Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 2 p-Block Elements – I Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements – I

12th Chemistry Guide p-Block Elements – I Text Book Questions and Answers

I. Choose the qorrect answer

1. An aqueous solution of borax is __________ .
a) neutral
b) acidic
c) basic
d) amphoteric
Answer:
c) basic

2. Boric acid is an acid because its molecule (NEET)
a) contains replaceable H+ ion
b) gives up a proton
c) combines with proton to form water molecule
d) accepts OH from water, releasing proton
Answer:
d) accepts OH from water, releasing proton

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

3. Which among the following is not a borane?
a) B2H6
b) B3H6
c) B4H10
d) none of these
Answer:
b) B3H6

4. Which of the following metals has the largest abundance in the earth’s crust?
a) Aluminium
b) Calcium
b) Magnesium
d) Sodium
Answer:
a) Aluminium

5. In diborane, the number of electrons that accounts for banana bonds is
a) six
b) two
c) four
d) three
Answer:
c) four

6. The element that does not show catenation among the following p-block elements is
a) Carbon
b) Silicon
c) Lead
d) germanium
Answer:
c) Lead

7. Carbon atoms in fullerene with formula C60 have
a) sp³ hybridised
b) sp hybridised
c) sp² hybridised
d) partially sp² and partially sp³ hybridised
Answer:
c) sp2 hybridised

8. Oxidation state of carbon in its hybrides
a) +4
b) -4
c) +3
d) +2
Answer:
a) +4

9. The basic structural unit of silicates is (NEET) (PTA – 1)
a) (SiO3)2-
b) (SiO4)2-
c) (SiO)
d) (SiO4)4-
Answer:
d) (SiO4)4-

10. The repeating unit in silicone is
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 1

11. Which of these is not a monomer for a high molecular mass silicone polymer?
a) Me3SiCl
b) PhSiCl3
c) MeSiCl3
d) Me2SiCl2
Answer:
a) Me3SiCl

12. Which of the following is not sp² hybridised?
a) Graphite
b) graphene
c) Fullerene
d) dry ice
Answer:
d) dry ice

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

13. The geometry at which carbon atom in diamond are bonded to each other is
a) Tetrahedral
b) hexagonal
c) Octahedral
d) None of these
Answer:
a) Tetrahedral

14. Which of the following statements is not correct?
a) Beryl is a cylic silicate
b) Mg2SiO4 is an orthosilicate
c) SiO44- is the basic structural unit of silicates
d) Feldspar is not aluminosilicate
Answer:
d) Feldspar is not aluminosilicate

15. Match items in Column-I with the items of Column-II and assign the correct code.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 2

Answer:
a) 2 1 4 3

16. Duralumin is an alloy of
a) Cu, Mn
b) Cu, AZ, Mg
c) AZ, Mn
d) AZ, Cu, Mn, Mg
Answer:
d) Al, Cu, Mn, Mg

17. The compound that is used in nuclear reactors as protective shields and control rods is
a) Metal borides
b) Metal oxides
c) Metal carbonates
d) Metal carbide
Answer:
a) Metal borides

18. The stability of +1 oxidation state increases in the sequence
a) AZ < Ga < In < TZ
b) TZ < In < Ga < Al
c) In < TZ < Ga < Al
d) Ga < In < AZ < TZ
Answer:
a) Al< Ga < In < TZ

II. Answer the following questions

Question 1.
Write a short note on anamolous properties of the first element of p-block.
Answer:
The following factors are resposible for the anamolous properties of the first elements of p-blick.
1. Small size of the first member
2. High ionisation enthalpy and high electronegativity.
3. Absence of d-orbitals in their valence shell.

First elemenl Property of First elements Other elements in the family
B Mettaloid Metals
C 1. Non-metal
2. It can form multiple bonds.
1. Metalloids – Si and Ge.
2. Other elements are metals.
3. It can’t form multiple bonds.
N 1. Non metal
2. It can form multiple bonds
3. Diamagnetic
1. Non metal – “P” Metalloids – As. Sb.
2. It cann’t form multiple bonds
O 1. Non metal and diatomic gas
2. It forms H-bonds
1. S, Se – non metals.
2. Te- metalloid and others are metals.
F 1. Non-metals
2. High electro­ negativity
3. Highly reactive.
1. Non-metals
2. Low reactive than ‘F’

Question 2.
Describe briefly allotropiam in p-block elements with specific reference to carbon.
Answer:

  • Some elements exist in more than one crystalline or molecular forms in the same physical state.
  • This phenomenon is called allotropism.
  • The different forms of an element are called allotropes.
  • Example: Carbon exists as diamond, graphite, graphene, fullerenes, carbon nanotubes

Question 3.
Give the uses of Borax.
Answer:

  1. Borax is used for the identification of coloured metal ions.
  2. In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
  3. It is also used as a flux in metallurgy and also acts as a good preservative.

Question 4.
What is catenation? Describe briefly the catenation property of carbon. (MARCH 2020)
Answer:
Catenation:
It is the phenomenon of an atom to form a strong covalent bond with the atoms of itself. Carbon shares the property of catenation to the maximum extent because it is small in size and can form pn-pn multiple bonds to itself. The following conditions are necessary for catenation.

  1. The valency of element is greater than or equal to two.
  2. Element should have the ability to bond with itself.
  3. The self-bond must be as strong as its bond with other elements.
  4. Kinetic inertness of catenated compound towards other molecules.
  5. Carbon possesses all the above properties and forms a wide range of compounds with itself.

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

Question 5.
Write a note on Fisher Tropsch synthesis. Fischer Tropsch synthesis: (PTA – 4)
Answer:
This is a reaction in which carbon monoxide reacts with hydrogen at a pressure less than 50 atm and temperature 500 – 700 K in presence of metal catalysts to give saturated and unsaturated hydrocarbons.
n CO + (2n+l) H2 → CnH2n+2 + nH2O
n CO + 2n H2 → CnH2n + nH2O

Question 6.
Give the structure of CO and CO2.
Answer:
Structure of CO:
Structure is linear.

Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 3

Structure of CO2:
Structure is linear.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 4

Question 7.
Give the uses of silicones.
Answer:

  1. Used for low temperature lubrication.
  2. Used in vacuum pumps.
  3. Used in high temperature oil baths.
  4. Used for making water proof cloths.
  5. Used as insulating material in electrical motor and other applicances.
  6. Mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals

Question 8.
Describe the structure of diborane. (PTA – 3)
Answer:

  • In diborane two BH2 units are linked by two bridged hydrogens, rherefore it has eight B-H bonds.
  • Diborane has only 12 valence electrons anc are- not sufficient to form normal covalen bonds.
  • The four terminal B-H bonds are norma covalent bonds. (2c 2e bond) (Totally 8e-s)
  • The remaining four electrons have to be used for the bridged bonds, ie two 3 centred B-H-B bonds utilise two electrons each.
  • Hence these bonds are 3c – 2e bonds.
    Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 5
  • The bridging hydrogen atoms are in a plane.
  • In diborane, boron is sp³ hybridised.
  • Three sp³ hybridised orbitals contain single electron and the fourth orbital is empty.
  • Two half filled sp³ hybridised orbitals of each boron overlap with two hydrogens to form four terminal 2C – 2e bonds.
  • One empty and one half filled sp³ hybridised orbital on each boron is left.
  • Empty sp³ hybridised orbital of one boron, overlaps with half filled sp³ hybridised orbital of the other boron and Is orbital of hydrogen to form two bridged 3C – 2e B-l 1-B bonds.

Question 9.
Write a short note on hydroboration.
Answer:

  • Diborane adds on to alkenes and alkynes in ether solvent at room temperature.
  • This reaction is known as hydroboration.
  • This is used in synthetic organic chemistry especially for anti Markovnikov addition.
    B2H6 + 6 RCH = CHR → 2B (CH2-CH2 R)3

Question 10.
Give one example for each of the following:
Answer:
i) icosogens
ii) tetragen
iii) pnictogen
iv) chalcogen

Group Name Example
i. Icosagens Boron
ii. Tetragens Carbon
iii. Pnictogen Nitrogen
iv. Chalcogens Oxy gen

Question 11.
Write a note on metallic nature of p-block elements.
Answer:

  • The tendency of an element to form a cation by losing electrons is known as electro positive or metallic character.
  • This character depends on the ionisation energy.
  • Generally on moving down a group ionisation energy decreases and hence the metallic character increases.
  • In p-block, the elements present in lower left part are metals, while the elements in the upper right part are non metals.
Group Non-metals Metalloids Metals
13 B Al, Ga, In, Tl
14 C Si, Ge Sn, Pb
15 N, P As, Sb Bi
16 O, S, Se Te, Po
17 F, Cl, Br, I
18 He, Ne, Ar, Kr, Xe

Question 12.
Complete the following reactions:
a) B(OH)3 + NH3
b) Na2B4O7 + H2SO4 + H2O →
c) B2H6 + 2NaOH + 2H2O →
d) B2H6 + CH3OH →
e) BF3 + 9H2O →
f) HCOOH+ H2SO4
g) SiCl4 + NH3
h) SiCl4 + C2H5OH →
I) B + NaOH →
j) H2B4O7 \(\underrightarrow { Red\quad hot } \)
Answer:
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 6
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 7

Question 13.
How will you identify borate radical? (PTA – 5)
Answer:

  • When boric acid or borate salt is heated with ethyl alcohol in presence of cone, sulphuric acid, an ester triaikvl borate is formed.
  • The vapour of this ester bums with a green edged flame.
  • This is ethyl borate test to identify borate radical,
    Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 8
    B(OC2H5)3 Ethyl borate (Green edged flame)

Question 14.
Write a note on zeolites. ( PTA – 2)
Answer:

  • Zeolites are three dimensional crystalline solids containing aluminium, silicon and oxvgen in their regular three dimensional frame work.
  • They are hydrated sodium alumino silicates.
  • General formula is
    Na2O.(Al2O3).x(SiO2).y(H2O)
    where x = 2 to 10; y = 2 to 6
  • Zeolites have porous structure in which the monovalent sodium ions and water molecules are loosely held.
  • Si and AI atoms are tetrahedrally coordinated with each other through shared oxygen atoms.
  • Zeolites are similar to Clay minerals but they differ in their crystalline structure.
  • Zeolites have a three dimensional crystalline structure looks like a honey comb consisting of a network of interconnected tunnels and cages.
  • Water molecules move freely In and out of these pores but the zeolite frame work remains rigid.
  • Another special aspect of this structure is that the pore/channel sizes are nearly uniform, allowing the crystal to act as a molecular sieve.
  • Zeolites are used in the removal of permanent hardness of water.

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

Question 15.
How will you convert boric acid to boron nitride? (PTA – 3)
Answer:
Fusion of urea with boric acid in an atmosphere of ammonia at 800 -1200 K gives boron nitride.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 9

Question 16.
A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducting agent (C). Identify (A), (B) and ( C) (PTA – 1)
Answer:
A hydride of 2nd period alkali metal (A) is LiH

Lithium hydride reacts with compound of boron (B) B2H6 to give reducing agent (C) lithium boro hydride.
∴ Compound B is diborane
Compound C is lithium boro hydride.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 10

Question 17.
A double salt which contains fourth period alkali metal (A) on heating at 500 K gives (B). Aqueous solution of (B) gives white precipitate with BaCl2 and gives a red colour compound with alizarin. Identify (A) and (B).
Answer:

  •  A double salt which contains fourth period alkali metal (A) is Potash alum
    K2SO4. Al2 (SO4)3.24H2O
  • (A) on heating at 500 K gives
    K2SO4.Al2(SO4)3 (B) which is burnt alum.

Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 11

Question 18.
CO is a reducing agent, justify with an example.
Answer:

  • CO is a strong reducing agent.
  • It reduces metallic oxides inlo melais.
    Example : 3CO + Fe2CO3 → 2Fe + 3CO2

III. Evaluate Yourself

Question 1.
Why group 18 elements are called inert gases? Write the general electronic configuraton of group 18 elements.
Answer:

  • These elements are gases.
  • Their outer electronic configuration is ns²np6 which is stable completely filled configuration.
  • So they are more stable and least reactive.
  • Hence they are called inert gases.

12th Chemistry Guide Chapter 2 p-Block Elements – I Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

1. The general electronic configuration of p-block elements is
a) ns¹
b) ns²
c) ns² np1-6
d) (n-1)s² np1-6
Answer:
c) ns² np1-6

2. p-block element consists of the groups
a) 1 & 2
b) 3 – 12
c) 13 – 17
d) 13 – 18
Answer:
d) 13 – 18

3. Group 18 elements are inert because of their
a) unstable incompletely filled orbitals
b) stable completely filled orbitals
c) half filled orbitals
d) stable nucleus
Answer:
b) stable completely filled orbitals

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

4. As we go down the group ionisation energy
a) decreases
b) increases
c) becomes constant
d) becomes zero
Answer:
a) decreases

5. As we go down the group metallic character
a) decreases
b) increases
c) becomes constant
d) becomes zero
Answer:
b) increases

6. As ionisation energy decreases, the metallic character of elements
a) decreases
b) increases
c) becomes constant
d) becomes zero
Answer:
b) increases

7. In p-block, metals are placed in
a) upper right part
b) middle part
c) lower left part
d) top of the group
Answer:
c) lower left part

8. In p-block, non-metals are placed in
a) upper right part
b) middle part
c) lower left part
d) bottom of the group
Answer:
a) upper right part

9. Which of the following factor is not responsible for the anamolous behaviour of the first member of each group in p-block elements?
a) small size
b) high ionisation enthalpy
c) outer electronic configuration
d) absence of d-orbitals
Answer:
c) outer electronic configuration

10. The correct order of catenation property in group 14 elements is
a) C << Si < Ge = Sn < Pb
b) C >> Si > Ge = Sn > Pb
c) C >> Si < Ge = Sn < Pb
d) C << Si » Ge = Sn > Pb
Answer:
b) C >> Si > Ge = Sn > Pb

11. The elements N, O, F readily forms hydrogen bonds due to their high
a) ionisation energy
b) electron affinity
c) electro negativity
d) atomic radius
Answer:
c) electro negativity

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

12. The most electro negative element is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
a) Flourine

13. The element with maximum electron affinity is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
b) Chlorine

14. The most reactive element among halogens is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
a) Flourine

15. The strongest oxidising agent among halogens is
a) Flourine
b) Chlorine
c) Bromine
d) Iodine
Answer:
a) Flourine

16. The important property shown by p-block elements is
a) complex formation
b) coloured ion formation
c) inert pair effect
d) metallic character
Answer:
c) inert pair effect

17. In 13th group Tl+1 ion is more stable than Tl3+ ion due to
a) high electronegatively
b) inert pair effect
c) high ionisation energy
d) stable electronic configuration
Answer:
b) inert pair effect

18. Diamond and graphite are ______ of carbon.
a) Isotopes
b) Isobars
c) Isomers
d) Allotropes
Answer:
d) Allotropes

19. The formula of Borax is
i) Na2B4O7.10H2O
ii) Na2[B4O5(OH)4].8H2O
iii) Na2[B4O5(OH)4].2H2O
a) (i) only
b) (i) & (ii) only
c) (i) & (iii) only
d) (iii) only
Answer:
b) (i) & (ii) only

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

20. Ortho boric acid on dehydration at 373K produces mainly (PTA – 3)
a) metaboric acid
b) boric anhydride
c) Boron metal and Oxygen
d) tetra boric acid
Answer:
a) metaboric acid

21. The formula of colemanite is
a) Na2B4O7
b) Na2B4O7.10H2O
c) Ca2B6O11
d) NaBO2
Answer:
c) Ca2B6O11

22. Which is used as moderator in nuclear reactors?
a) boron nitride
b) boron
c) borax
d) boric acid
Answer:
b) boron

23. The compound used in eye drops and antiseptics is
a) boron nitride
b) boric acid
c) sodium meta borate
d) boron tri oxide
Answer:
b) boric acid

24. The compound used as a flux in metallurgy is
a) boron nitride
b) boric acid
c) borax
d) boron tri oxide
Answer:
c) borax

25. Boric acid on heating at 413 K gives
a) meta boric acid
b) tetra boric acid
c) boric anhydride
d) borax
Answer:
b) tetra boric acid

26. In ethyl borate test the colour of the flame obtained is
a) red
b) yellow
c) blue
d) green
Answer:
d) green

27. On hydrolysis BF3 gives Boric acid and converted to fluroboric acid. The fluoroboric acid contains the species. (PTA – 6)
a) H+, F & BF3
b) H+ & [BF4]
c) [H BF3]+ & F
d) H+, B3+ & F
Answer:
b) H+ & [BF4]

28. In organic benzene is
a) diborane
b) borazole
c) borax
d) boric acid
Answer:
b) borazole

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

29. The formula of Inorganic benzene is
a) B3N3
b) B3N3H3
c) B3N3H6
d)B6N6H6
Answer:
c) B3N3H6

30. The most stable form of carbon is
a) graphite
b) diamond
c) fullerene
d) carbon nano tubes
Answer:
a) graphite

31. The formula of buckminster fullerene is
a) C32
b) C50
e) C60
d) C70
Answer:
c) C60

32. The number of six membered and five membered rings fused together respectively in buckminster fullerene is
a) 12 & 20
b) 20 & 12
c) 10 & 22
d) 22 & 10
Answer:
b) 20 & 12

33. Water gas is a mixture of
a) CO2 + H2
b) CO + H2O
c) CO + H2
d) CO + N2
Answer:
c) CO + H2

34. Producer gas is a mixture of
a) CO2 + H2
b) CO + H2O
c) CO + H2
d) CO + N2
Answer:
d) CO + N2

35. In the presence of light carbon monoxide reacts with chlorine to form a poisonous gas called
a) mustard gas
b) phosgene
c) phosphine
d) carbylamine
Answer:
b) phosgene

36. Fischer Tropsch synthesis is used for preparing
a) Silicones
b) Boranes
c) Hydrocarbons
d) Carbonyls
Answer:
c) Hydrocarbons

37. In metal carbonyls the oxidation state of metals is
a) 0
b) +1
c) +2
d) +3
Answer:
a) 0

38. The structure of CO molecule is
a) trigonal
b) tetrahedral
c) linear
d) square planar
Answer:
c) linear

39. The structure of CO2 molecule is
a) trigonal
b) tetrahedral
c) linear
d) square planar
Answer:
c) linear

40. The critical temperature of CO2 is
a) 21 °C
b) 31°C
c) 12°C
d) 13°C
Answer:
b) 31°C

41. When CO2 is dissolved in water, the solution is slightly
a) acidic
b) basic
c) amphoteric
d) neutral
Answer:
a) acidic

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

42. Which among the following is important for photo synthesis?
a) O2
b) N2
C) CO
d) CO2
Answer:
d) CO2

43. The water repellant property of silicones is due to the presence of
a) -OH group
b) -Si group
c) -R group
d) -Cl group
Answer:
c) -R group

44. The percentage of silicate minerals and silica present in earth’s crust is
a) 75
b) 85
c) 95
d) 100
Answer:
c) 95

45. The basic unit present in silicates is
a) SiO2
b) [SiO3]
c) [SiO4]2-
d) [SiO4]4-
Answer:
d) [SiO4]4-

46. Talc is an example of
a) Ino silicates
b) Phyllo silicates
c) Tecto silicates
d) Chain silicates
Answer:
b) Phyllo silicates

47. Quartz is an example of
a) Ino silicates
b) Phyllo silicates
c) Tecto silicates
d) Chain silicates
Answer:
c) Tecto silicates

48. The formula of Spodumene is
a) Sc2Si2O7
b) Li Ai(SiO3)2
c) [Be3 Al2(SiO3)6]
d) Be2SiO4
Answer:
b) Li Ai(SiO3)2

49. The silicate which is used in the removal of permanent hardness of water is
a) Feldspar
b) Quartz
c) Zeolites
d) Talc
Answer:
c) Zeolites

50. Thermodynamically the most stable form of carbon is (PTA – 4)
a) Diamond
b) Fullerenes
c) graphite
d) Nano tubes
Answer:
c) graphite

II. Pick the odd man out

1. W.r.t. their metallic character pick the odd man out.
a) Ge
b) Ga
c) B
d) As
Answer:
b) Ga – It is a metal while others are metalloids

2. W.r.t. their metallic character pick the odd man out
a) In
b)Pb
c) Cl
d) Bi
Answer:
c) Cl – It is a non metal while others are metals.

3. Pick the odd man out
a) Borax
b) Kernite
c) Colemanite
d) Bauxite
Answer:
d) Bauxite – It is an ore of aluminium others are ores of boron.

4. W.r.t. to hybridisation pick the odd man out.
a) Graphite
b) Diamond
c) Fullerene
d) Graphene
Answer:
b. Diamond – It is sp³ hybridised while others are sp² hybridised.

III. Assertion and Reason

i) Both A and R are correct, R explains A
ii) A is wrong but R is wrong
iii) A is wrong but R is correct
iv) Both A and R are correct but R does not explain A

1. Assertion (A) : Boron shows non metallic character.
Reason (R) : Atomic radius of boron is small and its nuclear charge is high.
Answer:
(i).Both A and R are correct, R explains A

2. Assertion (A) : As we move down Boron group the elements show less tendency to exhibit +1 oxidation state rather than +3. Reason (R) : As we move down Boron group the elements show inert pair effect.
Ans : (iii).A is wrong but R is correct

3. Assertion (A) : Graphite conducts electricity.
Reason (R) : In Graphite, successive carbon sheets are held together by weak Vander Waals force.
Answer:
(iv). Both A and R are corrrect but R does not explain A

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

4. Assertion (A) : Silicones are used for making water proofing clothes.
Reason (R) : In silicones the organic side groups which surrounds silicon make the molecule looks like an alkane.
Answer:
(i).Both A and R are correct, R explains A

IV. Choose the correct statement

1. i) Some of the p-block elements show negative oxidation states also.
ii) Halogens gain two electrons to give a stable halide ion.
iii) Inert gases have ns²np6 configuration and hence more stable.
iv) p-block elements have a general electronic configuration (n-1)s² np1-6
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correction:
ii) Halogens gain one electron to give a stable halide ion.
iv) p-block elements have a general electronic configuration ns² np1-6

2. i) Boron compounds are electron rich compounds.
ii) Boron does not react directly with hydrogen.
iii) Borax is sodium salt of metaboric acid.
iv) Boric acid is used as an antiseptic,
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
Answer:
c) (ii) & (iv)
Correction:
i) Boron compounds are electron deficient compounds.
iii) Borax is sodium salt of tetraboric acid.

3. i) In graphite carbon atoms are sp³ hybridised.
ii) A single planar sheet of graphite is known as graphene.
iii) In diamond each carbon atom is tetrahedrally surrounded by four other carbon atoms.
iv) Carbon nanotubes do not conduct electricity,
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correction:
i) In graphite carbon atoms are sp² hybridised.
iv) Carbon nanotubes conduct electricity.

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

4. i) Silicones are organo silicon polymers.
ii) Hydrolysis of R2SiCl2 yields complex cross linked polymer.
iii) Silicones are good thermal and electrical conductors.
iv) All silicones are water repellent,
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d)(i) & (iv)
Correction:
ii) Hydrolysis of R2SiCl2 yields a straight chain polymer.
iii) Silicones are good thermal and electrical insulators.

V. Choose the wrong statement

i) Boron is a metal.
ii) Nitrogen is a metalloid.
iii) Oxygen is a non metal.
iv) Antimony is a metalloid.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
a) (i) & (ii)
Correction:
i) Boron is a metalloid (or) non metal
ii) Nitrogen is a non metal.

2. i) Aluminium chloride is a Lewis acid.
ii) Alum is a double salt of potassium aluminium sulphate.
iii) Aluminium chloride is used as a styptic agent to arrest bleeding.
iv) Alum is used as a catalyst in Friedel Crafts reaction.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correction:
iii) Alum is used as a styptic agent to arrest bleeding.
iv) Anhydrous Aluminium chloride is used as a catalyst in Friedel Crafts reaction.

3. Which of the following statement about H3BO3 is not correct? (PTA – 5)
a) It is a strong tribasic acid
b) It is prepared by acidifying an aqueous solution of borax.
c) It is a layer structure in which planer BO3 units are joined by hydrogen bonds.
d) It does not act as proton donor but acts as a Lewis acid by accepting hydroxyl ion.
Answer:
a) It is a strong tribasic acid

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

4. i) Silicates which contain discrete [SiO4]4- units are called neso silicates.
ii) Beryl is an example for amphiboles.
iii) Spodumene is an example for phyllo silicates.
iv) Silicates which contain [Si7O7]6- ions are called Soro silicates,
a) (i) & (ii)
b) (ii) & (iv)
c) (ii) & (iii)
d) (i) & (iv)
Answer:
c) (ii) & (iii)
Correction:
ii) Beryl is an example for cyclic silicates.
iii) Spodumene is an example for chain silicates.

VI. Match the following
1.

Group No. Group Name
i 13 a) Pnictogens
ii 14 b) Chalcogens
iii 15 c) Inert gases
iv 16 d) Halogens
v 17 e) Icosagens
vi 18 f) Tetragens

Answer:

Group No. Group Name
i 13 e) Icosagens
ii 14 f) Tetragens
iii 15 a) Pnictogens
iv 16 b) Chalcogens
v 17 d) Halogens
vi 18 c) Inert gases

2.

1. Fluorine i) Identification of coloured metal ions
2. Borax ii) strong oxidising agent
3. Aluminium iii) chalgogens present in volcanic ashes
4. Sulphur iv) Most abundant element

Answer:

1. Fluorine ii) strong oxidising agent
2. Borax i) Identification of coloured metal ions
3. Aluminium iv) Most abundant element
4. Sulphur iii) chalgogens present in volcanic ashes

3.

Compound Uses
1. Boron a) Eye drops
2. Amorphous boron b) Pyrex glass
3. Boric acid c) Moderator
4. Boric oxide d) Rocket fuel igniter

Answer:

Compound Uses
1. Boron c) Moderator
2. Amorphous boron d) Rocket fuel igniter
3. Boric acid a) Eye drops
4. Boric oxide b) Pyrex glass

4.

Type of Example
1. Ortho silicates a) Quartz
2. Pyro silicates b) Asbestos
3. Cyclic silicates c) Mica
4. Chain silicates d) Thortveitite
5. Amphiboles e) Spodumene
6. Sheet silicates f) Phenacite
7. Tecto silicates g) Beryl

Answer:

Type of Example
1. Ortho silicates f) Phenacite
2. Pyro silicates d) Thortveitite
3. Cyclic silicates g) Beryl
4. Chain silicates e) Spodumene
5. Amphiboles b) Asbestos
6. Sheet silicates c) Mica
7. Tecto silicates a) Quartz

VII. 2 Marks questions

Question 1.
What are ‘p’-block elements? Write their general outer electronic configuration.
Answer:
The elements in which their last electron enters the ‘p’ orbital are called ‘p’-block elements.

  • They are placed in 13 -18 groups.
  • General outer electronic configuration is ns²np1-6.

Question 2.
How are the p-block elements classified.
Answer:

  • Based on the outer electronic configuration they are classified as 13 -18 group elements.
  • Based on the nature of the elements they are classified as non metals, metalloids and metals.

Question 3.
Aluminium (III) chloride is stable where as Thallium (III) chloride is unstable. Why? (PTA – 2)
Answer:

  • Due to inert pair effect, as we move down the 13th group ns² electrons remain inert and np¹ electron takes part in the reaction.
  • So Tl3+ ion is less stable and Tl+1 ion is more stable.
  • Hence AlCl3 is stable where as TICl3 is unstable and decomposes into TlCl.

Question 4.
How is boric acid prepared from borax?
Answer:
Boric acid can be extracted from borax by treating with HCl or H2SO4.
Na2B4O7 + 2HCl + 5H2O → 4H3BO3 + 2NaCl
Na2B4O7 + H2SO4 + 5H2O → 4H3BO3 + 2Na2SO4

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

Question 5.
How is boric acid prepared from Colemanite?
Answer:
When sulphur dioxide is passed through colemanite solution, boric acid is obtained.
Ca2B6O11 + 2SO2 + 9H2O → 2CaSO3 + 6H3BO3

Question 6.
What is the action of sodium hydroxide on boric acid?
Answer:
Boric acid reacts with sodium hydroxide to form sodium metaborate and sodium tetra borate.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 12

Question 7.
Write the action of water on diborane.
Answer:
Diborane reacts with water to form boric acid.
B2H6 + 6H2O → 2H3BO3 + 6H2

Question 8.
What is the action of NaOH on diborane.
Answer:
Diborane reacts with NaOH to form sodium meta borate.
B2H6 + 2NaOH + 2H2O → 2NaBO2 + 6H2

Question 9.
What is the action of air on diborane?
Answer:
At room temperature pure diborane does not react with air or oxygen.

But impure diborane reacts with air or oxygen to giveB203 along with large amount of heat.
B3H6 + 3O2 → B2O3 + 3H2O
∆H =-2165 KJ mol-1

Question 10.
How does diborane react with methyl alcohol?
Answer:
Diborane reacts with methyl alcohol to give trimethyl borate.
B2H6 + 6CH3OH → 2B(OCH3)3 + 6H2

Question 11.
How does diborane react with metal hydrides?
Answer:
When treated with metal hydrides, diborane forms metal boro hydrides.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 13

Question 12.
How does diborane react with ammonia at low temperature?
Answer:
When treated with excess ammonia at low temperature diborane gives diborane di ammonate.
3B2H6 + 6NH3 \(\underrightarrow { -153K } \) 3B2H6.2NH3

Question 13.
How is inorganic benzene prepared? (PTA – 1)
Answer:

  • On heating at higher temperatures with ammonia, diborane forms borazole or borazine.
  • Borazole or borazine is called as Inorganic benzene
    Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 14

Question 14.
BF3 acts as a Lewis acid. Give example.
Answer:
BF3 is an electron deficient compound and accepts electron pairs to form coordinate covalent bonds. Hence BF3 acts as a Lewis acid.
BF3 + NH3 → F3B ← NH3
BF3 + H2O → F3B ← OH2

Question 15.
Convert BF3 into hydro fluoro boric acid.
Answer:
On hydrolysis BF3 gives boric acid, which is converted into hydro fluoro boric acid.
4BF3 + 3H2O → H3BO3 + 3HBF4
3HBF4 (Hydro fluoro boric acid)

Question 16.
Write about McAfee process of manufacturing AlCl3.
Answer:
AlCl3 is obtained by heating a mixture of alumina and coke in a current of chlorine.
Al2O3 + 3C + 3Cl2 → 2AlCl3 + 3CO

Question 17.
Write the action of NaOH on AlCl3
Answer:
With excess of NaOH, AlCl3 gives sodium alumina te.
AlCl3 + 4NaOH → NaAlO2 + 2H2O + 3NaCl

Question 18.
Write the uses of aluminium chloride.
Answer:
1. Anhydrous AlCl3 is used as a catalyst in Friedel crafts reaction.
2. AlCl3 is used for the manufacture of petrol by cracking the mineral oils.
3. AlCl3 is used as a catalyst in the manufacture of dyes, drugs and perfumes.

Question 19.
What are alums? Give examples.
Answer:
1. Alum is a double salt of potassium aluminium sulphate.
2. Now a days the name alum is used for all the double salts with the formula
M’2 SO4 M”2 (SO4)3.24H2O
Where M’ is univalent metal ion or NH4+
M” is trivalent metal ion
Example: K2SO4.Al2(SO4)3.24H2O Potash alum
K2SO4.Cr2(SO4)3.24H2O Chrome alum

Question 20.
Aqueous solution of carbon di oxide is acidic. Why?
Answer:
Aqueous solution of carbon di oxide is slightly acidic as it forms carbonic acid which dissociates to give H+ ions.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 15

Question 21.
How is silicon tetra choride prepared?
Answer:
SiCl4 is prepared by passing dry chlorine over an intimate mixture of silica and carbon heating to 1675 K in a porcelain tube.
SiO2 + 2C + 2Cl2 → SiCl4 + 2CO
SiCl4 is prepared commercially by the reaction of silicon with hydrogen chloride gas above 600 K.
SiO + 4HCl → SiCl4 + 2H2

Question 22.
Write the uses of silicon tetra chloride.
Answer:
Silicon tetra chloride is used
i) In the production of semi conducting silicon.
ii) As a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

Question 23.
What is water gas equilibrium? (PTA – 5)
Answer:
Water gas equilibrium
The equilibrium involved in the reaction between carbon di oxide and hydrogen, has many industrial applications and is called water gas equilibrium.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 16

VIII. Three Marks questions

Question 1.
How is borax prepared from colemanite?
Answer:
When colemanite ore solution is boiled with sodium carbonate solution borax is obtained.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 17

Question 2.
Write the uses of boron.
Answer:
1. 5B10 absorbs neutrons, hence it is used as a 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. Boric acid and borax are used in eye drops, antiseptics, washing powders.
5. Boric oxide is used in the manufacture of pyrex glass.

Question 3.
Aqueous solution of borax is basic. Why?
Answer:
In hot water borax dissociates into boric acid and sodium hydroxide.
Na2B4O7 + 7H2O → 4H3BO3 + 2NaOH
Boric acid is a weak acid, whereas sodium hydroxide is a strong base.

As a result the resulting solution is basic.

Question 4.
What is the action of heat on borax?
Answer:
On heating borax loses its water of crystallisation first and then decomposes into sodium metaborate and boron trioxide.

Boron trioxide appears as transparent glassy beads.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 18

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

Question 5.
What is the action of heat on boric acid?
Answer:

Temperature Compound obtained
373 K Meta boric acid
413 K Tetra boric acid
Red hot Boric anhydride (glassy mass)

Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 19

Question 6.
Describe the structure of boric acid.
Answer:

  • Boric acid has a two dimensional structure.
  • It consists of [BO3]3- unit.
  • These unit are linked to each other by hydrogen bonds.

Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 20

Question 7.
Write the uses of boric acid.
Answer:
Boric acid is
1. Used in the manufacture of pottery glazes, glass, enamels and pigments.
2. Used as an antiseptic.
3. Used as an eye lotion.
4. Used as a food preservative.

Question 8.
How is diborane prepared?
Answer:

  • When sodium boro hydride in diglyme is reacted with iodine diborane is obtained.
    2NaBH4 + I2 → B2H6 + 2NaI + H2
  • On heating magnesium boride with Hcl, a mixture of volatile boranes are obtained.
    2Mg3B2 + 12HCl → 6MgCl2 + B4H10 + H2
    B4H10 + H2 → 2B2H6

Question 9.
Write the uses of diborane.
Answer:
Diborane is
1. Used as a high energy fuel for propellant.
2. Used as a reducing agent in organic

Question 10.
How is boron trifluoride prepared from boron trioxide?
Answer:
When boron trioxide is treated with calcium fluroide in presence of conc.sulphruic acid, boron trifluoride is obtained.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 21
When boron trioxide is reacted with carbon and fluorine, boron trifluoride is obtained.
B2O3 + 3C + 3F2 → 2BF3 + 3CO

Samacheer Kalvi 9th English Guide Prose Chapter 2 I Can’t Climb Trees Anymore

Question 11.
How is boron trifluoride prepared in the laboratory?
Answer:
In the laboratory pure BF3 is prepared by the. thermal decomposition of benzene, diozonium tetrafluro borate.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 22

Question 12.
How is potash alum prepared? (PTA – 4)
Answer:
Potash alum is prepared from alunite or alum stone.
When alunite is treated with excess of sulphuric acid, the aluminium hydroxide present is converted into aluminium sulphate.
A calculated quantity of potassium sulphate is added.
The solution is crystallised to obtain potash alum.
It is purified bv recrystallisation.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 23

Question 13.
Write the uses of alum.
Answer:
Alum is used
i) for the purification of water.
ii) for water proofing and textiles.
iii) in dyeing, paper and leather tanning industries.
iv) as a styptic agent to arrest bleeding.

Question 14.
Write the uses of carbon monoxide.
Answer:
Carbon monoxide is used
i) as a reducing agent and can reduce many metal oxides to metal.
ii) as an important ligand and forms metal carbonyls.
iii) a mixture of CO & H2 is called as water gas and a mixture of CO & N2 is called as producer gas. Both are used as important industrial fuels.

Question 15.
Write the uses of carbon dioxide.
Answer:
Carbon dioxide is used

  • to produce an inert atmosphere for chemical processing.
  • by plants in photosynthesis.
  • as fire extinguisher.
  • as a propellant gas.
  • in the production of carbonated beverages.
  • in the production of foam.

Question 16.
Write note on Boron Neutron Capture Therapy (BNCT).
Answer:

  • The affinity of Boron-10 for neutrons is the bases of this technique BNCT for treating patients suffering from brain tumours.
  • It is based on the nuclear reaction which occurs when Boron-10 is irradiated with low- energy thermal neutrons to give high linear energy a-particles and a Li particle.
  • Boron compounds are injected into a brain tumour patient and the compounds collect preferentially in the tumour.
  • The tumour area is then irradiated with / thermal neutrons and results in the release of an alpha particle.
  • This a-particle damages the tissue in the tumour each time a Boron-10 nucleus captures a neutron.
  • In this wav damage can be limited preferentially to the tumour, leaving the normal brain tissue less affected.
  • BNCT has been studied as a treatment for several other tumours of the head and neck, the breast the prostate, the bladder and the liver.

IX. Five Marks questions

Question 1.
How is higher boranes obtained from diborane.
Answer:
At high temperatures diborane forms higher boranes liberating hydrogen.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 24

Question 2.
Explain the allotropes of carbon.
Answer:

  • Carbon exists in many allotropic forms.
  • Graphite and diamond are the most common allotropes.
  • Graphene, fullerenes and carbon nano tubes are other important allotropes of carbon.

Graphite:

  • It is the most stable allotrope of carbon at normal temperature and pressure.
  • It is composed of flat two dimensional hexagonal sheets of sp² hybridised carbon atoms.
  • C-C bond length is 1.41 Å which is close to the C-C bond distance in benzene (1.40 Å )
  • Each carbon atom forms three sbonds with three neighbouring carbon atoms using three of its valence electrons and the fourth electron present in the unhybridised p-orbital forms a p-bond.
  • These pelectrons are delocalised over the entire sheet, hence graphite conducts electricity.
  • Successive carbon sheets at a distance of 3.40 Ao are held together by weak Vander Waals forces, hence graphite is soft, slippery and used as a lubricant.

Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 25

Diamond:

  • Carbon atoms in diamond are sp3 hybridised.
  • Each carbon is bonded tetra hedrally with four other carbon atoms by s-bonds with C-C bond length of 1.54 Å. Hence diamond is hard.
  • Since all the four valence electrons of carbon are involved in bonding and there is no free electrons, diamond is not a conductor.
  • Being the hardest substance, diamond is used for sharpening hard tools, cutting glasses, making bores and rock drilling.

Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 26

Fullerenes:

  • There are newly synthesised allotropes of carbon.
  • Unlike graphite and diamond these are discrete molecules of carbon like C32, C50, C60, C70, C76 …….
  • These have cage like structures.
  • Buck‘minster fullerene or bucky ball have a soccer ball like structure with the formula C60.
    It has a fused’ ring structure with 20 six membered rings and 12 five membered rings.
  • Each carbon is sp² hybridised and forms three π sbonds and one delocalised pbond giving aromatic character.
  • C-C bond distance is 1.44 Å and C=C bond distance is 1.38 Å.

Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 27

Carbon nano tubes:

  • This is another recently discovered allotropes of carbon.
  • They have graphite like tubes with fullerene ends.
  • Along the axis, carbon nano tubes are stronger than steel and conduct electricity.
  • They have many applications in nano scale electronics, Catalysis, polymers and medicine.

Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 28

Graphene:

  • It is a single planar sheet of graphite.
  • In this sp² hybridised carbon atoms are densely packed in a honey comb crystal lattice.

Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 29

Question 3.
Write about the preparation and structure of silicones.
Answer:

  • Silicones or poly siloxanes are organo silicon polymers.
  • Their general empirical formula is (R2SiO)
  • Since their empirical formula is similar to Ketones (R2CO) they are called as Silicones.
  • They may be linear or cross linked.
  • Due to their very high thermal stability they are called high-temperature polymers.

Types of Silicones:
i) Linear Silicones:
They are obtained by the hydrolysis and subsequent condensation of dialkyl or diaryl
a) Silicone rubbers:
These are bridged together by methylene or similar groups.
b) Silicone resins:
They are obtained by blending silicones with organic resins such as acrylic esters.

ii) Cyclic Silicones:
These are obtained by the hydrolysis of R2SiCl2.

iii) Cross linked Silicones:
These are obtained by the hydrolysis of RSiCl3.

Preparation:
Vapours of RCl or ArCl are passed over silicon at 570 K with copper catalyst gives R2SiCl2 (dialkyl dichloro silanes) or Ar2SiCl2 (diaryl dichloro silanes)
2RCl + Si \(\underrightarrow { Cu / 570K } \) R2SiCl2

Hydrolysis of R2SiCl2 gives a straight chain polymer which grows from both sides.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 30

Hydrolysis of mono alkyl trichloro silanes RSiCl3 gives a very complex cross linked polymer.

Linear silicones can be converted into cyclic or ring silicones when water molecules are removed from the terminal -OH groups.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 31

Question 4.
Explain various types of silicates.
Answer:
The mineral which contains silicon and oxygen in tetrahedral [SiO4]4- units linked together in different patterns are called silicates.
Samacheer Kalvi 12th Chemistry Guide Chapter 2 p-Block Elements – I 33