2SO4 → green solution
[X] and [Y] are:

(a)
Hint: Sulphides on treatment with cone H2SO4, produce SO2 gas, which reduces acidified K2Cr2O7 to green Cr2 (SO4)3 solution.

15. Which of the following chemical reactions depicts the oxidising nature of conc. H2SO4?

(a)

16. The correct order of reactivity of halogens is:
(a) F > Br > Cl > I
(b) F > Cl > Br > I
(c) I > Br > Cl > F
(d) Br > Cl > F > I
(b)

17. Which one of the following arrangements does not truly represent the property against it?
(a) Br2 < Cl2 < F2 Electronegativity
(b) Br2 < F2 < Cl2 Electron affinity
(c) Br2 < Cl2 < F2 Bond energy
(d) Br2 < Cl2 < F2 Oxidising power
Hint: F2 < Cl2 < Br2

18. Which products are expected from the disproportionation reaction of hypochlorus acid?
(a) HClO3 and Cl2O
(b) HClO2 and HClO3
(c) HCl and Cl2O
(d) HCl and HClO3
Hint- 3HOCl → 2HCl + HClO3

19. Among the halogens, the one which is oxidised by nitric acid is:
(a) F
(b) Cl
(c) Br
(d) I
(d)

20. Sea divers go deep in sea water with a mixture of the following gases?
(a) O2 and Ar
(b) O2 and He
(c) CO2 and Ar
(d) O2 and CO2
(b)

21. Among the following molecules:
(i) O2
(ii) XeOF4
(iii) XeF6
those having same number of lone pairs on Xe are:
(a) (i) and (ii) only
(b) (i) and (iii) only
(c) (ii) and (iii) only
(d) in all

22. As compared to nitrogen, oxygen is:
(a) less electronegative and less reactive
(b) more electronegative and less reactive
(c) more electronegative and more reactive
(d) less electronegative and more reactive
(c)

23. Among the following the pair in which two species are not isostructural is:
(a) SiF4 and SF4
(b) IO3 and XeO3
(c) BH4 and NH4+
(d) PF6 and SF6
(a)
Hint: SiF4 is tetrahedral while SF4 is square pyramidal.

24. Which one of the following statements is correct?
(a) The bond dissociation energy of fluorine is less than chlorine.
(b) Pene HBr can be prepared by treatment of NaBr with conc. H2SO4
(c) Hydrazine (N2H4) is a stronger base than NH3
(d) H2S is a weaker acid than H2O
(c)

25. Match the compounds in list I with list II.

Code:

(d)

26. Which of the following two are iso structural?
(a) XeF2, IF2
(b) NH3, BF3
(c) CO3-2, SO3-2
(d) PCl5, ICl5
(a)

27. Assertion (A): Nitrogen molecule is less reactive than molecular oxygen.
Reason (R): The bond length of N2 is shorter than that of oxygen.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
(a)

28. Assertion (A): HClO4 is a stronger acid than HClO3.
Reason (R): The oxidation state of chlorine in HClO4 is +5 and in HClO3 is +7.
(a) Both assertion and reason are true and reason is the correct«explanation of assertion.
(b) Both assertion and reason are true, but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
(c)
Hint: The oxidation state of chlorine in HClO4 is +7 and that in HClO5 is +5.

29. Which reactions are used in the preparation of halogen acid?

(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (iii) and (iv) only
(d) (i) and (iv) only
(c)

30. Iodine cannot form the ion:
(a) I+2
(b) I
(c) I+3
(d) I+
(a)

31. Xenon forms compounds with fluorine under different conditions. The known fluorides are:
(i) XeF
(ii) XeF2
(iii) XeF3
(iv) XeF4
(a) (i) and (iv) only
(b) (ii) and (iv) only
(c) (ii) and (iii) only
(d) (i) and (iii) only
(b)

## TN Board 12th Chemistry Important Questions Chapter 2 p-Block Elements – I

Students get through the TN Board 12th Chemistry Important Questions Chapter 2 p-Block Elements – I which is useful for their exam preparation.

## TN State Board 12th Chemistry Important Questions Chapter 2 p-Block Elements – I

Question 1.
Briefs outline the electronioconfiguration and oxidation state of block elements.

1. They have the general electronic configuration ns2, np1-6.
2. The elements of group 18 have completely filled ‘p’ orbitals and hence more stable than rest of the elements in the block.
3. They show variable oxidation state. The lower oxidation state corresponds to the loss of all the electrons and the higher oxidation state corresponds to the loss of both ‘s’ and ‘p’ electrons.
4. Coming down the group, the stability of the lower oxidation state increase and that of the higher oxidation state increases due to the inert pair effect.
5. Halogens have high electrons affinity values and have show -1 oxidation state.

Question 2.
Explain how does metallic character of ‘p’ block elements very down the group.
Metallic character depends on the tendency to lose an electron by the metal. The magnitude of ionisation energy decides whether an elements is a metal or not. Lower the value of ionisation energy, greater is the electropositive or matellic character. As the ionisation energy decreases down the group, metallic character increases down the group.

Question 3.
What is a metalloid? Name the metalloids present among ‘p’ block elements.
Metalloids are those elements which possess properties intermediate between the metals and non metals. Boron is group 13, silicon and germanium in group 14, arsenic and antimony in group 15, and tellurium in group 16 are metalloids.

Question 4.
The ionisations enthalpy decreases from Boron to aluminium, but from aluminium to thallium only a marginal increase is observed. Mention the cause for this observation.
This is due to the presence of inner d and f-electrons which has poor shielding effect compared to s and p-electrons. As a result, the effective nuclear charge on the valance electrons increases. The attraction between the nucleus and valence electron increases. This results in marginal increase in IE from Al to Ga.

Question 5.
Account for the trend in ionisation enthalpy of group 15/16/18 elefments.
In these groups, the ionisation enthalpy decreases, as we move down the group. Here poor shielding effect of d- and f-electrons are overcome by the increased shielding effect of the additional p-electrons.

Question 6.
How does electronegativity vary from boron to thallium?
The electronegativity first decreases from boron to aluminium and their show a marginal increase. This is due to the fact the increase in the nuclear charge is compensated by the poor shielding effect of ‘d’ and ‘p’ electrons which results in small increase in atomic size.

Question 7.
Mention the causes for the anomalous behaviour of the first element in each group.
The anomalous behaviour is due to its

1. small size
2. high electonegativity and ionisation enthalpy and
3. non-availibity of ‘d’ orbital to expand the valence beyond form.

Question 8.
The first member of each group differs from rest of the members in their properties. Explain the statement with an example.
The reason for this behaviour is due to its small size, high electronegativity and non availablity of ‘d’ orbitals for bond formations.
For example, in group 14 elements, the carbon has a tendency for catenation. Except silicon to some extent, this tendency is not shown by germanium, tin and lead. In nitrogen group, nitrogen is a diatomic gas while the rest its members are solids. Among the halogens, fluorine is the strongest oxidising agent.

Question 9.
Explain the term inert pair effect with a suitable example.
‘p’ block elements generally exhibit two oxidation states. The lower oxidation state is obtained by the loss of ‘np’ electrons and the higher oxidation state is obtained by the loss of both ‘ns’ and ‘np’ electrons. For example, corbon family elements exhibit +2 and +4 oxidation state.
As we come down the group, the stability of lower oxidation state increases while that of higher oxidation state decreases. For example,

i.e., Pb+2 is more stable than Pb+4
Sn+2 is more stable than Sn+4.
This means

PbCl4 is less stable than PbCl2. Hence PbCl4 (Pb+4 ions is a good oxidising agent).
The reason for the inert pair effect is with the increase in nuclear charge, the two electrons in ‘ns’ orbital are firmly bound to the nucleus and becomes ‘inert’ towards bond formation.

Question 10.
Name the allotropes of (i) phosphorus, (ii) tin, (iii) carbon, (iv) silicon, (v) sulphur.

 Element Allotrops (i) phosphorus White phosphorus, Red phosphorus, Scarlet phosphorus, violet and black phosphorus. (ii) carbon Diamond, graphite graphene, fullerenes carbon nanotubes. (iii) tin Grey tin, white tin, rhombic tin, sigma tin. (iv) silicon Amorphous and crystalline silicon. (v) Sulphur Rhombus sulphur and monoclinic sulphur.

Question 11.
Briefly detail the trend in physical and chemical properties of boron family.

1. They have the general electronic configuration of ns2 np1.
2. The atomic radius decreases from boron to aluminium due to the increase in nuclear charge and then show a marginal increase from aluminium to thallium. This is due to the screening effect of the ‘d’ and ‘f’ electrons which outweigh the increase in nuclear charge.
3. Down the group, the elements Ga, In and Tl sho w inert pair effect, i.e., The stability of +1 oxidation states increases while that of +3 oxidation state decreases.
4. Boron is a non-metal, while the rest are metals. Their compounds are electrons deficient and act as lewis acids (electron-pair acceptors).
5. Boron differs from the rest of its members in its properties due to its small size, high electronegativity and absence of ‘cf orbitals.
6. All the elements form hydrides, chlorides, oxides, and nitrides.
7. The oxides of boron and aluminium are amphoteric, while the rest are basic.

Question 12.
Give the chemical equation for the following reactions.

1. Boron combines with chromium at 1500K.
2. Boron trichloride reacts with tungsten in the presence of hydrogen at 1500K gaseous.
3. Boron trifluoride is treated with sodium hydride at 450K.
4. Boron and chlorine are Ideated.
5. Boron and nitrogen are heated at high temperatures.
2B + N2 → 2BN

Question 13.
Give one method of preparation of boric anhydride.
Boron is heated with oxygen at 900K. Boric anhydride is formed.

Question 14.
How does boron react with (i) conc.H2SO4 and (ii) conc.HNO3 give equations.
Boron being a non-metal is oxidised by these oxidising agents to its oxy acid boric and (H3BO3).

Question 15.
How is sodium borate formed from boron? Give equation.
Sodium borate is formed when boron is fused with sodium hydroxide.

Question 16.
Mention the uses of boron.

1. Boron has the capacity to absorb neutrons. Hence, its isotope $$10_{\mathrm{B}_{5}}$$ is used as moderator in nuclear reactors.
2. Amorphous boron is used as a rocket fuel igniter.
3. Boron is essential for the cell walls of plants.
4. Compounds of boron have many applications. For example, eye drops, antiseptics, washing powders etc., contains boric acid and borax. In the manufacture of Pyrex glass, the boric oxide is used.

Question 17.
How is borax prepared from colemanite? Give equation.
The aqueous solution of colemanite (Ca2B6O11) is boiled with sodium carbonate.

Question 18.
All aqueous solution of borax is alkaline in nature. Explain.
The aqueous solution of borax is alkaline due to hydrolysis. Along with boric acid, a strong base NaOH is formed. The pH of the resulting solution is greater than 7. Hence the aqueous solution of borax is alkaline.

Question 19.
What happens when borax is heated? Give equations.
On heating, it first loses its water of crystallisation and on further heating given transparent sodium metaborate (NaBO2).

Question 20.
Give the equation for the reaction of an aqueous solution of borax with (i) HCl and (ii) H2SO4.

Question 21.
Explain how boric acid behaves as a monobasic acid?
Monobasic acids usually release a proton. But boric acid accepts a hydroxyl group instead of donating the proton.

Question 22.
Give equations for the reactions of boric acid with sodium hydroxide.

Question 23.
What happens when boric acid is heated? Give equations.
On heating, the 373K, boric acid gives metaboric acid and at 413K, it gives tetraboric acid. When heated at red hot, it gives a glassy mass of boric anhydride.

Question 24.
What happens when
(i) Boric acid is heated with calcium fluoride in the presence of conc. H2SO4.
(ii) Heated with soda ash? Give equations.
(i) When boric acid is heated with calcium fluoride in the presence of cone. H2SO4 boron trifluoride is formed.

(ii) When heated with sodium carbonate (soda ash), it gives borax.

Question 25.
Give a brief account of the structure of boric acid.
Boric acid has a two-dimensional layered structure. It consists of [BO2]3- unit and these are linked to each other by hydrogen bonds.

Question 26.
Mention the uses of boric acid.

1. Boric acid is used in the manufacture of pottery glazes, glass, enamels and pigments.
2. It is used as an antiseptic and as an eye lotion.
3. It is also used as a food preservative.

Question 27.
Give two methods of preparation of diborane.

1. By the reaction of iodine with sodium borohydride in of diglyme (a solvent).
2. Heating magnesium boride with HC1 a mixture of volatile boranes is obtained.

Question 28.
Give equations for the reactions between diborane with (i) oxygen (ii) LiH (iii) NH3.

1. Pure diborane does not react with air or O2 at room temperature. But impure diborane gives B2O3. The reaction is exothermic.
2. When treated with lithium hydride, it gives lithium borohydride.
3. When treated with excess ammonia, at low temperatures, it gives diborane diammonate. On further heating it gives borazole.

Question 29.
What is inorganic benzene? How is it prepared? Write its structure.
Borazole is called inorganic benzene.
Preparation of borazole:
When treated with excess ammonia, at low temperatures, it gives diborane diammonate. On further heating it gives borazole.

Structure of borazole:

Question 30.
Mention the uses of diborane.

1. Diborane is used as a high energy fuel for propellant.
2. It is used as a reducing agent in organic chemistry.
3. It is used in welding torches.

Question 31.
How is boron trifluoride prepared? Give equations.
(i) By heating boron trioxide with calcium fluoride in the presence of cone. H2SO4.

(ii) By treating boron trioxide with carbon and fluorine.
B2O3 + 3C+ 3F2 → 2BF3 + 3CO
(iii) By thermal decomposition of benzene diazonium tetrafluoroborate.

Question 32.
Boron trifluoride is a lewis acid. Explain.
Boron trifluoride is an electron-deficient compound. It accepts a pair of electrons from donor atoms such as nitrogens in NH3 and forms a coordinate covalent bond.

Question 33.
Discuss the structure of boron trichloride.
Boron in boron trichloride is sp2 hybridised. The three sp2 hybrid orbitals orient in space in such a way they make an angle of 120°. Each of the sp2 hybrid orbitals contains are unpaired electron. These orbitals overlap with chlorine atoms containing one unpaired electron and form a σ bond. Thus BCl3 is a planar molecule.

Question 34.
How do you convert boron trifluoride to fluoroboric acid? Give equations.
Boron trifluoride, on hydrolysis, gives boric acid. This on treatment with hydrogen fluoride formed gives fluoro boric acid (HBF4).

Question 35.
Give two methods of preparation of aluminium chloride.

1. Aluminium when treated with hydrochloric acid, aluminium chloride is formed.
2Al + 6HCl → 2AlCl3 + 3H2
2. Aluminium hydroxide, gives aluminium chloride on treatment with dilute hydrochloric acid.
Al(OH)3 + 3HCl → AlCl3 + 3H20

Question 36.
How is aluminium chloride prepared by McAfee process?
Aluminium chloride is obtained by heating a mixture of alumina and coke in a current of chlorine.
Al2O3 + 3C + 3Cl2 → 2AlCl3 + 3CO2
On an industrial scale, it is prepared by chlorinating aluminium around 1000 K.

Question 37.
Explain why an aqueous solution of aluminium chloride is acidic?
An aqueous solution of aluminium chloride is acidic due to hydrolysis. A strong acid HCl is produced. The resultant solution has a pH less than 7. Hence it is acidic.
AlCl3 + 3H2O → Al(OH)3 + 3HCl

Question 38.
Complete and balance the following equations.

Question 39.
Mention the uses of aluminium chloride.

1. Anhydrous aluminium chloride is used as a catalyst in Friedel’s Crafts reactions.
2. It is used for the manufacture of petrol by cracking the mineral oils.
3. It is used as a catalyst in the manufacture of dyes, drugs and perfumes.

Question 40.
What are alums? Give examples.
Alums are double salts of potassium sulphate and aluminium sulphate. It has a general molecular formula M’2SO4.M”2(SO4)3. 24H2O
eg:
Potash alum: K2SO4. Al2(SO4)3. 24H2O
Sodium alum: Na2SO4. Al2(SO4)3. 24H2O
Ammonium alum: (NH4)2SO4. Al2(SO4)3. 24H2O
Chrome alum: K2SO4. Cr2(SO4)3. 24H2O

Question 41.
Mention the uses of alum.

1. It is used for the purification of water.
2. It is also used for waterproofing and textiles.
3. It is used in dyeing, paper and leather tanning industries.
4. It is employed as a styptic agent to arrest bleeding.

Question 42.
Give a brief account of the trends in properties of carbon group elements.

1. Carbon group elements have the general electronic configuration ns2 np2.
2. The atomic radius increases from carbon to silicon due to the increase in nuclear charge and the added electron enters to next higher energy level than carbon. Hence attraction between the nucleus and the added electron decrease. This accounts for the increase in atomic radius from carbon to silicon. There is a marginal increase in the atomic radius from silicon to tin. This is because even though there is an increase in nuclear charge, the shielding effect of inner ‘d’ and ‘f’ electrons increases. Both these factors are responsible for a small increase in the atomic radii. The decrease in atomic radius from tin to lead, due to an increase in the shielding effect of inner electrons, counterbalances the increase in nuclear charge.
3. A similar trend is followed in the case of the ionisation enthalpy of these elements.
4. Carbon is non-metal, white silicon and germanium are metalloids and tin and lead are metals.
5. Carbon has a greater tendency for catenation than the other elements of these groups. The catenation property decreases down the group.
6. All these elements exhibit +2 and +4 oxidation states. The +2 state becomes more stable down the group than the +4 state due to the inert pair effect.
7. Carbon, silicon, germanium and tin exhibit allotropy.
8. All these elements form oxides. The acidic character of the oxides decreases while that of the basic character increases down the group.
9. They form tetrahalides CCl4 is not readily hydrolysed, while the silicon tetrachloride can be readily hydrolysed. The tetrachlorides of other elements of this group are ionic.

Question 43.
Give a brief account of the allotropes of carbon with specific reference to their uses.
Carbon exists in different allotropic forms viz, graphite, diamond, fullerene, carbon nanotubes and graphene.

1. Graphite is soft and is a good conductor of electricity.
2. Diamond is hard and is used for sharpening hard tools, cutting glasses, making bones, and rock drilling.
3. Carbon nanotubes find many applications in nanoscale electrontes, catelysis, polymers and medicine.

Question 44.
Explain the structure of graphite.
Graphite has two-dimensional sheet-like structure. Each sheet consists of a hexagonal net of sp2 hybridised carbon of carbon atom with C—C bond distance closer to the C—C bond distance in benzene. Each carbon atom, by making use of its sp2 hybrid orbitals forms three covalent bonds with the neighbouring carbon atom. The fourth electron present in the unhybridized ‘p’ orbital of each carbon atom forms a π bond. The n electrons are delocalised over the entire sheet. This accounts for its electrical conductivity. The sheets are held together by weak vander waals forces and can be readily cleaned. This accounts for its lubricant properly.

Question 45.
Briefly explain the structure of the diamond.
The carbon atoms in diamond are sp3 hybridised and bonded to four neighbouring carbon atoms by σ bonds. This results in a tetrahedral arrangement around each carbon atom that extends to a three-dimensional space lattice. All four valance electrons of carbon are involved in bonding and there is no free electrons and have to diamond is a non-conductor of electricity.

Question 46.
Write a short note on fullerenes.
Fullerenes are newly synthesised allotropes of carbon. These allotropes are discrete molecules such as C32, C50, C60, C70, C76 etc. These molecules have cage-like structures. The C60 molecules have a soccer ball-like structure and are called buckminsterfullerene or buckyballs. It has a fused ring structure consists of 20 six-membered rings and 12 five-membered rings. Each carbon atom is sp2 hybridised and forms three σ bonds & a delocalised π bond giving aromatic character to these molecules.

Question 47.
How is carbon monoxide produced on an industrial scale?
By the reaction of carbon with air. The carbon monoxide formed will contain nitrogen gas also and the mixture of nitrogen and carbon monoxide is called producer gas.

The producer gas is then passed through a solution of copper(I) chloride under pressure which results in the formation of CuCl(CO).2H2O. At reduced pressures, this solution releases pure carbon monoxide.

Question 48.
Complete the following equations.
(i) HCOOH + H2SO4
(ii) CO + Fe2O3
(iii) CO + H2

Question 49.
Give the equation for the preparation of propanal by oxo process.

Question 50.
What are carbonyls? Give examples.
Carbonyl is a compound that contains carbon monoxide as a ligand. It is a complex compound formed by a transition metal and carbon monoxide. The transition metal in carbonyls exists in a zero oxidation state.
eg: Nickel tetra carbonyl Ni(CO)4
Iron pentacarbonyl Fe(CO)5
Chromium hexacarbonyl Cr(CO)6.

Question 51.
Mention the uses of carbon monoxide.

1. An equimolar mixture of hydrogen and carbon monoxide – ‘water gas and the mixture of carbon monoxide and nitrogen – producer gas is important industrial fuels.
2. Carbon monoxide is a good reducing agent and can reduce many metal oxides to metals.
3. Carbon monoxide is an important ligand and forms a carbonyl compound with transition metals.

Question 52.
How is carbon dioxide produced industrially?
It is produced by burning coke in excess of air.
2CO + O2 → 2CO2

Question 53.
Give an equation for the preparation of carbon dioxide in the laboratory.
CaCO3 + 2HCl → CaCl2 + H2O + CO2

Question 54.
Give a brief account of the properties of carbon dioxide.

1. It is an acidic oxide. It produces carbonic acid with water.
2. At high temperatures, it acts as a strong reducing agent. It is reduced to carbon by metals like Mg.
CO2 + Mg → 2MgO + C
3. It is forms water gas with hydrogen.
4. At temperatures above 3100 K, it undergoes decomposition.

Question 55.
Mention the uses of carbon dioxide.

1. Carbon dioxide is used to produce an inert atmosphere for chemical processing.
2. Biologically, it is important for photosynthesis.
3. It is also used as a fire extinguisher and as propellant gas.
4. It is used in the production of carbonated beverages and in the production of foam.

Question 56.
Give equations for the preparation of silicon tetrachloride from (i) silica and (ii) silicon.

1. SiCl4 from SiO2:
SiO2 + 2C + 2Cl2 → SiCl4 + 2CO
2. SiCl4 from Si:
Si + 4HCl → SiCl4 + 2H2

Question 57.
What is the actions of moisture on silicon tetrachloride? Give equations.
In moist air, silicon tetrachloride is hydrolysed to give silica and hydrochloric acid.
SiCl4 + 4H2O → 4HCl + Si(OH)4

Question 58.
What is the action of moist ether on silicon tetrachloride?
Silicon tetrachloride is hydrolysed to produce linear perchloro siloxanes. They have the repeating unit. [Cl.(SiCl2O)n.SiCl3 where n = 1 to 6]

Question 59.
Explain the terms alcoholysis and ammonolysis taking silicon tetrachloride as an example.
Alcoholysis:
The chlorine in silicon tetrachloride can be substituted by nucleophiles such as OH, OR, etc., using suitable reagents. For example, it forms silicic esters with alcohols.

Ammonialysis:
Similarly silicon tetrachloride undergoes ammonialysis to form chlorosilazanes.

Question 60.
Mention the uses of silicon tetrachloride.

1. Silicon tetrachloride is used in the production of semiconducting silicon.
2. It is used as a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

Question 61.
(i) What are silicones?
(ii) Mention the various types of silicons?
(iii) Give the various properties of silicones.
(i)

1. Silicones or polysiloxanes are organosilicon polymers with general empirical formula (R2SiO).
2. These silicones may be linear or cross-linked polymers.
3. Linear polymers contain as a repeating unit.
4. Complex cross-linked polymers have the structure.

(ii) Types of silicones: The following are the types of silicones.

1. Linear silicones
2. Cyclic silicones

(iii) Properties of silicones:

1. All silicones are water repellant.
2. They are thermal and electrical insulators.
3. Chemically they are inert.

Question 62.
Explain the formation of straight-chain or linear silicones.
The straight-chain or linear silicone is formed by the hydrolysis of dialkyl dichloro silanes.
(R2SiCl2 where R is an alkyl group).

This compound further reacts with another molecule of R2Si(OH)2 and form linear dialkyl chloro silanes.

Question 63.
Explain the formation of a complex cross-linked polymer with a suitable example.
The hydrolysis of mono alkylchloro silanes gives a complex cross-linked polymer.

When all the ‘OH’ groups are removed as water molecules, acyclic or ring silicones are obtained.

Question 64.
What are synthetic rubber and synthetic resins?
Synthetic rubber is silicones that are bridged together by methylene (—CH2—) or similar groups. Synthetic resins are obtained by blending silicones with organic resins such as acrylic esters.

Question 65.
What are silicates? Give examples for various types of silicates.
Silicates are minerals that contain [SiO4]-4 tetrahedra units linked together in different patterns.
Examples for other silicates: Phenacite BeSiO4 Olivine (Fe/Mg)2 SiO4.
Examples for phyllosilicates: Thortveitite Sc2Si2O7
Examples for cyclic silicates: Beryl [Bl3Al2(SiO3)6]
Examples for chain silicate(pyroxenes) spodumene: LiAl(SiO3)2
Example for double chain silicates or amphiboles: Asbestos.
Examples for sheet silicates: Talc, mica etc. Examples for three-dimensional silicates: Quartz, feldspar, zeolites etc.

Question 66.
What are ortho silicates?
Ortho silicates contain discrete [SiO4]-4 tetrahedral units. The silicon atom is at the centre of the tetrahedra and the four oxygen atoms occupy the comers of the tetrahedra.

In phenacite Be2SiO4, Be+2 ions are tetrahedrally surrounded by O-2 ions.

Question 67.
What are pyrosilicates? How are they formed?
Pyrosilicates are silicates containing [Si2O7]-2 units. They are formed by joining two [SiO4]-4 tetrahedral units by sharing one oxygen atom.

Question 68.
What are cyclic silicates? How are they formed?
Silicates that contain (SiO3)n2n- ions which are formed by linking three or more tetrahedral SiO44- units cyclically are called cyclic silicates. Each silicate unit shares two of its oxygen atoms with other units.

Question 69.
How are pyroxenes formed? [OR]
Explain the formation of chain silicates.
These silicates contain [(SiO3)n]2n- ions ‘ formed by linking ‘n’ number of tetrahedral [SiO4]4- units linearly. Each silicate unit shares two of its oxygen atoms with other units.

Question 70.
Explain the formation of sheet or phyllo silicates.
Silicates that contain (Si2O5)n2n- are called sheet or phyllo silicates. In these, Each [SiO4]4- tetrahedron unit shares three oxygen atoms with others and thus by forming two-dimensional sheets. These sheets of silicates form layered structures in which silicate sheets are stacked over each other. The attractive forces between these layers are very week, hence they can be cleaved easily just like graphite.

Question 71.
What are amphiboles? How are they formed?
Amphiboles are double-chain silicates that contain (Si4O11)n6n- ions. There are two types,

1. These sharing 3 vertices and
2. Those sharing only 2 vertices. Examples: asbestos.

Question 72.
Briefly explain the structure of three-dimensional silicates.
Silicates in which all the oxygen atoms of [SiO4]4- tetrahedra are shared with other tetrahedra to form a three-dimensional network are called three dimensional or tecto silicates. They have general formula (SiO2)n.
eg: Quartz These tecto silicates can be converted into Three-dimensional aluminosilicates by replacing [SiO4]4- units by [AlO4]5- units. E.g. Feldspar, Zeolites etc.,

1. Among the following pairs of elements which act as semiconductors?
(a) C and Si
(b) Si and Ge
(c) B and Al
(d) B and Si
(b) Si and Ge

2. Among ‘p’ block elements which group show +6 oxidations state?
(a) Icosagens
(b) Tetragens
(c) Prictogens
(d) chalcogens
(d) chalcogens
Hint: Chalcogens are group 16 elements which has a general electronic configuration ns2np4. The loss of all ns and np electrons results in +6 oxidation state.

3. Choose the correct statement:
(a) There is an increase in ionisation energy down the group as a result.
(b) All the elements in group 13 are metals.
(c) Boron and silicon exhibit diagonal relationship.
(d) Boron trifluoride is readily hydrolysed.
(c) Boron and silicon exhibit diagonal relationship.

4. Aluminium (III) chloride is stable where as thallium (III) chloride is highly unstable this is due to:
(a) inert pair effect
(b) increase in metallic character down the group
(c) decrease in metallic character down the group
(d) aluminium chloride is covalent
(a) inert pair effect

5. Which of the following is used as a moderator in nuclear reacters?
(a) $${ }_{6} \mathrm{C}^{14}$$
(b) $${ }_{5} \mathrm{B}^{10}$$
(c) $${ }_{5} \mathrm{N}^{14}$$
(d) $${ }_{8} \mathrm{O}^{17}$$
(b) $${ }_{5} \mathrm{B}^{10}$$

6. An aqueous solutions of borox is:
(a) acidic
(b) basic
(c) neutral
(d) is acidic as well as basic
(b) basic
Hint: Borox hydrolysis to give a strong base
Na2B4O7 + 7H2O → 4H3BO3 + 2NaOH

7. Borax Bead test is used to identify:
(a) boron
(c) onetal cations

8. Boric acid is:
(a) a weak mono basic acid
(b) a strong tetrabasic acid
(c) weak mono acidic base
(d) a diabasic acid
(a) a weak mono basic acid

9. Boric acid is heated to red hot. The product obtained is:
(a) metaboric acid
(b) pyroboric acid
(c) Boron trioxide
(d) all
(c) Boron trioxide
Hint:

10. Choose the incorrect statement with regard to boric acid:
(a) The structure consists of [BO3]-3 units linked to each other by hydrogen bonds.
(b) It is a monobasic acid as it releases a proton.
(c) It is a monobasis acid, as it accepts a hydroxyl ion.
(d) Boric acid is used as an antiseptic.
(b) It is a monobasic acid as it releases a proton.

11. Which is not true with respect to the structure of diborane?
(a) Each boron atoms, are sp3 hybridised
(b) The form B—H bonds are two centre two electron bonds.
(c) The B—H—B bonds are three centre – three electron bonds.
(d) The B—H—B bonds are three centre- two electron bonds.
(c) The B—H—B bonds are three centre – three electron bonds.

12. Producer gas is a mixture of:
(a) carbon monoxide and nitrogen
(b) carbon monoxide and hydrogen
(c) carbon dioxide and nitrogen
(d) carbon dioxide and hydrogen
(a) carbon monoxide and nitrogen

13. The hydrolysis of CH3SiCl3 yields:
(b) cyclic polymer
(c) linear polymer
(d) silicols

14. Phenacite is:
(a) Be3SiO4
(b) an orthosilicate
(c) both (a) and (b)
(d) a cyclic silicate
(c) both (a) and (b)

15. Choose the incorrect statement:
(a) pyroxenes contain (SiO3)n2- ions formed by sharing two of its oxygen atoms with other units.
(b) Pyroxene is also known as chain silicates.
(c) spodumore [LiAl(SiO3)3] is an example of pyroxene.
(d) Pyroxenes are silicates which contain [SiO4]-4 units, where all the oxygen atoms are shared with other [SiO4]-4 tetrahedra.
(d) Pyroxenes are silicates which contain [SiO4]-4 units, where all the oxygen atoms are shared with other [SiO4]-4 tetrahedra.

## TN Board 11th Bio Zoology Important Questions Chapter 12 Trends in Economic Zoology

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 12 Trends in Economic Zoology which is useful for their exam preparation.

## TN State Board 11th Bio Zoology Important Questions Chapter 12 Trends in Economic Zoology

Question 1.
What is economic zoology?
Economic Zoology is a branch of science that deals with economically useful animals. It involves the study of application of animals for human welfare.

Question 2.
Classify animals based on their economic importance.
Based on the economic importance, animals can be categorized as:

1. Animals for food and food products
2. Economically beneficial animals.
3. Animals of aesthetic importance
4. Animals for scientific research.

Question 3.
Write the relationship between the soil and earthworm.
Vermiculture is the process of using earthworms to decompose organic food waste, into a nutrient-rich material capable of supplying necessary nutrients which helps to sustain plant growth.

Question 4.
What is vermicompost?
Vermicompost is the compost produced by the action of earthworms in association with all other organisms in the compost unit.

Question 5.
Why earthworms has got the names of “farmer’s friends’ and biological indicators of soil fertility?
The disposal of solid wastes (bio-degradable and non-biodegradable) remains a serious challenge in most of the countries.
Earthworms play a vital role in maintaining soil fertility; hence these worms are called as “farmer’s friends”. These are also called as “biological indicators of soil fertility”. The reason is that they support bacteria, fungi, protozoans and a host of other organisms which are essential for sustaining a healthy soil.

Question 6.
What is vermicast?
The breakdown of organic matter by the activity of the earthworms and its elimination from its body is called vermicast.

Question 7.
How are earthworms divided into groups?
Earthworms are divided into two major groups. The first group, the humus formers, dwell on the surface and feed on organic matter. They are generally darker in colour. These worms are used for vermicomposting. The second group, the humus feeders, are burrowing worms that are useful in making the soil porous, and mixing and distributing humus throughout the soil.

Question 8.
Differentiate endemic and exotic species of earthworms with example.
There are different endemic (native) species of earthworms cultured in India for vermicomposting such as Periyonyx excavatus, Lampito mauritii, Octochaetona serrata. Some earthworm species have been introduced from other countries and called as exotic species eg: Eisenia fetida, Eudrilus eugeniae.

Question 9.
How can we prepare vermicompost bed for culturing earthworms as well as preparing vermicompost and vermiwash?

1. Vermicompost bed may be selected on upland or an elevated level as it prevents the stagnation of water.
2. A cement pit of 3 x 2 x 1m size(L x W x D) over ground surface using bricks. The size of pit may vary as per availability of raw materials. Cement pot or well rings are practically good. Provision should be made for excess water to drain. The vermibed should not be exposed to direct sunlight and hence shade may be provided. The first layer of vermibed contains gravel at about 5 cm in height, followed by coarse sand to a thickness of 3.5 cm, which will facilitate the drainage of excess water.
3. Earthworms collected from native soil prefer a layer of local soil in their compost beds. If local soil earthworms are used, add a layer of native loamy soil for about 15 cm on top of the gravel sand layer and introduce earthworms into it.
4. The unit can now be loaded with digested biomass or animal dung such as cow dung that has lost its heat. The number of earthworms to be introduced in an unit depends on the size of the vermibed prepared. Earthworms such as Periyonyx. excavatus, Eisenia fetida or Eudrilus Eugenie are introduced on the top. Jute bags or cardboards or broad leaves are used to cover the unit. As worms require moisture, water management is most important for the survival of earthworms. Too little or too much of water is not good for the worms.
5. Earthworms release their castings on the surface. One can start harvesting this from the surface on noticing the castings on the surface. It may take several days for the entire biomass to be composted depending on the amount of biomass. When all the compost is harvested, earthworms can be handpicked by creating small conical heaps of harvested compost and leaving in sunlight for a few hours. The earthworms then move down and settle at the bottom of the heap as a cluster. Earthworms from the lower layers of the compost can be recovered and the worms can be transferred to new composting units.
6. Vermiwash is a liquid collected after the passage of water through a column of vermibed. It is useful as a foliar spray to enhance plant growth and yield.

Question 10.
What is vermiwash? From where can we: obtained it. What is the use of it?
Vermiwash is a liquid collected after the passage of water through a column of vermibed. It is useful as a foliar spray to enhance plant growth and yield. It is obtained from the burrows or drilospheres formed by earthworms. Nutrients, plant growth promoter substances and some useful microorganisms are present in vermiwash.

Question 11.
What are tne enemies of earthworm?
Earthworm enemies include ants, springtails, centipedes, slugs, mites, certain beetle larvae, birds, rats, snakes, mice, toads, and other insects or animals which feed on worms. The earthworm has a number of internal parasites including numerous protozoa, some nematodes, and the larvae of certain flies.

Question 12.
Define sericulture?
Sericulture is an agro -based industry, the term which denotes commercial production of silk through silkworm rearing.

Question 13.
Write the three main components of sericulture.
Sericulture is an agro-based industry comprising three main components:

1. cultivation of food plants for the silkworms,
2. rearing of silkworms, and
3. reeling and spinning of silk.

Question 14.
Write down the various steps of development involved in the life cycle of Bombyx Mori.
Life cycle of Bombyx mori: The adult of Bombyx mori is about 2.5 cm in length and pale creamy white in colour. Due to heavy body and feeble wings, flight is not possible i by the female moth. This moth is unisexual in nature and does not feed during its very short life period of 2-3 days. Just after emergence, male moth copulates with female for about 2-3 hours and if not separated, they may die after few hours of copulating with female. Just after copulation, female starts egg laying which is completed in 1 24 hours. A single female moth lays 400 to 500 eggs depending upon the climatic conditions. Two types of eggs are generally found namely diapause type and non-diapause type. The diapause type is laid by silkworms inhabiting the temperate regions, whereas silkworms belonging to subtropical regions like India lay non-diapause type of eggs. The eggs after ten days of incubation hatch into larva called as caterpillar. The newly hatched caterpillar is about 3 mm in length and is pale, yellowish-white in colour. The caterpillars are provided with well developed mandibulate type of mouth-parts adapted to feed easily on the mulberry leaves.
After 1st, 2nd, 3 rd and 4th moultings caterpillars get transformed into 2nd, 3rd, 4th and 5th instars respectively. It takes about 21 to 25 days after hatching. The fully grown caterpillar is 7.5 cm in length. It develops salivary glands, stops feeding and undergoes pupation. The caterpillars stop feeding and move towards the comer among the leaves and secretes a sticky fluid through their silk gland. The secreted fluid comes out through spinneret (a narrow pore situated on the hypopharynx) and takes the form of long fine thread of silk which hardens on exposure to air and is wrapped around the body of caterpillar in the forms of a covering called as cocoon. It is the white coloured bed of the pupa whose outer threads are irregular while the inner threads are regular. The length of continuous thread secreted by a caterpillar for the formation • of cocoon is about 1000-1200 metres which requires 3 days to complete. The pupal period lasts for 10 to 12 days and the pupae cut through the cocoon and emerge into adult moth.

Question 15.
Classify the races of Bombyx mori on the basis of its moulting.
On the basis of the moults which they undergo during their larval life, B. mori is divided into three races – tri-moulters, tetra- moulters and penta-moulters.

Question 16.
Write few species of silk moth and the types of silk obtained from them.

 Species of silkmoth Types of silk Bombyx mori Mulberry silk Antheraea assamensis Muga silk Antheraea mylitta Tassar silk Attacus ricini Eri silk

Question 17.
What is Voltinism? Classify the races of Bombyx mori or mulberry worm according to this.
The number of broods raised per year is called voltinism. Three kinds of races are recognized in mulberry silkworm – univoltine (one brood only), bivoltine (two broods only), and multivoltine (more than two broods).

Question 18.
Define Moriculture? What for this is done?
Mulberry leaves are widely used as food for silkworm Bombyx mori and the cultivation of mulberry is called as Moriculture.

Question 19.
Name the improved varieties of mulberry plants used for planting.
Presently improved mulberry varieties like Victory 1, S36, G2, and G4 which can withstand various agro-climatic and soil conditions are used for planting.

Question 20.
How can we prepare a rearing house for silkworms, and how can it be reared?
A typical rearing house (6m x 4m x 3.5m) is constructed on an elevated place under shade to accommodate 100 dfls (disease-free layings). Space of lm should be provided surrounding the rearing house. Sufficient windows and ventilators should be provided for free circulation of air inside the rearing house. The windows and ventilators should be covered with a nylon nets to restrict the entry of uzi flies and other insects. Apart from the specified area of the rearing house; the following appliances such* as hygrometer, power sprayers, rearing stands, foam pads, wax-coated paraffin papers, nylon nets, baskets for keeping leaves, gunny bags, rotary or bamboo mountages and drier are needed for effective rearing of silkworms. The steps involved in the rearing process of silkworms are disinfection of rearing house, incubation of eggs, brushing, young larval rearing, and late age larval rearing. The selected healthy silk moths are allowed to mate for 4 hours. The female moth is then kept in a dark plastic bed, it lays about 400 eggs in 24 hours; the female is taken out, crushed, and examined for any disease, only certified disease-free eggs are reared for industrial purposes. The eggs are incubated in an incubator. The small larvae (caterpillars) hatch between 7-10 days. These larvae are kept in trays inside a rearing house at a temperature of about 20°C – 25°C. These are first fed on chopped mulberry leaves. After 4-5 days fresh leaves are provided. As the larvae grow, they are transferred to fresh leaves on clean trays, when fully grown they spin cocoons. Their maturity is achieved in about 45 days. At this stage, the salivary glands (silk glands) start secreting silk to spin cocoons.

Question 21.
What are various appliances used for rearing Silkworms?
The following appliances such as hygrometers, power sprayers, rearing stands, foam pads, wax-coated paraffin papers, nylon nets, baskets for keeping leaves, gunny bags, rotary or bamboo mountages and drier are needed for effective rearing of silkworms.

Question 22.
What are the steps involved in the rearing process of silkworms?
The steps involved in the rearing process of silkworms are disinfection of rearing house, incubation of eggs, brushing, young larval rearing, and late age larval rearing.

Question 23.
What is post cocoon processing? What are the steps involved?
The method of obtaining silk thread from the cocoon is known as post cocoon processing. This includes stifling and reeling.

Question 24.
Define stifling and Reeling.
The process of killing the cocoons is called stifling. The process of removing the threads from the killed cocoon is called reeling.

Question 25.
How the reeling process of silk thread is done?
For reeling silk the cocoons are gathered about 8-10 days after spinning had begun. The cocoons are first treated by steam or dry heat to kill the insect inside. This is necessary to prevent the destruction of the continuous fiber by the emergence of the moth. The cocoons are then soaked in hot water (95° -97°C) for 10-15 minutes to soften the gum that binds the silk threads together. This process is called cooking. The “cooked” cocoons are kept in hot water and the loose ends of the thread are caught by hand. Threads from several cocoons are wound together on spinning wheels (Charakhas) to form the reels of raw silk. Only about one-half of the silk of each cocoon is reliable, the remainder is used as a silk waste and formed into spun silk. Raw silk thus obtained is processed through several treatments to bring about the luster on the thread.

Question 26.
Tabulate some diseases which affect silkworms. Write their causative organisms also.

 Diseases Causative agent Pebrine Nosema bombycis, a protozoan Flacherie Streptococcus and Staphylococcus Grasserie Nuclear polyhedrosis vims called Baculovirus Muscardine Beauveria bassiana

Question 27.
What is apiculture or beekeeping.
Care and management of honey bees on a commercial scale for the production of honey is called Apiculture or Bee Keeping.

Question 28.
Write the names of well-recognized types of bees in the world.
There are five well-recognized types of bees in the world. They are Apis dorsata (Rock bee), Apis florea (Little bee), Apis iridic (Indian bee), Apis mellifera (European bee) and Apis Adamson (African bee).

Question 29.
What is a nuptial flight?
The Queen bee is a functional female bee present in each hive and feeds on Royal Jelly. The virgin queen bee mates only once in her life. During the breeding season in winter, a unique flight takes place by the queen bee followed by several drones. This flight is called “nuptial flight”.

Question 30.
Write an account on the worker bee?
Among the honey bees, workers are sterile females and smallest but yet function as the mainspring of the complicated machinery in. the colony. A worker bee lives in a chamber – called ‘Worker Cell’ and it takes about 21 days to develop from the egg to adult and its lifespan is about six weeks. Each worker has to perform different types of work in her lifetime. During the first half of her life, she becomes a nurse bee attending to indoor duties such as secretion of royal jelly, prepares bee-bread to feed the larvae, feeds the queen, takes care of the queen and drones, secretes beeswax, builds combs, cleans and fans the beehive.
Then she becomes a soldier and guards the beehive. In the second half of her life lasting for three weeks, she searches and gathers the pollen, nectar, propolis, and water.

Question 31.
Why drones are called ‘king of the colony’?
The drone is the functional male member of the colony which develops from an unfertilized egg. It lives in a chamber called a drone cell. Drones totally depend on workers for honey. The sole duty of the drone is to fertilize the virgin queen hence called “King of the colony”. During swarming (the process of leaving the colony by the queen with a large group of worker bees to form a new colony) the drones follow the queen, copulates, and dies after copulation.

Question 32.
Write the structure of a beehive.
The house of the honey bee is termed a beehive or comb. The hive consists of hexagonal cells made up of wax secreted by the abdomen of worker bees arranged in opposite rows on a common base. These hives are found hanging vertically from the rocks, buildings,s or branches of trees. The young stages of honey bees accommodate the lower and central cells of the hive called the brood cells. In Apis dorsata, the brood cells are similar in size and shape but in other species, brood cells are of three types viz., queen cell for queens, worker cell for workers, and drone cells for drones. The cells are intended for the storage of honey and pollen in the upper portion of the comb whereas the lower portions are for brood rearing.

Question 33.
Write the primary equipment of the Langstroth beehive.
The Langstroth beehive is made up of wood and consists of six parts.

1. The stand is the basal part of the hive on which the hive is constructed. The stands are adjusted to make a slope for rainwater to drain.
2. Bottom board is situated above the stand and forms the proper base for the hive. It has two gates, one gate functions as an entrance while the other acts as an exit.
3. The brood chamber is the most important part of the hive. It is provided with 5 to 10 frames arranged one above the other through which the workers can easily pass. The frame is composed of a wax sheet which is held in a vertical position up by a couple of wires. Every sheet of wax is known as Comb Foundation. The comb foundation helps in obtaining a regular strong worker brood cell comb which can be used repeatedly.
4. Super is also a chamber without cover and base. It is provided with many frames containing a comb foundation to provide additional space for the expansion of the hive.
5. The inner cover is a wooden piece used for covering the super with many holes for proper ventilation.
6. The top cover is meant for protecting the colonies from rains. It is covered with a sheet that is plain and sloping.

Question 34.
Write few lines about the following.

1. Queen Excluder,
2. Bee Gloves,
3. Bee Veil,
4. Smoker,
5. Bee brush,
6. Honey extractor.

1. Queen Excluder is utilized to prevent the entry of queen bees from the brood chamber into the super chamber.
2. Bee gloves are used by beekeepers for protecting their hands while inspecting the hives.
3. A bee veil is a device made of fine nettings to protect the bee-keeper from bee stings.
4. The smoker is used to scaring the bees during hive maintenance and honey collection by releasing smoke.
5. Bee brush is a large brush often employed to brush off bees from honeycombs particularly at the time of extraction.
6. Honey Extractor is a stainless-steel device that spins the combs rapidly to extract honey.

Question 35.
Give the economic importance of the chief products of beekeeping.
The chief products of the beekeeping industry are honey and bee wax.
Honey is the healthier substitute for sugar. The major constituents of honey are levulose, dextrose, maltose, other sugars, enzymes, pigments, ash, and water. It is an aromatic sweet material derived from the nectar of plants. It is a natural food, the smell and taste depend upon the pollen taken by the honey bee. It is used as an antiseptic, laxative, and as sedative. It is generally used in Ayurvedic and Unani systems of medicine. It is also used in the preparation of cakes, bread, and biscuits.
Bee wax is secreted by the abdomen of the worker bees at the age of two weeks. The wax is masticated and mixed with the secretions of the cephalic glands to convert it into a plastic resinous substance. The resinous chemical substance present in the wax is called propolis which is derived from pollen grains. The pure wax is white in color and the yellow color is due to the presence of carotenoid pigments. It is used for making candles, waterproofing materials polishes for floors, furniture, appliances, leather, and taps. It is also used for the production of comb foundation sheets in beekeeping and used in pharmaceutical industries.

Question 36.
Define Lac culture.
The culture of lac insect using techniques for the procurement of lac on large scale is known as Lac culture.

Question 37.
What is ‘swarming’ in lac insects?
The female develops very rapidly after fertilization and lays about 200 to 500 eggs. Eggs hatch into larvae after six weeks. The mass emergence of larvae from the egg in search of a host plant is called ‘swarming”.

Question 38.
Write an account of the life cycle of the ‘Lac’ insect with a simple sketch.

The female lac insect is responsible for the large-scale production of lac, which is larger than the male lac insect.
After copulation, the male insect dies. The female develops very rapidly after fertilization and lays about 200 to 500 eggs. Eggs hatch into larvae after six weeks. The mass emergence of larvae from the egg in search of a host plant is called ‘swarming’.
After settling on the host, the larvae start feeding continuously and the secretion of lac also starts simultaneously. Gradually the larvae become fully covered by lac. Then the larvae moult in their respective cells (chamber). The shapes of the cells are different for male and female insects, males are elongated whereas and the female is oval. The process of introducing lac insect on the host plant is called inoculation. Before inoculation, pruning of the host plant is done. The twigs having brood lac, i.e., lac insect about 20 cm in length are attached to fresh host plants. The lac insect then repeats its life cycle.

Question 39.
What is Aquaponics?
Aquaponics is a technique that is a combination of aquaculture (growing fish) and hydroponics (growing plants in non-soil media and nutrient-laden water).

Question 40.
What is the use of Aquaponics?
Aquaponics may also prevent toxic water runoff. It also maintains ecosystem balance by recycling the waste and excretory products produced by the fish.

Question 41.
What are the primary methods of aquaponic gardening used nowadays?
Deep water culture, media-based method, nutrient film technique, and aqua Vertica.

Question 42.
What is deep water culture?
Deepwater culture is otherwise known as the raft-based method. In this method, a raft floats in water. Plants are kept in the holes of the raft and the roots float in water. This method is applicable for larger commercial-scale systems. By this method, fast-growing plants are cultivated.

Question 43.
What is the media-based method?
The media-based method involves growing plants in inert planting media like clay pellets or shales. This method is applicable for the home and hobby scale systems. A larger number of fruiting plants, leafy green plants, herbs, and other varieties of plants can be cultivated.

Question 44.
What is the nutrient film technique?
The Nutrient Film technique involves the passage of nutrient-rich water through a narrow trough or PVC pipe. Plants are kept in the holes of the pipe to allow the roots to be in free contact within the water stream.

Question 45.
What is Aqua Vertica?
Aqua Vertica is otherwise known as vertical aquaponics. Plants are stacked on top of each other in tower systems. Water flows in through the top of the tower. This method is suitable for growing leafy greens, strawberries, and other crops that do not need supporting solid substratum to grow.

Question 46.
Give an account of the advantages of Aquaponic gardening?
Water conservation: No need for water discharge and recharge as the water is maintained by the recycling process.
Soil: Bottom soil may be loaded with freshwater. Microbes in water can convert the waste materials into usable forms like ammonia into nitrates which are used by the plants. Thus the soil fertility is maintained.
Pesticides: In this system use of pesticides is avoided and hence it is eco-friendly.
Weeds: Since the plants are cultured in confined conditions, the growth of weeds is completely absent. The utilization of nutrients by plants is high in this method.
Artificial food for fishes: ln this system plant waste and decays are utilized by fishes as food. So, the need for the use of supplementary feed can be minimized.
Fertilizer usage: Artificial or chemical fertilizers are not required for this system since the plants in the aquaponics utilize the nutrients from the fish wastes dissolved in water.

Question 47.
What is aquaculture?
Aquaculture is a branch of science that deals with the farming of aquatic organisms such as fish, molluscs, crustaceans, and aquatic plants.

Question 48.
Mention the classification of aquaculture?
On the basis of source, aquaculture can be classified into three categories. They are:

1. Freshwater aquaculture,
2. Brackishwater aquaculture,
3. Marine water aquaculture.

Question 49.
What is freshwater aquaculture?
Inland water bodies include freshwater bodies like rivers, canals, streams, lakes, flood plain wetlands, reservoirs, ponds, tanks, and other derelict water bodies and ponds constructed for freshwater aquaculture. The pH of the freshwater should be around neutral and salinity below 5 ppt (parts per thousand).

Question 50.
What is Brackishwater aquaculture? What are the fishes cultured here?
Brackish water fishes spend most of their life in river mouths (estuaries) backwaters, mangrove swamps, and coastal lagoons. Estuarine fish are more common in Bengal and Kerala. Culturing of animals in the water having salinity range 0.5 2 30 ppt are called as brackish water culture. Fishes cultured in brackish water are Milkfish (‘Chanos Chanos’), Sea bass (‘Koduva’), Grey mullet (‘Madavai ’), Pearl spots (‘Kari’meen) etc.

Question 51.
What is Mariculture? What are the fishes cultured here?
Culturing of animals in the water salinity ranges from 30 – 35% is called Mariculture. Some fishes like Chanos sp, Mugil cephalus are cultured here.

Question 52.
What is metahaline culture?
Culturing of animals in the salinity ranges from 36 – 40% is called Metahaline culture, eg: Brine shrimp (Artemia salina).

Question 53.
Write down the characteristics of cultivable fishes?
Characteristics of cultivable fishes. The special characteristic features of cultivable fishes are:

1. Fishes should have a high growth rate in a short period for culture.
2. They should accept a supplementary diet.
3. They should be hardy enough to resist some common diseases and infections of parasites.
4. Fishes proposed for polyculture should be able to live together without interfering or attacking other fishes.
5. They should have high conversion efficiency so that they can effectively utilize the food.

Question 54.
What are the types of cultivable fishes?
Types of cultivable fish: Cultivable fish are of 3 types.

1. Indigenous or native freshwater fishes (Major carps, Catla, Labeo, Clarias).
2. Saltwater fishes acclimatized for freshwater (Chanos, Mullet).
3. Exotic fishes or imported from other counties (Common carps).

Question 55.
Why carps have proved to be the best suited for culture in India?
Major carps have proved to be best suited for culture in India because of the carps.

1. Feed on zooplankton and phytoplankton, decaying weeds, debris, and other aquatic plants.
2. They can survive in turbid water with slightly higher temperatures.
3. Can tolerate O2 variations in water.
4. Can be transported from one place to another easily.
5. They are highly nutritive and palatable.

Question 56.
What are the external factors which affect fish culture?
External factors affecting fish culture: The factors that affect fish culture are temperature, light rain, water, flood, water current, turbidity of the water, pH hardness, salinity, and dissolved O2. Light and temperature also play an important role in fish breeding.

Question 57.
What are the two types of breeding among fishes?
Natural breeding and induced breeding.

Question 58.
Write about the induced breeding method in fishes?
Induced breeding: The fish seed is commonly collected from breeding grounds but does not guarantee that all fish seeds belong to the same species. Hence advanced techniques have been developed to improve the quality of fish seed by the artificial methods of fertilization and induced breeding. Artificial fertilization involves the removal of ova and sperm from female and male by artificial mechanical process and the eggs are fertilized. For artificial fertilization, the belly of mature female fish is held upward. Stripping is done with the thumb of the right hand from the anterior to the posterior direction for the ejection of eggs due to force. In this way, eggs are collected separately. Further, the male fish is- caught with its belly downwards. The milt of fish is striped and collected separately, and then the eggs are fertilized.
Induced breeding is also done by hypohydration (removal of the pituitary gland).
The gonadotropin hormone (FSH and LH) secreted by the pituitary gland influences the maturation of gonads and spawning in fishes. The pituitary gland is removed from a healthy mature fish. The pituitary extract is prepared by homogenizing in 0.3% saline or glycerine and centrifuged for 15 minutes at 8000 rpm.
The supernatant is injected intramuscularly at the base of the caudal fin or intra-peritonealy ‘ at the base of pectoral fin. Male and female fishes start to spawn (release of gametes) and eggs are fertilized. The fertilized eggs are removed from the spawning place and kept into hatching hapas.

Question 59.
What is composite fish farming or polyculture?
Few selected fishes belonging to different species are stocked together in proper proportion in a pond. This mixed farming is termed composite fish farming or polyculture.

Question 60.
What are the advantages of polyculture?

1. All available niches are fully utilized.
2. Compatible species do not harm each other.
3. No competition among different species is found.
4. Catla catla, Labeo rohita and Cirrhinus mrigala (surface feeder) are the commonly used fish species; for composite fish farming.

Question 61.
What are the different fishing methods carried out to harvest fish?
Different methods of fishing are carried out to harvest fish. These include Stranding, Angling, Traps, Dipnets, Cast nets, Gillnets, Drag nets, and purse nets.

Question 62.
What are exotic fishes? Give example.
The fishes imported into a country for fish culture are called exotic fishes and such fish culture is known as exotic fish culture. Examples of such exotic fishes introduced in India are Cyprinus carpio and Oreochromis mossambicus.

Question 63.
Write about some of the fish by-products in brief.

1. Fish oil is the most important fish by-product. It is derived from fish liver and from the fish body. Fish liver oil is derived from the liver which is rich in vitamin A and D, whereas fish body oil has a high content of iodine, not suitable for human consumption, but is used in the manufacture of laundry soaps, paints, and cosmetics.
2. Fish meal is prepared from fish waste after extracting oil from the fish. The dried wastes are used to prepare food for pigs, poultry, and cattle. The wastes obtained during the preparation of fish meals are widely used as manure.
3. Isinglass is a high-grade: collagen produced from dried air bladder or swim bladder of certain fishes viz. catfish and carps. The processed bladder which is dissolved in hot Water forms gelatin having adhesive property: It is primarily used for clarification of wine, beer, and vinegar.

Question 64.
What are the different types of prawn fishery?

1. Shallow water prawn fishery – located on the west coast restricted to shallow waters.
2. Estuaries and backwaters or saline lake prawn fishery – The area of production of prawns are the backwaters seen along the Western coast, Ennur, Pulicat, Ohilka lake, and Estuaries of Ganga and Brahmaputra rivers.
3. Freshwater prawn fishery – Prawns are caught from the rivers and lakes throughout India.
4. Marine prawn fishery – Most of the marine prawns are caught along the Indian coast belonging to the family Penaeidae.

Question 65.
Mention the names of some species of prawn?
Penaeus indicus, Penaeus monodon, Metapenaeus dobsoni and Macrobrachium rosenbergii.

Question 66.
How the freshwater prawn “Macrobrachium rosenbergii” is cultured?
Macrobrachium rosenbergii is commonly seen in rivers, fields, and low-saline estuaries. The prawn collected from ponds, rivers, and paddy fields is transferred to the tanks which are aerated. For fertilization, one pair Of prawn are kept in a separate tank. After mating; the eggs are laid. Spawning tanks of different sizes should be prepared with proper aeration. Temperature (240 C – 300 Q arid pH (7-8) should be maintained in the hatching tank. The eggs hatch into first and second-stage larva. Artificial feed is supplied. Young ones of 5cm length (60 days old) can be reared in fresh or slightly brackish water ponds and rice fields. Harvesting of prawns can be done twice a year.

Question 67.
How pearl formation occurs?
When a foreign particle accidentally enters into the space between the mantle and shell of the oyster, it adheres to the mantle. The mantle epithelium encloses it like a sac and starts to secrete concentric layers of nacre around it as a defensive mechanism. Nacre is secreted continuously by the epithelial layer of the mantle and is deposited around the foreign particle and over a period of time the formation of repeated layers of calcium carbonate makes them hard and glossy pearl. When the pearl enlarges the oyster dies. The shell is then carefully opened and the pearls are manually separated and graded.

Question 68.
How pearls are cultured artificially in the pearl industry?
Programming of Pearl Industry and Artificial Insertion of Nucleus: This can be achieved by an artificial device to insert the nucleus as the foreign particle in the shell of oyster has proved useful for the production of pearls in greater numbers.

1. Collection of oysters: Oysters are caught by special types of cages (84 x 54 x 20 cm) by covering a heavy wire frame with two-centimeter wire mesh. This cage is dipped into a sand-cement mixture providing a rough surface to the cages to which free-swimming spat get easily stuck up. These cages are suspended at a depth of 6 meters. From July to November, where spats are easily available. These collected oysters are now transferred to rearing cages.
2. Rearing of oysters: The collected oysters are stocked and reared in a special type of cage called a rearing cage. These cages are well protected from enemies of oysters like Octopus, Eel, Devil fishes, etc. The collected oysters are first cleaned and then placed into the culture cages for a period of about 10 to 20 days to recover from the strain due to excessive handling and for the physiological adjustment to the shallow water conditions.
3. Insertion of the nucleus: In this method, a piece of the mantle of living oyster is cut off and inserted together with a suitable nucleus inside the living tissue of another oyster. Following steps are taken for the insertion of the nucleus.
(a) Fitness of oysters for operation: The selected oysters for the insertion of the nucleus should be healthy and strong enough to overcome the stress during operation.
(b) Preparation of graft tissues: The piece of tissue which is inserted inside the mantle is called as ‘GRAFT’ tissue. The outer edges of these graft squares must be known because nacre secreting cells are found only on the outer surface of the mantle so it is essential to keep the outer surface in contact with the inserted nucleus.
(c) Preparation of nucleus: Any small particle may function as a nucleus to initiate the pearl formation but it is reported that calcareous nucleus is the best because the deposition of nacre was found to be more on the calcareous nucleus.
(d) Insertion of the nucleus: For the insertion of the nucleus, oysters are fixed in a desk clamp in the position of the right valve facing upward. Mantle folds are smoothly touched to expose the foot and the main body mass, followed by an incision into the epithelium of the foot and a slender channel into the main mass one graft tissue which functions as a bed for the nucleus.
(e) Post-operation care: Nucleated oysters are placed into cages and suspended into seawater and attached with floating rafts to a depth of 2 to 3 meters for about 6 to 7 days to recover from the shocks due to operation. This period of 6 to 7 days is known as the ‘Recovery period’. About 3000 to 3600 nucleated oysters are kept in different cages suspended in seawater at 2 to 3 meters depth for 3 to 6 years and undisturbed except at the time of clearing and inspection.
4. Harvesting of pearl: Pearls are harvested in the month of December to February which may slightly vary according to climatic conditions. After the completion of 3 years of the insertion of a nucleus, pearl oysters are harvested from the sea and the pearls are taken out from the shell.
5. Clearing of pearls: After taking out the pearls from the oyster’s shell they are washed properly, cleared with the soap solution.

Question 69.
Write the composition of pearl.
Composition of pearl: Pearl comprises water, organic matter, calcium carbonate, and residue.

1. Water: 2 – 4%
2. Organic matter: 3.5-5.9%
3. Calcium Carbonate: 90%
4. Residue: 0.1 – 0.8%, Carbonate: 90%.

Question 70.
What is meant by Animal husbandry?
Animal husbandry is the practice of breeding and raising livestock cattle like cows, buffaloes, and goats and birds, etc., that are useful to human beings.

Question 71.
What are the parameters to be taken into account to maintain dairy and poultry farms?
Parameters such as adequate ventilation, temperature, sufficient light, water, and proper housing accommodation should be taken into account to maintain dairy and poultry farms.

Question 72.
Write the objectives of animal breeding?
Objectives of Animal breeding:

1. To improve the growth rate.
2. Enhancing the production of milk, meat. Egg etc.
3. Increasing the quality of animal products.
4. Improved resistance to diseases.
5. Increased reproductive rate.

Question 73.
What is outcrossing?
Outcrossing: It is the breeding between unrelated animals of the same breed but, having no common ancestry. The offspring of such a cross is called outcross. This method is suitable for breeding animals below average in productivity.

Question 74.
What is artificial insemination?
Artificial insemination: Artificial insemination is a technique in which the semen collected from the male is injected into the reproductive tract of the selected female. Artificial insemination is the economical measure where fewer bulls are required and maximum use can be made of the best sire.

Question 75.
What is thawing?
Thawing means to melt or become liquid. When the semen collected for artificial insemination is taken to far-off places/stored for a long time in the frozen condition it should be brought to room temperature slowly before use. This process is called thawing.

Question 76.
What is MOET? Write when this method is followed?
Multiple ovulation embryo transfer technology (MOET): It is another method of propagation of animals with desirable traits. This method is applied when the success rate of crossing is low even after artificial insemination. In this method, Follicle-stimulating hormone (FSH) is administered to cows for inducing follicular maturation and superovulation. Instead of one egg per cycle, 6-8 eggs can be produced by this technology. The eggs are carefully recovered non-surgically from the genetic mother and fertilized artificially. The embryos at 8-32 celled stages are recovered and transferred to a surrogate mother. For*another round -of ovulation, the same genetic mother is utilized. This technology can be applied to cattle, sheep, and buffaloes.

Question 77.
How are cattle classified?
Cattles are classified under three groups based on the purpose they serve to man. They are:

1. Dairy breeds or Milk breeds: They are high milk yielders with extended lactation, eg: Sindhi, Gir, Sahiwal, Jersy, Brown Swiss, Holstein cattle.
2. Draught purpose breeds: Bullocks are good for draught purposes, eg: Kangayam, Malvi.
3. Dual Purpose breeds: Cows are meant for yielding more milk and bullocks are used for better drought purposes, eg: Ongole, Hariana.

Question 78.
Mention the names of the diseases of cattle.
The main diseases of dairy cattle are rinderpest, foot and mouth disease, cowpox, hemorrhagic fever, anthrax.

Question 79.
What is poultry fanning?
Poultry farming is essential for the purpose of meat, eggs, and feather production.

Question 80.
Some of the chicken breeds are of egg layers and are mainly farmed for the production of eggs. Name the egg layers and write their characters.
Egg layers: These are farmed mainly for the production of eggs.
Leghorn: This is the most popular commercial breed in India and originated in Italy. They are small, compact with a single comb and wattles with white, brown, or black color. They mature early and begin to lay eggs at the age of 5 or 6 months. Hence these are preferred in commercial farms. They can also thrive well in dry areas.
Chittagong: It is the breed chiefly found in West Bengal. They are golden or light yellow colored. The beak is long and yellow in color. Ear lobes and wattles are small and red in color. They are good egg layers and are delicious.

Question 81.
Write the characters of white Plymouth rock.
White Plymouth rock: They have white plumage throughout the body. It is commonly used in broiler production. This is an American breed. It is a fast-growing breed and well suitable for growing intensively in confined farms.

Question 82.
Write the special characteristic features of Aseel?
Aseel: This breed is white or black in color. The hens are not good egg layers but are good in the incubation of eggs. It is found in all states of India. Aseel is noted for its pugnacity, high stamina, and majestic gait, and dogged fighting qualities. Although poor in productivity, this breed is well-known for its meat qualities.

Question 83.
Given an example for the ornamental breed of ‘ chicken and write its features.
Silkie: It is a breed of chicken that has a typical fluffy plumage, which is said to feel like silk and satin. The breed has numerous additional special characters, such as black skin and bones, blue earlobes, and five toes on each foot, while the majority of chickens only have four. They are exhibited in poultry shows and come out in various colors. Silkies are well recognized for their calm, friendly temperament. Silkie chicken is especially simple to maintain as pets.

Question 84.
What are the different methods of poultry farming?
The types of poultry farming are Free-range farming. Organic method, Yarding method, Battery cage method, and Furnished cage method.

Question 85.
Explain the steps involved in rearing chicken?
There are some steps involved in rearing chicken.
Selection of the best layer: An active intelligent-looking bird, with a bright comb, not obese should be selected.
Selection of eggs for hatching: Eggs should be selected very carefully. Eggs should be fertile, medium-sized, dark brown shelled and freshly laid eggs are preferred for rearing. Eggs should be washed, cleaned, and dried.
Incubation and hatching: The maintenance of newly laid eggs in optimum condition till hatching is called incubation. The fully developed chick emerges out of the egg after an incubation period of 21 – 22 days. There are two types of incubation namely natural incubation and artificial incubation. In the natural incubation method, only a limited number of eggs can be incubated by a mother hen. In artificial incubation, more eggs can be incubated in a chamber (Incubator).
Brooding: Caring and management of young chicks for 4-6 weeks immediately after hatching is called brooding. It can also be categorized into two types namely natural and artificial brooding.
The housing of Poultry: To protect the poultry from sun, rain, and predators it is necessary to provide housing to poultry. Poultry house should be moisture-proof, rat proof and it should be easily cleanable and durable.
Poultry feeding: The diet of chicks should contain an adequate amount of water, carbohydrates, proteins, fats, vitamins, and minerals.

Question 86.
What are the main products and byproducts of poultry farming?

1. The main products of poultry farming are eggs and meat.
2. The feathers of poultry birds are used for making pillows and quilts. Droppings of poultry can be used as manure in fields.
3. A number of poultry byproducts like blood-meal, feather meal, poultry by-product meal, and hatchery by-product meal are used as good sources of nutrients for meat producing animals and poultry.

Question 87.
Name the diseases that affect poultry animals.
Ranikhet, Coccidiosis, and Fowlpox are some common poultry diseases.

Question 88.
Write the benefits of keeping poultry. Benefits of Poultry farming:
The advantages of poultry farming are-

1. It does not require high capital for construction and maintenance of poultry farming:
2. It does not require a big space.
3. It ensures a high return of investment within a very short period of time.
4. It provides fresh and nutritious food and has a huge global demand.
5. It provides employment opportunities for the people.

Question 89.
Mention the native breeds and exotic breeds of duck.
The native one includes Indian Runner and Syhlet meta. The exotic breeds include Muscori, Pekin, Aylesbury and Campbell.

Question 90.
Ramu has reared about 60 to 70 ducks of his own. In what way he is benefited from duck farming.
Advantages of duck farming: They can be reared in small backyards where water is available and needs less care and management as they are very hardy. They can adapt themselves to all types of environmental conditions and are breed for feed efficiency, growth rate, and resistance to diseases.

Question 91.
Why should we breed animals?
Through animal breeding, improved breeds of animals can be produced by improving their genotype through selective breeding.

Question 92.
How can you identify healthy cattle?
Healthy cattle appear bright, alert, and active in their movement with a shiny coat.

Question 93.
Define Hapa?
Hapa is a cage-like, rectangular, or square net impoundment placed in a pond for holding fish for various purposes. They are made of fine mesh netting material.

Question 94.
What is drilosphere?
Drilospfiere is the part of the soil influenced by earthworm secretions, burrowing, and castings.

Question 95.
What is a biological indicator?
The biological indicator refers to organisms, species or communities whose characteristics show the presence of specific environmental conditions.

1. A branch of science that deals with economically useful animals is called:
(a) apiculture
(b) horticulture
(c) economic zoology
(d) aquaculture
(c) economic zoology

2. ……….. is the primary goal of vermiculture.
(a) Vermiwash
(b) Vermi compost
(c) Voltinism
(d) Manure production
(b) Vermi compost

3. ………… is called biological indicators of soil fertility.
(a) Silkworm
(b) Honeybee
(c) Earthworm
(d) Lac insect
(c) Earthworm

4. ………. are the earthworms, which dwell on the surface and feed on organic matter.
(a) Humus formers
(b) Humus feeders
(c) Vermi composers
(d) Soil improvers
(a) Humus formers

5. Lampito mauritii are the ……….. species of earthworms.
(a) endemic species
(b) extinct species
(c) exotic species
(d) notho species
(a) endemic species

6. ……… is a liquid collected after the passage of water through a column of vermibed.
(a) Hydrophonics
(b) Vermiwash
(c) Vermicompost
(d) Mucilage
(b) Vermiwash

7. Vermiwash is obtained from the …………. formed by earthworms
(a) compost unit
(b) bunds
(c) drilospheres
(d) soil
(c) drilospheres

8. Which of the following statement is not correct.
The main components of sericulture
(a) Cultivation of food plants
(b) The mass emergence of larvae from the egg in search of host plant is called swarming
(c) Rearing of worms
(d) Reeling and spinning of silk.
(b) The mass emergence of larvae from the egg in search of host plant is called swarming

9. The ‘Muga Silk’ is produced from the ………. species of silkmoth.
(a) Bombyx mori
(b) Antheraea mylitta
(c) Antheraea assamensis
(d) Attacus ricini
(c) Antheraea assamensis

10. The preferred food leaves of the silkworm species Attacus ricini is:
(a) mulberry
(b) champa
(c) castor
(d) aijun
(c) castor

11. Which of the following state in which Mulberry silk is produced?
(b) Nagaland
(c) West Bengal
(d) Assam

12. A single female moth of silkworm lays …………. number of eggs depending upon the climatic conditions.
(a) 40-50
(b) 400-500
(c) 1-24
(d) 2-3
(b) 400-500

13. The caterpillars of silkworm developed …………. type of mouth parts adapted to feed easily on the mulberry leaves.
(a) maxillary
(b) mandibulate
(c) cutter
(d) sickle
(b) mandibulate

14. Name the species that produces large amount of silk is:
(a) Attacus ricini
(b) Bombyx mori
(c) Antheraea mylitta
(d) Antheraea assamensis
(b) Bombyx mori

15. The cultivation of mulbeny is Called:
(a) sericulture
(b) apiculture
(c) horticulture
(d) moriculture
(d) moriculture

16. The method of obtaining silk thread from the cocoon is known as
(a) pre-cocoon processing
(b) post-cocoon processing
(c) reeling
(d) cooking
(b) post-cocoon processing

17. The process of killing the cocoon is called:
(a) reeling
(b) stifling
(c) cooking
(d) rearing
(b) stifling

18. The process of removing die threads from the killed cocoon is called:
(a) reeling
(b) stifling
(c) cooking
(d) bundling
(a) reeling

19. Pebrine is a dangerous disease caused by:
(a) Nosema bombycis
(b) Streptococcus
(c) Baculovirus
(d) Staphylococcus
(a) Nosema bombycis

20. Worker bee lives in a chamber called:
(a) brood cell
(b) drone cell
(c) worker cell
(d) sub cells
(c) worker cell

21. Match the correct pair:

 1. Metabolic process (a) Liquid from vermibed 2. Classification of animals (b) Embryologist 3. Early development stage (c) Physiologist 4. Vermitech Taxonomists 5. Vermiwash (e) Bio-remediation of soil

(a) 1 -(d), 2-(c), 3-(e), 4-(b), 5-(a)
(b) 1-(c), 2-(d), 3-(b), 4-(a), 5-(e)
(c) 1-(e), 2-(a), 3-(d), 4-(c), 5-(b)
(d) 1-(a), 2-(b), 3-(c), 4-(d), 5-(e)
(b) 1-(c), 2-(d), 3-(b), 4-(a), 5-(e)

22. The nectar collected from the flowers and is stored in the stomach region of the worker bee, is transformed into honey by the enzyme called:
(a) enterokinase
(b) peptidase
(c) invertase
(d) transferase
(c) invertase

23. The process of leaving the colony by the queen with a large group of worker bees to form a new colony is called:
(a) nupital flight
(b) migration
(c) swarming
(d) flocking
(c) swarming

24. Match the following:

 1. Apis dorsata (a) Little bee 2. Apis florea (b) Rock bee 3,Apisindica (c) European bee 4. Apis mellifera (d) African bee 5. Apis adamsoni (e) Indian bee

(a) 1 -(a), 2-(b), 3-(d), 4-(e), 5-(c)
(b) 1 -(c), 2-(d), 3-(b), 4-(a), 5-(e)
(c) 1-(b), 2-(a), 3-(e), 4-(c), 5-(d)
(d) 1 -(d), 2-(c), 3-(a), 4-(b), 5-(e)
(c) 1-(b), 2-(a), 3-(e), 4-(c), 5-(d)

25.. ……….. is the most important part of the Langstroth bee hive.
(a) Inner cover
(b) Top cover
(c) Bottom board
(d) Brood chamber
(d) Brood chamber

26. ………. is a device made of fine nettings to protect the bee keeper form bee sting.
(a) Bee gloves
(b) Bee veil
(c) Bee brush
(d) Comb foundation
(b) Bee veil

27. Which of the following statement is incorrect regarding honey?
(a) It is an aromatic sweet material
(b) It is used as an antiseptic and as a sedative
(c) The resinous chemical substance present is called propolis
(d) It is generally used in Ayurvedic and unani systems of medicines
(c) The resinous chemical substance present is called propolis

28. ……….. is secreted by the abdomen of the worker bees at the age of two weeks.
(a) Beewax
(b) Honey
(c) Lac
(d) Pollen
(a) Beewax

29. Lac insect Tachardia lacca is previously known as:
(a) Bombyx mori
(b) Apis florea
(c) Laccifer Lacca
(d) Penaeus Indicus
(c) Laccifer Lacca

30. Match the following:

 1. S36,G2,G4 (a) Spinning wheels 2.Voltinism (b) Silk road 3. Gharakhas (c) Number of broods 4. Verini compost (d) Mulberry varieties 5. 7000 mile (e) Plant nutrients

(a) 1-(a), 2-(b), 3-(d), 4-(e), 5-(c)
(b) 1-(d), 2-(c), 3-(a), 4-(e), 5-(b)
(c) 1 -(b), 2-(d), 3-(a), 4-(c), 5-(e)
(d) 1-(b), 2-(a), 3-(d), 4-(e), 5-(c)
(b) 1-(d), 2-(c), 3-(a), 4-(e), 5-(b)

31. Which one of the following is not related to Lac insect:
(a) Lac insect sucks plant juices, grows and secretes Lac
(b) The quality of lac depends on the quality of the host plant
(c) After copulation, the male insect dies
(d) It is a functional female which feeds on royal jelly.
(d) It is a functional female which feeds on royal jelly.

32. The “seed lac” is sub dried and then melted to produce:
(a) Ari lac
(b) Stick lac
(c) Shellac
(d) Cerelac
(c) Shellac

33. Assertion: Aquaponics is a technique which is a combination of aquaculture and hydroponics.
Reason: In this system plant waste and decays are utilized by fishes as food
(a) Assertion and reason is correct but not related.
(b) Assertion and reason is incorrect but related.
(c) Assertion and reason is correct but related.
(d) Assertion and reason is incorrect but not related.
(c) Assertion and reason is correct but related.

34. Match the following:

 1. Ari lac (a) Acacia nilotica 2. Stick lac (b) Acacia catechu 3. Khair (c) Schleichera oleosa 4. Kamvelai (d) Immature harvesting 5. Karangalli (e) Cut from the host plant

(a) 1 -(b), 2-(c), 3-(d), 4-(e), 5-(a)
(b) 1-(e), 2-(d), 3-(c), 4-(b), 5-(a)
(c) l-(e), 2-(b), 3-(a), 4-(d), 5-(c)
(d) 1 -(d), 2-(c), 3-(b), 4-(a), 5-(c)
(d) 1 -(d), 2-(c), 3-(b), 4-(a), 5-(c)

35. Aqua vertica is otherwise known as
(a) Vertical aquaponics
(b) Plants are kept in PVC pipe holes
(c) Growing plants in clay pellets
(d) Raft based method.
(a) Vertical aquaponics

36. Choose the correct statement from the below
(a) Use of pesticides is avoided in aquaphonics
(b) Growth of weeds is more in aquaphonics
(c) Estuarine fishes are more common in Tamilnadu and Pondichery
(d) Exotic fishes are native breeds.
(a) Use of pesticides is avoided in aquaphonics

37. Cultivable fishes like tilapia, trout, koi, gold fish, bass, etc are cultured in aquaphonics – say this statement is true or false.
true

38. Match the following:

 1. Artemia (a) Benchijal 2. Catla (b) High quality pearls 3. Spawn collecting net (c) Indigenous freshwater fishes 4. High grade collagen (d) Brine shrimp 5. Pinctada (e) Isinglass

(a) 1-(e), 2-(d), 3-(a), 4-(b), 5-(c)
(b) 1-(d), 2-(c), 3-(a), 4-(e), 5-(b)
(c) 1-(b), 2-(a), 3-(c), 4-(e), 5-(d)
(d) 1-(e), 2-(c), 3-(d), 4-(a), 5-(b)
(b) 1-(d), 2-(c), 3-(a), 4-(e), 5-(b)

39. The spawn collecting net is called:
(a) Benchijal
(b) Hapas
(c) Fishnet
(d) Dipnets
(a) Benchijal

40. Management of fish farm includes
(a) Fish seed – nursery pond- stocking pond – harvesting
(b) Breeding pond – fish seed-hatching pit – nursery pond – rearing pond – stocking pond –harvesting
(c) Breeding pond – nursery pond- rearing pond- stocking pond- harvesting.
(d) Fish seed – hatching pit- rearing pond-stocking pond-harvesting
(b) Breeding pond – fish seed-hatching pit – nursery pond – rearing pond – stocking pond –harvesting

41. The newly hatched fries of fishes are transported from the hatching happa to nursery ponds where they grow into:
(a) Hatchling
(b) Larva
(c) Fingerlings
(c) Fingerlings

42. Say which statement of the following is not correct.
(a) Selected fishes of different species are stocked together in a pond is called composite fish farming
(b) Catla, Rohita, are the commonly used fishes for composite fish farming.
(c) The fishes imported into the country are called exotic fishes.
(d) Parasitic infestations and microbial infections cannot be found and treated among fishes.
(d) Parasitic infestations and microbial infections cannot be found and treated among fishes.

43 is the fish product rich in Vitamin A and D.
(a) Fish meal
(b) Fish oil
(c) Ising glass
(d) Dried fish
(b) Fish oil

44. ……….. are the aquatic crustacean their flesh contains more glycogen protein with low fat content.
(a) Prawns
(b) Oysters
(c) Crayfish
(d) Crabs
(a) Prawns

45. ……….. is commonly seen in rivers and low saline estuaries.
(a) Metapenaeus dobsoni
(b) Panaeus indicus
(c) Panaeus monodon
(d) Macrobrachium rosenbergii
(d) Macrobrachium rosenbergii

46. High quality of pears are obtained from pearl oysters of;
(a) Lamellidens
(b) Olympia oyster
(d) Sydney rock oyster

47. …………. breed is the smallest breed of cow.
(a) Brown Swiss
(b) Jersey
(c) Sindhi
(d) Vechur
(d) Vechur

48. The most popular commercial chicken breed in India is ………. which thrive well in dry areas.
(a) Aseel
(b) White Plymouth rock
(c) Brahma
(d) Leghorn
(d) Leghorn

49. Fast growing chicken breed is:
(a) Chittagong
(b) White Plymouth rock
(c) Wyandotte
(d) Karaknath
(b) White Plymouth rock

50. ……….. is a type of chicken breed known for its pugnacity.
(a) Aseel
(b) Brahma
(c) Busra
(d) Leghorn
(a) Aseel

51. Which is the ornamental breed of chicken of the following?
(a) Chittagong
(b) Leghorn
(c) Aseel
(d) Silkie
(d) Silkie

52. Match the following:

 1. Egg layers (a) White Plymouth rock 2. Broiler type (b) Silkie 3. Dual purpose breed (c) Leghorn 4. Game breeds (d) Brahma 5. Ornamental breeds (e) Aseel

(a) 1-(b), 2-(a), 3-(e), 4-(c), 5-(d)
(b) 1 -(a), 2-(c), 3-(b), 4-(e), 5-(d)
(c) 1 -(d), 2-(e), 3-(c), 4-(b), 5-(a)
(d) 1 -(c), 2-(a), 3-(d), 4-(e), 5-(b)
(d) 1 -(c), 2-(a), 3-(d), 4-(e), 5-(b)

53. Which of the following statement is true?
(a) The most common and commercially farmed birds are chicken, ducks, goose, patridges, Guinea fowl, Emu, etc.
(b) Broiler type chicken are well known for its slow growth and hard quality of meat
(c) Aseel hens are good egg layers, but are not good in incubation of eggs
(d) Silkies are well recognized for its majestic look and fighting qualities.
(a) The most common and commercially farmed birds are chicken, ducks, goose, patridges, Guinea fowl, Emu, etc.

## TN Board 11th Bio Zoology Important Questions Chapter 11 Chemical Coordination and Integration

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 11 Chemical Coordination and Integration which is useful for their exam preparation.

## TN State Board 11th Bio Zoology Important Questions Chapter 11 Chemical Coordination and Integration

Question 1.
What are hormones?
Hormones (hormone means to excite) are chemical messengers released into the blood and circulated as chemical signals and act specifically on certain organs or tissues called target organs Or target tissues.

Question 2.
Hormones secreted by the endocrine glands remain in the blood or destroyed?
The hormones secreted do not remain permanently in the blood but are converted by the liver into inactive compounds and excreted by the kidneys.

Question 3.
Differentiate Exo and endocrine glands for example.

 Exocrine gland Endocrine gland The exocrine glands secrete enzymes, saliva, and sweat and have ducts that carry their substances to the membrane surfaces. The endocrine glands, called ductless glands produce hormones and lack ducts; they release their hormone to the surrounding tissue fluid. The hormones circulate around the body and eventually reach the target organs. eg: salivary gland and gastric gland. eg: Pituitary, thyroid.

Question 4.
What are the functions of the hypothalamus?
Hypothalamus maintains homeostasis, blood pressure, body temperature, the cardio and fluid-electrolyte balance of the body. As part of the limbic system, it influences various emotional responses.

Question 5.
List out the names of hormones secreted by the anterior lobe of the pituitary?
The anterior lobe of the pituitary secretes six tropic hormones like growth hormone (GH), thyroid-stimulating hormone (TSH), adrenocorticotropic hormone (ACTH), follicle-stimulating hormone (FSH), luteinizing hormone (LH), luteotropic hormone (LTH).

Question 6.
Write the functions of growth hormones.
Growth hormone (GH) is also known as somatotropic hormone (STH) or Somatotropin.

1. Growth hormone promotes the growth of all the tissues and metabolic processes of the body.
2. It influences the metabolism of carbohydrates, proteins, and lipids and increases the rate of protein biosynthesis in the cells.
3. It stimulates chondrogenesis (cartilage formation), osteogenesis (bone formation) and helps in the retention of minerals like nitrogen, potassium,
phosphorus, sodium, etc., in the body.
4. GH increases the release of fatty acid from adipose tissue and decreases the rate of glucose utilization for energy by the cells.
5. Thus it conserves glucose for glucose-dependent tissues, such as the brain.

Question 7.
Write the negative feedback system operated in TSH.
TSH is a glycoprotein hormone, which stimulates the thyroid gland to secrete Tri-iodothyronine (T3) and thyroxine (T4). TSH secretion is regulated by a negative feedback mechanism. Its release from the anterior pituitary is induced by the thyrotropin-releasing hormone (TRH). When thyroxine levels in the blood increase, TRH acts on both the pituitary and hypothalamus to inhibit TSH secretion.

Question 8.
Show how ACTH secretion is regulated by a negative feedback mechanism.
ACTH is a peptide hormone that stimulates the adrenal cortex to secrete glucocorticoids and mineralocorticoids. It stimulates melanin synthesis in melanocytes, induces the release of fatty acids from adipose tissues, and stimulates insulin secretion. ACTH secretion is regulated by a negative feedback mechanism.

Question 9.
What are the functions of FSH over gonads?
Follicle-stimulating hormone FSH is a glycoprotein hormone that regulates the functions of the gonads (ovary and testis). In males, FSH along with androgens acts on the germinal epithelium of seminiferous tubules and stimulates the production and release of sperms (spermatogenesis). In females, FSH acts on the ovaries and brings about the development and maturation of graffian follicles.

Question 10.
Write the impact of LH in both males and females?
Luteinizing hormone (LH) is a glycoprotein hormone which is also known as interstitial cell-stimulating hormone (ICSH). In males, ICSH acts on the interstitial cells of the testis to produce the male sex hormone, testosterone. In females, LH; along with FSH matures the ovarian follicles. LH independently induces ovulation, maintains the corpus luteum, and promotes synthesis and release of ovarian hormones. FSH and LH are collectively referred to as gonadotropins. FSH and LH are not produced during childhood. The secretion of FSH and LH starts only during the prepubertal period.

Question 11.
When the LTH have more impact on females? Why this hormone is named so?
Luteotropic hormone (LTH) is also called luteotropin or lactogenic hormone or prolactin or mammotropin. It is a protein hormone that stimulates milk secretion after childbirth in females. High prolactin secretion during lactation suppresses LH secretion and ovulation since it induces the corpus luteum hence named as luteo tropic hormone.

Question 12.
Why ADH is called antidiuretic hormone? What is the condition if this secretion is less?
Vasopressin or antidiuretic hormone (ADH) is a peptide hormone that promotes reabsorption of water and electrolytes by distal tubules of the nephron and thereby reduces loss of water through urine. Hence it is called an antidiuretic hormone. It also causes constriction of blood vessels when released in large amounts and increases blood pressure. ADH deficiency causes Diabetes insipidus which induces the production of a large amount of urine.

Question 13.
Which hormone is called “rapid birth hormone”? Why? What are its functions?’
Oxytocin (means quick birth) is a peptide hormone that stimulates vigorous contraction of the smooth muscles of the uterus during child birth and ejection of milk from the mammary glands.

Question 14.
Name some Glycoprotein hormones.

1. Thyroid-stimulating hormone.
2. Follicle-stimulating hormone.
3. Luteinizing hormone.

Question 15.
Name some peptide hormones.

1. Growth hormone.
3. Vasopressin.

Question 16.
What are the two amino acids in a sequence which make such a difference in the action of oxytocin and vasopressin?

1. Phenylalanine and Arginine in vasopressin.
2. Isoleucine and Leucine in oxytocin.

Question 17.
How light has its effect on the production of melatonin?
Melatonin is secreted at night, Light falling on the retina of the eye decreases melatonin production.

Question 18.
Circadian rhythm is the 24-hour cycle of biological activities associated with natural periods of light and darkness. Example sleep-wake cycle, body temperature, appetite, etc.

Question 19.
Name the three hormones of the thyroid gland?
The follicular cells of the thyroid gland secrete two hormones namely tri-iodothyronine (T3) and thyroxine or tetra-iodothyronine (T4). The parafollicular cells or ‘C’ cells of the thyroid gland secrete a hormone called thyrocalcitonin.

Question 20.
What are the different functions of thyroxine?
Thyroxine regulates the basal metabolic rate (BMR) and body heat production. It stimulates protein synthesis and promotes growth. It is essential for the development of the skeletal and nervous systems. Thyroxine plays an important role in maintaining blood pressure. It reduces serum cholesterol levels, Optimum levels of thyroxine in the blood are necessary for gonadial functions.

Question 21.
Mention the functions of thyrocalcitonin?
TCT is a polypeptide hormone, which regulates blood calcium and phosphate levels. It reduces the blood calcium level and opposes the effects of the parathyroid hormone.

Question 22.
What are the two types of cells in the parathyroid gland? What are its functions?
The parathyroid gland is composed of two types of cells, the chief cells, and oxyphil cells. The chief cells secrete parathyroid hormone (PTH) and the functions of oxyphil cells are not known.

Question 23.
Write about the requirement of iodine in the formation of thyroxine hormone?
To produce normal quantities of thyroxine, about 1 mg/Week of iodine is required. To prevent iodine deficiency common table salt is iodized with 1 part sodium iodide to every 1,00,000 parts of sodium chloride.

Question 24.
Which hormone is called “hypercalcemic hormone”? Why?
The secretion of PTH is controlled by calcium levels in the blood. It increases the blood calcium level by stimulating osteoclasts to dissolve the bone matrix. As result calcium and phosphate are released into the blood. PTH enhances the reabsorption of calcium and excretion of ‘phosphates by the renal tubules and promotes activation of vitamin D to increase calcium absorption by intestinal mucosal cells.

Question 25.
Write the names of four hormones of the thymus gland.
The thymus gland secretes four hormones such as thymulin, thymosin, thymopoietin, and thymic humoral factor.

Question 26.
What is the primary function of the thymus? How much it is important to humans?

1. The primary function of the thymus is the production of immunocompetent TV lymphocytes which provide cell-mediated immunity.
2. Due to degeneration of the thymus gland, thymosin level decreases, as a result, the immunity of old age people becomes weak and causes sickness.

Question 27.
Write the three zones of the adrenal cortex of the adrenal gland.
Zona glomerulosa, Zona fasciculata, and Zona reticularis.

Question 28.
Which part of the pancreas is endocrine in nature? What are the types of cells present there and write its functions also?
The islets of Langerhans secrete hormones like insulin and glucagon. The human pancreas has one to two million islets of langerhans.
In each islet about 60% of cells are beta cells, 25% cells are alpha cells and 10% cells are delta cells. The alpha cells secrete glucagon, the beta cells secrete insulin and delta cells secrete somatostatin.

Question 29.
What are the functions of glucocorticoids?
Glucocorticoids stimulate gluconeogenesis, lipolysis, and proteolysis (the life-saving activity). Cortisol is a glucocorticoid involved in maintaining cardiovascular and kidney functions. It produces anti-inflammatory reactions and suppresses the immune response. It stimulates RBC production. It is also known as the stress combat hormone.

Question 30.
What are the functions of mineralocorticoids?
Mineralocorticoids regulate the water and electrolyte balance of our bodies. Aldosterone stimulates the reabsorption of sodium and water and eliminates potassium and phosphate ions through excretion, thus it helps in maintaining electrolytes, osmotic pressure, and blood pressure. Adrenal androgen plays a role in hair growth in the axial region, pubis, and face during puberty.

Question 31.
How can we say insulin is a hypoglycemic hormone?
Insulin is a peptide hormone and plays an important role in glucose homeostasis. It’s main effect is to lower blood glucose levels by increasing the uptake of glucose into the body cells, especially muscle and fat cells. Insulin also inhibits the breakdown of glycogen to glucose, the conversion of amino acids or fats to glucose, so insulin is rightly called a hypoglycemic hormone.

Question 32.
Why insulin is given as an injection to diabetic patients and not by oral consumption?
Because these are digested by digestive enzymes.

Question 33.
Define the following terms.
(i) Glygogenolysis, (ii) Gluconeogenesis.

1. Glucagon is a potent hyperglycaemic hormone that acts on the liver and promotes the breakdown of glycogen to glucose (Glygogenolysis).
2. Synthesis of glucose from lactic acid and from non-carbohydrate molecules (gluconeogenesis).

Question 34.
Write the functions of testosterone?
Under the influence of FSH and LH, testosterone initiates maturation of male reproductive organs, and the appearance of secondary sexual characters, muscular growth, growth of facial and axillary hair, masculine voice, and male sexual behavior.
It enhances the total bone matrix and plays a stimulating role in the process of spermatogenesis.

Question 35.
What is the function of the female hormones such as oestrogen and progesterone?
Oestrogen is responsible for the maturation of reproductive organs and the development of secondary sexual characters at puberty. Along with progesterone, oestrogens promote breast development and initiate cyclic changes during the menstrual cycle.
Progesterone prepares the uterus for implantation of the fertilized ovum. It decreases uterine contraction during pregnancy and stimulates the development of mammary glands and milk secretion. It is responsible for premenstrual changes in the uterus and is essential for the formation of the placenta.

Question 36.
Why myxoedema is caused? What are its symptoms?
Hyposecretion of the thyroid in adults causes myxoedema. It is otherwise called Gull’s disease. This disease is characterized by decreased mental activity, memory loss, slowness of movement, speech, and general weakness of body, dry coarse skin, scarce hair, puffy appearance, disturbed sexual function, low BMR, poor appetite, and subnormal body temperature.

Question 37.
What is the other name for Grave’s disease? Why it is caused? What are the symptoms of Grave’s disease?
Grave’s disease also called thyrotoxicosis or exophthalmic goiter. This disease is caused due to hypersecretion of the thyroid. It is characterized by enlargement of the thyroid gland, increased BMR (50% – 100%), elevated respiratory and excretory rates, increased heartbeat, high BP, increased body temperature, protrusion of the eyeball, and weakness of eye muscles, and weight loss.

Question 38.
Differentiate exophthalmic goitre from endemic goitre.

 Exophthalmic goitre Endemic goitre This disease is caused due to hypersecretion of the thyroid. It is caused due to hyposecretion of thyroxine. Increased BMR. elevated respiratory and excretory rates. Fall in serum thyroxine level increased TSH secretion.

Question 39.
What is tetany? Why is it caused? What are its symptoms?
Tetany is caused due to the hyposecretion of parathyroid hormone (PTH). Due to hyposecretion of PTH serum calcium level decreases (Hypocalcemia), as a result, serum phosphate level increases. Calcium and phosphate excretion levels decrease. Generalized convulsion, locking of jaws increased heartbeat rate, increased body temperature, muscular spasm are the major symptoms of tetany.

Question 40.
What are the symptoms of hyperparathyroidism?
Demineralization of bone, cyst formation, softening of bone, loss of muscle tone, general weakness, renal disorders are the symptoms of hyperparathyroidism.

Question 41.
What are the symptoms of Addison’s disease? Why is it caused?
Addison’s disease is caused due to hyposecretion of glucocorticoids and mineralocorticoids from the adrenal cortex. Muscular weakness, low BP., loss of appetite, vomiting, hyperpigmentation of the skin, low metabolic rate, subnormal temperature, reduced blood volume, weight loss are the symptoms that occur in Addison’s disease. Reduced aldosterone secretion increases urinary excretion of NaCl. and water and decreases potassium excretion leading to dehydration.

Question 42.
Write the symptoms of Cushing’s syndrome.
Obesity of the face and trunk, redness of the face, hand, feet, thin skin, excessive hair growth, loss of minerals from bone (osteoporosis) systolic hypertension are features of Cushing’s syndrome. Suppression of sexual function like atrophy of gonads is the other symptom of Cushing’s Syndrome.

Question 43.
What will be the effect of consuming synthetic soft drinks?
The branded soft drinks damage our endocrine system. While consuming soft drinks, the sugar level increases in the blood which leads to elevated insulin secretion to reduce the blood glucose level. The elevated insulin level diminishes immunity and causes obesity, cardiovascular disorders, etc.

Question 44.
What are the effects of consuming alcohol in due course?
Alcohol consumption has a widespread effect on the endocrine system. Alcohol impairs the regulation of blood glucose levels, reduces testosterone levels and increases the risk of osteoporosis.

Question 45.
Write the characteristic features, of the peptide hormones?
Peptide hormones cannot cross the phospholipid cell membrane and bind to the receptors on the exterior cell surface. They are being transported to the Golgi, which is the site of modification. It acts as a first messenger in the cell. Hormones on binding to their receptors do not enter the target cell but generate the production of second messengers such as cyclic AMP (cAMP), which in turn regulates cellular metabolism. This is catalyzed by the enzyme adenylate cyclase. The interaction between the hormone at the surface and the effect brought out by cAMP within the cell is known as a signaling cascade. At each step, there is a possibility of amplification.

1. One hormone molecule may bind to multiple receptor molecules before it is degraded.
2. Each receptor may activate several adenylate cyclases each of which make much c AMP.
3. Thus there is more signal after each step. The actions of cAMP are terminated by phosphodiesterases.

The effect of peptide hormones like insulin, glucagon, somatotropin is usually short-lived. because they work through a second messenger system.

Question 46.
Give an account on steroid hormones?
Steroid hormones can easily cross the cell membrane and bind to their receptors, which are intracellular or intranuclear. Upon binding to the receptors, they pair up with another receptor – hormone complex (dimerize). This dimer tan then binds to DNA and alters its transcription. The effect of steroid hormones t such as aldosterone, oestrogen, FSH is long-lived, as they alter the amount of mRNA and protein in a cell.

Question 47.
Brief the characters of ammo acid-derived hormones?
Amino acid-derived hormones are derived from one or two amino acids with a few additional modifications. Thyroid hormone is synthesized from tyrosine and includes the addition of several iodine atoms. Epinephrine an amino acid derivative may function through the second messenger system like peptide hormones or they may actually enter the cell and function like steroid hormones.

Question 48.
Why steroid component drugs be avoided?
The abuse of anabolic steroids can cause serious health problems like high BP, heart diseases, liver damage, cancer, stroke, and blood clots. Other side effects of steroid use include nausea, vomiting, ligament and tendon injuries, headache, joint pain, muscle cramps, diarrhea, sleep problem etc.

Question 49.
Define BMR?
Basal Metabolic Rate (BMR) is the amount of energy needed to keep the body at rest.

Question 50.
Naturally occurring amines that function as neurotransmitters. They are characterized by a catechol group in which an amine group is attached, eg: Epinephrine and norepinephrine.

Question 51.
What is the Limbic system?
It is a collection of special Structures located in the. middle of the brain. lt is also known as the paleomammalian brain. It controls emotions, behavior motivation of long-term memory, and olfaction.

Question 52.
Define acidosis.
A condition characterized by lower blood pH, due to the increase of keto acids (ketosis).

Question 53.
What are the bright future opportunities for endocrinologists?
An endocrinology career is a medical career that involves studying hormones and their effects on the human body. They investigate and find new ways Of treatment of hormonal imbalance. Just like doctors endocrinologists speak with patients about their medical history and share current findings. They also study the test result of patients and advise them for treatment.

Question 54.
Write any five hypos and hyperactivity, of the endocrine gland and its related disorders.
Dwarfism, gigantism acromegaly cretinism, and myxodema.

1. Which of the following pair is exclusive endocrine gland:
(a) Thyroid and parathyroid
(b) Hypothalamus and pituitary
(c) Thymus and heart
(d) Thyroid and pancreas
(a) Thyroid and parathyroid

2. Which is considered as a neuroendocrine gland?
(a) Thymus
(b) Pineal gland
(c) Hypothalamus
(d) gastro intestinal tract
(c) Hypothalamus

3. Choose the correct partial, endocrine gland
(a) Placenta
(b) Pineal body
(c) Pituitary
(d) Parathyroid
(a) Placenta

4. The posterior pituitary is connected with the hypothalamus by a nerve bundle called:
(a) Hypophyseal portal system
(b) Dendrites of the neurohypophysis
(c) Axons of neurphypophysis
(d) Hypothalamic-hypophyseal axis.
(d) Hypothalamic-hypophyseal axis.

5. The pituitary gland is located in:
(a) Rathke’s pouch
(b) Sella turcica
(c) Behind the third ventride of the brain
(d) Just above the heart
(b) Sella turcica

6. Pituitary gland is connected to the hypothalamic region of the brain by a stalk called:
(a) Isthmus
(b) Infundibulum
(c) Optic chiasma
(d) Mamillary body
(b) Infundibulum

7. The anterior lobe of the pituitary originates from the embryonic invagination of pharyngeal epithelium called:
(a) Pharyngeal pouches
(b) Sella turcica
(c) Acinus
(d) Rathke’s pouch
(d) Rathke’s pouch

8. The anterior lobe of pituitary secret …………. hormone in lower animals only
(a) Oxytocin
(b) Melanocyte stimulating hormone
(c) Antidiuretic hormone
(d) Leuteotrophic hormone
(b) Melanocyte stimulating hormone

9. The function of melanocyte-stimulating hormone is:
(a) Stimulates cartilage formation
(b) Induces ovulation
(c) Induces pigmentation in the skin
(d) Stimulates insulin secretion
(c) Induces pigmentation in the skin

10. Adrenricorticotrophic hormone secretion is regulated by:
(a) Stimulation
(b) Negative feedback system
(c) Neurotransmitters
(d) Blood
(b) Negative feedback system

11. ………… hormone is otherwise called antidiuretic hormone.
(a) Oxytocin
(b) Luteinizing hormone (LH)
(c) ACTH
(d) Vasopressin
(d) Vasopressin

12. ………… gland is commonly called the “Master gland” of the body.
(a) Pineal gland
(b) Thymus
(c) Pituitary gland
(c) Pituitary gland

13. ………. gland is also called hypothalamus cerebri
(a) Pituitary
(c) Pancreas
(d) Thyroid
(a) Pituitary

14. Pineal gland secretes …………. hormone which plays a central role in the regulation of circadian rhydim of bur body.
(a) Melanin.
(b) Melatonin
(c) Melanocyte stimulating hormone
(d) Mineralocorticoids
(b) Melatonin

15. ………… is the largest endocrine gland of our body.
(c) Thyroid
(d) Pancreas
(c) Thyroid

16. The two lateral lobes of the thyroid gland is connected by a median tissue mass called:
(a) Acini
(c) Isthmus
(d) Elastic cartilage
(c) Isthmus

17. ………… is essential for the normal synthesis of thyroid hormones.
(a) Calcium
(b) Potassium
(c) Iodine
(d) Magnesium
(c) Iodine

18. ………….. is a genetic disease and is not caused by iodine or thyroxine deficiency.
(a) Exophthalmic goitre
(b) Endemic goitre
(c) Simple goitre

19. The ……….. cells of parathyroid secretes parathyroid hormone.
(a) Oxyphil cells
(b) cuboidal cells
(c) Chief-cells
(d) Islet cells
(c) Chief-cells

20. ……….. hormone is called hypercalcemic hormone.
(a) Thymic humoral factor
(b) Parathormone
(c) Catecholamines
(d) Cortisol
(b) Parathormone

21. ………. endocrine glands are called suprarenal glands.
(a) Pineal
(b) Pituitary
(d) Testis

22. Which pair is called catecholamines
(a) Thymulin, thymosin

23. ………. is also known as stress combat hormone.
(a) Aldosterone
(b) Cortisol
(c) Thymopoietin
(d) Insulin
(b) Cortisol

24. The secretions of the……… gland is called ‘3F’ hormones.
(b) Pancreatic
(c) Medulla of the kidney
(d) Duodenal

25. …………. hormone is called hypoglycemic hormone.
(a) Glucagon
(b) Testosterone
(c) Insulin
(d) Thyroxine
(c) Insulin

26. ……….. is called hyperglycemic hormone.
(a) Glucagon
(b) Insulin
(c) Oestrogen
(c) Thyroxine
(a) Glucagon

27. The half-life period of insulin in plasma is:
(a) 6 minutes
(b) 8 minutes
(c) 10 minutes
(d) 16 minutes
(a) 6 minutes

28. The testis is composed of Leydig cells that secretes several male sex hormones collectively called:
(a) Oestrogen
(b) Progesterone
(c) Androgen
(d) Renin
(c) Androgen

29. Formation of RBC is called:
(a) Erythropoiesis
(b) Gluconeogenesis
(c) Immune response
(d) Ossification
(a) Erythropoiesis

30. In the heart, cardiocytes on the atrial wall’s secretes an important peptide hormone called
(a) Antidiuretic hormone
(b) Atrial natriuretic factor
(c) Cholecystokinin
(d) Inhibiting hormone
(b) Atrial natriuretic factor

31. ………… secreted by the proximal tubules of nephron which promotes calcium and phosphorous absorption from intestine and accelerates bone formation.
(a) Calcitonin
(b) Secretin
(c) Calcitriol
(d) Cholecystokinin
(c) Calcitriol

(a) Gigantism
(b) myxoedema
(c) Acromegaly
(d) Goitre
(c) Acromegaly

33. In infants hypothyroidism causes:
(a) Myxoedema
(b) Cretinism
(c) Tetany
(d) Acromegaly
(b) Cretinism

(b) Cushing’s syndrome
(c) Diabetes mellitus
(d) Diabetes insipidus

35. …………. is known as insulin-dependent diabetes.
(a) Type I
(b) Type II
(c) Type I and II
(d) None of the above
(a) Type I

36. Myxodema is otherwise called:
(a) Simple goitre
(c) Cretinism
(d) Gull’s disease
(d) Gull’s disease

37. Normal blood glucose level before fasting is:
(a) 70 – 110mg/dl
(b) 80 – 120mg/dl
(c) 110- 140mg/dl
(d) 50 – 110mg/dl
(a) 70 – 110mg/dl

38. Normal blood glucose level after food is:
(a) 110 – 140mg/dl
(b) 80 – 120mg/dl
(c) 140 – 260mg/dl
(d) 220 – 300mg/dl
(a) 110 – 140mg/dl

39. Human insulin is produced artificially by method.
(a) Gene targetting
(b) Integrated DNA technology
(c) Genome editing
(d) Recombinant DNA technology
(d) Recombinant DNA technology

40. ……….. is secreted by juxtaglomerular cells which increases blood pressure when angiotensin is formed in blood.
(a) Gastrin
(b) Renin
(c) Secretin
(d) Pepsin
(b) Renin

## TN Board 11th Bio Zoology Important Questions Chapter 10 Neural Control and Coordination

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 10 Neural Control and Coordination which is useful for their exam preparation.

## TN State Board 11th Bio Zoology Important Questions Chapter 10 Neural Control and Coordination

Question 1.
What are the basic functions of the neural systems in higher animals?
The neural system of higher animals is well developed and performs the following basic functions:

1. Sensory functions: It receives sensory input from the internal and external environments.
2. Motor functions: It transmits motor commands from the brain to the skeletal and muscular systems.
3. Autonomic functions: Reflex actions.

Question 2.
What are the divisions of the human neural system?
The human neural system is divided into two, the central neural system (CNS) and the peripheral neural system.

Question 3.
What are the divisions of neurons based on their function?
There are three functional classes of neurons. They are the afferent neurons that take sensory impulses to the Central Neural system (CNS) from the sensory organs; the efferent neurons that carry motor impulses from the CNS to the effector organs; and interneurons that lie entirely within the CNS between the afferent and efferent neurons.

Question 4.
Write the functions performed by Neuroglia.
Neuroglia performs several functions such as providing nourishment to the surrounding neurons; involving, the memory process; repairing the injured tissues due to their dividing and regenerating capacity; and acting as phagocyte cells to engulf the foreign particles at the time of any injury to the brain.

Question 5.
Describe the structure of a neuron, which is the functional unit of the nervous system with a neat sketch.

A neuron is a microscopic structure composed of three major parts namely the cell body (soma), dendrites, and axon. The cell body is the spherical part of the neuron that contains all the cellular organelles as a typical cell (except centriole). The plasma membrane covering the neuron is called neurilemma and the axon is axolemma. The repeatedly branched short fibers coming out of the cell body are called dendrites, which transmit impulses towards the cell body. The cell body and the dendrites contain cytoplasm and granulated endoplasmic reticulum called Nissl’s granules.
An axon is a long fiber, that arises from a cone-shaped area of the cell body called the Axon hillock and ends at the branched distal end.
Axon hillock is the place where the nerve impulse is generated in the motor neurons.
The axon of one neuron branches and forms connections with many other neurons. An axon contains the same organelles found in the dendrites and cell body but lacks Nissl’s granules and Golgi apparatus.
The axon, particularly of peripheral nerves is surrounded by Schwann cells to form the myelin sheath, which acts as an insulator.
Myelin sheath is associated only with the axon; dendrites are always non-myelinated. Schwann cells are not continuous along the axon; so there are gaps in the myelin sheath between adjacent Schwann cells. These gaps are called Nodes of Ranvier. Large myelinated nerve fibers conduct impulses rapidly, whereas non-myelinated fibers conduct impulses quite slowly.
Each branch at the distal end of the axon terminates into a bulb-like structure called a synaptic knob which possesses synaptic vesicles filled with neurotransmitters. The axon transmits nerve impulses away from the cell body to the interneural space or to a neuromuscular junction.

Question 6.
Classify neurons according to their structural difference with a simple diagram.
The neurons are divided into three types based on the number of axons and dendrites they possess.

1. Multipolar neurons have many processes with one axon and two or more dendrites. They are mostly interneurons.
2. Bipolar neurons have two, processes with one axon and one dendrite. These are found in the retina of the eye, inner ear, and the olfactory area of the brain.
3. Unipolar neurons have a single short process and one axon. Unipolar neurons are located in the ganglia of cranial and spinal nerves.

Question 7.
What are the two phases of transmission of nerve impulses?
The transmission of impulse involves two main phases:

1. Resting membrane potential,
2. Action membrane potential.

Question 8.
Write about the “resting membrane potential” state of impulse transmission.
Resting membrane Potential: The electrical potential difference across the plasma membrane of a resting neuron is called the resting potential during which the interior of the cell is negative due to greater efflux of K+ outside the cell than Na+ influx into the cell. When the axon is not conducting any impulses i.e. in resting condition, the axon membrane is more permeable to K+ and less permeable to Na+ ions, whereas it remains impermeable to negatively charge protein ions.

The axoplasm contains a high concentration of K+ and negatively charged proteins and a low concentration of Na+ ions. In contrast, fluid outside the axon (ECF) contains a low concentration of K+ and a high concentration of Na+, and this forms a concentration gradient. This ionic gradient across the resting membrane is maintained by ATP driven Sodium-Potassium pump, which exchanges 3Na+ outwards for 2K+ into the cells. In this state, the cell membrane is said to be polarized. In neurons, the resting membrane potential ranges from – 40mV to – 90mV, and its normal value is – 70mV. The minus sign indicates that the inside of the cell is negative with respect to the outside.

Question 9.
Write briefly about the different phases of action membrane potential.
Action membrane potential: An action potential occurs when a neuron sends information down an axon, away from the cell body. It includes the following phases, depolarization, repolarization, and hyperpolarization.
Depolarization – Reversal of polarity: When a nerve fiber is stimulated, sodium voltage-gated opens and makes the axolemma permeable to Na+ ions; meanwhile the potassium voltage gate closes. As a result, the rate of flow of Na+ ions into the axoplasm exceeds the rate of flow of K+ ions to the outside fluid [ECF]. Therefore, the axolemma becomes positively charged inside and negatively charged outside. This reversal of electrical charge is called Depolarization.
During depolarization, when enough Na+ ions enter the cell, the action potential reaches a certain level, called threshold potential [-55mV], The particular stimulus. which is able to bring the membrane potential to the threshold is called threshold stimulus.
The action potential occurs in response to a threshold stimulus but does not occur at subthreshold stimuli. This is called the all or none principle. Due to the rapid influx of Na+ ions, the membrane potential shoots rapidly up to +45mV which is called the Spike potential.

Repolarisation [Falling Phase]: When the membrane reaches the spike potential, the sodium voltage-gated closes and potassium voltage-gated opens. It checks the influx of Na+ ions and initiates the efflux of K+ ions which lowers the number of positive ions within the cell. Thus, the potential falls back towards the resting potential. The reversal of membrane potential inside the axolemma to negative occurs due to the efflux of K+ ions. This is called Repolarisation.
Hyperpolarization: If repolarization becomes more negative than the resting potential -70 mV to about -90 mV, it is called Hyperpolarization. During this, K+ ion gates are more permeable to K+ even after reaching the threshold level as it closes slowly; hence called Lazy gates. The membrane potential returns to its original resting state when K+ ion channels close completely. During hyperpolarization, the Na+ voltage gate remains closed.

Question 10.
Is there any speed difference among the neurons? Why there is such a difference?
The conduction speed of a nerve impulse depends on the diameter of the axon. The greater the axon’s diameter, the faster is the conduction. The myelinated axon conducts the impulse faster than the non-myelinated axon.

Question 11.
What is ‘saltatory conduction?
The voltage-gated Na+ and K+ channels are concentrated at the nodes of Ranvier. As a result, the impulse jumps node to node, rather than traveling the entire length of the nerve fiber. This mechanism of conduction is called Saltatory Conduction.

Question 12.
How nerve impulse is transmitted in the synaptic region of the neurons?

The junction between two neurons is called a Synapse through which a nerve impulse is transmitted. The first neuron involved in the synapse forms the pre-synaptic neuron and the second neuron is the post-Synaptic neuron. A small gap between the pre and postsynaptic membranes is called Synaptic Cleft that forms a structural gap and a functional bridge between neurons. The axon terminals contain synaptic vesicles filled with neurotransmitters. When an impulse [action potential] arrives at the axon terminals, it depolarizes the presynaptic membrane, opening the voltage-gated calcium channels. The influx of calcium ions stimulates the synaptic vesicles towards the pre-synaptic membrane and fuses with it. In the neurilemma, the vesicles release their neurotransmitters into the synaptic cleft by exocytosis. The released neurotransmitters bind to their specific receptors on the post¬synaptic membrane, responding to chemical signals. The entry of the ions can generate a new potential in the post-synaptic neuron, which may be either excitatory or inhibitory. Excitatory post-synaptic potential causes depolarization whereas inhibitory post-synaptic potential causes hyperpolarization of postsynaptic membrane.

Question 13.
What are the three meninges of the brain? Where are they present.
The brain is located in the cranial cavity and is covered by three cranial meninges.

1. The outer thick layer is Duramater which lines the inner surface of the cranial cavity.
2. The median thin layer is the Arachnoid mater which is separated from the dura mater by a narrow subdural space.
3. The innermost layer is Piamater which is closely adhered to the brain but separated from the arachnoid mater by the subarachnoid space.

Question 14.
Write an account of the structure and function of the forebrain.
Fore Brain, comprises the following regions: Cerebrum and Diencephalon. The cerebrum is the ‘seat of intelligence’ and forms the major part of the brain. The cerebrum consists of an outer cortex, inner medulla, and basal nuclei. The superficial region of the cerebrum is called the cerebral cortex, which looks grey due to the presence of unmyelinated nerve cells. The cerebral cortex consists of the neuronal cell body, dendrites, associated glial, and blood vessels. The surface of the cerebrum shows many convolutions (folds) and grooves. The folds are called gyri (singular gyrus); the shallow grooves between the gyri are called sulci (singular sulcus) and deep grooves are called fissures. These sulci and gyri increase the surface area of the cerebral cortex. Several sulci divide the cerebrum into eight lobes: a pair of frontals, parietals, temporals, and occipital lobes.

A median longitudinal fissure divides the cerebrum longitudinally into two cerebral hemispheres. A transverse fissure separates the cerebral hemispheres from the cerebellum. The hemispheres are connected by a tract of nerve fibers called the corpus callosum. The cerebral cortex has three functional areas namely sensory areas that occur in the parietal, temporal, and occipital lobes of the cortex. They receive and interpret the sensory impulses. The motor area of the cortex which controls voluntary muscular movements lies in the posterior part of the frontal lobes. The areas other than sensory and motor areas are called Association areas that deal with integrative functions such as memory, communications, learning, and reasoning. Inner to the cortex is the medulla which is white in color and acts as a nerve tract between the cortex and the diencephalon.
Diencephalon consists largely of the following three paired structures.
The epithalamus forms the roof of the diencephalon and it is a non-nervous tissue. The anterior part of the epithalamus is vascular and folded to form the choroid plexus. Just behind the choroid plexus, the epithalamus forms a short stalk that ends in a rounded body called the pineal body which secretes the hormone, melatonin which regulates the sleep and wake cycle.
Thalamus is composed of grey mater which serves as a relay center for impulses between the spinal cord, brain stem, and cerebrum. Within the thalamus, information is sorted and edited and plays a key role in learning and memory. It is a major coordinating center for sensory and motor signaling.
Hypothalamus forms the floor of the diencephalon. The downward extension of the hypothalamus, the infundibulum connects the hypothalamus with the pituitary gland. The hypothalamus contains a pair of small rounded bodies called mammillary bodies that are involved in olfactory reflexes and emotional responses to odor. Hypothalamus maintains homeostasis and has many centers which control the body temperature, urge for eating, and drinking. It also contains a group of neurosecretory cells that secrete the hypothalamic hormones. Hypothalamus also acts as the satiety center.
Functions of brain lobes

 Structure Functions Frontal Behaviour, Intelligence, Memory, Movement Parietal Language, Reading, Sensation Temporal Speech, Hearing, Memory Occipital Visual processing

Question 15.
What is corpora quadrigemina? What is its function?
The dorsal portion of the midbrain consists of four rounded bodies called corpora quadrigeminal which acts as a reflex center for vision and hearing.

Question 16.
Given an account of the different areas of the hindbrain region.
Hindbrain: Rhombencephalon forms the hindbrain. It comprises of cerebellum, pons varolii and medulla oblongata. The cerebellum is the second largest part of the brain. It consists of two cerebellar hemispheres and a central worm-shaped part, the vermis. The cerebellum controls and coordinates muscular movements and body equilibrium. Any damage to the cerebellum often results in uncoordinated voluntary muscle movements.
Pons varoli lies infront of the cerebellum between the midbrain and the medulla oblongata. The nerve fibers in the pons Varolii form a bridge between the two cerebellar hemispheres and connect the medulla oblongata with the other region of the brain. The respiratory nuclei found in the pons cooperate with the medulla to control respiration.
Medulla oblongata forms the posterior-most part of the brain. It connects the spinal cord with various parts of the brain. It receives and integrates signals from the spinal cord and sends it to the cerebellum and thalamus. The medulla contains vital centers that control cardiovascular reflexes, respiration, and gastric secretions.

Question 17.
What are the fluid-filled spaces of the brain called? Write few lines about its position.
Ventricles of the Brain: The brain has four hollow, fluid-filled spaces. The C-shaped space found inside each cerebral hemisphere forms the lateral ventricles I and II which are separated from each other by a thin membrane called the septum pellucidum. Each lateral ventricle communicates with the narrow III ventricle in the diencephalon through an opening called interventricular foramen (foramen of Monro). The ventricle III is continuous with the ventricle IV in the hindbrain through a canal called the aqueduct of Sylvius.

Question 18.
Brief an account on the G.S. of the spinal cord with a simple sketch.

In the cross-section of the spinal cord, there are two indentations: the posterior median sulcus and the anterior median fissure. Although there might be slight variations, the cross-section of the spinal cord is generally the same
throughout its length. In contrast to the brain, the grey matter in the spinal cord forms an inner butterfly-shaped region surrounded by the outer white matter. The grey matter consists of neuronal cell bodies and their dendrites, interneurons, and glial cells. White matter consists of bundles of nerve fibers. In the center of the grey matter, there is a central canal that is filled with CSF. Each half of the grey matter is divided into a dorsal horn, a ventral horn, and a lateral horn.
The dorsal horn contains cell bodies of interneurons on which afferent neurons terminate. The ventral horn contains cell bodies of the efferent motor neurons supplying the skeletal muscle. Autonomic nerve fibers, supplying cardiac and smooth muscles and exocrine glands, originate from the cell bodies found in the lateral horn. In the white matter, the bundles of nerve fibers form two types of tracts namely ascending tracts which carry sensory impulses to the brain, and descending tracts which carry motor impulses from the brain to the spinal nerves at various levels of the spinal cord. The spinal cord shows two enlargements, one in the cervical region and another one in the lumbosacral region. The cervical enlargement serves the upper limb and lumbar enlargement serves the lower limbs.

Question 19.
Write the anatomical structure of the spinal cord.
The spinal cord is a long, slender, cylindrical nervous tissue. It is protected by the vertebral column and surrounded by the three membranes as in the brain. The spinal cord extends from the brain stem into the vertebral canal of the vertebral column up to the level of 1st or 2nd lumbar vertebra. So the nerve roots of the remaining nerves are greatly elongated to exit the vertebral column at their appropriate space. The thick bundle of elongated nerve roots within the lower vertebral canal is called the cauda equina (horse’s tail) because of its appearance.

Question 20.
In what way the grey arid white matter of the brain and the spiral cord differs.
The superficial region of the cerebrum is called the cerebral cortex, which looks grey due to the presence of unmyelinated nerve cells. The cerebral cortex consists of the neuronal cell body, dendrites, associated glial, and blood vessels. Inner to the cortex is the medulla which is white in color and acts as a nerve tract between the cortex and the diencephalon. In contrast to the brain, the grey matter in the spinal cord forms an inner butterfly-shaped region surrounded by the outer white matter. The grey matter consists of neuronal cell bodies and their dendrites, interneurons, and glial cells. White matter consists of bundles of nerve fibers. In the center of the grey matter, there is a central canal that is filled with CSF.

Question 21.
On touching a hot pan, the hand is withdrawn rapidly without our willingness. What is this action called? Which part is controlling this action? How it is occurs?
On touching a hot pan, the hand is withdrawn rapidly. This is called reflex action.
The spinal cord remains as a connecting functional nervous structure in between the brain and effector organs. But sometimes when a very quick response is needed, the spinal cord can affect motor initiation as the brain and brings about an effect. This rapid action by the spinal cord is called reflex action. It is a fast, involuntary, unplanned sequence of actions that occurs in response to a particular stimulus. The nervous elements involved in carrying out the reflex action constitute a reflex arc or in other words, the pathway followed by a nerve impulse to produce a reflex action is called a reflex arc.

Question 22.
Draw the schematic sketch of functional components of a reflex arc.
Sensory Receptor: It is a sensory structure that responds to a specific stimulus.
Sensory Neuron: This neuron takes the sensory impulse to the grey (afferent) matter of the spinal cord through the dorsal root of the spinal cord.
Interneurons: One or two interneurons may f serve to transmit the impulses from the d sensory neuron to the motor neuron.
Motor Neuron: It transmits impulses from CNS to the effector organ.
Effector Organs: It may be a muscle or gland which responds to the impulse received.
Sensory organ → Sensory of afferent neuron → Grey matter of the spinal cord → Intermediary or relay neuron → efferent or motor neuron → effector organ.

Question 23.
Write about the types of reflexes.
There are two types of reflexes. They are –

1. Unconditional reflex is an inborn reflex for an unconditioned stimulus. It does not need any past experience, knowledge, or training to occur; eg: Blinking of an eye when a dust particle about to fall into it, sneezing, and coughing due to foreign particles entering the nose or larynx.
2. A conditioned reflex is a response to a stimulus that has been acquired by learning. This does not naturally exist in animals. Only an experience makes it a part of the behavior, eg: The excitement of the salivary gland on seeing and smelling food. The conditioned reflex was first demonstrated by the Russian physiologist Pavlov in his classical conditioning experiment in a dog. The cerebral cortex controls the conditioned reflex.

Question 24.
What is the peripheral neural system? What are its components?

1. Peripheral Neural System (PNS) consists of all nervous tissue outside the CNS.
2. Components of PNS include nerves, ganglia, enteric plexuses, and sensor receptors.

Question 25.
Write out cranial nerves rich are arising from the brain coming under the peripheral neural system.
Cranial nerves: There are 12 pairs of cranial nerves, of which the first two pairs arise from the forebrain and the remaining 10 pairs from the midbrain. Other than the Vagus nerve, which extends into the abdomen, all cranial nerves serve the head and face.

Question 26.
How many spinal nerves emerged from the spinal cord and how are they named?
Spinal nerves:- 31 pairs of spinal nerves emerge out from the spinal cord through spaces called the intervertebral foramina found between the adjacent vertebrae. The spinal nerves are named according to the region of the vertebral column from which they originate.

1. Cervical nerves (8 pairs)
2. Thoracic nerves (12 pairs)
3. Lumbar nerves (5 pairs)
4. Sacral nerves (5 pairs)
5. Coccygeal nerves (1 pair)

Each spinal nerve is a mixed nerve containing both afferent (sensory) and efferent (motor) fibers. It originates as two roots: 1) a posterior dorsal root with a ganglion outside the spinal cord and 2) an anterior ventral root with no external ganglion.

Question 27.
What is the somatic neural system? What are its functions?
The somatic neural system (SNS or voluntary neural system) is the part of the peripheral neural system associated with the voluntary control of body movements via skeletal muscles. The sensory and motor nerves that innervate striated muscles form the somatic neural system. Major functions of the somatic neural system include voluntary movement of the muscles and organs and reflex movements.

Question 28.
What are the components of the autonomic neural system?
An autonomic neural system comprises the following components:
Preganglionic neuron: Whose cell body is in the brain or spinal cord; its myelinated axon exits the CNS as part of cranial or spinal nerve and ends in an autonomic ganglion.
Autonomic ganglion: Consists of the axon of preganglionic neuron and cell bodies of, postganglionic neuron.
Postganglionic neuron: Conveys nerve impulses from autonomic ganglia to visceral effector organs.

Question 29.
Name the neural systems coming under the autonomic system.
The autonomic neural system consists of the Sympathetic – neural system and Parasympathetic neural system.

Question 30.
Write down the differences which differentiate sympathetic nerves from the parasympathetic neural system.

 Sympathetic Neural system (SNS) Parasympathetic Neural system (PNS) SNS originates in the thoracic and lumbar regions of the spinal cord. PNS originates in the cranial region of the brain and the sacral region of the spinal cord. Sympathetic ganglia are linked up to form a chain. Its ganglia remain isolated. Preganglionic fibers are short and postganglionic fibers are long. Preganglionic fibers are long and postganglionic fibers are short. Noradrenaline is produced at the terminal ends of the postganglionic fibers at the effector organs. Hence the system is adrenergic. Acetylcholine is produced at the terminal ends of the postganglionic fibers at the effector organs. Hence the system is cholinergic. Active during stressful conditions preparing the body to face them. Active during relaxing times restoring normal activity after stress. The overall effect is excitatory and stimulating. The overall effect is inhibitory. It is considered as the flight or fight system. It is considered as ‘The Rest and Digest System’ or ‘The Feed and Breed System’.

Question 31.
Name the senses that occur in our brain.
Sensation [awareness of the stimulus] and perception [interpretation of the meaning of the stimulus] occur in the brain.

Question 32.
Classify receptors based on their location.
Receptors are classified based on their location:

1. Exteroceptors are located at or near the surface of the body. These are sensitive to external stimuli and receive sensory inputs for hearing, vision, touch, taste, and smell.
2. Interoceptors are located in the visceral organs and blood vessels.
They are sensitive to internal stimuli. Proprioceptors are also a kind of interoceptors. They provide information about the position and movements of the body. These are located in the skeletal muscles, tendons, joints, ligaments, and in connective tissue coverings of bones and muscles.

Question 33.
What are the muscles involved in placing the eyeball held its position in the orbit of the skull?
The eye is the organ of vision; located in the orbit of the skull and held in its position with the help of six extrinsic muscles. They are superior, inferior, lateral, medial rectus muscles, superior oblique, and inferior oblique muscles.

Question 34.
What are the functions of the extrinsic muscle of the eye?
The extrinsic muscles aid in the movement of the eyes and they receive their nerve innervation from III, IV, and VI cranial nerves.

Question 35.
What are the functions of the eyelids?
The eyelids protect the eyes from excessive light and foreign objects and spread lubricating secretions over the eyeballs.

Question 36.
What are the glands present related to the eyes? What is its function?
Sebaceous glands at the base of the eyelashes are called ciliary glands which secrete a lubricating fluid into the hair follicles. Lacrymal glands, located in the upper lateral region of each orbit, secrete tears.

Question 37.
Write briefly about the structure of L.S of the eye of humans.
The conjunctiva is a thin, protective mucous membrane found lining the outer surface of the eyeball.
The eye has two compartments, the anterior and posterior compartments. The anterior compartment has two chambers, the first one lies between the cornea and iris and the second one lies between the iris and lens. These two chambers are filled with a watery fluid called aqueous humor. The posterior compartment lies between the lens and retina and it is filled with a jelly-like fluid called vitreous humor that helps to retain the Spherical nature of the eye. The eye lens is transparent and biconvex, made up of long columnar epithelial cells called lens fibers.
The eyeball is spherical in nature. The wall of the eyeball consists of three layers: fibrous Sclera, vascular Choroid, and sensory Retina.
The outer coat is composed of dense non-vascular connective tissue. It has two regions: the anterior cornea and the posterior sclera. The cornea is a non-vascular transparent coat formed of stratified squamous epithelium.
Sclera forms the white of the eye and protects the eyeball. Posteriorly the sclera is innervated by the optic nerve.
The choroid is a highly vascularized pigmented layer that nourishes all the eye layers and its pigments absorb light to prevent internal reflection.
Anteriorly the choroid thickens to form the ciliary body and iris. Iris is the colored portion of the eye lying between the cornea and lens. The aperture at the center of the iris is the pupil through which the light enters the inner chamber.
Iris is made of two types of muscles the dilator papillae (the radial muscle) and the sphincter papillae (the circular muscle).In the bright light, the circular muscle in the iris contract; so that the size of the pupil decreases and less light enters the eye. In the dim light, the radial muscle in the iris contract; so that the pupil size increases and more light enters the eye. Smooth muscle present in the ciliary body is called the ciliary muscle which alters the convexity of the lens for near and far vision. The ability of the eyes to focus objects at varying distances is called accommodation which is achieved by suspensory ligament, ciliary muscle, and ciliary body. The suspensory ligament extends from the ciliary body and helps to hold the lens in its upright position. The ciliary body is provided with blood capillaries that secrete a watery fluid called aqueous humor that fills the anterior chamber.
The retina forms the innermost layer of the eye. The neural retina layer contains three types of cells: photoreceptor cells – cones and rods, bipolar cells, and ganglion cells. The yellow flat spot at the center of the posterior region of the retina is called macula lutea which is responsible for sharp detailed vision. A small depression present in the center of the yellow spot is called fovea centralis which contains only cones. The optic nerves and the retinal blood vessels enter the eye slightly below the posterior pole, which is devoid of photoreceptors; hence this region is called a blind spot.

Question 38.
Write the mechanism involved in the vision process.
When light enters the eyes, it gets refracted by the cornea, aqueous humor, and lens and it is focused on the retina and excites the rod and cone cells. The photopigment consists of Opsin, the protein part, and Retinal, a derivative of vitamin A. Light induces dissociation of retinal from opsin and causes the structural changes in opsin. This generates an action potential in the photoreceptor cells and is transmitted by the optic nerves to the visual cortex of the brain, via bipolar cells, ganglia, and optic nerves, for the perception of vision.

Question 39.
Arun cannot able to see the nearby objects clearly. What is his problem with his eye called? How this condition occurs and how can it be rectified?
Myopia.
The affected person can see the nearby objects but not the distant objects. This condition may result due to an elongated eyeball or thickened lens; so that the image of a distant object is formed in front of the yellow spot. This error can be corrected using a concave lens that diverges the entering light rays and focuses it on the retina.

Question 40.
What is long-sightedness? What is the condition of the eye in this case? How can it be rectified?
Hypermetropic (long-sightedness): the affected person can see only the distant objects clearly but not the objects nearby. This condition results due to a shortened – eyeball and thin lens; so the image of the closest object is converged behind the retina. This defect can be overcome by using a convex lens that converges the entering light rays on the retina.

Question 41.
Differentiate the rod and cone cells of the eyes.

 Rod cells Cone cells Rods are responsible for vision in dim light. The cones are responsible for color vision and work best in bright light. The pigment present in the rods is rhodopsin, formed of a protein scotopsin and retinal (an aldehyde of vitamin A). The pigment present in the cones is photopsin, formed of opsin protein and retinal. There are about 120 million rod cells. There may be 6-7 million cone cells. Rods are predominant in the extra fovea region. Cones are concentrated in the fovea region.

Question 42.
Write few lines about the eye lens.
The eye lens is transparent and biconvex, made up of long columnar epithelial cells called lens fibers. These cells are accumulated with the proteins called crystalline.

Question 43.
What is style?
Infection of ciliary glands of the eye by bacteria causes a painful, pus-filled swelling called a Stye.

Question 44.
What is the reason for getting “Madras eye” in human eyes?
Dilation and congestion of the blood vessels of the conjunctiva due to local irritation or infection are the cause of bloodshot eye (conjunctivitis – commonly called Madras eye).

Question 45.
What are the visual pigments present in the cones for color vision?
Visual pigments for color vision are

1. The red cones having the visual pigment, Erythropsin is sensitive to long-wavelength close to 560 nm.
2. The green cones having the pigment, chloropsin is sensitive to the medium wavelength of 530 nm.
3. The blue cones having the pigment, cyanopsin is sensitive to a short-wavelength of 420 nm.

Question 46.
Write the different types of refractive errors that occur in the eye with suitable diagrams?
Myopia (nearsightedness): The affected person can see the nearby objects but not the distant objects. This condition may result due to an elongated eyeball or thickened lens; so that the image of a distant object is formed in front of the yellow spot. This error can be corrected using a concave lens that diverges the entering light rays and focuses them on the retina.
Hypermetropia (long-sightedness): The affected person can see only the distant objects clearly but not the objects nearby. This condition results due to a shortened eyeball and thin lens; so the image of the closest object is converged behind the retina. This defect can be overcome by using a convex lens that converges the entering light rays on the retina.
Presbyopia: Due to aging, the lens loses elasticity and the power of accommodation. Convex lenses are used to correct this defect.
Astigmatism is due to the rough (irregular) curvature of the cornea or lens. Cylindrical glasses are used to correct this error.
Cataract: Due to the changes in the nature of the protein, the lens becomes opaque. It can be corrected by surgical procedures.

Question 47.
Name the parts of the organ of equilibrium involved in the following functions.

1. Linear movement of the body – Maculae
2. Changes in the body position – Perilymph and endolymph
3. Rotational movement of the head – Semicircular canals

Question 48.
Write the anatomy of the ear.
Anatomically, the ear is divided into three regions: the external ear, the middle ear, and the internal ear.
The external ear consists of the pinna, external auditory meatus, and eardrum. The pinna is the flap of elastic cartilage covered by skin. It collects the sound waves. The external auditory meatus is a curved tube that extends up to the tympanic membrane [the ear drum]. The tympanic membrane is composed of connective tissues covered with skin outside and with mucus membrane inside.
There are very fine hairs and wax-producing sebaceous glands called ceruminous glands in the external auditory meatus. The combination of hair and ear wax [cerumen] helps in preventing dust and foreign particles from entering the ear.

The middle ear is a small air-filled cavity in the temporal bone. It is separated from the external ear by the eardrum and from the internal ear by a thin bony partition; the bony partition contains two small membrane-covered openings called the oval window and the round window.
The middle ear contains three ossicles: malleus [hammer bone], incus [anvil bone], and stapes [stirrup bone] which are attached to one another. The malleus is attached to the tympanic membrane and its head articulates with the incus which is the intermediate bone lying between the malleus and stapes. The stapes is attached to the oval window in the inner ear. The ear ossicles transmit sound waves to the inner ear. A tube called Eustachian tube connects the middle ear cavity with the pharynx. This tube helps in equalizing the pressure of air on either side of the eardrum.
The inner ear is the fluid-filled cavity consisting of two parts, the bony labyrinth, and the membranous labyrinths. The bony labyrinth consists of three areas: cochlea, vestibule, and semicircular canals. The cochlea is a coiled portion consisting of 3 chambers namely: scala vestibuli and scala tympani- these two are filled with perilymph; and the scala media is filled with endolymph. At the base of the cochlea, the scala vestibule ends at the ‘oval window’ whereas the scala tympani ends at the ‘round window’ of the middle ear. The chambers scala vestibuli and scala media are separated by a membrane called Reisner’s membrane whereas the scala media and scala tympani are separated by a membrane called Basilar membrane.
Organ of Corti The organ of Corti is a sensory ridge located on the top of the Basilar membrane and it contains numerous hair cells that are arranged in four rows along the length of the basilar membrane. Protruding from the apical part of each hair cell is hair-like structures known as stereocilia. During the conduction of sound waves, stereocilia makes a contact with the stiff gel membrane called the tectorial membrane, a roof-like structure overhanging the organ of Corti throughout its length.

Question 49.
Give an account bn the mechanism of hearing.
Sound waves entering the external auditory meatus fall on the tympanic membrane. This causes the eardrum to vibrate, and these vibrations are transmitted to the oval window through the three auditory ossicles. Since the tympanic membrane is 17-20 times larger than the oval window, the pressure exerted on the oval window is about 20 times more than that on the tympanic membrane. This increased pressure generates pressure waves in the fluid of perilymph. This pressure causes the round window to alternately bulge outward and inward meanwhile the basilar membrane along with the organ of Corti move up and down. These movements of the hair alternately open and close the mechanically gated ion channels in the base of hair cells and the action potential is propagated to the brain as sound sensation through the cochlear nerve.

Question 50.
Write clown the various defects of the ear.
Deafness may be temporary or permanent. It can be further classified into conductive deafness and sensory-neural deafness. Possible causes for conductive deafness may be due to

1. The blockage of the ear canal with earwax,
2. Rupture of the eardrum.
3. Middle ear infection with fluid accumulation.
4. Restriction of ossicular movement.

In sensory-neural deafness, the defect may be in the organ of the Corti or the auditory nerve or in the ascending auditory pathways or auditory cortex.

Question 51.
Explain, the structure of the “Organ of equilibrium”.
Balance is part of a sense called proprioception, which is the ability to sense the position, orientation, and movement of the body. The organ of balance is known as the vestibular system which is located in the inner ear next to the cochlea. The vestibular system is composed of a series of fluid-filled sacs and tubules. These sacs and tubules contain endolymph and are kept in the surrounding perilymph. These two fluids, perilymph, and endolymph, respond to the mechanical forces, during changes occurring in body position and acceleration.
The utricle and saccule are two membranous sacs, found nearest the cochlea, and contain equilibrium receptor regions called maculae that are involved in detecting the linear movement of the head. The maculae contain the hair cells that act as mechanoreceptors. These hair cells are embedded in a gelatinous otolithic membrane that contains small calcareous particles called otoliths.
The canals that lie posterior and lateral to the vestibule are semicircular canals; they are anterior, posterior, and lateral canals oriented at right angles to each other. At one end of each semicircular canal, at its lower end has a swollen area called the ampulla. Each ampulla has a sensory area known as crista ampullar which is formed of sensory hair cells and supporting cells. The function of these canals is to detect the rotational movement of the head.

Question 52.
Write down the receptors present in the skin.
Some of the sensory receptors present in the skin are –

1. Tactile Merkel disc is a light touch receptor lying in the deeper layer of the epidermis.
2. Hair follicle receptors are light-touch receptors lying around the hair follicles.
3. Meissner’s corpuscles are small light pressure receptors found just beneath the epidermis in the dermal papillae. They are numerous in hairless skin areas such as fingertips and soles of the feet.
4. Pacinian corpuscles are the large egg-shaped receptors found scattered deep in the dermis and monitoring vibration due to pressure. It allows detecting different textures, temperatures, hardness, and pain.
5. Ruffini endings that lie in the dermis respond to continuous pressure.
6. Krause end bulbs are thermoreceptors that sense temperature.

Question 53.
What is vitiligo? What are its symptoms?
Vitiligo (Leucoderma) is a condition in which the melanin pigment is lost from areas of the skin, causing white patches, often with no clear cause. Vitiligo is not contagious. It can affect people of any age, gender, or ethnic group. The patches appear when melanocytes fail to synthesis melanin pigment.

Question 54.
What is spike potential?
Due to the rapid influx of Na+ ions, the membrane potential shoots rapidly up to +45mV in the neuron, which is called the spike potential.

Question 55.
Define threshold potential and threshold stimulus.
During depolarization, when enough Na+ ions enter the cell, the action potential reaches a certain level, called threshold potential [-55mV]. The particular stimulus which is able to bring the membrane potential to the threshold is called threshold stimulus.

Question 56.
Define the following.
(i) Leakage channels.
(ii) Ligand-gated channels.
(iii) Voltage-gated channels.

1. Leakage Channels are ionic channels that remain open all the time.
2. Ligand-gated channels are chemically gated channels that open or close in response to chemical stimuli.
3. Voltage-gated channels are mechanically gated channels that open in response to a physical stimulus in the form of vibration such as touch and pressure.

Question 57.
What are actions take place during Ligand-gated channels?
Ligand-gated channels are located between the presynaptic membrane of the first axon and postsynaptic membrane of the cell body of the second neuron [i.e. dendrites and cell bodies]. The neurotransmitter acetylcholine opens ligand channels that allow Na+ and Ca++ ions to diffuse inward and K+ ions diffuse outward.

Question 58.
How potential difference across the axolemma is maintained during leakage channels.
K+ leakage channels are more in number than the Na+ leakage channels. Sarcolemma has greater permeability to K+ ions than Na+ ions. These ions keep moving continuously maintain the potential difference across the axolemma.

Question 59.
Name the two types of voltage-gated channels.
There are two types of voltage-gated channels.

1. Sodium voltage-gated channel
2. Potassium voltage-gated channel.

1. The structural and functional units of the neural system are:
(a) nephrons
(b) neurons
(c) neuroglia
(d) intemeurons
(b) neurons

2. The non-nervous special cells which forms the supporting cells of the nervous tissue are called:
(a) afferent neurons
(b) efferent neurons
(c) neuroglia
(d) dendrons
(c) neuroglia

3. The plasma membrane covering the neuron is called:
(a) cell wall
(b) primary wall
(c) neurilemma
(d) axolemma
(c) neurilemma

4. The area from where the axon arises from the cell body of the neuron is called:
(a) Schwann cells
(b) Node of ranvier
(c) Nissil body
(d) Axon hillock
(d) Axon hillock

5. Nissil’s granules are absent in the ……….. area of the neuron.
(a) dendrites
(b) cell body
(c) axon
(d) myelin sheath
(c) axon

6. The longest cells in the human body are the:
(a) nephrons
(b) axons
(c) dendrons
(d) neurons
(d) neurons

7. ………… is the longest axon in the human body.
(a) Vagus nerve
(b) Cervical nerve
(c) Sciatic nerve
(d) Sacral nerve
(c) Sciatic nerve

8. The axon of the peripheral nerves is surrounded by:
(a) nodes of ranvier
(b) nissil’s bodies
(c) axon hillock
(d) schwann cells
(d) schwann cells

9. The gaps in the myelin sheath of the axon in between the adjacent Schwann cells are called:
(a) synaptic vesicle
(b) synaptic knob
(c) nodes of ranvier
(d) neuromuscular junction
(c) nodes of ranvier

10. In neurons, the normal value of resting membrane potential is:
(a) -70mV
(b) -40m V
(c) -50mV
(d) -17mV
(a) -70mV

11. When a nerve fiber is in the stimulated stage the following action will result:
(a) Sodium voltage-gated opens
(b) Potassium voltage-gated opens
(c) Sodium voltage-gated closes
(d) Ligand-gated opens
(a) Sodium voltage-gated opens

12. If repolarization becomes more negative than the resting potential -70mV to about -90mV, it is called:
(a) depolarization
(b) repolarization
(c) hyperpolarization
(d) hyperpolarization
(c) hyperpolarization

13. Nerve impulses travel at the speed of:
(a) 1-3 m/s
(b) 1-300 m/s
(c) 2-250 m/s
(d) 1-310 m/s
(b) 1-300 m/s

14. The junction between two neurons is called a:
(a) synapse
(b) nodes of Ranvier
(c) synaptic cleft
(d) septum pellucidum
(a) synapse

15. A small gap between the pre and postsynaptic membrane is called:
(a) synaptic cleft
(b) synaptic vesicle
(c) synaptic knob
(d) synapse
(a) synaptic cleft

16. The synaptic vesicles of the axon terminal is filled with:
(a) neurotransmitters
(b) cerebrospinal fluid
(c) plasma
(d) mucus
(a) neurotransmitters

17. The outer dura mater and the median Arachnoid membranes of the. the brain has separated from each other by means of space called:
(a) dural space
(b) arachnoid space
(c) subarachnoid space
(d) subdural space
(d) subdural space

18. ……….. part of the brain is called “Seat of intelligence” and forms the major part of the brain.
(a) Cerebrum
(b) Cerebellum
(c) Diencephalon
(d) Hypothalamus
(a) Cerebrum

19. The cerebral hemispheres of the cerebrum are connected by a tract of nerve fibers called:
(a) infundibulum
(b) corpus callosum
(c) cauda equina
(d) choroid plexus
(b) corpus callosum

20. ……….. serve as a relay center for impulses between the spinal cord, brainstem, and cerebrum.
(a) Hypothalamus
(b) Pineal body
(c) Thalamus
(d) Infundibulum
(c) Thalamus

21. ……….. is the major coordinating center for sensory and motor signaling.
(a) Olfactory bulbs
(b) Brainstem
(c) Thalamus
(c) Thalamus

22. The hypothalamus contains a pair of the small rounded bodies called ……….. which are involved in olfactory reflexes and emotional responses to odor.
(b) mamillary bodies
(c) foramen of Monro
(d) hippocampus
(b) mamillary bodies

23. ………. system of our body is called the “emotional brain.”
(a) Neural system
(b) Muscular system
(c) Sensory receptor system
(d) Limbic system
(d) Limbic system

24. The four rounded bodies in the dorsal portion of the midbrain is called:
(b) mamillary bodies
(c) olfactory bulbs
(d) cauda equina

25. ……….. is the reflex center of the brain for vision and hearing.
(a) Cerebral peduncles
(b) Septum pellucidum
(d) Choroid plexus

26. ………… is the second largest part of the brain.
(a) Cerebrum
(b) Medulla oblongata
(c) Cerebellum
(d) Pons varoli
(c) Cerebellum

27. ………… part of the brain controls and coordinates the muscular movements and body equilibrium.
(a) Cerebrum
(b) Hypothalamus
(c) Vermis
(d) Cerebellum
(d) Cerebellum

28. Cardiovascular reflexes, respiration, and gastric secretions are controlled by:
(a) Medulla oblongata
(b) Interventricular foramen.
(c) Cerebral hemispheres
(d) Cerebellum
(a) Medulla oblongata

29. The 1st and the 2nd ventricles of the brain are communicated with the 3rd ventricle through an opening called:
(a) aqueduct of Sylvius
(b) foramen of Monro
(c) intervertebral foramina
(d) pupil
(b) foramen of Monro

30. The thick bundle of elongated nerve roots within the lower vertebral canal is called the:
(a) choroid plexus
(b) cauda equina
(c) peripheral neural, system
(d) intervertebral foramina
(b) cauda equina

31. There are ……….. pairs of cranial nerves that arise from the brain.
(a) 11
(b) 12
(c) 13
(d) 14
(b) 12

32. There are ……….. pairs of spinal nerves emerge out from the spinal cord.
(a) 30
(b) 29
(c) 31
(d) 13
(c) 31

33. The spinal nerves emerge out from the spinal cord through spaces called:
(a) foramen of Monro
(b) intervertebral foramina
(c) aqueduct of sylvium
(d) synapse
(b) intervertebral foramina

34. …………. located in the upper lateral region of each orbit of secrete tears.
(a) Sebaceous glands
(b) Lacrymal glands
(c) Meibomian glands
(d) Oil glands
(b) Lacrymal glands

35. The excess of aqueous humor in an eye drains out through:
(a) foramen of Monro
(b) canal of schlemm
(c) fovea centralis
(d) macula lutea
(b) canal of schlemm

36. ……….. layer of an eye is the highly vascularized pigmented layer.
(a) Sclera
(b) Choroid
(c) Retina
(d) Eyelids
(b) Choroid

37. The normal value of intraocular pressure of an eye is:
(a) 16 mmHg
(b) 61 mmHg
(c) 22 mmHg
(d) 6 mmHg
(a) 16 mmHg

38. The convexity of the lens of an eye for near and far vision is altered by:
(a) sphincter muscle
(b) ciliary muscle
(c) suspensory ligaments
(d) dilator papillae
(b) ciliary muscle

39. ………. is responsible for sharp detailed vision of eye, which is present in the centre of the posterior region of retina.
(a) Macula lutea
(b) Fovea centralis
(c) Pupil
(d) Canal of Schlemm
(a) Macula lutea

40. The pigmenf present in the cone cells of an eye is:
(a) anthocyanin
(b) carotenoids
(c) phytochrome
(d) photopsin
(d) photopsin

41. Due to aging, the lens of an eye looses its elasticity and the power of accommodation, what is the condition called?
(a) Astigmatism
(b) Myopia
(c) Presbyopia
(d) Hypermetropia
(c) Presbyopia

42. There are about ………. million-rod cells in the human eye.
(a) 1200
(b) 120
(c) 12
(d) 102
(b) 120

43. Match:

 1. Myopia (a) Lens loses elasticity 2. Hypermetropia (b) The lens becomes opaque 3. Presbyopia (b) Rough curvature of cornea or lens 4. Astigmatism (c) Short sightedness 5. Cataract (d) Long sightedness

(a) 1 -(d), 2-(e), 3-(a), 4-(c), 5-(b)
(b) 1-(e), 2-(d),3-(c), 4-(a), 5-(b)
(c) 1-(b), 2-(a),3-(c), 4-(d),5-(e)
(d) 1-(d), 2-(a), 3-(e), 4-(c), 5-(b)
(a) 1 -(d), 2-(e), 3-(a), 4-(c), 5-(b)

44. The wax glands present in an ear is called:
(a) sebaceous glands
(b) meibomian glands
(c) oil glands
(d) ceruminous glands
(d) ceruminous glands

45. The hair cells of the basilar membrane of the inner ear has hair-like projections called:
(a) cilia
(b) flagella
(c) stereocilia
(d) ciliary epithelium
(c) stereocilia

46. The intensity of sound is measured in:
(a) mV
(b) decibels
(c) kelvin
(d) mole
(b) decibels

47. The receptors for taste and smell are called:
(a) chemoreceptors
(b) mechanoreceptors
(c) tactile receptors
(d) phono receptors
(a) chemoreceptors

48. The tongue has many small projections called ……….. which give the tongue an abrasive feel.
(a) papillae
(b) dilator papillae
(c) sphincter papillae
(d) ruffini endings
(a) papillae

49. ………. are the small light pressure receptors found numerous in hairless skin areas such as fingertips and soles of the feet.
(a) Pacinian corpuscles
(b) Meissner’s corpuscles
(c) Krause end bulbs
(d) Tactile Merkel disc
(b) Meissner’s corpuscles

50. ………… are the thermoreceptors found on the skin that sense temperature.
(a) Gustatory epithelial cells
(b) Olfactory receptor cells
(c) Krause end bulbs
(d) Pacinian corpuscles
(c) Krause end bulbs

51. Match the following:

 1. Nerve cell (a) Neurilemma, axolemma 2. Plasma membrane (b) Nissl’s granules 3. Cytoplasm (c) Neuron 4. Endoplasmic reticulum (d) Retina of the eye 5. Bipolar neurons (e) Neuroplasm

(a) 1-(d), 2-(a), 3-(c), 4-(e), 5-(b)
(b) 1-(a), 2-(c), 3-(d), 4-(b), 5-(e)
(c) 1-(d), 2-(e), 3-(c), 4-(a), 5-(b)
(d) 1-(c), 2-(a), 3-(e), 4-(b), 5-(d)
(d) 1-(c), 2-(a), 3-(e), 4-(b), 5-(d)

## TN Board 11th Bio Zoology Important Questions Chapter 9 Locomotion and Movement

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 9 Locomotion and Movement which is useful for their exam preparation.

## TN State Board 11th Bio Zoology Important Questions Chapter 9 Locomotion and Movement

Question 1.
What is locomotion?
The movement of organisms from one place to another in search of food, shelter, mate and escape from predators is called locomotion.

Question 2.
What are the different types of movements that occur in the cells of the human body?
The different types of movements that occur in the cells of our body are amoeboid, ciliary, flagellar and muscular movements.

1. Amoeboid- movement: Cells such as macrophages exhibit amoeboid movement for engulfing pathogens by pseudopodia formed by the streaming movement of the cytoplasm.
2. Ciliary movement: This type of movement occurs in the respiratory passages and genital tracts which are lined by ciliated epithelial cells.
3. Flagellar movement: This type of movement occurs in the cells which are having flagella or whip-like motile organelle. The sperm cells show flagellar movement.
4. Muscular movement: The movement of hands, legs, jaws, tongue are caused by the contraction and relaxation of the muscle which is termed as the muscular movement.

Question 3.
Write the names of the types of muscles?
The muscles are classified into three types, namely skeletal, visceral and cardiac muscles.

Question 4.
Why skeletal muscles are called’ voluntary muscles?
Skeletal muscles control voluntary actions such as walking, running, swimming, writing hence termed as voluntary muscles.

Question 5
Differentiate epimysium, perimysium, and endomysium.
The connective tissue covering the whole muscle is the epimysium, the covering around each fascicle is the perimysium and the muscle fibre is surrounded by the endomysium.

Question 6.
How does myoglobin serve as a reservoir of oxygen?
Myoglobin is a red-coloured respiratory pigment of the muscle fibre. It is similar to haemoglobin and contains the iron group that has an affinity towards oxygen and serves as the reservoir of oxygen. Glycosomes are the granules of stored glycogen that provide glucose during the period of muscle fibre activity.

Question 7.
Differentiate anisotropic bands and isotropic bands.

 Anisotropic band Isotropic band A – bands are called anisotropic. I – bands are called isotropic bands. They are dark bands. They are light bands.

Question 8.
Define neuromuscular junction.
Muscle contraction is initiated by a nerve impulse sent by the central nervous system (CNS) through a motor neuron. The junction between the motor neuron and the sarcolemma of the muscle fibre is called the neuromuscular junction or motor endplate.

Question 9.
Write the schematic presentation of muscle contraction?

Question 10.
Write the two primary types of muscle contractions?
There are two primary types of muscle contractions. They are isotonic contraction and isometric contraction. The types of contractions depend on the changes in the length and tension of the muscle fibres at the time of its contraction.

Question 11.
Differentiate isotonic and isometric contraction for example.

 Isotonic contraction Isometric contraction In isotonic contraction, the length of the muscle changes but the tension remains constant. Here, the force produced is unchanged, eg: lifting dumbbells and weightlifting. In isometric contraction, the length of the muscle does not change but the tension of the muscle changes. Here, the force produced is changed, eg: pushing against a wall, holding a heavy bag.

Question 12.
What are the different types of skeletal muscles? On which basis the classification is done.
The muscle fibres can be classified on the basis of their rate of shortening, either fast or slow and the way in which they produce the ATP needed for contraction, either oxidative or glycolytic. Fibres that contain numerous mitochondria and have a high capacity for oxidative phosphorylation are classified as oxidative fibres. The oxidative fibres are termed red muscle fibres. Fibres that contain few mitochondria but possess a high concentration of glycolytic enzymes and large stores of glycogen are called glycolytic fibres. They are termed white muscle fibres.

Question 13.
Write an account of three types of skeletal muscles.
Skeletal muscle fibres are further classified into three types based on the above classification. They are slow – oxidative fibres, fast – oxidative fibres and fast – glycolytic fibres.

1. Slow – oxidative fibres have low rates of myosin ATP hydrolysis but have the ability to make large amounts of ATP. These fibres are used for prolonged, regular activity such as long-distance swimming. Long-distance runners have a high proportion of these fibres in their leg muscles.
2. Fast-oxidative fibres have high myosin ATPase activity and can make large amounts of ATP. They are particularly suited for rapid actions.
3. Fast-glycolytic fibres have myosin ATPase activity but cannot make as much ATP as oxidative fibres, because of their source of ATP in glycolysis. These fibres are best suited for rapid, intense actions, such as short sprints at maximum speed.

Question 14.
Name the two divisions of the human skeletal system.
In human beings, the skeletal system is made up of 206 bones and cartilages. It is grouped into two principal divisions – the axial skeleton and the appendicular skeleton.

Question 15.
Write down the functions of the skeletal system.

1. Support: It forms a rigid framework and supports the weight of the body against gravity.
2. Shape: It provides and maintains the shape of the body.
3. Protection: It protects the delicate internal organs of the body.
4. Acts as reservoir: It stores minerals such as calcium and phosphate. Fat (Triglyceride) is stored in the yellow bone marrow and represents a source of stored energy for the body.
5. Locomotion: It acts as a lever along with the muscles attached to it.
6. Strength: It can withstand heavyweight and absorbs mechanical shock.
7. As a haemopoietic tissue: Red and White blood cells are produced in the bone marrow of the ribs, spongy bones of vertebrae and extremities of long bones.

Question 16.
Give an account of the structures of the skull.
The skull is composed of two sets of bones, cranial and facial bones. It consists of 22 bones of which 8 are cranial bones and 14 are facial bones. The cranial bones form the hard protective outer covering of the brain and called the brain box. The capacity of the cranium is 1500 cm3. These bones are joined by sutures that are immovable. They are paired parietal, paired temporal and individual bones such as the frontal, sphenoid, occipital and ethmoid.

The large hole in the temporal bone is the external auditory meatus. In the facial bones maxilla, zygomatic, palatine, lacrimal, nasal are paired bones whereas mandible or lower jaw and vomer are unpaired bones. They form the front part of the skull. A single U-shaped hyoid bone is present at the base of the buccal cavity. It is the only bone without any joint. Each middle ear contains three tiny bones- malleus, incus and stapes collectively are called ear ossicles. The upper jaw is formed of the maxilla and the lower jaw is formed of the mandible. The upper jaw is fused with the cranium and is immovable. The lower jaw is connected to the cranium by muscles and is movable. The most prominent openings in the skull are the orbits and the nasal cavity. The foramen magnum is a large opening found at the posterior base of the skull. Through this opening, the medulla oblongata of the brain descends down as the spinal cord.

Question 17.
Write about the bones forming the vertebral column.
The vertebral column is also called the backbone. It consists of 33 serially arranged vertebrae which are interconnected by cartilage known as an intervertebral disc. The vertebral column extends from the base of the skull to the pelvis and forms the main framework of the trunk. The vertebral column has five major regions. They are the Cervical, Thoracic, Lumbar, Sacrum (5 sacral vertebrae found in the infant which are fused to form one bone in the adult) and Coccyx (4 coccygeal vertebrae found in the infant which are fused to form one bone in the adult).
Each vertebra has a central hollow portion, the neural canal, through which the spinal cord passes. The first vertebra is called the atlas and the second vertebra is called the axis. Atlas is articulated with the occipital condyles.
The vertebral column protects the spinal cord, supports the head and serves as the point of attachment for the ribs and musculature of the back.

Question 18.
What is the sternum? What is its importance?
The sternum is a flat bone on the midventral line of the thorax. It provides space for the attachment of the thoracic ribs and abdominal muscles.

Question 19.
Write the features of the pectoral girdle.
The upper limbs are attached to the pectoral girdles. These are very light and allow the upper limbs a degree of mobility not seen anywhere else in the body. The girdle is formed of two halves. Each half of the pectoral girdle consists of a clavicle or collar bone and a scapula. The scapula is a large, thin, triangular bone situated in the dorsal surface of the ribcage between the second and seventh ribs. It has a slightly elevated ridge called, the spine which projects as a flat, expanded process called the acromion. The clavicle articulates with this process. Below the acromion is a depression called the glenoid cavity which articulates with the head of the humerus to form the shoulder joint. Each clavicle is a long slender bone with two curvatures that lies horizontally and connects the axial skeleton with the appendicular skeleton.

Question 20.
Write about the bones of the upper limb.
The upper limb consists of 30 separate bones and is specialized for mobility. The skeleton of the arm, the region between the shoulder and elbow is the humerus. The head of the humerus articulates with the glenoid cavity of the scapula and forms the shoulder joint. The distal end of the humerus articulates with the two forearm bones the radius and ulna. The forearm is the region between the elbow and the wrist. The Olecranon process is situated at the upper end of the ulna which forms the pointed portion of the elbow. The hand consists of carpals, metacarpals and phalanges.
Carpals, the wrist bones, 8 in number are arranged in two rows of four each. The anterior surface of the wrist has a tunnel-like appearance, due to the arrangement of carpals with the ligaments. This tunnel is termed a carpal tunnel.
Metacarpals, the palm bones are 5 in number and phalanges the bones of the digit are 14 in number.

Question 21.
Differentiate false ribs from floating ribs.

 False ribs Floating ribs The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but joined with the cartilaginous (hyaline cartilage) part of the seventh rib. These are called ‘false ribs’ or vertebra-chondral ribs. The last 11th and 12th pairs of ribs are not connected ventrally. Therefore, they are called as ‘floating ribs’ or vertebral ribs. Thoracic vertebrae, ribs and sternum together form the ribcage.

Question 22.
Write about the pelvic girdle briefly.
The pelvic girdle is a heavy structure specialized for weight-bearing. It is composed of two hip bones called coxal bones that secure the lower limbs to the axial skeleton. Together, with the sacrum and coccyx, the hip bones form the basin like the bony pelvis. Each coxal bone consists of three fused bones, ilium, ischium and pubis. At the point of fusion of ilium, ischium, and pubis a deep hemispherical socket called the acetabulum is present on the lateral surface of the pelvis. It receives the head of the femur or thigh bone at the hip joint and helps in the articulation of the femur. Ventrally the two halves of the pelvic girdle meet and form the pubic symphysis containing fibrous cartilage. The ilium is the superior flaring portion of the hip bone. Each ilium forms a secure joint with the sacrum posteriorly. The ischium is a curved bar of bone. The V-shaped pubic bones articulate anteriorly at the pubic symphysis. The pelvis of males is deep and narrow with larger heavier bones and the female is shallow, wide and flexible, in nature, and this helps during pregnancy which is influenced by female hormones.

Question 23.
Write an account on the bones of the lower limb.
The lower limb consists of 30 bones that carry the entire weight of the erect body and is subjected to exceptional forces when we jump or run. The bones of the lower limbs are thicker and stronger than the upper limbs.
The three segments of each lower limb are the thigh, the leg or the shank and the foot.
The femur is the single bone of the thigh. It is the largest, longest and strongest bone in the body. The head of the femur articulates with the acetabulum of the pelvis to form the hip joint. Two parallel bones, the tibia and fibula, form the skeleton of the shank. A thick, triangular patella forms the knee cap, which protects the knee joint anteriorly and improves the leverage of thigh muscles acting across the knee. The foot includes the bones of the ankle, the tarsus, the metatarsus and the phalanges or toe bones. The foot supports our body weight and acts as a lever to propel the body forward while walking and running.
The tarsus is made up of seven bones called tarsals. The metatarsus consists of five bones called metatarsals. The arrangement of the metatarsals is parallel to each other. There are 14 phalanges in the toes which are smaller than those of the fingers.

Question 24.
Differentiate carpals and metacarpals from tarsus and metatarsus.

 Carpals and metacarpals Tarsus and Metatarsus The hand consists of carpals and metacarpals. The foot includes the bones called tarsus and metatarsus. Carpals are the wrist bones that are eight in number and are arranged in two rows of four each. The tarsus is made up of seven bones called tarsals. Metacarpals are five in number. Metatarsals are five bones in number.

Question 25.
Given an account of the structure of a long bone with a neat diagram.
Atypical long bone has a diaphysis, epiphyses (singular-epiphysis) and membranes. A tubular diaphysis or shaft forms the long axis of the bone. It is constructed of a thick collar of compact bone that surrounds a central medullary cavity or marrow cavity. The epiphyses are the bone ends. Compact bone forms the exterior of epiphyses and their interior contains spongy bone with red marrow. The region where the diaphysis and epiphyses meet is called the metaphysis. The external surface of the entire bone except the joint surface is covered by a double-layered membrane called the periosteum. The outer fibrous layer is dense irregular connective tissue. The inner osteogenic layer consists of osteoblasts (bone-forming cells) which secrete bone matrix elements and osteoclasts (bone-destroying cells). In addition, there are primitive stem cells, osteogenic cells, that give rise to the osteoblasts. The periosteum is richly supplied with nerve fibres, lymphatic vessels and blood vessels. Internal bone surfaces are covered with a delicate connective tissue membrane called the endosteum. The endosteum covers the trabeculae of spongy bone and lines the canals that pass through the compact bone. It also contains both osteoblasts and osteoclasts. Between the epiphysis and diaphysis epiphyseal plate or growth, the plate is present.

Question 26.
(i) What are joints?
(ii) Write its importance.
(i) The joints are points of contact between bones.
(ii) Joints are essential for all types of movements performed by the bony parts of the body.

Question 27.
Write about the different types of joints with examples.

1. Fibrous joints or Synarthroses: They are immovably fixed joints in which no movement between the bones is possible. Sutures of the flat skull bones are fibrous joints.
2. Cartilaginous joints or Amphiarthroses: They are slightly movable joints in which the joint surfaces are separated by cartilage and slight movement is only possible.

Question 28.
Write about the different disorders of the muscular system.
Myasthenia gravis: An autoimmune disorder affecting the action of acetylcholine at the neuromuscular junction leading to fatigue, weakening and paralysis of skeletal muscles. Acetylcholine receptors on the sarcolemma are blocked by antibodies leading to weakness of muscles. When the disease progresses, it can make chewing, swallowing, talking and even breathing difficult.
Tetany: Rapid muscle spasms occur in the muscles due to deficiency of parathyroid hormone resulting in reduced calcium levels in the body.
Muscle fatigue: Muscle fatigue is the inability of a muscle to contract after repeated muscle contractions. This is due to lack of ATP and accumulation of lactic acid by the anaerobic breakdown of glucose will lead to rigour Mortis.
Atrophy: A decline or cessation of muscular activity results in the condition called atrophy which results in the reduction in the size of the muscle and makes the muscle become weak, which occurs with lack of usage as in chronic bedridden patients.
Muscle pull: Muscle pull is actually a muscle tear. A traumatic pulling of the fibres produces a tear known as a sprain. This can occur due to sudden stretching of muscle beyond the point of elasticity. Back pain is a common problem caused by muscle pull due to improper posture with static sitting for long hours.
Muscular dystrophy: The group of diseases collectively called muscular dystrophy are associated with the progressive degeneration of skeletal muscle fibres, weakening the muscles and leading to death from lung or heart failure. The most common form of muscular dystrophy is called Duchene Muscular Dystrophy (DMD).

Question 29.
Differentiate muscle fatigue from muscle pull.
Muscle fatigue is the inability of a muscle to contract after repeated muscle contractions. Muscle pull is actually a muscle tear. A traumatic pulling of the fibres produces a tear known as a sprain.

Question 30.
What is Duchene muscular dystrophy?
The progressive degeneration of skeletal muscle fibres, weakening the muscles and leading to death from lung or heart failure. The most common form of muscular dystrophy is called Duchene Muscular Dystrophy (DMD).

Question 31.
Write briefly about the disorders of the skeletal system.
Arthritis and osteoporosis are the major disorders of the skeletal system.
(i) Arthritis: Arthritis is an inflammatory (or) degenerative disease that damages, the joints. There are several types of arthritis.
(a) Osteoarthritis: The bone ends of the knees and other freely movable joints wear away as a person ages. The joints of the knees, hip, fingers and vertebral column are affected.
(b) Rheumatoid arthritis: The synovial membranes become inflamed and there is an accumulation of fluid in the joints. The joints swell and become extremely painful. It can begin at any age but symptoms usually emerge before the age of fifty.
(c) Gouty arthritis or gout: Inflammation of joints due to accumulation of uric acid crystals or inability to excrete it. It gets deposited in synovial joints.
(ii) Osteoporosis: It occurs due to deficiency of vitamin D and hormonal imbalance. The bone becomes soft and fragile. It causes rickets in children and osteomalacia in adult females. It can be minimized with adequate calcium intake, vitamin D intake and regular physical activities.

Question 32.
Define gout.
Gouty arthritis or gout: Inflammation of joints due to accumulation of uric acid crystals or inability to excrete it. It gets deposited in synovial joints.

Question 33.
What is osteoporosis?
Osteoporosis occurs due to deficiency of vitamin D and hormonal imbalance. The bone becomes soft and fragile. It causes rickets in children and osteomalacia in adult females. It can be minimized with adequate calcium intake, vitamin D intake and regular physical activities.

Question 34.
What are the four basic categories of exercises?
Endurance or aerobic activities increase the breathing and heart rate. They keep the circulatory system healthy and improve overall fitness. Strength exercises make the muscles stronger. They help to stay independent and carry out everyday activities such as climbing stairs and carrying bags.
Balance exercises help to prevent falls which is a common problem in older adults. Many strengthening exercises also improve balance.
Flexibility exercises help to stretch body muscles for more freedom of joint movements.

1. The sperm cells show ………. movement.
(a) amoeboid movement
(b) ciliary movement
(c) flagellar movement
(d) muscular movement
(c) flagellar movement

2. Muscles are made of cells called:
(a) mucocytes
(b) myocytes
(c) monocytes
(d) chondrocytes
(b) myocytes

3. Skeletal muscle is attached to the bone by a bundle of collagen fibres known as:
(a) fascicle
(b) myofibrils
(c) tendon
(d) sarcomere
(c) tendon

4. The cytoplasm of the muscle fibre is called:
(a) sarcoplasm
(b) protoplasm
(c) leucoplast
(d) mycoplasma
(a) sarcoplasm

5. ………… is red coloured respiratory pigment of the muscle fibre.
(a) Haemoglobin
(b) Haemocyanin
(c) Myoglobin
(d) phycoerythrin
(c) Myoglobin

6. The functional unit of the skeletal muscle is:
(a) fascicle
(b) epimysium
(c) myofibrils
(d) sarcomere
(d) sarcomere

7. Contraction of the muscle depends on the presence of contractile proteins such as
(a) actin and troponin
(b) troponin and tropomyosin
(c) myosin and tropomyosin
(d) actin and myosin
(d) actin and myosin

8. Sliding filament theory was proposed by:
(a) Huxley and Rolf
(b) Niels Bohr
(c) Max Planck
(d) Bill Nye
(a) Huxley and Rolf

9. The secretions in the neuromuscular junction is:
(a) melatonic
(b) acetylcholine
(c) hormones
(d) neurotransmitter
(b) acetylcholine

10. The skeletal system is derived from the embryonic:
(a) ecotoderm
(b) myotome
(c) mesoderm
(d) endoderm
(c) mesoderm

11. The number of facial bones are:
(a) 26
(b) 25
(c) 14
(d) 15
(c) 14

12. The appendicular skeleton consists of ………. number of bones
(a) 206
(b) 126
(c) 134
(d) 203
(b) 126

13. The cervical vertebra supporting the head is:
(a) axis
(b) Atlas
(c) sacral
(d) lumbar
(b) Atlas

14. The ………… pairs of ribs in rib cage are called floating ribs.
(a) 8th and 9th
(b) 10th and 12th
(c) 11th and 12th
(d) first seven pairs
(c) 11th and 12th

15. The head of the humerus bone articulates with the ………… of the pectoral girdle to form the shoulder joint.
(a) acetabulum
(b) glenoid cavity
(c) orbicularis oris
(d) scapula
(b) glenoid cavity

16. The process which is situated at the upper end of the ulna, which forms the pointed portion of the elbow is called:
(a) patella
(b) olecranon process
(c) diaphysis
(d) ischium
(b) olecranon process

17. The head of the femur fits into the cavity called:
(a) glenoid fossa
(b) occipital condyle
(c) marrow cavity
(d) acetabulum
(d) acetabulum

18. The bone-forming cells are named as:
(a) osteoclasts
(b) Atlas
(c) osteoblasts
(d) mutated cells
(c) osteoblasts

19. The sutures of the flat skull bones are of ………… type of joints
(a) cartilaginous joints
(b) synovial joints
(c) fibrous joints
(c) fibrous joints

20. The joint between the humerus and the pectoral is of ……….. type.
(a) cartilaginous joint
(c) ball and socket joint
(d) pivot joint
(c) ball and socket joint

21. Knee joint is of ……….. type of joint.
(a) pivot joint
(b) ball and socket joint
(d) hinge joint
(d) hinge joint

22. Carpal tunnel syndrome is mostly seen among:
(a) athlete’s
(b) software professionals
(c) women
(d) weight lifters
(b) software professionals

23. Muscle fatigue is due to lack of:
(a) acetylcholine
(c) ATP
(d) Lactic acid
(c) ATP

24. Inflammation of joints due to accumulation of uric acid crystals is called:
(a) Gout
(b) Muscular atrophy
(c) Tetany
(d) Arthritis
(a) Gout

25. ………… exercises keep the circulatory system healthy and improve overall fitness.
(a) Endurance
(b) Balance
(c) Strength
(d) Flexibility
(a) Endurance

26. ………… is applied to study skeletal muscle physiology, anatomy and pathology.
(a) Diffusion tensor imaging
(b) Lithotripsy
(c) Angioplasty
(d) Computed tomography.
(a) Diffusion tensor imaging

27. Match the correct pair.

 (i) Cranium (a) 12 (ii) Facial bones (b) 5 (iii) Thoracic vertebra (c) 8 (iv) Lumbar vertebra (d) 7 (v) Tarsal bones (e) 14

(a) (i)-(c), (ii)-(e), (iii)-(a); (iv)-(b), (v)-(d)
(b) (i)-(e), (ii)-(c), (iii)-(b), (iv)-(a), (v)-(d)
(c) (i)-(d), (ii)-(b), (iii)-(e), (iv)-(a), (v)-(c)
(d) (i)-(b), (ii)-(c), (iii)-(e), (iv)-(d), (v)-(a)
(a) (i)-(c), (ii)-(e), (iii)-(a); (iv)-(b), (v)-(d)

28. Match:

 (i) Muscle contraction (a) cockroach (ii) Exoskeleton (b) Man (iii) Endoskeleton (c) Skull (iv) Axial skeleton (d) CNS (v) Foramen magnum (e) 80 bones

(a) (i)-(d), (ii)-(a), (iii)-(b), (iv)-(e), (v)-(c)
(b) (i)-(b), (ii)-(d), (iii)-(c), (iv)-(e), (v)-(a)
(c) (i)-(e), (ii)-(c), (iii)-(d), (iv)-(a), (v)-(b)
(d) (i)-(d), (ii)-(e), (iii)-(b), (iv)-(a), (v)-(c)
(a) (i)-(d), (ii)-(a), (iii)-(b), (iv)-(e), (v)-(c)

29. Match:

 (i) Pivot joints (a) Knee joints (ii) Gliding joints (b) Between Atlas and axis (iii) Hinge joints (c) Between humems and pectoral girdle (iv) Saddle joints (d) Between carpals (v) Ball and socket (e) Between carpals and joint metacarpals

(a) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c), (v)-(e)
(b) (i)-(d), (ii)-(c), (iii)-(e), (iv)-(a), (v)-(b)
(c) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(e), (v)-(c)
(d) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(e), (v)-(d)
(c) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(e), (v)-(c)

## TN Board 11th Bio Zoology Important Questions Chapter 8 Excretion

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 8 Excretion which is useful for their exam preparation.

## TN State Board 11th Bio Zoology Important Questions Chapter 8 Excretion

Question 1.
Mention the three homeostatic processes of kidneys.
Three homeostatic processes namely, osmotic regulation, ionic regulation, and nitrogen excretion.

Question 2.
Define the following terms, (i) Osmotic regulation, (ii) Ionic regulation, (iii) Nitrogen excretion.

1. Osmotic regulation is the control of tissue osmotic pressure which acts as a driving force for the movement of water across biological membranes.
2. Ionic regulation is the control of the ionic composition of body fluids.
3. The process by which the body gets rid of the nitrogenous waste products of protein metabolism is called excretion. Nitrogen excretion is the pathway by which animals excrete ammonia, the toxic nitrogenous end product of protein catabolism.

Question 3.
Differentiate osmoconformers from osmoregulators.

 Osmoconformers Osmoregulators Osmoconformers are able to change their internal osmotic concentration with changes in the external environment, eg: marine molluscs and sharks. Osmoregulators maintain their internal osmotic concentration irrespective of their external osmotic environment, eg: Otters.

Question 4.
What are stenohaline and euryhaline animals?
The stenohaline animals can tolerate only narrow fluctuations in the salt concentration (example Goldfish), whereas the euryhaline animals are able to tolerate wide fluctuations in the salt concentrations, eg: Artemia, tilapia, and salmons.

Question 5.
Define excretion.
The process by which the body gets rid of the nitrogenous waste products of protein metabolism is called excretion.

Question 6.
What are the different waste products of protein metabolism?
Waste products of protein metabolism are trimethylamine oxide (TMO) in marine teleosts, guanine in spiders, hippuric acid, allantoin, allantoic acid, ornithuric acid, creatinine, creatine, purines, pyrimidines, and pterines.

Question 7.
Name the three major nitrogenous waste materials.
The major nitrogenous waste products are ammonia, urea, and uric acid.

Question 8.
What are ammonoteles, uricoteles and ureoteles.

1. Animals that excrete most of its nitrogen in the form of ammonia are called ammonoteles. Many fishes, aquatic amphibians and aquatic insects are ammOnotelic.
2. Reptiles, birds, land snails and insects excrete uric acid crystals, with a minimum loss of water and are called uricoteles. In terrestrial animals, less toxic urea and uric acid are produced to conserve water.
3. Mammals and terrestrial amphibians mainly excrete urea and are called ureoteles.

Question 9.
The animal kingdom shows a wide variety of excretory structures. Name some of the excretory organs and in which animals they are present as well as its functional aspect.

1. Protonephridia are excretory structures with specialized cells in the form of flame cells (cilia) in Platyhelminthes (tapeworm) and solenocytes (flagella) in Amphioxus.
2. Nematodes have rennette cells, Metanephridia are the tubular excretory structures in annelids and molluscs.
3. Malpighian tubules are the excretory structures in most insects.
4. Antennal glands or green glands perform an excretory function in crustaceans like prawns.

Question 10.
What are the structures involved in the excretory system of humans? Write the structure of the kidney with a neat diagram.
The excretory system in humans consists of a pair of kidneys, a pair of meters, urinary bladder, and urethra. Kidneys are reddish-brown, bean-shaped structures that lie in the superior lumbar region between the levels of the last thoracic and third lumber vertebra close to the dorsal inner wall of the abdominal cavity. The right kidney is placed slightly lower than the left kidney. Each kidney weighs an average of 120-170 grams. The outer layer of the kidney is covered by three layers of supportive tissues namely, renal fascia, perirenal fat capsule, and fibrous capsule.

Question 11.
Describe the internal anatomy of the kidney with a neat diagram of L.S. of the kidney.
The longitudinal section of the kidney shows an outer cortex, inner medulla, and pelvis. The medulla is divided into a few conical tissue masses called medullary pyramids or renal – pyramids. The part of the cortex that extends in between the medullary pyramids in the renal columns of Bertini. The center of the inner concave surface of the kidney has a notch Called the renal hilum, through which the ureter, blood vessels, and nerves innervate. Inner to the hilum is a broad funnel-shaped space called the renal pelvis with a projection called calyces. The renal pelvis is continuous with the ureter once it leaves the hilum. The walls of the calyces, pelvis, and ureter have smooth muscles which contract rhythmically. The calyces collect the urine and empties it into the ureter, which is stored in the urinary bladder temporarily. The urinary bladder opens into the urethra through which urine is expelled out.

Question 12.
Give an account of the structure of the nephron with a suitable sketch.
Each kidney has nearly one million complex tubular structures called nephrons. Each nephron consists of a filtering corpuscle called the renal corpuscle (malpighian body) and a renal tubule. The renal tubule opens into a longer tubule called the collecting duct.
The renal tubule begins with a double-walled cup-shaped structure called the Bowman’s capsule, which encloses a ball of capillaries that delivers fluid to the tubules, called the glomerulus. The bowman’s capsule and the glomerulus together constitute the renal corpuscle. The endothelium of the glomerulus has many pores (fenestrae). The external parietal layer of the Bowman’s capsule is made up of simple squamous epithelium and the visceral layer is made of epithelial cells called podocytes. The podocytes end in foot processes which cling to the basement membrane of the glomerulus. The openings between the foot processes are called filtration slits.

The renal tubule continues further to form the proximal convoluted tubule [PCT] followed by a U-shaped loop of Henle (Henle’s loop) that has a thin descending and a thick ascending limb. The ascending limb continues as a highly coiled tubular region called the distal convoluted tubule [DCT], The DCT of many nephrons opens into a straight tube called the collecting duct. The collecting duct runs through the medullary pyramids in the region of the pelvis. Several collecting ducts fuse to form a papillary duct that delivers urine into the calyces, which opens into the renal pelvis.

Question 13.
Write about the capillary bed of the nephrons.
The capillary bed of the nephrons: First capillary bed of the nephron is the glomerulus and the other is the peritubular capillaries.
The glomerular capillary bed is different from other capillary beds in that it is supplied by the afferent and drained by the efferent arteriole. The efferent arteriole that comes out of the glomerulus forms a fine capillary network around the renal tubule called the peritubular capillaries. The efferent arteriole serving the juxtamedullary nephron forms bundles of a long straight vessel called vasa recta and runs parallel to the loop of Henle. Vasa recta are absent or reduced in cortical nephrons.

Question 14.
Name the steps involved in urine formation.
Urine formation involves three main processes namely, glomerular filtration, tubular reabsorption, and tubular secretion.

Question 15.
Describe the process of glomerular filtration.
Blood enters the kidney from the renal artery, into the glomerulus. Blood is composed of large quantities of water, colloidal proteins, sugars, salts, and nitrogenous end products. The first step in urine formation is the filtration of blood that takes place in the glomerulus. This is called glomerular filtration which is a passive process. The fluid that leaves the glomerular capillaries and enters the Bowman’s capsule is called the glomerular filtrate. The glomerular membrane has a large surface area and is more permeable to water and small molecules present in the blood plasma. Blood enters the glomerulus faster with greater force through the afferent arteriole and leaves the glomerulus through the efferent arterioles, much slower. This force is because of the difference in sizes between the afferent and efferent arteriole (afferent arteriole is wider than efferent arteriole) and glomerular hydrostatic pressure which is around 55mm Hg.
Kidneys produce about 180l glomerular filtrate in 24 hours. The molecules such as water, glucose, amino acids, and nitrogenous substances pass freely from the blood into the glomerulus. Molecules larger than 5nm are barred from entering the tubule. Glomerular pressure is the chief force that pushes water and solutes out of the blood and across the filtration membrane. The glomerular blood pressure (approximately 55 mmHg) is much higher than in other capillary beds. The two opposing forces are contributed by the plasma proteins in the capillaries. These include colloidal osmotic pressure (30 mmHg) and capsular hydrostatic pressure (15 mmHg) due to the fluids in the glomerular capsule. The net filtration pressure of 10 mmHg is responsible for renal filtration.

The effective glomerular pressure of 10 mmHg results in ultrafiltration. Glomerular filtration rate (GFR) is the volume of filtrate formed min-1 in all nephrons (glomerulus) of both kidneys. In adults, the GFR is approximately 120-125 mL/min.

Question 16.
The glomerulus of the nephron filtered the blood, and it is called glomerular filtrate. While it is passed through the different segments of the nephron, certain substances are selectively reabsorbed. What is the substance reabsorbed and in which segment? Explain.

1. Tubular reabsorption: This involves the movement of the filtrate back into the circulation. The volume of filtrate formed per day is around 170-180 L and the urine released is around 1.5 L per day, i.e„ nearly 99% of the glomerular filtrate that has to be reabsorbed by the renal tubules as it contains certain substances needed by the body. This process is called selective reabsorption. Reabsorption takes place by the tubular epithelial cells in different segments of the nephron either by active transport or passive transport, diffusion, and osmosis.
2. Proximal convoluted Tubule (PCT): Glucose, lactate, amino acids, Na+, and water in the filtrate are reabsorbed in the PCT. Sodium is reabsorbed by active transport through the sodium-potassium (Na+ – K+) pump in the PCT. Small amounts of urea and uric acid are also reabsorbed.
3. Descending limb of Henle’s loop: It is permeable to water due to the presence of aquaporins, but not permeable to salts. Water is lost in the descending limb, hence Na+ and Cl get concentrated in the filtrate.
4. Ascending limb of Henle’s loop: It is impermeable to water but permeable to solutes such as Na+, Cl and K+.
5. The distal convoluted tubule: It recovers water and secretes potassium into the tubule. Na+, Cl and water remain in the filtrate of the DCT. Most of the reabsorption from this point is dependent on the body’s needs and is regulated by hormones. Reabsorption of bicarbonate (HCO3) takes place to regulate the blood pH. Homeostasis of K+ and Na+ in the blood is also regulated in this region.
6. Collecting duct: It is permeable to water, secretes K+ (potassium ions are actively transported into the tubule), and reabsorbs Na+ to produce concentrated urine.

Question 17.
How ADH by means of a feedback system regulates the function of kidneys.
The functioning of kidneys is efficiently monitored and regulated by hormonal feedback control mechanisms involving the hypothalamus, juxtaglomerular apparatus, and to a certain extent the heart. Osmoreceptors in the hypothalamus are activated by changes in blood volume, body fluid volume, and ionic concentration. When there is excessive loss of fluid from the body or when there is an increase in blood pressure, the osmoreceptors of the hypothalamus respond by stimulating the neurohypophysis to secrete the antidiuretic hormone (ADH) or vasopressin (positive feedback). ADH facilitates the reabsorption of water by increasing the number of aquaporins on the cell surface membrane of the distal convoluted tubule and collecting duct. This increase in aquaporins causes the movement of water from the lumen into the interstitial cells, thereby preventing excess loss of water by diuresis. When you drink excess amounts of your favorite juice, osmoreceptors of the hypothalamus are no longer stimulated and the release of ADH is suppressed from the neurohypophysis (negative feedback) and the aquaporins of the collecting ducts move into the cytoplasm. This makes the collecting ducts impermeable to water and the excess fluid flows down the collecting duct without any water loss. Hence dilute urine is produced to maintain the blood volume.

Question 18.
What is diabetes insipidus? How is it called? What are the symptoms?

1. Vasopressin secretion is controlled by positive and negative feedback mechanisms.
2. Defects in ADH receptors or the inability to secrete ADH leads to a condition called diabetes insipidus.
3. Characterized by excessive thirst and excretion of large quantities of dilute urine resulting in dehydration and fall in blood pressure.

Question 19.
What is the role of ANF in kidneys? How it acts antagonistically to the renin-angiotensin system?
Atrial Natriuretic factor: Excessive stretch of cardiac atrial cells cause an increase in blood flow to the atria of the heart and release Atrial Natriuretic Peptide or factor (ANF) travels to the kidney where it increases Na+ excretion and increases the blood flow to the glomerulus, acting on the afferent glomerular arterioles as a vasodilator or on efferent arterioles as a vasoconstrictor. It decreases aldosterone release from the adrenal cortex and also decreases the release of renin, thereby decreasing angiotensin II. ANF acts antagonistically to the renin-angiotensin system, aldosterone, and vasopressin.

Question 20.
What are the clinical diagnostic results of testing urine in case of diabetes mellitus?
The presence of glucose (glucosuria) and ketone bodies (ketonuria) in the urine are indications of diabetes mellitus.

Question 21.
In what situation there is a chance for the production of hypotonic urine.
Hypotonic urine is formed when the osmotic pressure of the body fluid is decreased due to water retention or solute loss when ADH secretion is lowered. If you drink a large volume of water without eating anything salty, the total body fluid volume increases quickly and the osmolarity decreases. The kidneys increase the volume of urine excreted.

Question 22.
Name and the role played by the other organs in excretion apart from kidneys.

1. Apart from kidneys, organs such as lungs, liver, and skin help to remove wastes.
2. Our lungs remove large quantities of carbon dioxide (181/day) and significant quantities of water every day.
3. The liver secretes bile containing substances like, bilirubin and biliverdin, cholesterol, steroid hormones,, vitamins and drugs which are excreted out along with the digestive wastes.
4. Sweat and sebaceous glands in the skin eliminate certain wastes through their secretions.
5. The sweat produced by the sweat glands primarily helps to cool the body and secondarily excretes Na+ and Cl, small quantities of urea and lactate. Sebaceous glands eliminate certain substances like sterols, hydrocarbons, and waxes through sebum that provides a protective oily covering for the skin.
6. Small quantities of nitrogenous wastes are also excreted through saliva.

Question 23.
Write some disorders caused in relation to the excretory system in humans.
Urinary tract infection: Female’s urethra is very short and its external opening is close to the anal opening, hence improper toilet habits can easily carry faecal bacteria into the urethra. The urethral mucosa is continuous with the urinary tract and the inflammation of the urethra (urethritis) can ascend the tract to cause bladder inflammation (cystitis) or even renal inflammation (pyelitis or pyelonephritis). Symptoms include dysuria (painful urination), urinary urgency, fever, and sometimes cloudy or blood-tinged urine. When the kidneys are inflamed, back pain and severe headache often occur. Most urinary tract infections can be treated by antibiotics.
Renal Failure (Kidney Failure): Failure of the kidneys to excrete wastes may lead to accumulation of urea with a marked reduction in the urine output. Renal failure is of two types, Acute, and chronic renal failure. In acute renal failure, the kidney stops its function abruptly, but there are chances for recovery of kidney functions. In chronic renal failure, there is a progressive loss of .function of the nephrons which gradually decreases the function of kidneys.
Uremia: Uremia is characterized by an increase in urea and other non-protein nitrogenous substances like uric acid and creatinine in the blood. The normal urea level in human blood is about 17-30mg/100mL of blood. The urea concentration rises to 10 times normal levels during chronic renal failure.
Renal calculi: Renal calculi, also called renal stone or nephrolithiasis, is the formation of hard stone-like masses in the renal tubules of the renal pelvis. It is mainly due to the accumulation of soluble crystals of salts of sodium oxalates and certain phosphates.
This results in severe pain called “renal colic pain” and can cause scars in the kidneys. Renal stones can be removed by techniques like pyleothotomy or lithotripsy.
Glomerulonephritis: It is also called Bright’s disease and is characterized by inflammation of the glomeruli of both kidneys and is usually due to post-streptococcal infection that occurs in children. Symptoms are haematuria, proteinuria, salt and water retention, oliguria, hypertension, and pulmonary oedema.

Question 24.
Why is females, urinary infection is more easy and common than in males?
The Female’s urethra is very short and its external opening is close to the anal opening, hence improper toilet habits can easily carry faecal bacteria into the urethra. The urethral mucosa is continuous with the urinary tract and the inflammation of the urethra (urethritis) can ascend the tract to cause bladder inflammation (cystitis) or even renal inflammation.

Question 25.
Write about the hemodialysis process with schematic diagram, (or) Write about an artificial kidney used in the dialysis process.
A dialyzing machine or an artificial kidney is connected to the patient’s body. A dialyzing machine consists of a long cellulose tube surrounded by the dialyzing fluid in a water bath. The patient’s blood is drawn from a convenient artery and pumped into the dialyzing unit after adding an anticoagulant like heparin. The tiny pores in the dialysis tube allow small molecules such as glucose, salts, and urea to enter the water bath, whereas blood cells and protein molecules do not enter these pores. This stage is similar to the filtration process in the glomerulus.
The dialyzing liquid in the water bath consists of a solution of salt and sugar in correct proportion in order to prevent the loss of glucose and essential salts from the blood. The cleared blood is then pumped back to the body through a vein.

Question 26.
What is kidney transplantation? When and how it can be done?
It is the ultimate method for the correction of acute renal failure. This involves the transfer of a healthy kidney from one person (donor) to another person with kidney failure. The donated kidney may be taken from a healthy person who is declared brain dead or from sibling or close relatives to minimize the chances of rejection by the immune system of the host. Immunosuppressive drugs are usually administered to the patient to avoid tissue rejection.

Question 27.
What are aquaporins?
Aquaporins are Water-permeable channels (membrane transport proteins) that allow water to move across the epithelial cells in relation to the osmotic difference from the lumen to the interstitial fluid.

Question 28.
What is osmolarity? How is it expressed?
Osmolarity – (The solute concentration of a solution of water is known as the osmolarity of the solution, expressed as milliosmoles /liter (mOsm/L).

Question 29.
Describe how the countercurrent multiplier mechanism helps in the formation of concentrated urine in humans.

Formation of concentrated urine is accomplished by kidneys using counter current mechanisms. The major function of Henle’s loop is to concentrate Na+ and Cl. There is low osmolarity near the cortex and high osmolarity towards the medulla. This osmolarity in the medulla is due to the presence of the solute transporters and is maintained by the arrangement of the loop of Henle, collecting duct, and vasa recta. This arrangement allows the movement of solutes from the filtrate to the interstitial fluid. At the transition between the proximal convoluted tubule and the descending loop of Henle, the osmolarity of the interstitial fluid is similar to that of the blood – about 300mOsm. Ascending and descending limbs of Henle, create a counter-current multiplier (interaction between flow of filtrate through the limbs of Henle’s and JMN) by active transport. Figure (a) shows, the countercurrent multiplier created by the long loops of Henle of the JM nephrons which creates a medullary osmotic gradient. As the fluid enters the descending limb, water moves from the lumen into the interstitial fluid, and the osmolarity decreases. To counteract this dilution the region of the ascending limb actively pumps solutes from the lumen into the interstitial fluid and the osmolarity increases to about 1200mOsm in the medulla. This mismatch between water and salts creates an osmotic gradient in the medulla. The osmotic gradient is also due to the permeability of the collecting duct to urea.
The vasa recta maintain the medullary osmotic gradient via countercurrent exchanger (the flow of blood through the ascending and descending vasa recta blood vessels) by passive transport. Figure (b) shows the counter current, exchanger where the vasa recta preserve the medullary gradient while removing reabsorbed Water and solutes. This system does not produce an osmotic gradient, but protects the medulla by removal of excess salts from the interstitial fluid and removing reabsorbed water. The vasa recta leaves the kidney at the junction between the cortex and medulla. The interstitial fluid at this point is iso-osmotic to the blood. When the blood leaves the efferent arteriole and enters the vasa recta the osmolarity in the medulla increases (1200mOsm) and results in passive uptake of solutes and loss of water. As the blood enters the cortex, the osmolarity in the blood decreases (300mOsm), and the blood loses solutes and gains water to form concentrated urine (hypertonic). Human kidneys can produce urine nearly four times concentrated than the initial filtrate formed.

1. Animals that can tolerate only narrow fluctuations in salt concentration are called:
(a) Osffioconformers
(b) Euryhaline
(c) Stenohaline
(d) Osmoregulators
(c) Stenohaline

2. Which is euryhaline animal of the following?
(a) Tilapia
(b) gold fish
(c) sharks
(d) otters
(a) Tilapia

3. The waste products of protein metabolism in spider is:
(a) Hippuric acid
(b) Quanine
(c) Allantonic
(d) Creatine
(b) Quanine

4. Mammals and terrestrial amphibians mainly excrete urea and are called:
(a) Uricoteles
(b) Ammonoteles
(c) Ureoteles
(d) Gunotelic
(c) Ureoteles

5. Each kidney of a human weighs about ………… gms.
(a) 120-280
(b) 120-270
(c) 130-170
(d) 120-170
(d) 120-170

6. The primitive kidneys of invertebrates are called:
(a) protonephridia
(b) Mesonephridia
(c) Meganephridia
(d) malpighian tubules
(a) protonephridia

7. (i) Flame cells (a) Nematodes
(ii) Solenocytes (b) Platyhelminthes
(iii) Renfiette cells (c) Amphioxus
(iv) Malpighian tubes (d) Insects
(a) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d)
(b) (i)-(c), (ii)-(b), (iii)-(d), (iv)-(a)
(c) (i)-(a), (ii)-(d), (iii)-(b), (iv)-(c)
(d) (a)-(d), (b)-(c), (c)-(a), (d)-(b)
(b) (i)-(c), (ii)-(b), (iii)-(d), (iv)-(a)

8. The granular cells of the juxta glomerular apparatus secrete an enzyme called:
(a) thymin
(b) renin
(d) angiotensin
(d) angiotensin

9. The process of release of urine from the bladder is called:
(a) micturition
(b) malnutrition
(c) holozoic nutrition
(d) pasteurization
(a) micturition

10. The formation of hard stone-like masses in the renal tubules is called:
(a) uremia
(b) gall stones
(d) renal calculi
(d) renal calculi

11. The renal stone is formed mainly by the accumulation of:
(a) salts of sodium oxalates and certain phosphates
(b) calcium and sodium
(c) creatinine
(d) potassium and phosphate
(a) salts of sodium oxalates and certain phosphates

12. Renal stones can be removed by techniques like:
(a) dialysis
(b) medication
(c) lithotripsy
(d) holistic therapy
(c) lithotripsy

13. Urine is a transparent yellowish fluid, because of the presence of the pigment called;
(a) chlorocruorirt
(b) cytochrome
(c) urochrome
(d) haemocyanin
(c) urochrome

14. Animals that excrete of its nitrogen in the form of ammonia are called:
(a) ammonoteles
(b) uricoteles
(c) urioteles
(d) aminoteles
(a) ammonoteles

15. Animals such as reptiles, birds and insects excrete,uric acid crystals with a minimum ross of water called:
(a) guanoteles
(b) uricoteles
(c) urioteles
(d) aminoteles
(b) uricoteles

16. The excretory organ of platyhelminthes is:
(a) solenocytes
(b) flame cells
(c) rennette cells
(d) green glands
(b) flame cells

17. ……… are the excretory structures in insects.
(a) Flame cells
(b) Antennal glands
(c) Malpighian tubules
(d) Solenocytes
(c) Malpighian tubules

18. ………… is the structural and functional unit of kidneys.
(a) Neurons
(b) Malpighian tubules
(c) Nephrons
(d) Flame cells
(c) Nephrons

19. Absense of ………… in Amphibians and freshwater fishes, produce dilute urine.
(a) Glomerulus
(b) Node of Ranvier
(c) Nephrons
(d) Henle’s loop
(d) Henle’s loop

20. The ureter, blood vessels and nerves innervate the kidney through:
(a) ureter
(b) cortex
(c) medulla
(d) hilum
(d) hilum

21. The biosynthesis of urea takes place in:
(a) blood
(b) kidney
(c) liver
(d) brain
(c) liver

(a) diabetes mellitus
(b) diabetes insipidus
(c) Grave’s disease
(d) renal failure
(b) diabetes insipidus

23. Increase in amount of urea in blood is called:
(a) nephritis
(b) Uremia
(c) Renal calculi
(d) cystitis
(b) Uremia

24. ……….. is not a symptom of Renal failure.
(a) Increase in urea in blood
(b) Formation of hard stone like masses in renal tubules
(c) inflammation of the glomeruli
(d) all the above
(d) all the above

## TN Board 11th Bio Zoology Important Questions Chapter 7 Body Fluids and Circulation

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 7 Body Fluids and Circulation which is useful for their exam preparation.

## TN State Board 11th Bio Zoology Important Questions Chapter 7 Body Fluids and Circulation

Question 1.
In what way the circulatory system helps our body?
The circulatory system helps to maintain the homeostasis of the body fluids and body temperature (heat exchange). The homeostatic regulation of the cardiovascular system maintains blood flow, or perfusion, to the heart and brain.

Question 2.
Why vasovagal syncope (fainting) occurs?
Signals from the nervous system cause a sudden decrease in blood pressure, and the individual faints from lack of oxygen to the brain.

Question 3.
Name the three types of extracellular fluids.
The three types of extracellular fluids are the interstitial fluid or tissue fluid (surrounds the cell), the plasma (fluid component of the blood) and lymph.

Question 4.
What are the four main types of plasma proteins?
The four main types of plasma proteins synthesized in the liver are albumin, globulin, prothrombin and fibrinogen.

Question 5.
Write the functions of the plasma protein.
Albumin maintains the osmotic pressure of the blood. Globulin facilitates the transport of ions, hormones, lipids and assists in immune function. Both Prothrombin and Fibrinogen are involved in blood clotting.

Question 6.
What happens to the constituents of blood immediately after a meal?
The composition of plasma is not always constant. Immediately after a meal, the blood in the hepatic portal vein has a very high concentration of glucose as it is transporting glucose from the intestine to the liver where it is stored. The concentration of the glucose in the blood gradually falls after some time as most of the glucose is absorbed.

Question 7.
If too much protein consumed more than the normal need, what will happen to the excess of protein consumed?
If too much protein is consumed, the body cannot store the excess amino acids formed from the digestion of proteins. The liver breaks down the excess amino acids and produces urea. Blood in the hepatic vein has a high concentration of urea than the blood in other vessels namely, the hepatic portal vein and hepatic artery.

Question 8.
What is the name of the pigment present in RBC? Write its important role.
The red colour of the RBC is due to the presence of a respiratory pigment, haemoglobin dissolved in the cytoplasm. Haemoglobin plays an important role in the transport of respiratory gases and facilitates the exchange of gases with the fluid outside the cell.

Question 9.
What is haematocrit?
Erythropoietin is a hormone secreted by the kidneys in response to low oxygen and helps in the differentiation of stem cells of the bone marrow to erythrocytes (erythropoiesis) in adults. The ratio of red blood cells to blood plasma is expressed as Haematocrit.

Question 10.
Write about the red cells of the blood.

Red blood cells are abundant than other blood cells. There are about 5 million to 5.5 millions of RBC mm”3 of blood in a healthy man and 4.5 – 5.0 million RBC mnr3 in healthy women. The RBCs are very small with a diameter of about 7pm (micrometer). The structure of RBC is shown in Figure. The red colour of the RBC is due to the presence of a respiratory pigment, haemoglobin dissolved in the cytoplasm. Haemoglobin plays an important role in the transport of respiratory gases and facilitates the exchange of gases with the fluid outside the cell (tissue fluid). The biconcave shaped RBCs increases the surface area to volume ratio, hence oxygen diffuses quickly in and out of the cell. The RBCs are devoid of the nucleus, mitochondria, ribosomes and endoplasmic reticulum. The absence of these organelles accommodates more haemoglobin thereby maximising the oxygen-carrying capacity of the cell. The average life span of RBCs in a healthy individual is about 120 days after which they are destroyed in the spleen (graveyard/cemetery of RBCs) and the iron component returns to the bone marrow for reuse. Erythropoietin is a hormone secreted by the kidneys in response to low oxygen and helps in the differentiation of stem cells of the bone marrow to erythrocytes (erythropoiesis) in adults. The ratio of red blood cells to blood plasma is expressed as Haematocrit

Question 11.
Give an account of different types of white blood cell with suitable diagrams.

White blood cells (leucocytes) are colourless, amoeboid, nucleated cells devoid of haemoglobin and other pigments. Approximately 6000 to 8000 per cubic mm of WBCs are seen in the blood of an average healthy individual. The different types of WBCs are shown in Figure. Depending on the presence or absence of granules, WBCs are divided into two types, granulocytes and agranulocytes. Granulocytes are characterised by the presence of granules in the cytoplasm and are differentiated in the bone marrow. The granulocytes include neutrophils, eosinophils and basophils.
Neutrophils are also called heterophils or polymorphonuclear (cells with 3-4 lobes of nucleus connected with delicate threads) cells which constitute about 60% – 65% of the total WBCs. They are phagocytic in nature and appear in large numbers in and around the infected tissues.
Eosinophils have a distinctly bilobed nucleus and the lobes are joined by thin strands. They are non – phagocytic and constitute about 2 – 3% of the total WBCs. Eosinophils increase during certain types of parasitic infections and allergic reactions.
Basophils are less numerous than any other type of WBCs constituting 0.5% – 1.0% of the total number of leucocytes. The cytoplasmic granules are large-sized but fewer than eosinophils. The nucleus is large-sized and constricted into several lobes but not joined by delicate threads. Basophils secrete substances such as heparin, serotonin and histamines. They are also involved in inflammatory reactions.
Agranulocytes are characterised by the absence of granules in the cytoplasm and are differentiated in the lymph glands and spleen. These are of two types, lymphocytes and monocytes. Lymphocytes constitute 28% of WBCs. These have a large round nucleus and a small amount of cytoplasm. The two types of lymphocytes are B and T cells. Both B and T cells are responsible for the immune responses of the body. B cells produce antibodies to neutralize the harmful effects of foreign substances and T cells are involved in cell-mediated immunity.
Monocytes (Macrophages) are phagocytic cells that are similar to mast cells and have kidney shaped nucleus. They constitute 1 – 3% of the total WBCs. The macrophages of the central nervous system are the ‘microglia’, in the sinusoids of the liver they are called ‘Kupffer cells’ and in the pulmonary region they are the ‘alveolar macrophages.

Question 12.
Which type of WBC increase in its amount during allergic reactions? Write a few words about it.

1. Eosinophils increase during certain types of parasitic infections and allergic reactions.
2. Eosinophils have a distinctly bilobed nucleus and the lobes are joined by thin strands. They are non – phagocytic and constitute about 2 – 3% of the total WBCs.

Question 13.
Write a few lines about the two types of lymphocytes.
The two types of lymphocytes are B and T cells. Both B and T cells are responsible for the immune responses of the body. B cells produce antibodies to neutralize the harmful effects of foreign substances and T cells are involved in cell-mediated immunity.

Question 14.
Write short note on platelets.
Platelets are also called thrombocytes that are produced from megakaryocytes (special cells in bone marrow) and lack nuclei. Blood normally contains 1, 50,000 – 3, 50,000 platelets mm-3 of blood. They secrete substances involved in the coagulation or clotting of blood. The reduction in platelet number can lead to clotting disorders that result in excessive loss of blood from the body.

Question 15.
Tabulate the different types of blood group in human with the antigen and antibodies.
Distribution of antigens and antibodies in different blood groups

 Blood group Agglutinogens (antigens) on the RBC Agglutinin (antibodies) in the plasma A A Anti B B B Anti A AB AB No antibodies O No antigens Anti A and Anti B

Question 16.
How can we identify the ‘+ve’ or’-ve’ factor in the blood groups?
Rh factor is a protein (D antigen) present on the surface of the red blood cells in the majority (80%) of humans. This protein is similar to the protein present in Rhesus monkey, hence the term Rh, Individuals who carry the antigen D on the surface of the red blood cells are Rh” (Rh-positive) and the individuals who do not carry antigen D, are Rh (Rh-negative).

Question 17.
What is erythroblastosis foetalis? How is it caused? Can it be treated?
Rh factor compatibility is also checked before blood transfusion. When a pregnant woman is Rh and the foetus is Rh+ incompatibility (mismatch) is observed. During the first pregnancy, the Rh antigens of the foetus does not get exposed to the mother’s blood as both their blood are separated by the placenta. However, a small amount of the foetal antigen becomes exposed to the mother’s blood during the birth of the first child. The mother’s blood starts to synthesize D antibodies. But during subsequent pregnancies, the Rh antibodies from the mother (Rh) enters the foetal circulation and destroys the foetal RBCs. This becomes fatal to the foetus because the child suffers from anaemia and jaundice. This condition is called erythroblastosis foetalis. This condition can be avoided by the administration of anti D antibodies (Rhocum) to the mother immediately after the first childbirth.

Question 18.
Give a brief account of the general structure of blood vessels.
The vessels carrying the blood are of three types; they are arteries, veins and capillaries. These vessels are hollow structures and have complex walls surrounding the lumen. The blood vessels in humans are composed of three layers, tunica intima, tunica media and tunica externa. The inner layer, tunica intima or tunica interna supports the vascular endothelium, the middle layer, tunica media is composed of smooth muscles and an extracellular matrix that contains a protein, elastin. The contraction and relaxation of the smooth muscles result in vasoconstriction and vasodilation. The outer layer, tunica externa or tunica adventitia is composed of collagen fibres.

Question 19.
Describe the process of clotting blood.

The clotting process begins when the endothelium of the blood vessel is damaged and -the connective tissue in its wall is exposed to the blood. Platelets adhere to collagen fibres in the connective tissue and release substances that form the platelet plug which provides emergency protection against blood loss. Clotting factors released from the clumped platelets or damaged cells mix with clotting factors in the plasma. The protein called prothrombin is converted to its active form called thrombin in the presence of calcium and vitamin K. Thrombin helps in the conversion of fibrinogen to fibrin threads. The threads of fibrins become interlinked into a patch that traps blood cell and seals the injured vessel until the wound is healed. After some time fibrin fibrils contract, squeezing out a straw-coloured fluid through a meshwork called serum (Plasma without fibrinogen is called serum).

Question 20.
How blood in the blood vessels are liquid in nature without clotting?
Heparin is an anticoagulant produced in small quantities by mast cells of connective tissue which prevents coagulation in small blood vessels.

Question 21.
What are anastomoses?
These are connections of one blood vessel (arteries) with another blood vessel. They provide an alternate route of blood flow if the original blood vessel is blocked. For example, Arteries in the joints contain numerous anastomoses. This allows blood to flow freely even if one of the arteries closes during bending of the joints.

Question 22.
What are coronary blood vessels? Why did they call so?
Blood vessels that supply blood to the cardiac muscles with all nutrients and remove wastes are the coronary arteries and veins. Heart muscle is supplied by two arteries namely the right and left coronary arteries. These arteries are the first branch of the aorta. Arteries usually surround, the heart in the manner of a crown, hence called the coronary artery.

Question 23.
What is the Law of Laplace? Where the law can be used in the biological system?
The Law of Laplace is used to understand the structure and function of blood vessels and the heart. Laplace law states that the tension in the walls of the blood vessel is proportional to the blood pressure and vessel radius. Blood vessels such as the aorta that is subjected to high pressures have thicker walls than the arterioles that are subjected to low pressures.

Question 24.
Why vigorous exercise after a meal can cause indigestion or abdominal cramps?
When exercising vigorously, blood is rerouted from the digestive organ (food or no food) to the capillary beds of the skeletal muscles where it is needed immediately. This rerouting explains why vigorous exercise after a meal can cause indigestion or abdominal cramps.

Question 25.
What is incomplete double circulation?
Amphibians have two auricles and one ventricle and no interventricular septum whereas reptiles except crocodiles have two auricles and one ventricle and an incomplete interventricular septum. Thus mixing of oxygenated and deoxygenated blood takes place in the ventricles. This type of circulation is called incomplete double circulation.

Question 26.
Write an account on the L.S. of heart with a simple sketch.
The heart is divided into four chambers, the upper two small auricles or atrium and the lower two large ventricles. The two auricles are separated by inter auricular septum and the two ventricles are separated by the interventricular septum The separation of chambers avoids mixing of oxygenated and deoxygenated blood. The auricle communicates with the ventricle through an opening called auriculo ventricular aperture which is guarded by the auriculo ventricular valves. The opening between the right atrium and the right ventricle is guarded by the tricuspid valve (three flaps or cusps), whereas* a bicuspid (two flaps or cusps) or mitral valve guards the opening between the left atrium and left ventricle. The valves of the heart allow the blood to flow only in one direction, i.e., from the atria to the ventricles and from the ventricles to the pulmonary artery or the aorta. These valves prevent the backward flow of blood. The opening of the right and left ventricles into the pulmonary artery and aorta are guarded by aortic and pulmonary valves and are called semilunar valves. The opening and closing of the semilunar valves are achieved by the chordae tendinae. The chordae tendinae are attached to the lower end of the heart by papillary muscles.

Question 27.
What does the following term define?

1. Heartbeat,
2. Systole,
3. Diastole,
4. ‘lub’ and ‘dub’ sound of the heart,
5. Cardiac cycle.

1. Rhythmic contraction and expansion of the heart are called heartbeat.
2. The contraction of the heart is called systole.
3. The relaxation of the heart is called diastole.
4. The first heart sound (lub) is associated with the closure of the tricuspid and bicuspid valves whereas the second heart sound (dub) is associated with the closure of the semilunar valves.
5. The events that occur at the beginning of the heartbeat and last until the beginning of the next beat is called the cardiac cycle.

Question 28.
Write about the series of event takes place in the cardiac cycle.
The series of events takes place in a cardiac cycle.
PHASE 1: Ventricular diastole – The pressure in the auricles increases than that of the ventricular pressure. AV valves are open while the semilunar valves are closed. Blood flows from the auricles into the ventricles passively.
PHASE 2: Atrial systole – The atria contracts while the ventricles are still relaxed. The contraction of the auricles pushes the maximum volume of blood to the ventricles until they reach the end-diastolic volume (EDV). EDV is related to the length of the cardiac muscle fibre. The more the muscle is stretched, greater the EDV and the stroke volume.
PHASE 3: Ventricular systole (isovolumetric contraction) – The ventricular contraction forces the AV valves to close and increases the pressure inside the ventricles. The blood is then pumped from the ventricles into the aorta without a change in the size of the muscle fibre length and ventricular chamber volume (isovolumetric contraction).
PHASE 4: Ventricular systole (ventricular ejection) – Increased ventricular pressure forces the semilunar valves to open and blood is ejected out of the ventricles without backflow of blood. This point is the end of the systolic volume (ESV).
PHASE 5: (Ventricular diastole) – The ventricles begin to relax, pressure in the arteries exceeds ventricular pressure, resulting in the closure of the semilunar valves. The heart returns to phase 1 of the cardiac cycle.

Question 29.
Define cardiac output. How can it be calculated?
The amount of blood pumped out by each ventricle per minute is called cardiac output(CO).
It is a product of heart rate (HR) and stroke volume (SV). Heart rate or pulse is the number of beats per minute.
Pulse pressure = systolic pressure – diastolic pressure.
Stroke volume (SV) is the volume of blood pumped out by one ventricle with each beat. SV depends on ventricular contraction.
CO = HR x SV
SV represents the difference between EDV (amount of blood that collects in a ventricle during diastole) and ESV (volume of blood remaining in the ventricle after contraction).
SV = EDV – ESV

Question 30.
Explain how the Frank-Starling effect protects the heart from an abnormal increase in blood volume.
According to the Frank-Starling law of the heart, the critical factor controlling SV is the degree to which the cardiac muscle cells are stretched just before they contract. The most important factor stretching the cardiac muscle is the amount of blood returning to the heart and distending its ventricles, venous return. During vigorous exercise, SV may double as a result of venous return. The heart’s pumping action normally maintains a balance between cardiac output and venous return. Because the heart is a double pump, each side can fail independently of the other. If the left side of the heart fails, it results in pulmonary congestion and if the right side fails, it results in peripheral congestion. Frank-Starling effect protects the heart from an abnormal increase in blood volume.

Question 31.
What is blood pressure? What are its types? How is it caused?

1. Blood pressure is the pressure exerted on the surface of blood vessels by the blood.
2. There are two types of pressure, systolic pressure and diastolic pressure.
3. Systolic pressure is the pressure in the arteries as the chambers of the heart contracts. Diastolic pressure is the pressure in the arteries when the heart chambers relax.

Question 32.
What is orthostatic hypotension?
The primary reflex pathway for homeostatic control of mean arterial pressure is the baroreceptor reflex. The baroreceptor reflex functions every morning when you get out of bed. When you are lying flat the gravitational force is evenly distributed. When you stand up, gravity causes blood to pool in the lower extremities. The decrease in blood pressure upon standing is known as orthostatic hypotension.

Question 33.
What is ECG? How ECG graph helps the activity of the heart during one cardiac cycle?
Electrocardiogram (ECG): An electrocardiogram (ECG) records the electrical activity of the heart over a period of time using electrodes placed on the skin, arms, legs and chest.
It records the changes in electrical potential across the heart during one cardiac cycle.
A normal ECG shows 3 waves designated as P wave, QRS complex and T wave.

P Wave (atrial depolarisation): It is a small upward wave and indicates the depolarization of the atria. This is the time taken for the excitation to spread through the atria from the SA node. Contraction of both atria lasts for around 0.8 – 1.0 sec.
PQ Interval (AV node delay): It is the onset of the P wave to the onset of the QRS complex. This is from the start of depolarization of the atria to the beginning of ventricular depolarization. It is the time taken for the impulse to travel from the atria to the ventricles (0.12 – 0.21 sec). It is the measure of AV conduction time.
QRS Complex (ventricular depolarisation): No separate wave for atrial depolarization in the ECG is visible. Atrial depolarization occurs simultaneously with ventricular depolarization. The normal QRS complex lasts for 0.06 – 0.09 sec. The QRS complex is shorter than the P wave because depolarization spreads through the Purkinje fibres. Prolonged QRS wave indicates delayed conduction through the ventricle, often caused due to ventricular hypertrophy or due to a block in the branches of the bundle of His.
ST-Segment: It lies between the QRS complex and the T wave. It is the time during which all regions of the ventricles are completely depolarised and reflects the long plateau phase before repolarisation. In the heart muscle, prolonged depolarization is due to retardation of K+ efflux and is responsible for the plateau. The ST segment lasts for 0.09 sec.
T wave (ventricular depolarisation): It represents ventricular depolarization. The duration of the T wave is longer than the QRS complex because repolarisation takes place simultaneously throughout the ventricular depolarisation.

Question 34.
Show the double circulation by way of the diagrammatic sketch.

Question 35.
In what way systemic circulation is different from pulmonary circulation incomplete double blood circulation?
In the systemic circulation, the oxygenated blood entering the aorta from the left ventricle is carried by a network of arteries, arterioles and capillaries to the tissues. The deoxygenated blood from the tissue is collected by venules, veins and vena cava and emptied into the right atrium. In pulmonary circulation, the blood from the heart (right ventricle) is taken to the lungs by the pulmonary artery and the oxygenated blood from the lungs is emptied into the left auricle by the pulmonary vein.

Question 36.
How cardiac activity is regulated?
The type of heart in human is myogenic because the heartbeat originates from the muscles of the heart. The nervous and endocrine systems work together with paracrine signals (metabolic activity) to influence the diameter of the arterioles and alter the blood flow. Neuronal control is achieved through the autonomic nervous system (sympathetic and parasympathetic). Sympathetic neurons release norepinephrine and the adrenal medulla releases epinephrine. The two hormones bind to p – adrenergic receptors and increase the heart rate. The parasympathetic neurons secrete acetylcholine that binds to muscarinic receptors and decreases the heartbeat.

Question 37.
Write about the following disorders of the circulatory system.
(a) Hypertension
(b) Coronary heart disease.
(c) Angina pectoris.
(d) Stroke
(e) Myocardial infarction.
(f) Rheumatoid heart disease.
Hypertension is the most common circulatory disease. The normal blood pressure in man is 120/80 mm Hg. In cases when the diastolic pressure exceeds 90 mm Hg and the systolic pressure exceeds 150 mm Hg persistently, the condition is called hypertension. Uncontrolled hypertension may damage the heart, brain and kidneys.
Coronary heart disease occurs when the arteries are lined by atheroma. The build-up of atheroma contains cholesterol, fibres, dead muscle and platelets and is termed Atherosclerosis. The cholesterol-rich atheroma forms plaques in the inner lining of the arteries making them less elastic and reduces the blood flow. Plaque grows within the artery and tends to form blood clots, forming coronary thrombus. A thrombus in a coronary artery results in a heart attack.
Angina pectoris (ischemic pain in the heart muscles) is experienced during the early stages of coronary heart disease. Atheroma may partially block the coronary artery and reduce the blood supply to the heart. As a result, there is tightness or choking with difficulty in breathing. This leads to angina or chest pain. Usually, it lasts for a short duration of time.
Stroke is a condition when the blood vessels in the brain bursts, (Brain haemorrhage) or when there is a block in the artery that supplies the brain, (atherosclerosis) or thrombus. The part of the brain tissue that is supplied by this damaged artery dies due to a lack of oxygen (cerebral infarction).
Myocardial infarction (Heart failure): The prime defect in heart failure is a decrease in cardiac muscle contractility. The Frank-Starling curve shifts downwards and towards the right such that for a given EDV, a failing heart pumps out a smaller stroke volume than a normal healthy heart.
When the blood supply to the heart muscle or myocardium is remarkably reduced it leads to the death of the muscle fibres. This condition is called a heart attack or myocardial infarction. The blood clot or thrombosis blocks the blood supply to the heart and weakens the muscle fibres. It is also called Ischemic heart disease due to a lack of oxygen supply to the heart muscles. If this persists it leads to chest pain or angina. Prolonged angina leads to death of the heart muscle resulting in heart failure.
Rheumatoid Heart Disease: Rheumatic fever is an autoimmune disease that occurs 2-4 weeks after throat infection usually a streptococcal infection. The antibodies developed to combat the infection cause damage to the heart. Effects include fibrous nodules on the mitral valve, fibrosis of the connective tissue and accumulation of fluid in the pericardial cavity.

Question 38.
What is CPR? Which case is more useful?
CPR is a life-saving procedure that is done at the time of emergency conditions such as when a person’s breath or heartbeat has stopped abruptly in case of drowning, electric shock or heart attack. CPR includes rescue of breath, which is achieved by mouth to mouth breathing, to deliver oxygen to the victim’s lungs by external chest compressions which helps to circulate blood to the vital organs. CPR must be performed within 4 to 6 minutes after cessation of breath to prevent brain damage or death.

Question 39.
How varicose veins are formed? Where are they usual located?
Varicose veins The veins are so dilated that the valves prevent the backflow of blood. The veins lose their elasticity and become congested. Common sites are legs, rectoanal regions (haemorrhoids), the oesophagus and the spermatic cord.

Question 40.
What is embolism?
Embolism is the obstruction of the blood vessel by an abnormal mass of materials such as a fragment of the blood clot, bone fragment or an air bubble. An embolus may lodge in the lungs, coronary artery or liver and leads to death.

Question 41.
What is an aneurysm? How much it is dangerous?
Aneurysm the weakened regions of the wall of the artery or veins bulges to form a balloon-like sac. An unruptured aneurysm may exert pressure on the adjacent tissues or may burst to cause a massive haemorrhage.

Question 42.
Mention the three layers of blood vessels.
The blood vessels in humans are composed of three layers, tunica intima, tunica media and tunica externa.

1. The average blood volume in an adult is:
(a) 5000 ml
(b) 50000 ml
(c) 50 ml
(d) 5000 l
(a) 5000 ml

2. Liver receives oxygenated blood from the heart through:
(a) hepatic portal vein
(b) hepatic vein
(c) hepatic artery
(d) Renal artery
(c) hepatic artery

3. ………… plasma protein is responsible for maintaining the osmotic pressure of blood.
(a) Globulin
(b) Prothrombin
(c) Fibrinogen
(d) Albumin
(d) Albumin

4. ………… plasma proteins are involved in blood clotting.
(a) Albumin and prothrombin
(b) Globulin and albumin
(c) Prothrombin and fibrinogen
(d) Globulin and fibrinogen.
(c) Prothrombin and fibrinogen

5. Healthy man has ………. millions of RBC in mm-3 of blood.
(a) 5.0-5.5
(b) 5.0-6
(c) 4.5-5.0
(d) 5.5-5.6
(a) 5.0-5.5

6. Healthy women has ……….. millions of RBC in mm-3 of blood.
(a) 5.0 – 5.5
(b) 5.5 – 6.0
(c) 4.5 – 5.5
(d) 4.5 – 5.0
(d) 4.5 – 5.0

7. Human RBC has ………… number of nucleus.
(a) 3
(b) 2
(c) 0
(d) 1
(c) 0

8. The average life span of RBCs in a healthy individual is:
(a) 102 days
(b) 201 days
(c) 120 days
(d) 21 days.
(c) 120 days

9. After the life span of RBCs are over they get destroyed in the:
(a) kidneys
(b) bone marrow
(c) thymus
(d) spleen
(d) spleen

10. Differentiation of stem cells of the bone marrow to erythrocytes in adults is called:
(a) erythropoiesis
(b) apoptosis
(c) coagulation
(d) embolus
(a) erythropoiesis

11. ………….. are the types of WBCs which are involved in inflammatory reactions.
(a) Neutrophils
(b) Basophils
(c) Eosinophils
(d) Monocytes
(b) Basophils

12. The macrophages of the liver are called:
(a) microglia
(b) alveolar macrophages
(c) kupffer cells
(d) lymphocytes
(c) kupffer cells

13. Blood normally contains ……….. platelets mm-3 of blood.
(a) 1,00,000 – 2,00,000
(b) 1,50,000 – 3,50,000
(c) 3,50,000 – 4,00,000
(d) 2,50,000 – 3,50,000
(b) 1,50,000 – 3,50,000

14. The Rh factor protein is otherwise called ………… present on the surface of the red blood cells, in the majority of human.
(a) B – antigen
(b) D – antigen
(c) C – antigen
(d) A – antigen
(b) D – antigen

15. The capillaries doesn’t posses the ………… layer of the blood vessel.
(a) tunica media
(b) tunica intima
(c) tunica Externa
(d) epicardium
(a) tunica media

16. Complete double circulation is seen in ………….
(a) amphibians
(b) crocodiles
(c) fishes
(d) annelids
(b) crocodiles

17. ………… valve of the heart is otherwise called mitral valve.
(a) Semilunar valves
(b) Tricuspid valves
(c) Bicuspid valves
(d) Atrial valve
(c) Bicuspid valves

18. The heart sounds can be heard through …………
(a) electrocardiogram
(b) sphygmomanometer
(c) haemocytometer
(d) stethoscope
(d) stethoscope

19. The amount of blood pumped out by each ventricle per minute is called:
(a) pulse pressure
(b) cardiac output
(c) stroke volume
(d) heart rate.
(b) cardiac output

20. The volume of blood pumped out by one ventricle with each beat is called:
(a) Cardiac output
(b) Heartbeat
(c) Stroke volume
(d) systolic pressure
(c) Stroke volume

21. Stroke volume depends on:
(a) Ventricular contraction
(b) Ventricular relaxation
(c) Atrial contraction
(d) Atrial relaxation
(a) Ventricular contraction

22. Blood pressure of man is measured by using:
(a) Electrocardiogram
(b) Sphygmomanometer
(c) Cytometer
(d) Computed tomography
(b) Sphygmomanometer

23. Normal blood pressure in man is:
(a) 120/80 mm Hg
(b) 130/70 mm Hg
(c) 120/60 mm Hg
(d) 150/60 mm Hg
(a) 120/80 mm Hg

24. The QRS complex wave of ECG lasts for ………… seconds.
(a) 0.05 -0.1
(b) 0.06 – 0.09
(c) 0.8
(d) 0.12-0.21
(b) 0.06 – 0.09

25. The blood clot inside the blood vessel is called:
(a) thrombus
(b) embolus
(c) atheroma
(d) plaques
(b) embolus

26. Cells found in the lymphatics are called:
(a) basophils
(b) neutrophils
(c) lymphocytes
(d) monocytes
(c) lymphocytes

27. Plasma without fibrinogen is called:
(a) Plasma protein
(b) Serum
(c) Fibrin
(d) atheroma
(b) Serum

28. The damaged tissues and dead leucocytes are removed from the body in the forms of:
(a) pus
(b) blood
(c) serum
(d) plasma