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

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

Samacheerkalvi new Reduced and new syllabus chemistry Guide volume 1,2 Tamil Medium and English Medium full guide book back Questions and answer guide solution available here.

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

Question 1

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

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) pa³tially 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


  • The following factors are resposible for the anamolous properties of the first elements of p-blick.
  1. Small size of the first member
  1.  High ionisation enthalpy and high electronegativity.
  1.  Absence of d-orbitals in their valence shell
First elemenl Property of First elements Other elements in the family
B row1 col 2 Metals
C 1. Non-metal 2. It can form multiple bonds.. row2 col 3
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'

  • 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.
  •  Carbon exists as diamond, graphite, graphene, fullerenes, carbon nanotubes
  • Used for the identification of coloured metal ions (Borax bead test)
  • Manufacture of optical and borosilicate glass, enamels and glazes for pottery.
  • Flux in metallurgy.
  • Good preservative.
  • Catenation is an ability of an element to form chain of atoms.
  • The conditions for catenation are
  • Carbon possesses all the above properties and shows catenation.
  • Carbon forms a wide range of compounds with itself and with other elements such as H, O, N, S and halogens.
  • 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.

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

Group name Example
i. Icosagens. Boron
ii. Tetragens Corban
iii.pnictogen Nitrogen
iv. Chalcogens Oxygen
Group Non metals Metallodios 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.


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

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) Mg₃SiO₄ 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.

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:

Question 2.

Describe briefly allotropiam in p-block elements with specific reference to carbon.

Answer:

Example:

Question 3.

Give the uses of Borax.

Answer:

Question 4.

What is catenation? Describe briefly the catenation property of carbon. (MARCH 2020)

Answer:

i) The valency of the element is greater than or equal to two.

ii) The element should have an ability7 to bond with itself.

iii) The self bond must be as strong as its bond with other elements.

iv) Kinetic inertness of catenated compound towards other molecules.

Question 5.

Write a note on Fisher Tropsch synthesis. Fischer Tropsch synthesis: (PTA – 4)

Answer:

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

Structure of CO₂:

Structure is linear.

Question 7.

Give the uses of silicones.

Answer:

Question 8.

Describe the structure of diborane. (PTA – 3)

Answer:

Question 9.

Write a short note on hydroboration.

Answer:

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

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.


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:

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,

  • 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.

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.

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.

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 BaCl₂and 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.

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 ns²np⁶which is stable completely filled configuration.
  • So they are more stable and least reactive.
  • Hence they are called inert gases.

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