Category: Important Questions

Students get through the TN Board 12th Chemistry Important Questions Chapter 6 Solid State which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 6 Solid State

Answer the following questions.

Question 1.
Why are solids rigid?
Answer:
In solids, the constituent particles (atoms, molecules, or ions) are not free to move but can only oscillate about their mean position due to strong interatomic or intermolecular forces. This imparts rigidity.

Question 2.
Classify the following as amorphous or crystalline solids.
Answer:
Polyurethane, naphthalene, benzoic acid Teflon, potassium nitrate, cellophane, polyvinyl chloride fiberglass, copper.
Amorphous solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, and fiberglass. Crystalline solids: Naphthalene, benzoic acid, potassium nitrate, and copper.

Question 3.
Explain why glass is considered an amorphous solid.
Answer:
Like liquids, it has a tendency to flow, though very slowly. It does not have a sharp melting point. It is isotropic.

Question 4.
Classify the following solids in different categories based on the nature of intermolecular forces operating in potassium sulfate, tin, benzene, urea, ammonia water, zinc sulfide, graphite, rubidium, argon, silicon carbide.
Answer:
Ionic solids: potassium sulfate, zinc sulfide. Covalent solids: graphite, silicon carbide. Molecular solids: benzene, urea, ammonia water, argon.
Metallic solids: rubidium, tin.

Question 5.
Ionic solids conduct electricity in the molten state but not in the solid state. Explain.
Answer:
In ionic solids, free ions are present in a molten state and they move towards the oppositely charged electrode under the influence of electric current. In a solid state, these ions are fixed in their lattice position and cannot move under the influence of electric current.

Question 6.
Briefly outline the properties of covalent solids.
Answer:

1. They have a high melting point.
2. They are poor thermal and electrical conductors.

Question 7.
What are covalent solids? Give example.
Answer:
In covalent solids, the constituents (atoms) are bound together in a three-dimensional network entirely by covalent bonds. eg; Diamond, silicon carbide, etc.

Question 8.
What are molecular solids? Give examples.
Answer:
In molecular solids, the constituents are neutral molecules. They are held together by weak Van der Waals forces. Generally, molecular solids are soft and they do not conduct electricity, eg: I₂, graphite.

Question 9.
Explain various types of molecular solids for example.
Answer:
Molecular solids are classified into (i) non-polar molecular solids, (ii) polar molecular solids, (iii) hydrogen-bonded molecular solids.

1. In nonpolar molecular solids, the constituent molecules are held together by weak dispersion or London forces, eg: naphthalene, anthracene.
2. In polar molecular solids, the constituent molecules are formed by polar covalent bonds, eg solid CO₂, solid NH₃.
3. In hydrogen-bonded molecular solids, the constituent molecules are held together by hydrogen bonds, eg: ice (H₂O), glucose, urea, etc.

Question 10.
Explain the types of the force of attraction that exist (i) nonpolar molecular solids, (ii) polar molecular solids, (iii) hydrogen-bonded molecular solids.
Answer:

1. In non-polar molecular solids, the constituent particles are held together by weak dispersion or London forces.
2. In polar molecular solids, the constituent molecules are held by covalent bonds. They are attracted to each other by relatively strong dipole-dipole attraction.
3. In hydrogen-bonded molecular solids, the molecules are held together by hydrogen bonds.

Question 11.
What are the characteristics of metallic solids? What type of attractive forces exists among the constituent particles?
Answer:
The constituents of a metallic solid are held together by a metallic bonding. The lattice points are occupied by metals ion and the electron’s charge bond encloses the metal ion.
They are hard, possess a high melting point, They conduct heat and electricity. They possess metallic luster.

Question 12
Give examples for metallic solids.
Answer:
Metal and metal alloys, eg: Cu, Fe, Zn, Ag, Au, Cu, Zn, etc.

Question 13.
Define the following: (i) Lattice point, (ii) Crystal lattice, (iii) Unit cell.
Answer:

1. Lattice point: Each lattice point represents one constituent particle of a solid. This may be an atom, molecule, or ion.
2. The crystal lattice is a three-dimensional arrangement of identical points in the space which represents how the constituent particles (atoms, molecules, or ions) are arranged in a crystal.
3. Unit cell A unit cell is the smallest portion of a space lattice which when repeated in different directions generates the entire lattice.

Question 14.
What are the characteristics of a unit cell?
Answer:
A unit cell is characterized by the three edge lengths or lattice constants a, b, and c and the angle between the edges α, β, and γ.

Question 15.
Briefly explain how constituent particles are arranged in (i) simple cubic, (ii) body-centered cubic, and (iii) face-centered cubic unit cells.
Answer:

1. In the simple cubic unit cell, each comer is occupied by identical atoms or ions or molecules. And they touch along the edges of the cube, do not touch diagonally. The coordination number of each atom is 6.
2. In a body-centered cubic unit cell, each comer is occupied by an identical particle, and in addition to that one atom occupies the body center. Those atoms which occupy the comers do not touch each other, however, they all touch the one that occupies the body center. Hence, each atom is surrounded by eight nearest neighbors and the coordination number is 8.
3. In a face-centered cubic unit cell, identical atoms lie at each comer as well as in the center of each face. Those atoms in the comers touch those in the faces but not each other. The coordination number is 2.

Question 16.
Distinguish between primitive and nonprimitive unit cells.
Answer:

Primitive unit cells	Nonprimitive unit cells
A unit cell that contains one lattice point is called a primitive unit cell. In a primitive unit cell, constituent particles are present only at the comers of a unit cell.	In the nonprimitive unit cells, the constituent particles are present not only at the comer but also in the centers of the face and body of the unit cell.

Question 17.
Define coordination number? What is the coordination number of each constituent particle present at (i) simple cubic, (ii) body-centered cubic, and (iii) face-centered cubic unit cells?
Answer:
The number of constituent particles that surround an atom/ion in a crystal lattice is called the coordination number.

1. The coordination number of each atom in a simple cubic unit cell is 6.
2. The coordination number of each atom in a body-centered cubic unit cell is 8.
3. The coordination number of an atom is a face-centered cubic unit cell is 2.

Question 18.
Explain how will you calculate the number of particles in a unit cell.
Answer:

1. A point that lies at the comer of a unit cell is shared among the eight-unit cells and therefore, only one eight (\(\frac{1}{8}\)th) of each point lies within the unit cell.
2. A point along the edge is shared by 4 unit cells and only \(\frac{1}{4}\)th of it lies within any one cell.
3. A face-centered point is shared by 2 unit cells and only \(\frac{1}{2}\) of it is present in a given unit cell.
4. A body-centered point lies entirely within the unit cell and contributes one complete point to the cell.
   
   Total number of particles in an unit cell (z) = (\(\frac{1}{8}\) × occupied by corners) + \(\frac{1}{4}\) × (occupied by edge centres) + \(\frac{1}{2}\) × (occupied by 24 face centres) + one occupied by body centre.

Question 19.
Calculate the number of particles present in (i) simple cubic (ii) body-centered and (iii) face-centered unit cells.
Answer:

1. In simple cubic unit cell particles are present only at the comers of the cube and this is shared by 8 other unit cells. Hence, the contribution of the point at the comers to the given unit cell is
   ∴ Z = 8 × \(\frac{1}{8}\) = 1
   i.e, A simple cubic unit cell has a single constituent (atom/molecule/ion) unit per unit cell.
2. In the body-centered cubic unit cell, comers, as well as the center of the unit cell, are occupied.
   ∴ Z = 8 × \(\frac{1}{8}\) + 1 = 2
   Hence body-centered cubic (bcc) has 2 constituent units per unit cell.
3. In face-centered cubic unit cell (fee) comers as well as face centers are occupied.
   ∴ Z = 8 × \(\frac{1}{8}\) + 6 + 2 = 4
   Hence, the face-centered cubic unit cell has four constituent units per unit cell.
   [Note: Remember that the number of atoms per unit cell is in the same ratio as the stoichiometry of the compound. Hence, it helps to predict the formula of the compound.]

Question 20.
A compound formed by elements A and B has a cubic structure in which A atoms are at the comers and B atoms are at face centers. Derive the formula of the compound.
Answer:
As ‘A’ atoms are present at the 8 comers of the cube, therefore a number of atoms of A in the unit cell = 8 × \(\frac{1}{8}\) = 1.
As ‘B’ atoms are present at the face centers of the 6 faces of the cube, therefore, the number
of atoms of B in the unit cell = \(\frac{1}{2}\) × 6 = 3.
∴ Ratio of atoms A : B = 1 : 3
∴ The formula of the compound is AB₃.

Question 21.
A cubic solid is made up of two elements X and Y. Atoms ‘Y’ are present at the comers of the cube and atoms X at the body center. What is the formula of the compound? What are the coordination numbers of X and Y?
Answer:
As Y atoms are present at 8 comers, of the cube, the number of atoms of Y in the unit cell
= \(\frac{1}{8}\) × 8 = 1
As atoms X are present at the body center, therefore, the number of atoms of X in the unit cell = 1.
∴ Ratio of atoms X : Y = 1 : 1
Hence the formula of the compound is XY.
The coordination number of each X and Y = 8.

Question 22.
A compound formed by elements X and Y crystallizes in the cubic structure where Y atoms are at the comers of the cube and X atoms are at alternate faces. What is the formula of the compound?
Answer:
As there are 8 Y atoms are at the comers of the cube and the contribution of each
= \(\frac{1}{8}\) therefore no. of Y atoms per unit cell
= 8 × \(\frac{1}{8}\) = 1
These can only be 2 × atoms an alternate forces.
As contribution of each of them = \(\frac{1}{2}\).
Therefore number of X atom per unit cell
= 2 × \(\frac{1}{2}\) = 1
Ratio of atom X : Y = 1 : 1
Hence the formula of the compound is 1 : 1

Question 23.
Give the expression to find out the inter planar distance (d) between two successive planes of atoms.
Answer:

Where n is the number of planes,
λ is the wavelength of X-ray used,
θ is the angle of diffraction.
Using these values, the edge of the unit cell can be calculated.

Question 24.
X-rays of wavelength 1.54Å strike a crystal and are observed to be deflected at an angle of 22.5°. Assuming that n = 1, calculate the spacing between the planes of atoms that are responsible for this reflection.
Answer:
Applying Bragg equation,

Question 25.
Derive an expression to find the density of a crystal from lattice parameters.
Answer:
Using the edge length of a unit cell, we can calculate the density (ρ) of the crystal by considering a cubic unit cell as follows.

substitute (3) in (2)
mass of the unit cell = n × \(\frac{\mathrm{M}}{\mathrm{N}_{\mathrm{A}}}\) …(4)
For a cubic unit cell, all the edge lengths are equal i.e, a = b = c
Volume of the unit cell = a × a × a = a³ …(5)

Equation (6) contains four variables namely ρ, n, M, and a. If any three variables are known, the fourth one can be calculated.

Question 26.
The edge length of a unit cell is 408 pm. Its density is 10.6 g cm^-3. Predict whether the metal is body-centered, or face-centered, or simple cubic.
Answer:

The value of Z indicates that the metal has a fee structure.

Question 27.
An element crystallizes in bcc structure the edge length of its unit cell is 288 pm. The density of the crystal is 7.2g cm^-3. What is the atomic mass of the element?
Answer:

Question 28.
A metallic element exists as a body-centered cubic lattice. Each edge of the unit cell is 2.88 pm. The density of the metal is 7.2g cm^-3. How many atoms and unit cells are there in 100 g of the atom?
Answer:
Let the atomic mass of the element be M.

Question 29.
The density of a face-centered cubic clement is 6.25 g cm^-3. Calculate the length of the unit cell. (Atomic mass of the element = 60.2 AMU).
Answer:
Given Z = 4; M = 60.2 amu; ρ = 625 g cm^-3
N_A = 6.023 × 10²³
Applying the formula:

Question 30.
KBr has NaCl type structure. What is the distance between K⁺ and Br^– in KBr, if the density is 2.75 g cm^-3?
Answer:
Edge length ‘a’ of the unit cell is calculated as

Question 31.
Derive an expression to calculate the packing efficiency in a simple cubic arrangement.
Answer:
Let us calculate the packing efficiency in simple cubic arrangement,

Let us consider a cube with an edge length ‘a’ as shown in fig.
Volume of the cube with edge length a is = a × a × a = a³
Let V is the radius of the sphere. From the figure, a = 2r ⇒ r = \(\frac{a}{2}\)
∴ Volume of the sphere with radius ‘r’

In a simple cubic arrangement, the number of spheres that belongs to a unit cell is equal to one
∴ Total volume occupied by the spheres in sc unit cell = 1 × \(\left(\frac{\pi a^{3}}{6}\right)\) ……(2)
Dividing (2) by (3)

i.e., only 52.31% of the available volume is occupied by the spheres in simple cubic packing, making inefficient use of available space and hence minimizing the attractive forces.

Question 32.
Show that the packing efficiency in the face-centered cubic unit cells is 74%.
Answer:
The cubic close packing is based on the face-centered cubic unit cell. Let us calculate the packing efficiency in the fcc unit cell.


The total number of spheres belongs to a single fcc unit cell is 4

Question 33.
Give the relationship between the nearest neighbor distance (d) and the edge (a) of the unit cell of a cubic crystal.
Answer:
For simple cubic: d = a
For face centered cubic: d = \(\frac{a}{\sqrt{2}}\) = 0.707a
For body centered cubic:
d = \(\frac{\sqrt{3}}{2} a\)

Question 34.
Give the relationship between atomic radius (r), (which is d/2 for crystals of pure substances) and edge (a) of the unit cell of a cubic crystal.
Answer:
For simple cubic: r = \(\frac{a}{2}\)

Question 35.
Xenon crystallizes in face-centered cubic lattice and the edge of the unit cell is 620 pm. What is the nearest neighbor distance and what is the radius of the xenon atom?
Answer:
Here, a = 620 pm; d = 2 ; r = ?

Question 36.
CsCl has bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl.
Answer:

The bcc arrangement of CsCl is shown in fig. Where black circle is Cs⁺ ion and colored circles are Cl^– ions. The aim is to find half of the body diagonal AE. If the edge of the unit cell is ‘a’.

Question 37.
If the radius of the atom is 75 pm and the lattice type is body-centered cubic, what is the edge of the unit cell?
Answer:

Question 38.
The radius of an atom of an element is 500 pm. If it crystallizes as fee lattice, what is the length of the side of the unit cell?
Answer:

Question 39.
A solid AB has a CsCl type structure. The edge length of the unit cell is 404 pm. Calculate the distance of the closest approach between A⁺ / B^– ions.
Answer:
The distance of the closest approach is equal to the distance between nearest neighbors (d). As CsCl has a bcc lattice.

Question 40.
What is the radius ratio? Mention its importance.
Answer:
In ionic solids, the ratio of the radius of the cation to the radius of an anion is known as the radius ratio.

The radius ratio of an ionic solid gives the structural arrangement of ionic solids.

Question 41.
What is the meaning of the term imperfection in solids?
Answer:
Imperfection refers to the departure from the perfect periodic arrangement of atoms, ions, or molecules in the structure of crystalline solids.

Question 42.
What are the types of lattice imperfections found in crystals?
Answer:

1. Stoichiometric defect i.e., Schottky defect and Frenkel defect.
2. Nonstoichiometric defect viz metal excess, metal deficiency, and impurity defects.

Question 43.
What are interstitials in a crystal?
Answer:
Atoms or ions that fill the normal vacant interstitial voids in a crystal are known as interstitials.

Question 44.
What is the Schottky defect?
Answer:
If an equal number of cations and anions are missing from their lattice sites, the defect is known as the Schottky defect.

Question 45.
What is Frenkel’s defect?
Answer:
When some ions (usual cation) are missing from the lattice sites and they occupy the interstitial sites. So that, the electro-neutrality, as well as stoichiometry are maintained, it is called Frenkel defect.

Question 46.
Which crystal defect lowers the density of the solid?
Answer:
Schottky defect.

Question 47.
Which crystal defect in crystals does not alter the density of a relevant solid?
Answer:
Frenkel defect.

Question 48.
Which point does defect increase the density of the solid?
Answer:
Schottky defect.

Question 49.
Which point defect in the crystal increases the density of the solid?
Answer:
Interstitial defect.

Question 50.
Name one solid in which both Frenkel and Schottky defects occur.
Answer:
Silver Bromide.

Question 51.
Why does the Frenkel defect does not change the density of AgCl crystal?
Answer:
Because in Frenkel defect, no ion is missing from the lattice sites. Therefore, no change in density occurs.

Question 52.
What are F. Centres?
Answer:
The free-electron trapped in the anion vacancy is called F.Centres.

Question 53.
What are non-stoichiometric defects?
Answer:
If as a result of imperfections in the crystal, the ratio of the cation to the anions becomes different from that indicated by their ideal chemical formula, then the defect is termed a non-stoichiometric defect.

Question 54.
Why does table salt, NaCl appear yellow in color?
Answer:
Yellow color in sodium chloride is due to metal excess defect due to which unpaired electrons occupy anionic sites. These sites are called F. Centres. These electrons absorb energy from the visible region for excitation which makes the crystal yellow.

Question 55.
Why is FeO (s) not formed in stoichiometric composition?
Answer:
In the crystal of FeO, some of Fe⁺² cations are replaced by Fe⁺³ cations. Three Fe⁺² cations are replaced by two Fe⁺³ cations to make up for the loss of positive charge. Eventually, there would be fewer atoms of the metal as compared to stoichiometric proportion.

Question 56.
How would you account for the following?

1. Frenkel defects are not found in alkali metal halides.
2. Schottky defects lower the density of the solid.
3. Impurity doped silicon is a semiconductor.

Answer:

1. This is because alkali metal ions have larger sizes that cannot fit into interstitial sites.
2. As the number of ions decreases as a result of the Schottky defect, the mass decreases whereas the volume remains the same.
3. This is due to additional electrons or the creation of holes and doping with impurity. Creation of hole results in p-type semiconductor and creation of electron results in the n-type semiconductor.

Question 57.
Briefly explain the metal excess defect.
Answer:
The metal excess defect arises due to the presence of more metal ions as compared to anions. Alkali metal halides NaCl, KCl show this type of defect. The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the excess metal ions (or) by the presence of extra cation and electrons present in the interstitial position.

Question 58.
Briefly explain metal deficiency defects.
Answer:
Metal deficiency defect arises due to the presence of fewer cations than the anions. This defect is observed in a crystal in which, the cations have variable oxidation states.

Question 59.
Briefly explain the impurity defect.
Answer:
A general method of introducing defects in ionic solids is by adding impurity ions. If the impurity ions are in a different valance states from that of the host, vacancies are created in the crystal lattice of the host. For example, the addition of CdCl₂ to silver chloride yields solid solutions where the divalent cation Cd²⁺ occupies the position of Ag⁺. This will disturb the electrical neutrality of the crystal. In order to maintain the same, a proportional number of Ag⁺ ions leaves the lattice. This produces a cation vacancy in the lattice, such kinds of crystal defects are called impurity defects.

Choose the correct answer.

1. Which one of the following is a covalent crystal?
(a) Rock salt
(b) Ice
(c) Quartz
(d) Dry ice
Answer: (c)
Hint: Rock salt is an ionic solid, while dry ice and ice are molecular solids.

2. Total volume of atoms present in a face centered cubic unit cell of a metal is:
(r = atomic radius)

Answer: (b)
Hint: Total no. of atoms in fee unit cell = 4

3. The fraction of the total volume occupied by the atoms present in a simple cube is:

Answer: (b)
In a simple cube, no. of atoms / unit cell = 8 × \(\frac{1}{8}\) = 1

4. Three elements A, B and C crystallise into a cubic solid lattice. Atoms A occupies the comers, B atoms, the cubic centres, and atoms C, the edges. The formula of the compound is:
(a) ABC
(b) ABC₂
(c) ABC₃
(d) ABC₄
Answer: (c)
Hint:
Atoms A per unit cell = 8 × \(\frac{1}{8}\) = 1
Atom B per unit cell = 1
Atoms C per unit cell = 12 × \(\frac{1}{4}\) = 3
Ratio A : B : C = 1 : 1 : 3.
Hence the formula is ABC₃.

5. An alloy of copper, silver and gold is found to have copper forming the simple cubic close packed lattice. If silver atoms occupy the comers and gold atoms are present at the body centres, the formula of the alloy will be:
(a) Cu₃ Ag Au
(b) Cu Ag₃ Au
(c) Cu₄ Ag₂ Au
(d) Cu Ag Au.
Answer: (b)
Hint:
Cu atoms per unit cell (present at the comer) = 8 × \(\frac{1}{8}\) = 1
Ag atoms per unit cell (present at the face centres) = 6 × \(\frac{1}{2}\) = 3
Au atoms per unit cell (present at body centre) = 1
Hence the formula is Cu Ag₃ Au.

6. A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y^–) will be:
(a) 275.1 pm
(b) 322.5 pm
(c) 241.5 pm
(d) 165.7 pm
Answer: (c)
Hint: NaCl has a face centered cubic structure. Cl^– ions and Na⁺ ions are present in the octahedral voids. Hence for such a solid, radius of the cation is = 0.414 × radius of the anion.

7. The number of atoms in 100g of the fcc crystal with density = 10 g/cm³, and the cell edge equal to 200 pm is equal to:
(a) 5 × 10²⁴
(b) 5 × 10²⁵
(c) 6 × 10²³
(d) 2 × 10²⁵
Answer:
(a)
Hint:

Thus 12g of the atom contain = 6.023 × 10²³ atom.
∴ 100g of the atom will contain

8. Which of the following statements is not true about amorphous solids?
(a) On heating they may become crystalline at certain temperature.
(b) They may become crystalline or keeping for long time.
(c) Amorphous solids can be moulded by heating.
(d) They are anisotropic in nature.
Answer:
(d)

9. The sharp melting point of crystalline solids is due to:
(a) a regular arrangement of constituent particles observed over a short distance in the crystal lattice.
(b) a regular arrangement of constituent particles observed over a long distance in the crystal lattice.
(c) same arrangement of constituent particles in different directions.
(d) different arrangement of constituent particles in different directions.
Answer:
(b)

10. Iodine molecules are held in the crystal lattice by:
(a) London forces
(b) dipole – dipole interaction
(c) covalent bonds
(d) columbic forces
Answer:
(a)

11. Which of the following is a network solid?
(a) SO₂ (solid)
(b) I₂
(c) diamond
(d) H₂O (ice)
Answer:
(c)

12. Which of the following solids is not an electrical conductor?
(i) Mg(s)
(ii) TiO(s)
(iii) I₂ (s)
(iv) H₂O (ice)
(a) (i) only
(b) (ii) only
(c) (iii) and (iv)
(d) (ii), (iii) and (iv)
Answer:
(c)

13. The lattice sites in a pure crystal cannot be occupied by:
(a) molecule
(b) ion
(c) electron
(d) atom
Answer:
(c)
Hint: Electrons can occupy only in the interstitial sites.

14. Cations are present in interstitial sites in:
(a) Frankel defect
(b) Schottky defect
(c) Valency defect
(d) Metal deficiency defect
Answer:
(a)

15. Schottky defect is obtained in crystal when:
(a) Some cations move from their lattice sites to interstitial sites.
(b) equal number of cations and anions are missing from the lattice.
(c) Some lattice sites are occupied by electrons.
(d) Some impurity is present in the lattice.
Answer:
(b)

16. The total number of tetrahedral void in the face centered unit cell is:
(a) 6
(b) 8
(c) 10
(d) 12
Answer:
(b)
Hint: No. of atoms per unit cell in fee = 4
No. of tetrahedral void: 2 × 4 = 8.

17. Which of the following statements is not true about hexagonal close packing?
(a) The coordination number is 12.
(b) It has 74% packing efficiency.
(c) Tetrahedral voids of the second layer are covered by spheres of the third layer.
(d) In this arrangement spheres of the fourth layer are exactly aligned with those of the first.
Answer: (d)
Hint: In hcp arrangement ABAB type and not ABC ABC type. Hence (d) is not true.

18. What is the coordination number in square close packed structure in two dimensions?
(a) 2
(b) 3
(c) 4
(d) 6
Answer:
(c)
Hint: The arrangement is AA AA type. In this arrangement each sphere is touching four other spheres touching the particular sphere, a square is formed. Hence, the coordination number is 4.

19. The correct order to packing efficiency in different type of unit cells is:
(a) fee < bee < simple cubic
(b) fee > bcc > simple cubic
(c) fee < bcc > simple cubic
(d) bcc < fee > simple cubic
Answer:
(b)
Hint: fcc = 74%, bcc = 68%; simple cubic = 52.4%

20. In cubic close packing, the unit cell has:
(a) 4 tetrahedral voids each of which is shared by adjacent unit cells. .
(b) 4 tetrahedral voids within the unit cell.
(c) 8 tetrahedral voids each of which is shared by four adjacent unit cells.
(d) 8 tetrahedral voids within the unit cells.
Answer:
(d)
Hint: ccp = hcp
No. of atoms per unit cell in fcc unit cell = 4.
∴ No. of tetrahedral voids = 2 × 4 = 8

21. KCl crystallises in the same type of lattice as does NaCl.
Given that \(r_{\mathrm{Na}^{+}} / r_{\mathrm{Cl}^{-}}\) is 0.55 and \(r_{\mathrm{K}^{+}} / r_{\mathrm{Cl}^{-}}\) is 0.74, Calculate the ratio of the side of the unit cell of KCl and NaCl.
(a) 1.123
(b) 0.891
(c) 1.414
(d) 0.414
Answer:
(a)

22. Ice crystallises in a hexagonal lattice having a volume of the unit cell is 132 × 10 cm³. It density of ice at the given temperature is 0.92 g cm^-3, then number of H₂O molecules per unit cell is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d)
Hint:

24. The edge length of the face-centered cubic unit cell is 508 pm. If the radius of the cation is 110 pm, the radius of the anion will be:
(a) 144 pm
(b) 288 pm
(c) 618 pm
(d) 398 pm
Answer:
(a)
Hint: For the face-centered cubic unit cell (eg: NaCl),
the edge length = 2 × distance between cation and anion
= 2 (r₊ + r_–)
2(r₊ + r_–) = 508 pm
given r₊ = 110 pm
2(110 + r_–) = 508 pm
r_– = 144 pm

25. Which of the following statements is not correct?
(a) The fraction of the total volume occupied by the atoms in a primitive cell is 0.48
(b) Molecular solids are generally volatile.
(c) The number of carbon atoms in a unit cell of a diamond is 4.
(d) The number of brave lattices in which a crystal can be categorized it.
Answer:
(c)

26. Match the entities with a column I with appropriate entities in column II and choose the correct option using the code given.

(a) (A) – r; (B) – p, r, s; (C) – r, (D) – q
(b) (A) – p,r; (B) – s; (C) – r; (D) – q
(c) (A) – r; (B) – p,s; (C) – q; (D) – r
(d) (A) – r, (B) – s; (C) – r; (D) – p
Answer:
(a)

27. Match the entities with column I with appropriate entities in column II and choose the correct option using the code given.

a = cell edge
d = nearest neighbour distance
r = radius
(a) (A) – q; (B) – r; (C) – p; (D) – s
(b) (A) – q; (B) – p; (C) – r; (D) – 5
(c) (A) – q; (B) – s; (C) – q; (D) – p
(d) (A) – q; (B) – s; (C) – r; (D) – p
Answer:
(c)

28. Match the entities with column I with appropriate entities in column II and choose the correct option using the code given.

(a) (A) – s; (B) – q; (C) – r; (D)-p
(b) (A) – q; (B) – p; (C) – r; (D) – s
(c) (A) – r; (B) – s; (C) – q; (D) – p
(d) (A) – s; (B) – r; (C) – p; (D) – q
Answer:
(a)

29. Assertion (A): Hexagonal close packing is more closely packed than cubic close packing.
Reason (R): Hexagonal close-packing has a coordination number 12 whereas the cubic close packing has a coordination number 8.
(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, and 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.
Answer:
(d)
Hint:
Correct assertion: Hexagonal close packing and cubic close packing are equally close-packed with a packing efficiency of 74%.
Correct reason: Both have the coordination number 12.

30. Assertion (A): Frankel defects are shown by silver halides.
Reason (R): Ag⁺ ion is smaller in size.
(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, and 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.
Answer:
(a)

Students get through the TN Board 12th Chemistry Important Questions Chapter 5 Coordination Chemistry which is useful for their exam preparation.

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

Answer the following questions.

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]
Answer:
(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)]^–
Answer:
(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]
Answer:
(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).
Answer:
(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.
Answer:
(i) Hexaammine cobalt (III) chloride.
(ii) Pentaammine chloridocobalt (III) chloride.
(iii) Potassium hexacyano ferrate (III).
(iv) Potassium trioxalato ferrate (III).
(v) Potassium tetrachlorido palladate (II).
(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)
Answer:
(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]
Answer:
(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.
Answer:
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
Answer:
(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.
Answer:
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
Answer:
(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.
Answer:
(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
Answer:
(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.
Answer:
Square planar complexes: In square planar complexes of the form [MA₂B₂]^n± and [MA₂BC]^n± (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]^n± where xy is a bidentate ligand with two different coordinating atoms also shows cis-trans isomerism. Square planar complex of the form [MABCD]^n± 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.
Answer:
(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.
Answer:
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.
Answer:
Octahedral complex of the type [MA₃B₃]^n± 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]
Answer:
(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.
Answer:
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]
Answer:
(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.
Answer:
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]⁺
Answer:

Question 23.
Write all the geometrical isomers of [Pt (NH₃) BrCl Py] and how many of these exhibit optical isomerism?
Answer:
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)]⁺
Answer:
(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]^–.
Answer:

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:
Answer:

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]⁺
Answer:
(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.
Answer:
(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.
Answer:
(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?
Answer:
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
Answer:
(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?
Answer:
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.
Answer:
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?
Answer:
(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.
Answer:
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?
Answer:
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]
Answer:
(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
Answer:
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
Answer:
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
Answer:
(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)
Answer:
(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.
Answer:
(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
Answer:
(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?
Answer:
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?
Answer:
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.
Answer:
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 e_g{d_x²-y², d_z²) orbitals lie in the path of approaching ligands these orbitals experience greater forces of repulsion than those of t_2g (d_xy, d_yz and d_xz) 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 Δ_o. (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 Δ_o below the average ‘d’ orbital energy and two ‘d’ orbitals lie at an energy 3/5 Δ_o above the average energy. The energy gap between t_2g and e_g sets also denoted by 10Dq.
The energy of t_2g orbitals is 4Dq less than that of hypothetical degenerate ‘d’ orbitals and that of e_g orbitals is 6Dq above that of hypothetical degenerate ‘d’ orbitals: Thus, t_2g set loses energy equal to 0.4 Δ_o or 4 Dq white e_g sets gain energy equal to 0.6 Δ_o 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)
Answer:
(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).
Answer:

Question 49.
Briefly discuss the crystal field splitting in tetrahedral complexes.
Answer:
The three orbitals i.e., t_2g orbitals are close to approaching ligands. As a result of this, the t_2g electrons suffer more repulsions than e_g electrons. The energy of t_2g orbitals increases more than e_g orbitals. The splitting is shown below:

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

Question 50.
Briefly outline crystal field splitting in square planar complexes.
Answer:
As the lobes of d_x²-y² point towards the ligands, this orbital has the highest energy. The lobes of d_xy orbital lie between the ligands but are coplanar with them this orbital is the next highest energy. The lobes of d_z² 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 d_z² orbital is the next highest in energy. The lobes of d_x² and d_y² 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 Δ_o decide the actual configuration of ‘d’ orbitals in a coordination entity?
Answer:
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 (t_2g and e_g) of orbitals is called crystal field splitting energy Δ_o for an octahedral field.
If Δ_o < p (pairing energy), the fourth electron enters one of the e_g orbitals giving the configuration t_2g³ e_g¹, thereby forming high spin complexes. Such ligands for which Δ_o < p are called weak field ligands.
If Δ_o > p, the fourth electron pair up in one of the t_2g orbital giving the configuration t_2g⁴ e_g⁰, thus forming low spin complexes. Such ligands for which Δ_o > 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.
Answer:
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 Δ_o CFSE (Δ_o) is called weak field ligands whereas those with a large value of CFSE(Δ_o) are called strong field ligands.

Question 53.
How does the oxidation state of a metal ion influence or affect the crystal field splitting energy?
Answer:
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 t_2g and e_g orbitals in octahedral complexes in a weak ligand field.
Answer:
Under the influence of weak ligands the energy difference, Δ_o 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 t_2g and e_g sets occurs according to Hund’s rule.
When the ligands are weak, the first three electrons numbered as 1,2,3 go to t_2g set those numbered 4, 5 go to e_g set, those numbered 6, 7, 8 go to t_2g set and the remaining two electrons numbered 9 and 10 will occupy e_g set. i.e.,
In complexes of weak ligands, Δ_o is less than P(P is called pairing energy which is the energy required to pair two electrons in the same orbital). Δ_o is the octahedral crystal field splitting energy, tends to force as many electrons into t_2g set while P tends to prevent the electrons to pair up in the t_2g 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.
Answer:
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 t_2g³ e_g² i.e., all the electrons are unpaired.
In the other hand CN^– ion is a strong ligand. The distribution of electrons is t_2g⁵ e_g⁰.
i.e., and it has one unpaired electron.

Question 56.
Why do compounds having similar geometry have a different magnetic moment?
Answer:
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 t_2g and e_g sets in an octahedral complex in the presence of a strong field ligand.
Answer:
Under the influence of strong ligands, the energy difference between t_2g and e_g sets is relatively high and thus the distribution of ‘d’ electrons in t_2g and e_g sets does not obey Hund’s rule. The first electrons numbers 1, 2, 3, 4, 5, 6 will go to t_2g set remaining four electrons numbered 7, 8, 9, 10 will go to e_g set.

For these complexes, Δ_o 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
Answer:
(i) F^– is a weak field ligand and cobalt is in a +3 oxidation state. It has d⁶ configuration. The electron distribution is t_2g⁴ e_g².

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., t_2g⁵ e_g².

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 t_2g 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.
Answer:
With weak field ligand, Δ_o < P, the electronic configuration (of cobalt (III)) i.e., d⁶ will be t_2g⁴ e_g² and it has 4 unpaired electrons and is paramagnetic. With strong field ligands, Δ_o > P, the electron distribution is t_2g⁶ e_g⁰. It has no unpaired electron and hence diamagnetic.

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

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 Δ_o and P.
Answer:
For d⁶ configuration, three electrons will first enter t_2g. Now, if P < Δ_o electrons will pair up in t_2g orbitals giving the configuration t_2g⁶ e_g⁰. If Δ_o< P, two electrons will first enter into e_g orbitals and one will pair up in t_2g, giving the configuration t_2g⁴ e_g².

Question 62.
Arrange the following ligands in the increasing order of crystal field splitting power.
Answer:
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
Answer:
(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 t_2g and e_g sets follow Hund’s rule.

i.e., t_2g⁴ e_g². 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 t_2g and e_g orbitals take place according to Hund’s rule.

i.e., t_2g⁴ e_g². 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 t_2g and e_g sets take place not in accordance with Hund’s rule. All the t_2g 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.
Answer:

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.
Answer:
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 t_2g⁶ e_g⁰. 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., t_2g⁴ e_g². It has 4 unpaired electron and hence paramagnetic.

Question 66.
CO is stronger ligand than NH₃ for many metals. Explain giving appropriate reason.
Answer:
Ligands such as CO, CN^–, NO⁺ have empty π orbitals which overlap with the filled d orbitals (t_2g) of transition metals forming π bonds (back bonding). These π interactions increase the value of Δ_o. 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?
Answer:
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?
Answer:
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?
Answer:
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.
Answer:
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.
Answer:
The equilibrium is

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

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.
Answer:

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.
Answer:
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.
Answer:

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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:

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.
Answer:
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.

Choose the correct answer.

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

Answer:
(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:

Answer:
(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)
Answer:
(a)

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

Answer:
(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:
(a) linkage isomerism
(b) ionisation isomerism
(c) coordination isomerism
(d) no isomerism
Answer:
(d)

6. Which of the following species is not expected to be a ligand?
(a) NO
(b) NH₄⁺
(c) NH₂CH₂CH₂NH₂
(d) Co
Answer:
(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
Answer:
(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.
Answer:
(d)
Hint:

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

Answer:
(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.
Answer:
(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:

Answer:
(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

Answer:
(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)
Answer:
(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 Δ_O?

Answer:
(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
Answer:
(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
Answer:
(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
Answer: (c)
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.
Answer:
(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?

Answer:
(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 Δ_o
(b) -1.8 Δ_o
(c) -1.66 Δ_o + P
(d) -1.2 Δ_o
Answer:
(a)
Hint: d⁴ octahedral complex has the configuration t_2g³ e_g¹
CFSE = (-0.4x + 0.6y) Δ_o
x = no. of electron in t_2g
y = no. of electron in e_g
CFSE = [-0.4 x 3+ 0.6 x 1] Δ_o
= [-1.2 + 0.6] Δ_o
= – 0.6 Δ_o

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.
Answer:
(c)

22. The magnitude of crystal field stabilisation energy (CFSE or Δ_t) 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 Δ_t, 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 (Δ_t) by further 2/3.
(c) both (a) and (b) are correct
(d) both (a) and (b) are wrong.
Answer:
(c)

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

(At. No: Zn = 30, Sc = 21, Ti = 22; Cr = 24) ,
Answer:
(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.
Answer:
(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^–
Answer:
(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)
Answer:
(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
Answer:
(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:

Answer:
(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.
Answer:
(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.
Answer:
(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.

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

Answer the following questions.

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.
Answer:

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.
Answer:
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.
Answer:
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.
Answer:
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
VO₄^-3(vanadate ion):
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.
Answer:

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.
Answer:
(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.
Answer:
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.
Answer:
(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?
Answer:
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 t_2g 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.
Answer:
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?
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
(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?
Answer:
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.
Answer:

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?
Answer:
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?
Answer:
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?
Answer:
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.
Answer:

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?
Answer:
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.
Answer:
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.
Answer:
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).
Answer:
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.
Answer:
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?
Answer:
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?
Answer:
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.
Answer:

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.
Answer:

Question 32.
Explain how transition metals and their compounds act as catalysts.
Answer:
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.
Answer:

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?
Answer:
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.
Answer:
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.
Answer:
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, ZrH_1.92, Mn₄N etc., The elements that occupy the metal lattice provide them new properties.

Question 37.
Mention the properties of interstitial compounds.
Answer:

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.
Answer:
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.
Answer:

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.
Answer:
(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.
Answer:
(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
Answer:
(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^–.
Answer:
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.
Answer:
It decomposes to give green Cr₂O₃ and oxygen.

Question 45.
Discuss the action of alkali on potassium dichromate.
Answer:
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 p^H on a solution of potassium dichromate?
Answer:
In aqueous solution

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

Question 47.
Draw the structure of the dichromate ion.
Answer:

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

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

Answer:

Question 50.
Describe the chromyl chloride test.
Answer:
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.
Answer:

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.
Answer:
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?
Answer:
(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.
Answer:
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.
Answer:
MnO₄^– + 2H₂O + 3e → MnO₂ + 4OH^–

Question 56.
What is Bayer’s reagent? Mention its use.
Answer:
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.
Answer:
(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.
Answer:

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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
+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.
Answer:

Question 65.
Compare the chemistry of actinoids with that of lanthanoids with special reference to
(i) electronic configuration
(ii) Oxidation state
(iii) Atomic and ionic radii
(iv) Chemical reactivity
Answer:
(i) The general electronic configuration of lanthanoids is [Xe]⁵⁴ 4f^1-14 5d^0-1 6s² where as that of actinoids in [Rn]⁸⁶ 5f^1-14 6d^0-1 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?
Answer:
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?
Answer:
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.
Answer:
(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.
Answer:
(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.
Answer:
(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.
Answer:
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

Choose the correct answer.

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
Answer:
(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.
Answer:
(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.
Answer:
(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.
Answer:
(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) \(\frac{1}{5}\)
Answer:
(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
Answer:
(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(t_2g ) 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
Answer:
(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)
Answer:
(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.
Answer:
(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.
Answer:
(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)
Answer:
(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)
Answer:
(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.
Answer:
(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.
Answer:
(c) The +4 oxidation state of Ce is not known in solution.

15. MnO₄^– reacts with Br^– in alkaline p^H to give:
(a) BrO₃^–, MnO₂
(b) Br₂, MnO₄^-2
(c) Br₂, MnO₂
(d) BrO^–, MnO₄^-2
Answer:
(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.
Answer:
(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 B_μ) is approximately:
(a) 4
(b) 1
(c) 5
(d) 6
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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
Answer:
(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.
Answer:
(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.
Answer:
(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.
Answer:
(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)
Answer:
(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.
Answer:
(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
Answer:
(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
Answer:
(a) Ni⁺², Ti⁺³
Hint: Ni⁺² = [Ar] 3d⁸ t_2g⁶ e_g². 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
Answer:
(d) potassium manganate

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

Answer the following questions.

Question 1.
Briefly account for the trend in atomic radius of elements in the nitrogen family.
Answer:
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.
Answer:
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.
Answer:
In both cases, nitrogen is formed.

Question 4.
Explain why nitrogen is inert at room temperature.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:

Question 8.
How is ammonia manufactured?
Answer:
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.
Answer:
(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.
Answer:

Question 11.
Give an example for a reducing property of ammonia. [OR] Ammonia is a reducing agent. Give an example to prove this statement.
Answer:
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?
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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:
Answer:
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.
Answer:
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.
Answer:
(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.
Answer:
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.

Lead, under similar condition, give lead nitrate and nitric oxide.

(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.
Answer:

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.

Answer:

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

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

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.
Answer:

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.
Answer:
(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.
Answer:

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.
Answer:

Question 31.
Explain the structure of white phosphorus.
Answer:
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.
Answer:
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.
Answer:
(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.
Answer:

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

Question 36.
How phosphine is formed from (i) White phosphorus, (ii) Calcium phosphide, (iii) Phosphorus acid, (iv) Phosphonium iodide. Give equations.
Answer:
(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.
Answer:

(iii) PH₃ + 4Cl₃ → PCl₅ + 3HCl

Question 38.
Give examples for (i) basic nature and (ii) reducing property of phosphine.
Answer:
(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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
(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.
Answer:
The ‘OH’ group in alcohols and acids are replaced by ‘Cl’.

Question 44.
Explain the structure of phosphorus trichloride.
Answer:
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.
Answer:

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

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

Question 48.
Mention the uses of phosphorus pentachloride.
Answer:
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.
Answer:
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.
Answer:
(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?
Answer:
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?
Answer:
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.
Answer:
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.
Answer:
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
Answer:

Question 56.
Complete and balance the following equations:

Answer:

Question 57.
Briefly explain the trend in physical properties of group 16 elements.
Answer:
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.
Answer:
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.
Answer:
(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?
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
Ozone oxidises potassium iodide to iodine.

Question 65.
Explain how ozone can be estimated quantitatively.
Answer:
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.
Answer:

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?
Answer:
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.
Answer:
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.
Answer:
(i) Burning sulphur in air.
S + O₂ → SO₂

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

Question 71.
With an example, prove sulphur dioxide in an acidic oxide.
Answer:
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.
Answer:
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.
Answer:
(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.
Answer:

Question 75.
Explain the use of sulphur dioxide as a bleaching agent.
Answer:
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.
Answer:
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.
Answer:

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?
Answer:
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?
Answer:
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.
Answer:
(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.

Answer:

Question 82.
Give a brief account of the action of sulphuric acid on metals.
Answer:
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?
Answer:
(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.
Answer:
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.
Answer:

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.

Answer:

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

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?
Answer:
(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.

Answer:

Question 90.
Briefly outline the manufacture of chlorine by
(i) Electrolytic process, (ii) Deacon’s process.
Answer:
(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.
Answer:

Question 92.
What happens when (i) chlorine is treated with excess ammonia (ii) ammonia is treated with an excess of chlorine? Give equation.
Answer:
(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.
Answer:
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.
Answer:
(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.
Answer:
Chlorine bleaches by oxidation

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

Question 96.
Complete and balance the equation:

Answer:

Question 97.
How is bleaching powder prepared? Give equation.
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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.
Answer:
(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.
Answer:
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.
Answer:
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.
Answer:

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.
Answer:
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?
Answer:
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.
Answer:
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?
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
Moist hydrofluoric acid (not dry) rapidly react with silica and glass.

Question 113.
What is etching? Explain with an example.
Answer:
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.
Answer:
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.
Answer:

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.
Answer:
(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.
Answer:
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?
Answer:

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.
Answer:
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.
Answer:

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.
Answer:
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.
Answer:

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

Question 124.
Name the addition compounds formed by the Xenon difluoride.
Answer:
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).
Answer:

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).
Answer:
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).
Answer:
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).
Answer:
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).
Answer:
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).
Answer:
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.
Answer:

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.
Mention the uses of radon.
Answer:

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.

Choose the correct answer.

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

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

5. The structural formula of hypo phosphorus acid is:

Answer:
(a)

6. Which of the following oxides is the most acidic?
(a) P₂O₅
(b) N₂O₅
(c) Sb₂O₃
(d) AS₂O₃
Answer:
(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°
Answer:
(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
Answer:
(b)

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

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

Answer:
(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?

Answer:
(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
Answer:
(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
Answer: (c)
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
Answer: (d)
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
Answer:
(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
Answer:
(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
Answer: (d)

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
Answer:
(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
Answer:
(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
Answer:
(c)

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

Code:

Answer:
(d)

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

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

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

Answer the following questions.

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

1. They have the general electronic configuration ns², np^1-6.
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.
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
‘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.
Answer:

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.
Answer:

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.
Answer:

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.
Answer:
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.
Answer:
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.
Answer:
Sodium borate is formed when boron is fused with sodium hydroxide.

Question 16.
Mention the uses of boron.
Answer:

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.
Answer:
The aqueous solution of colemanite (Ca₂B₆O₁₁) is boiled with sodium carbonate.

Question 18.
All aqueous solution of borax is alkaline in nature. Explain.
Answer:
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.
Answer:
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.
Answer:

Question 21.
Explain how boric acid behaves as a monobasic acid?
Answer:
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.
Answer:

Question 23.
What happens when boric acid is heated? Give equations.
Answer:
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.
Answer:
(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.
Answer:
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.
Answer:

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.
Answer:

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.
Answer:

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.
Answer:
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.
Answer:

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.
Answer:
(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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:

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?
Answer:
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?
Answer:
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.
Answer:

Question 39.
Mention the uses of aluminium chloride.
Answer:

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.
Answer:
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.
Answer:

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.
Answer:

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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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 →
Answer:

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

Question 50.
What are carbonyls? Give examples.
Answer:
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.
Answer:

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?
Answer:
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.
Answer:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

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

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.
Answer:

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.
Answer:

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.
Answer:
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?
Answer:
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.
Answer:
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.
Answer:

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.
Answer:
(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
3. Cross-linked 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.
Answer:
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.
Answer:
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?
Answer:
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.
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
Silicates that contain (SiO₃)_n^2n- ions which are formed by linking three or more tetrahedral SiO₄^4- 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.
Answer:
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.
Answer:
Silicates that contain (Si₂O₅)_n^2n- 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?
Answer:
Amphiboles are double-chain silicates that contain (Si₄O₁₁)_n^6n- 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.
Answer:
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.,

Choose the correct answer.

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
Answer:
(b) Si and Ge

2. Among ‘p’ block elements which group show +6 oxidations state?
(a) Icosagens
(b) Tetragens
(c) Prictogens
(d) chalcogens
Answer:
(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.
Answer:
(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
Answer:
(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) \({ }_{8} \mathrm{O}^{17}\)
Answer:
(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
Answer:
(b) basic
Hint: Borox hydrolysis to give a strong base
Na₂B₄O₇ + 7H₂O → 4H₃BO₃ + 2NaOH

7. Borax Bead test is used to identify:
(a) boron
(b) borate radical
(c) onetal cations
(d) boride radical
Answer:
(b) borate radical

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

13. The hydrolysis of CH₃SiCl₃ yields:
(a) complex cross linked polymer
(b) cyclic polymer
(c) linear polymer
(d) silicols
Answer:
(a) complex cross linked polymer

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

15. Choose the incorrect statement:
(a) pyroxenes contain (SiO₃)_n^2- 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.
Answer:
(d) Pyroxenes are silicates which contain [SiO₄]^-4 units, where all the oxygen atoms are shared with other [SiO₄]^-4 tetrahedra.

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

Answer the following.

Question 1.
What is economic zoology?
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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.
Answer:
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?
Answer:

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?
Answer:
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?
Answer:
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?
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:

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.
Answer:
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?
Answer:
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.
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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.
Answer:
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?
Answer:
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.
Answer:

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.
Answer:
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.
Answer:
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?
Answer:
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?
Answer:
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’?
Answer:
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.
Answer:
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.
Answer:
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.

Answer:

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.
Answer:
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.
Answer:
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?
Answer:
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.
Answer:

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?
Answer:
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?
Answer:
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?
Answer:
Deep water culture, media-based method, nutrient film technique, and aqua Vertica.

Question 42.
What is deep water culture?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
Natural breeding and induced breeding.

Question 58.
Write about the induced breeding method in fishes?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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.
Answer:
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.
Answer:

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?
Answer:

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?
Answer:
Penaeus indicus, Penaeus monodon, Metapenaeus dobsoni and Macrobrachium rosenbergii.

Question 66.
How the freshwater prawn “Macrobrachium rosenbergii” is cultured?
Answer:
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?
Answer:
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?
Answer:
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.
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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.
Answer:
The main diseases of dairy cattle are rinderpest, foot and mouth disease, cowpox, hemorrhagic fever, anthrax.

Question 79.
What is poultry fanning?
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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.
Answer:
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?
Answer:
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?
Answer:
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?
Answer:

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.
Answer:
Ranikhet, Coccidiosis, and Fowlpox are some common poultry diseases.

Question 88.
Write the benefits of keeping poultry. Benefits of Poultry farming:
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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?
Answer:
Healthy cattle appear bright, alert, and active in their movement with a shiny coat.

Question 93.
Define Hapa?
Answer:
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?
Answer:
Drilospfiere is the part of the soil influenced by earthworm secretions, burrowing, and castings.

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

Choose the correct answer.

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

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

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

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

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

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

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

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
Answer:
(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
Answer:
(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
Answer:
(b) Bombyx mori

15. The cultivation of mulbeny is Called:
(a) sericulture
(b) apiculture
(c) horticulture
(d) moriculture
Answer:
(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
Answer:
(b) post-cocoon processing

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

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

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

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

29. Lac insect Tachardia lacca is previously known as:
(a) Bombyx mori
(b) Apis florea
(c) Laccifer Lacca
(d) Penaeus Indicus
Answer:
(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)
Answer:
(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.
Answer:
(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
Answer:
(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.
Answer:
(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)
Answer:
(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.
Answer:
(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.
Answer:
(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.
Answer:
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)
Answer:
(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
Answer:
(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
Answer:
(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
(d) Tadpoles
Answer:
(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.
Answer:
(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
Answer:
(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
Answer:
(a) Prawns

45. ……….. is commonly seen in rivers and low saline estuaries.
(a) Metapenaeus dobsoni
(b) Panaeus indicus
(c) Panaeus monodon
(d) Macrobrachium rosenbergii
Answer:
(d) Macrobrachium rosenbergii

46. High quality of pears are obtained from pearl oysters of;
(a) Lamellidens
(b) Olympia oyster
(c) Pinctada
(d) Sydney rock oyster
Answer:
(c) Pinctada

47. …………. breed is the smallest breed of cow.
(a) Brown Swiss
(b) Jersey
(c) Sindhi
(d) Vechur
Answer:
(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
Answer:
(d) Leghorn

49. Fast growing chicken breed is:
(a) Chittagong
(b) White Plymouth rock
(c) Wyandotte
(d) Karaknath
Answer:
(b) White Plymouth rock

50. ……….. is a type of chicken breed known for its pugnacity.
(a) Aseel
(b) Brahma
(c) Busra
(d) Leghorn
Answer:
(a) Aseel

51. Which is the ornamental breed of chicken of the following?
(a) Chittagong
(b) Leghorn
(c) Aseel
(d) Silkie
Answer:
(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)
Answer:
(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.
Answer:
(a) The most common and commercially farmed birds are chicken, ducks, goose, patridges, Guinea fowl, Emu, etc.