CLS Aipmt 16 17 XI Che Study Package 3 SET 2 Chapter 11

Chapter

11

The p-Block Elements Solutions SECTION - A Objective Types Questions 1.

On moving down the group 13 density (1) Increases

(2)

Decreases

(3) First decreases then increases

(4)

Remains same

Sol. Answer (1) Density =

Mass Volume

On moving down the group atomic mass increases as well as due to addition of shells, there is an increase in volume. But mass increases at a rate faster than volume. 2.

High melting point of boron is due to its existance as (1) Small covalent molecule

(2)

Giant covalent molecule

(3) Giant covalent solid

(4)

Giant ionic molecule

Sol. Answer (3) Boron exists as a giant covalent solid due to which its melting point is high. 3.

Elements of group 13 mainly form covalent compounds because (1) Size of ions is small

(2)

Sum of three ionization energies is very high

(3) Electronegativity values are high

(4)

All of these

Sol. Answer (4) Due to small size of ions, high electronegative values and high sum of the three ionization energies, the elements of group 13 form covalent compounds. 4.

Oxidation state shown by group 13 elements is (1) +1 and +3

(2)

+1, +2 and +3

(3)

+2, +3 and +4

(4)

+1 and +4

Sol. Answer (1) Boron shows +1 and +3 oxidation states. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

40 5.

The p-Block Elements

Solution of Assignment (Set-2)

Which one of the following elements of group 13 can react with alkali solutions to give H2 gas? (1) Boron

(2)

Aluminium

(3)

Gallium

(4)

All of these

Sol. Answer (4) Boron, aluminium and gallium are all elements of group 13. 6.

Lewis acid character of boron trihalides is as follows (1) BI3 > BBr3 > BCl3 > BF3

(2)

BF3 > BCl3 > BBr3 > BI3

(3) BCl3 > BF3 > BBr3 > BI3

(4)

BI3 > BBr3 < BF3 < BCl3

Sol. Answer (1) Acidic strength is inversely proportional to back-bonding as back-bonding decreases from BF3 to BI3, hence the Lewis acidic strength will be BF3 < BCl3 < BBr3 < BI3. 7.

Dimer Al2Cl6 is formed because (1) Al is electron rich (2) Aluminium is having lone pair of electron (3) Aluminium forms coordinate bonds with chlorine to complete its octet (4) Aluminium donates lone pair to form bridge

Sol. Answer (3)

Cl

Cl

Cl

Al

Al

Cl Cl Cl Aluminium forms coordinate bonds with chlorine to complete its octet. 8.

When we heat borax strongly then it will yield the following compound (1) NaBO2

(2)

B2O3

(3)

Na2B4O7

(4)

Both (1) & (2)

(3)

3

(4)

4

Sol. Answer (4)

Pt Loop

Na2B4O7.10H2O

Na2B4O7 + 10H2O Anhyd Borax

 NaBO2 + B2O3 Glassy bead

9.

–

B(OH)3 accept how many OH ions? (1) 1

(2)

2

Sol. Answer (1)

H

O

B

O O

H H

+ H2O

H+[B(OH)4]–

It does not lose 3H+ but accepts one lone pair from H2O. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

The p-Block Elements

41

10. Boric acid is having a polymeric type structure because of its (1) Basic nature

(2)

Acidic nature

(3)

Hydrogen bonds

(4)

Co-ordinate bonds

Sol. Answer (3) Boric acid is layered structure. B(OH)3 units are joined by hydrogen bonds and form two dimensional sheet. 11. Select the incorrect statement for B2H6 3

(1) It contains B—B ionic bond

(2)

Each boron is sp hybridised

(3) It has two types of hydrogen bonds

(4)

All of these

(3)

Four bonds

Sol. Answer (1)

H

H

H B

B H

H

H

It does not contain B–B bonds. 12. In diborane each boron forms (1) Two bonds

(2)

Three bonds

(4)

Five bonds

(4)

Aluminium carbide

Sol. Answer (3) In diborane each boron forms four bonds. 13. Which one of the following compounds has similar structure to that of graphite? (1) Boron nitride

(2)

Boron carbide

(3)

Aluminium oxide

Sol. Answer (1) Boron nitride has graphite-like structure. 14. The number of sigma and pi bonds present in inorganic benzene are respectively (1) 3, 12

(2)

12, 3

(3)

3, 3

(1) Resistance to corrosion

(2)

Poor conductivity

(3) Heaviness

(4)

All of these

(4)

12, 12

Sol. Answer (2) Borazine B3N3H6 is called inorganic benzene.

H H

H

N

B

B

H

N B

N

H

H

H

H

H Borazine

+ N

– B

N+ – B

–B N+

H

H

H

15. Aluminium is used for making alloys because of its

Sol. Answer (1) Due to the lightness and resistance to corrosion aluminium is used for making alloys. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

42

The p-Block Elements

Solution of Assignment (Set-2)

16. Group 14 elements have general electronic configuration 2

(1) ns

(2)

2

4

ns np

(3)

2

6

ns np

(4)

2

2

ns np

Sol. Answer (4) General electronic configuration of group 14 elements is ns2np2. 17. Tendency of carbon for catenation is because carbon–carbon atom bond energy is (1) Low

(2)

High

(3)

Zero

(4)

Negative

Sol. Answer (2) Due to high carbon-carbon atom bond energy, the tendency of carbon for catenation is generally high. 18. All elements except carbon have tendency to show maximum covalency of six (1) Due to absence of vacant d-orbitals

(2)

Due to presence of vacant d-orbitals

(3) Due to presence of partially filled d-orbitals

(4)

Due to presence of completely filled d-orbitals

Sol. Answer (2) All elements except carbon have tendency to show maximum covalency of six due to the presence of vacant d-orbitals. 19. Most abundant metal by mass in earth crust is (1) Silicon

(2)

Germanium

(3)

Aluminium

(4)

Arsenic

Sol. Answer (3) Next to silica and oxygen, aluminium is the most widely distributed element. It is present to the extent of 7.3 percent in the earth's crust. 20. Which one of the following elements is a metalloid? (1) Carbon

(2)

Germanium

(3)

Lead

(4)

All of these

Sol. Answer (2) Carbon – Non-metal

(On moving down the group the metallic character increases)

Germanium – Metalloid Lead – Metal 21. On moving down the group, acidic nature of oxides of group 14 (1) Decreases

(2)

Increases

(3) Remains same

(4)

Increases then decreases

Sol. Answer (1) The acidic nature decreases with increase of atomic number of the oxides. CO2 and SiO2 are acidic and GeO2, SnO2 & PbO2 are amphoteric. Basic character of both mono and dioxide increases down the group. 22. Which one of the following elements forms double or triple bond involving p-p bonding? (1) Carbon

(2)

Silicon

(3)

Germanium

(4)

Tin

Sol. Answer (1) Carbon atoms form double or triple bonds involving p-p bonding. Carbon shows anomalous behaviour due to its smaller size, higher electronegativity, higher ionization enthalpy and unavailability of d-orbitals. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

The p-Block Elements

43

23. Allotropy is due to (1) Difference in the number of atoms in the molecules (2) Difference in the arrangement of atoms in the molecules in the crystal (3) Difference in chemical properties (4) All of these Sol. Answer (2) Allotropy : The existence of a chemical element in two or more forms, which may differ in the arrangements of atoms in crystalline solids. 24. In diamond, carbon have 3

(1) sp hybridisation

(2)

sp hybridisation

(3)

2

sp hybridisation

(4)

3 2

sp d hybridisation

Sol. Answer (1) In diamond each carbon is joined to other four carbon tetrahedrally and carbon-carbon bond length is 1.54Å and the bond angle is 109°28' having sp3 hybridisation on each carbon. 25. Graphite has (1) 2-D sheet structure (2) Van der Waals forces between different layers 2

(3) sp hybridised carbon linked with other three carbon atoms in hexagonal planar structure (4) All of these Sol. Answer (4) In graphite each carbon is ‘sp2’ hybridised. It has a layered structure. These layers are attracted by van der Walls forces. 26. In graphite, the bond is (1) Ionic

(2)

Covalent

(3)

Co-ordinate

(4)

Metallic

Sol. Answer (2) In graphite a carbon atom is attached to another carbon atom i.e. the two atoms bonded together are of equal electronegativity. Thus, a covalent bond is formed. 27. Which one of the following is properties of CO gas? (1) It is a colourless gas

(2)

It is an odourless gas

(3) It is a neutral oxide

(4)

All of these

Sol. Answer (4) Carbon monoxide is a colourless, odourless gas. It is also a neutral oxide as it does not form salts when reacted with acids or bases. 28. Carbonic acid is a (1) Weak tribasic acid

(2)

Weak dibasic acid

(3) Strong tribasic acid

(4)

Strong dibasic acid

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44

The p-Block Elements

Solution of Assignment (Set-2)

Sol. Answer (2) Carbon dioxide on reaction with water gives a weak dibasic acid called carbonic acid which dissociates as – +   H 2CO 3 (aq) + H 2O(l)   HCO 3 (aq) + H 3O (aq) 2– +   HCO 3 – (aq) + H 2O(l)   CO 3 (aq) + H 3O (aq)

It is called dibasic because it has two hydrogen ions to donate to a base in an acid base reaction. 29. Organosilicon polymers containing Si—O—Si linkage is called (1) Silicates

(2)

Silicones

(3)

Glass

(4)

Silica

Sol. Answer (2) Silicones are polymers having

30.

Si

O

n

monomer.

SiO44  ion has geometry (1) Triangular

(2)

Tetrahedral

(3) Pentagonal bipyramidal

(4)

Linear

Sol. Answer (2) Ortho silicates (SiO4–4) –

O

Si –

O

(Tetrahedral geometry) –

O

–

O

SECTION - B Objective Type Questions 1.

Boron compounds behave as lewis acids because of their (1) Acidic nature

(2)

Covalent nature

(3) Ionisation energy

(4)

Electron deficient nature

Sol. Answer (4) Boron compounds, especially the hydrides are electron deficient compounds which can accept a lone pair of electrons hence behave as Lewis acids. 2.

The compound that is not a lewis acid is (1) BF3

(2)

AlCl3

(3)

PCl3

(4)

SnCl4

Sol. Answer (3) In PCl3 molecule, P-atom has a ‘lone-pair’ of electrons. This lone pair of electrons can be donated to electron deficient species (Lewis acid). PCl3 can act as a Lewis base. 3.

Borax is (1) Na2[B4O5(OH)4].8H2O

(2)

Na2[B4O5(OH)6].7H2O

(3) Na2[B4O3(OH)8].6H2O

(4)

Na2[B4O2(OH)10].5H2O

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Solution of Assignment (Set-2)

The p-Block Elements

45

Sol. Answer (1) Na2B4O710H2O / Na2[B4O5(OH)4]8H2O. 4.

In Diborane, the incorrect statement is (1) All 6 B–H bond are on same plane (2) 4 B–H bonds are on the plane and two B–H bonds above and below the plane (3) It is the 12 e– species (4) Two BH3 are attached with three centre electron pair bond

Sol. Answer (1) Diborane (B2H6)

H H

H 120° B 97°

B H

H H

Number of atoms present in same plane = 6. 5.

On strong heating, boric acid yields (1) B

(2)

B2H6

(3)

B2O3

(4)

BO2

(4)

B + HNO3 

Sol. Answer (3) 100 C H 3BO 3   HBO 2 + H 2O 160 C 4HBO 3   H 2B 4O 7 + H 2O

Red hot H 2B 4O 7   2B 2O 3 + H 2O

6.

In which of the following reaction boron does not act as reducing agent? (1) B + CO2 

(2)

B + Mg 

(3)

B + SiO2 

Sol. Answer (2) B + Mg  In the above reaction boron oxidises Mg to Mg2+, hence it acts as an oxidising agent. 7.

Which of the following statement is correct? (1) Boron and aluminium halides behave as Lewis acids (2) Al forms [AlF6]3– ion but B does not form [BF6]3– ion (3) The p – p back bonding occurs in the halides of boron and not in those of aluminium (4) All of these

Sol. Answer (4) Boron and aluminium halides behave as Lewis acid. Al forms [AlF6]3– ion but B does not form [BF6]3– ion. The p-p back bonding occurs in the halides of boron and not in those of aluminium. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

46 8.

The p-Block Elements

Solution of Assignment (Set-2)

Aluminium chloride in acidified aqueous solution forms (1) Tetrahedral [Al(H2O)4]3+ ion

(2)

Octahedral [Al(H2O)4]3+ ion

(3) Tetrahedral [Al(H2O)6]3+ ion

(4)

Octahedral [Al(H2O)6]3+ ion

Sol. Answer (4) Aluminium chloride in acidified aqueous solution forms octahedral [Al(H2O)6]3+ ion.

9.

Na2B4O7

740°C



2NaBO2 + B2O3 X + Y Transparent

Z + CuO(s)  Cu(BO2)2 Blue Bead

The 'Z' will be (1) X

(2)

Y

(3) Mixture of X & Y in 2 : 1 ratio

(4)

Mixture of X & Y in 1 : 2 ratio

Sol. Answer (2)

Na2B4O7

740°C



2NaBO2 + B2O3 Glass bead

CuO + B2O3

Cu(BO2)2

CuO + NaBO2

NaCuBO3

Blue bead

10. Reaction of ammonia with diborane gives initially B2H6.2NH3 which can also be written as (1) [BH2(NH3)2]+[BH4]–

(2)

[BH4]+[BH2(NH3)2]–

(3)

[BH3NH3]+[BH4]–

(4)

[B2N2H6]+[H3]–

Sol. Answer (1) Boron hydride reacts with excess ammonia to given B2H6.2NH3 which is white ionic solid and consists of (H3N  BH2  NH3)+ and BH4–. 11. Diborane can't be obtained from (1) Na2B4O7 + HCl

(2)

NaBH4 + I2

(3)

BF3 + LiAlH4

(4)

BF3 + NaH

Sol. Answer (1) Na2B4O7 + 10H2O + 2HCl  4H3BO3 + 2NaCl + 5H2O 12. White fumes appear around the bottle of anhydrous AlCl3 due to (1) Decomposition of AlCl3

(2)

Hydrolysis of AlCl3 liberating H2 gas

(3) Hydrolysis of AlCl3 liberating Cl2 gas

(4)

Hydrolysis of AlCl3 liberating HCl gas

Sol. Answer (4) AlCl3 + 3H2O  Al(OH)3 + 3HCl Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

The p-Block Elements

47

13. Number of hydroxyl group attached to Boron in Borax are (1) Four

(2)

Five

(3)

Six

(4)

Ten

Sol. Answer (1) Structure of Borax

OH B O

O HO

+1

2Na

– B

– B O

O

OH

.8H2O

O B

OH 14. The correct match is (1) C60 – Buckminster fullerene

(2)

Na2B4O7.4H2O – Kernite

(3) Borazole – B3N3H6

(4)

All of these

(1) B(OH)3 – basic

(2)

SnO, PbO – amphoteric

(3) GeO2 – basic

(4)

PbO2 – only acidic

Sol. Answer (4) C60 – Buckminster fullerene Na2B4O7.4H2O – Kernite Borazole – B3N3H6 15. The correct match is

Sol. Answer (2) SnO, PbO are amphoteric oxides. The acidic nature decreases with increase of atomic number of oxides. 16. C – O bond length is maximum in (1) CH3CHO

(2)

CO2

(3)

CO

(4)

CO32

Sol. Answer (4) Due to small size of carbon and repulsions along lp-bp, bp-bp in CO32–. The bond length C–O will be maximum.

Bond Order 

2–

O O

C

1 Bond Length

O

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48

The p-Block Elements

Solution of Assignment (Set-2)

17. Dry ice is composed of (1) Solid He

(2)

Solid CO2

(3)

Solid SO2

(4)

Solid C6H6

(1) Decreases from top to bottom

(2)

Increases from top to bottom

(3) Does not change gradually

(4)

Metallic character is not seen

Sol. Answer (2) Dry ice composed of solid CO2. 18. The metallic character of group 14

Sol. Answer (2) The atomic radii increases down the group and ionization energies decreases. According to periodic trends the metallic character of group 14 increases down the group. 19. In carbon family the tendency to show +2 oxidation state increases in order of (1) Ge < Sn < Pb

(2)

Pb < Sn < Ge

(3)

Sn < Ge < Pb

(4)

Sn < Pb < Ge

Sol. Answer (1) Due to inert pair effect, the stability of +4 oxidation state decreases down the group while the stability of +2 oxidation state increases. 20. Which one of the following is correct statement of fullerenes –C60? (1) Fullerenes are made by heating of graphite in an electric arc in the presence of Hydrogen (2) Fullerenes are the only impure form of carbon due to presence of dangling bonds (3) Both (1) & (2) (4) It contains twenty six-membered rings and twelve five membered rings Sol. Answer (4) According to the structure of fullerene which has a football-like structure. It contains 20 hexagonal sixmembered ring and 12 pentagonal five-membered ring. 21. The mixture of CO & H2 is known as (1) Water gas or producer gas

(2)

Water gas or synthesis gas

(3) Synthesis gas or producer gas

(4)

Producer gas

Sol. Answer (2)

H2O + C

H2 + CO

( H = +131 kJ/mol)

(steam) (Red hot fuel) Water gas or synthesis gas

22. When SiCl4 is allowed to undergo hydrolysis it gives (1) SiO2 – Silicic acid

(2)

Si(OH)4 – Silicic acid

(3) Si(OH)Cl3 – Silicic acid

(4)

SiCl4 do not undergo hydrolysis

Sol. Answer (2) SiCl4 + H2O

Si(OH) 4 + 4HCl (Silicic acid)

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Solution of Assignment (Set-2)

The p-Block Elements

49

23. Which statement is correct for carbon family? (1) Tin mainly occurs as Cassiterite, SnO2 (2) Silicon is the third most abundant element on earth's crust (3) Only two isotopes of carbon are present C12 and C13 (4) Germanium is most abundant than other members of carbon family Sol. Answer (1) Tin mainly occurs as cassiterite (SnO2). 24. p–p multiple bond is seen in (1) Mostly carbon

(2)

All carbon family member

(3) Sn but not in carbon

(4)

Boron family and not in carbon family

Sol. Answer (1) Carbon has smaller size, high electronegativity; higher ionization enthalpy and unavailability of d-orbitals due to which it can form only p–p bond. 25. A + CO  CO2 B + CO  CO2 X + O2  CO2 A, B & X respectively are (1) CH4, Carbon, Fe2O3

(2)

Fe2O3, ZnO, CH4

(3) Fe2O3, CH4, ZnO

(4)

HCOOH, Carbon, CH4

(3)

Octahedral

Sol. Answer (2) Fe2O3 + 3CO  2Fe + 3CO2 ZnO + CO  CO2 + Zn CH4 + 2O2  CO2 + 2H2O 26. The geometry of SiCl4 is (1) Tetrahedral

(2)

Square planar

(4)

Planar triangular

Sol. Answer (1)

Cl Si Cl

Cl (Tetrahedral geometry)

Cl

27. The silicates which contain discrete tetrahedral units are (1) Sheet silicates

(2)

Ortho silicates

(3) Three dimensional silicates

(4)

Pyrosilicate

Sol. Answer (2) –

O

Si –

O

(Tetrahedral geometry) –

O

–

O

The orthosilicates contain discrete tetrahedral units. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

50 28.

The p-Block Elements

Solution of Assignment (Set-2)

Cu Powder CH3Cl  Si  

x

2H2O y 

570 K

y & x respectively are (1) (CH3)2SiCl2, (CH3)2Si(OH)2

(2)

(CH3)2Si(OH)2, (CH3)2SiCl2

(3) SiCl4, Si(OH)4

(4)

Si(OH)4, SiCl4

Sol. Answer (2) Silicone polymers Si + 2CH3Cl  (CH3)2SiCl2 (CH3)2SiCl2 + 2H2O  (CH3)2Si(OH)2 + 2HCl 29. Hydrolysis of dimethyldichloro silane; (CH3)2SiCl2 followed by condensation polymerisation yields straight chain polymer of

(1)

O

O

O

Si

Si

O

O

CH3 (3)

O

Sol. Answer (3)

Si CH3

O

(2)

Si O

Si

Si

CH3 CH3

CH3 O

O

O

(4)

O

Si

Si

O

CH3 CH3

CH3

Polymerisation of silicones

R HO

Si

R OH + HO

R

Si

R OH + HO

R

Si

R OH

Polymerisation –H2O

R

Si R

R O

Si R

R O

Si

O

R

Silicon

30. Silicons are (1) Water repelling in nature

(2)

With high thermal stability

(3) With high dielectric strength

(4)

All of these

Sol. Answer (4) Silicons are water repellants, good electrical insulators, stable towards heat, non-toxic resistant to chemicals. 31. Which one is correct statement for zeolite? (1) They are alumino silicates (2) Hydrated zeolites are used as ion exchangers in hardening of soft water (3) ZSM-5 is used to convert gasoline to alcohol (4) All of these Sol. Answer (1) If Al atoms replace few silicon atoms in 3D-network of SiO2, overall structure is known as aluminosilicates acquires a negative charge. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

The p-Block Elements

51

32. The correct statement is (1) Diamond is covalent yet it has high melting point (2) [SiF6]2– is known whereas [SiCl6]2– is not (3) SiO only exist at high temp. (4) All of these Sol. Answer (4) Due to extended covalent bonding, diamond is hard to break six large Cl– ions cannot be accommodated around silicon ion SiO exists at only high temperature. 33. (i) SiO2 + NaOH ? (ii) SiO2 + HF ? The products of (ii) & (i) respectively are (1) H2SiF6, SiO44–

(2)

SiF4, Na2SiO3

(3)

Na2SiO3, SiF4

(4)

Na2SiO4, H2SiF6

(3)

Methane

(4)

Propyne

(3)

C3O2

(4)

CO

Sol. Answer (2) SiO2 + 2NaOH  Na2SiO3 + H2O SiO2 + 4HF  SiF4 + 2H2O 34. Calcium carbide on hydrolysis gives (1) Ethylene

(2)

Acetylene

Sol. Answer (2) CaC2 + 2H2O  Ca(OH)2 + C2H2 (Acetylene) 35. Carbon suboxide has the formula (1) H2CO3

(2)

C2O3

Sol. Answer (3) Carbon suboxide or tricarbon dioxide, C3O2 or O=C=C=C=O.

SECTION - C Previous Years Questions 1.

The stability of +1 oxidation state among Al, Ga, In and Tl increases in the sequence (1) TI < In < Ga < Al

(2)

In < Tl < Ga < Al

(3) Ga < In < Al < Tl

(4)

Al < Ga < In < Tl

[Re-AIPMT-2015]

Sol. Answer (4) Due to INERT PAIR EFFECT group oxidation state less by 2 becomes more stable going down the group. Group oxidation state of group 13 is +3.  Stability of +1 oxidation state would be Al < Ga < In < Tl Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

52 2.

The p-Block Elements

Solution of Assignment (Set-2)

Which of the following structure is similar to graphite? (1) B

(2)

B4C

[NEET-2013] (3)

B2H6

(4)

BN

Sol. Answer (4) One B atom and one N atom together have the same number of valency electron as two carbon atom. Thus, BN has almost the same structure as graphite, with sheets made up of hexagonal rings of alternate B and N atoms joined together. 3.

Which of these is not a monomer for a high molecular mass silicone polymer? (1) Me2SiCl2

(2)

Me3SiCl

(3)

PhSiCl3

[NEET-2013] (4)

MeSiCl3

Sol. Answer (2) Linear chain silicone formed by hydrolysis of Me2SiCl2 followed by condensation. Cross link silicone form by hydrolysis of Me3SiCl3. (MeSiCl is used to stop chain length) HOH Me 3SiCl   Me 3SiOH

Me 3SiOH + HOSiMe 3   Me 3Si – O – OSiMe 3 4.

Which of these is least likely to act as a Lewis base ? (1) F–

(2)

BF3

[NEET-2013] (3)

PF3

(4)

CO

Sol. Answer (2) 5.

The basic structural unit of silicates is (1) SiO44 

(2)

SiO32 

[NEET-2013] (3)

SiO24

SiO–

(4)

Sol. Answer (1) –

O

Si –

O 6.

[SiO4]4– –

O

O–

Which of the following is electron-deficient? (1) (SiH3)2

(2)

(BH3)2

[NEET-2013] (3)

PH3

(4)

(CH3)2

Sol. Answer (2) 7.

Which of the following species contains three bond pairs and one lone pair around the central atom ? [AIPMT (Prelims)-2012] (1) NH2–

(2)

PCl3

(3)

H2O

(4)

BF3

Sol. Answer (2) 8.

When Cl2 gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from [AIPMT (Prelims)-2012] (1) Zero to –1 and zero to +3

(2)

Zero to +1 and zero to –3

(3) Zero to +1 and zero to –5

(4)

Zero to –1 and zero to +5

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Solution of Assignment (Set-2)

The p-Block Elements

53

Sol. Answer (4) This reaction is 0

5

–1

6NaOH + 3Cl 2   NaCl + NaClO 3 + 3H 2O

hot & conc.

In Cl2, the oxidation number of Cl is O but in NaClO3 the oxidation number of Cl is +5 and – 1 in NaCl. 9.

Which of the following is least likely to behave as Lewis base? –

(2)

(1) OH

H2O

(3)

NH3

[AIPMT (Prelims)-2011] (4)

BF3

Sol. Answer (4) 10. Name the type of the structure of silicate in which one oxygen atom of [SiO4]4– is shared? [AIPMT (Prelims)-2011] (1) Three dimensional

(2)

Linear chain silicate (3)

Sheet silicate

(4)

Pyrosilicate

Sol. Answer (4) –

–

O

–

O

O

Si

Si –

O

O

–6

O– O

–

(Si2O7 ) Pyrosilicates

11. Which of the following oxide is amphoteric ? (1) SiO2

(2)

CO2

[AIPMT (Mains)-2011] (3)

SnO2

(4)

CaO

Sol. Answer (3) CaO – Basic CO2 and SiO2 – Acidic SnO2 – Amphoteric SnO2 + 4HCl  SnCl2 + 2H2O SnO2 + 2NaOH  Na2SnO3 + H2O 12. Which one of the following molecular hydrides acts as a Lewis acid ? (1) NH3

(2)

H2O

(3)

B2H6

[AIPMT (Prelims)-2010] (4)

CH4BF3

Sol. Answer (3) Since borane has vacant p orbitals in central atom hence it can act as Lewis acid. 13. The tendency of BF3, BCl3 and BBr3 to behave as Lewis acid decreases in the sequence [AIPMT (Prelims)-2010] (1) BCl3 > BF3 > BBr3

(2)

BBr3 > BCl3 > BF3

(3)

BBr3 > BF3 > BCl3

(4)

BF3 > BCl3 > BBr3

Sol. Answer (2) Acidic strength is inversely proportional to back bonding, as back bonding decreases from BF3 to BI3. Hence Lewis acidic strength will increase as BF3 < BCl3 < BBr3 < BI3 14. Which of the following molecules acts as a Lewis acid? (1) (CH3)2O

(2)

(CH3)3P

[AIPMT (Prelims)-2009] (3)

(CH3)3N

(4)

(CH3)3B

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54

The p-Block Elements

Solution of Assignment (Set-2)

Sol. Answer (4) 15. Which one of the following anions is present in the chain structure of silicates ? (1) SiO44 

(2)

Si2O76 

(3)

 SiO  2 3

n

[AIPMT (Prelims)-2007] (4)

 Si O  2

2 5

n

Sol. Answer (4) 16. Which of the following oxidation states are the most characteristic for lead and tin respectively ? [AIPMT (Prelims)-2007] (1) +2, +2

(2)

+4, +2

(3)

+2, +4

(4)

+4, +4

Sol. Answer (3) Lower oxidation state (+2) is more stable as we move down the group. 17. Al2O3 can be converted to anhydrous AlCl3 by heating

[AIPMT (Prelims)-2006]

(1) Al2O3 with HCl gas

(2)

Al2O3 with NaCl in solid state

(3) A mixture of Al2O3 and carbon in dry Cl2 gas

(4)

Al2O3 with Cl2 gas

Sol. Answer (3) Al2O3 + 3C + 3Cl2  2AlCl3 + 3CO 18. Which of the following is the most basic oxide ? (1) Al2O3

(2)

Sb2O3

[AIPMT (Prelims)-2006] (3)

Bi2O3

(4)

SeO2

Sol. Answer (3) 19. Which of the following is not isostructural with SiCl4 ? (1) SCl4

(2)

SO24

[AIPMT (Prelims)-2006] (3)

PO34

(4)

NH4

Sol. Answer (1) 20. In which of the following molecules are all the bonds not equal? (1) ClF3

(2)

BF3

(3)

AlF3

[AIPMT (Prelims)-2006] (4)

NF3

Sol. Answer (1) 21. Which of the following is the electron deficient molecule? (1) B2H6

(2)

C2H6

(3)

[AIPMT (Prelims)-2005] PH3

(4)

SiH4

Sol. Answer (1) Boranes have vacant ‘p’ orbitals in central boron atom. 22. Which statement is wrong? (1) Feldspars are not aluminosilicates (2) Beryl is an example of cyclic silicate (3) Mg2SiO4 is orthosilicate (4) Basic structural unit in silicates is the SiO44  tetrahedron Sol. Answer (1) Feldspar : These are aluminosilicates of two types : (a) Orthoclase feldspar (eg.) KAlSi3O8 (b) Plagioclase feldspar (eg.) NaAlSi3O8 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

The p-Block Elements

55

23. Carbon and silicon belong to (IV) group. The maximum coordination number of carbon in commonly occurring compounds is 4, whereas that of silicon is 6. This is due to (1) Availability of low lying d-orbitals in silicon

(2)

Large size of silicon

(3) More electropositive nature of silicon

(4)

Both (2) & (3)

Sol. Answer (1) Due to non-availability of d-orbitals in carbon and availability of low lying d-orbitals in silicon, the difference in coordination number is observed. 24. Which of the following statements about H3BO3 is not correct? (1) It has a layer structure in which planar BO3 units are joined by hydrogen bonds (2) It does not act as proton donor but acts as a Lewis acid by accepting hydroxyl ion (3) It is a strong tribasic acid (4) It is prepared by acidifying an aqueous solution of borax Sol. Answer (3)

O

H

O

B

O

H

+ H2O

H+[B(OH)4]–

H

It is a weak monobasic Lewis acid. 25. Aluminium (III) chloride forms a dimer because aluminium (1) Belongs to 3rd group

(2)

Can have higher coordination number

(3) Cannot form a trimer

(4)

Has high ionization energy

Sol. Answer (2) AlCl3 achieves stability by forming a dimer

Cl

Cl Al

Cl

Cl Al

Cl

Cl

26. Boron compounds behave as Lewis acids, because of their (1) Ionisation property

(2)

Electron deficient nature

(3) Acidic nature

(4)

Covalent nature

Sol. Answer (2) Boron compounds, especially the hydrides are electron deficient compounds which can accept a lone pair of electrons hence behave as Lewis acids. 27. In graphite, electrons are (1) Localised on each C-atom

(2)

Localised on every third C-atom

(3) Delocalised within the layer

(4)

Present in anti-bonding orbital

Sol. Answer (3) In graphite all electrons get delocalised in one layer and form -bond. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

56

The p-Block Elements

Solution of Assignment (Set-2)

28. In borax bead test which compound is formed? (1) Orthoborate

(2)

Metaborate

(3)

Double oxide

(4)

Tetraborate

Sol. Answer (2) In borax bead test metal metaborates are formed  Na 2B 4O 7   2NaBO 2 + B 2O 3



Sodium metaborate



CuO + B 2O 3   Cu(BO 2 ) 2



Cupric metaborate



29. Which one of the following statements about the zeolite is false? (1) They are used as cation exchangers (2) They have open structure which enables them to take up small molecules (3) Zeolites are aluminosilicates having three dimensional network (4) Some of the SiO44– units are replaced by AlO45– and AlO69– ions in zeolites Sol. Answer (4) Fact 30. The straight chain polymer is formed by (1) Hydrolysis of (CH3)2 SiCl2 followed by condensation polymerisation (2) Hydrolysis of (CH3)3 SiCl followed by condensation polymerisation (3) Hydrolysis of CH3 SiCl3 followed by condensation polymerisation (4) Hydrolysis of (CH3)4 Si by addition polymerisation Sol. Answer (1) Hydrolysis of (CH3)2 SiCl2 followed by condensation polymerisation forms a straight chain polymer. Reactions(CH3)2SiCl2 + 2H2O  (CH3)2Si(OH)2 + 2HCl n(CH3)2Si(OH)2  [–(CH3)2SiO–]n + nH2O 31. The metal oxide which cannot be reduced to metal by carbon is (1) Fe2O3

(2)

Al2O3

(3)

PbO

(4)

ZnO

Sol. Answer (2) Carbon + Metal oxide  Metal + Carbon dioxide According to the reactivity series, carbon cannot reduce the oxides of reactive metals such as potassium, sodium, calcium, magnesium and aluminium. 32. Oxalic acid on heating with conc. H2SO4 gives (1) CO only

(2)

CO2 only

(3)

CO2 + H2O

(4)

CO + CO2 + H2O

Sol. Answer (4)

COOH + H2SO4 COOH Oxalic acid

CO + CO2 + H2O

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Solution of Assignment (Set-2)

The p-Block Elements

57

33. Chemical formula of phosgene is (1) COCl2

(2)

CaOCl2

(3)

CaCO3

(4)

COCl

Sol. Answer (1) Phosgene is COCl2

SECTION - D Assertion-Reason Type Questions 1.

A : Borazine is more reactive than benzene. R : Borazine is isostructural with benzene.

Sol. Answer (2) Borazine is considerably more reactive than benzene B3N3H6 + 3HCl  B3N3H9Cl3 Addition reactions occur readily Structure of borazine is iso-structural with benzene

H H

H

N

H

B

N

B

B N

H

H

H

H

H

H H Benzene

H Borazine

2.

A : In Diborane containing eight –B–H bonds only four B–H bonds are on the plane. R : Boron in B2H6 is sp2 hybridised.

Sol. Answer (3)

H

H B

H

H B

H

H

Total B–H bonds = 8 The bridge bonded hydrogens are out of the plane. Hybridization of boron is sp3. 3.

A : All the oxides of boron family with the general formula M2O3 are basic. R : From B2O3 to Tl2O3 basic character decreases.

Sol. Answer (4) Boron compounds are acidic in nature. Basic character increases with atomic number of central atom. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

58 4.

The p-Block Elements

Solution of Assignment (Set-2)

A : When borax is strongly heated it forms transparent glassy bead. R : Borax is the other name for sodium tetraborate decahydrate.

Sol. Answer (2)

Na2B4O7.10H2O

Pt Loop

Na2B4O7 + 10H2O  NaBO2 + B2O3 Glassy bead

Borax is also called sodium tetraborate decahydrate. 5.

A : CBr4 is thermally more stable than CI4. R : C–Br bond energy is more than C–I.

Sol. Answer (1) Smaller size of carbon makes it difficult to accommodate large anions. Due to this bond energy increases for small anions. 6.

A : Boric acid is weak monobasic acid. R : Boric acid give one H+ ion.

Sol. Answer (3) H3BO3 is a weak monobasic acid. It reacts with water molecule and releases H+ ion. Three of its OH– groups are not ionisable. 7.

A : Al forms [AlF6]3–. R : It is octahedral complex.

Sol. Answer (2) Al has vacant d-orbitals and can expand its octet in [AlF6]3–.

F F

F 3–

Al F

F

Octahedral geometry

F 8.

A : Anhydride of carbonic acid is CO2. R : Carbonic acid is dibasic.

Sol. Answer (2) Anhydride of carbonic acid is CO2 and carbonic is dibasic as it gives 2H+ on ionization. 9.

A : CaC2 is interstitial carbide. R : Calcium ions are present in the Interstices.

Sol. Answer (4) Interstitial carbides are formed by transition metals of the periodic group IV B, V B and VI B. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-2)

The p-Block Elements

59

10. A : Fullerene is the purest allotrope of carbon. R : They have smooth structure without dangling bonds. Sol. Answer (1) Fullerene is the purest allotrope of carbon. They have smooth structure without dangling bonds. 11. A : GeCl4 is easily hydrolysed by water. R : Central atom can accommodate lone pair of e – from oxygen atom of water molecules in GeCl4. Sol. Answer (1) GeCl4 is easily hydrolysed by water because the central atom can accommodate lone pair of oxygen atom of water molecule of GeCl4. 12. A : Carbon has maximum tendency of catenation among group 14. R : C – C bond strength is very strong. Sol. Answer (1) Catenation in carbon is due to its small size and high electronegativity. Catenation is mostly seen in carbon is because of the high bond strength. 13. A : Oxides of carbon in higher oxidation state is more acidic than in lower oxidation state. R : Both CO2 and CO can exist. Sol. Answer (2) Oxides of carbon in higher oxidation state is more acidic than in lower oxidation state. Both CO2 and CO can exist. 14. A : Heavier elements of 14th group do not form p – p bonds. R : Atomic orbital of heavier elements are too large and do not have effective overlapping. Sol. Answer (1) Heavier elements of 14th group do not form p – p bonds because the atomic orbital of heavier elements are too large and do not have effective overlapping. 15. A : Carbon shows anomalous behaviour in group Gp-14. R : Carbon has maximum covalency of 4. Sol. Answer (2) Carbon shows anomalous behaviour because of small size, high electronegativity, high ionization enthalpy. Carbon has the maximum covalency of 4.

  

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