p-Block Elements Test

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p-Block Elements Test - 10

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Question 1
1 / -0

The Ge (II) is a powerful reducing agent while Pb (IV) is a powerful oxidising agent. This is because

A
the inert pair effect is more pronounced in Pb than in Ge

B
the ionisation energy of Pb is less than that of Ge

C
the ionic radii of Pb²⁺ and Pb⁴⁺ are larger than those of Ge²⁺ and Ge⁴⁺

D
Pb (IV) is more electropositive than Ge (II)

Solution

Ge has an electronic configuration: [Ar], 3d¹⁰, 4s², 4p²
Pb has an electronic configuration: [Xe], 4f¹⁴, 5d¹⁰, 6s², 6p²
As is evident from the electronic configuration, the inert pair effect is more pronounced in Pb due to poor shielding by the d- and f-electrons, and hence the +4 oxidation state in lead is less stable and Pb (IV) is a powerful oxidising agent.
On the other hand, Ge (II) is a strong reducing agent because it can easily be oxidised to Ge(IV).
Question 2
1 / -0

Which of the following oxides of Group 14 are amphoteric in nature?

A. CO₂
B. SiO₂
C. SnO₂
D. PbO₂

A
A, B, C

B
A and B

C
B and C

D
B, C and D

Solution

SnO₂ is amphoteric in nature.

SnO₂ + 4H⁺ → Sn⁺⁴ + 2H₂O

SnO₂ + 2H₂O + 2OH^-→ [Sn(OH)₆]^2-

SiO₂ is amphoteric in nature.

2NaOH + SiO₂ → Na₂SiO₃ + H₂O

4HF + SiO₂ → SiF₄ + 2H₂O
Question 3
1 / -0

Which of the following pairs of compounds are incorrectly matched with their chemical formulae?

(A) Butter of tin - SnCl₄.5H₂O (Stannic chloride)
(B) Sugar of lead - 2PbCO₃.Pb(OH)₂ (Basic lead carbonate)
(C) Carborundum - SiC (Silicon carbide)
(D) Water glass - Na₂SiO₃ (Sodium silicate)
(E) White lead - Pb(CH₃COO)₂ (Lead acetate)

A
(A), (C) and (D)

B
(C) and (E)

C
(A), (B) and (E)

D
(B) and (E)

Solution

Sugar of lead is lead acetate, i.e. Pb(CH₃COO)₂.
White lead is basic lead carbonate, i.e. 2PbCO₃.Pb(OH)₂.
Question 4
1 / -0

White metallic tin buttons get converted to grey powder. This transformation is related to

A
a change in the crystalline structure of tin

B
a change in the partial pressure of oxygen in air

C
the interaction with water vapour present in the humid air

D
the interaction with nitrogen present in air at low temperature

Solution

At 13.2 °C (about 56 °F) or below, pure tin transforms from the silvery, ductile, metallic allotrope white α-tin to brittle, nonmetallic, β-form grey tin with a diamond cubic structure.

Tin was the major metal used to make buttons in the French army's uniforms. When the Napoleon's army took on the Russian's, the severe Russian temperatures approached -30°C, which led to the disintegration of the crystalline form of tin causing fatal maladies in the soldiers.
Question 5
1 / -0

Consider the following statements:

(A) Graphite is a good conductor of electricity due to the availability of delocalised electrons.
(B) Carbon can form C=S bond because C has the ability to form dπ-dπ bond.
(C) Both diamond and boron nitride have similar structures.
(D) Silicon carbide has three-dimensional structures with each silicon and carbon being tetrahedrally surrounded by four atoms of the same kinds.

Which of the given statements are incorrect?

A
(A) and (B)

B
(C) and (D)

C
(A) and (D)

D
(B) and (D)

Solution

(B) Carbon does not form dπ-dπ bond because of the absence of d-orbital.
(D) Silicon carbide has three-dimensional structures with each silicon and carbon being tetrahedrally surrounded by four atoms of the different kinds. In each case. carbon is tetrahedrally is surrounded by four silicon atoms and vice versa.
Question 6
1 / -0

Which of the following statements is incorrect?

A
Diamond is unaffected by concentrated acids, but graphite reacts with hot concentrated HNO₃, forming mellitic acid C₆(COOH)₆.

B
COCl₂ is phosphine gas.

C
C₃O₂ (carbon suboxide) is a foul-smelling gas.

D
CO is toxic because it has the ability to form a complex with haemoglobin in blood.

Solution

COCl₂ is phosgene which is a toxic gas. Phosphine is PH₃.

Mellitic acid, also called graphitic acid or benzenehexacarboxylic acid, can be formed by reaction of graphite with hot and concentrated HNO₃.
C₃O₂ (carbon suboxide) is a foul-smelling lacrimatory (tear-stimulating) gas produced by the dehydration of malonic acid, CH₂(COOH)₂, with P₄O₁₀.

CO is toxic because it has the ability to form a complex with haemoglobin in blood, and hence can deprive the body of oxygen.
Question 7
1 / -0

To prepare a silicon polymer containing 10 Si-O-Si linkages, the number of R₂Si(OH)₂ units required is

A
8

B
9

C
10

D
11

Solution

To prepare a silicon polymer containing 'n' Si-O-Si linkages, the total number of chain-building units, R₂Si(OH)₂ required is (n - 1). Thus, the number of R₂Si(OH)₂ units needed is 9.

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p-Block Elements Test - 10

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Question 1

the inert pair effect is more pronounced in Pb than in Ge

the ionisation energy of Pb is less than that of Ge

the ionic radii of Pb2+ and Pb4+ are larger than those of Ge2+ and Ge4+

Pb (IV) is more electropositive than Ge (II)

Solution

Question 2

A, B, C

A and B

B and C

B, C and D

Solution

Question 3

(A), (C) and (D)

(C) and (E)

(A), (B) and (E)

(B) and (E)

Solution

Question 4

a change in the crystalline structure of tin

a change in the partial pressure of oxygen in air

the interaction with water vapour present in the humid air

the interaction with nitrogen present in air at low temperature

Solution

Question 5

(A) and (B)

(C) and (D)

(A) and (D)

(B) and (D)

Solution

Question 6

Diamond is unaffected by concentrated acids, but graphite reacts with hot concentrated HNO3, forming mellitic acid C6(COOH)6.

COCl2 is phosphine gas.

C3O2 (carbon suboxide) is a foul-smelling gas.

CO is toxic because it has the ability to form a complex with haemoglobin in blood.

Solution

Question 7

8

9

10

11

Solution

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the ionic radii of Pb²⁺ and Pb⁴⁺ are larger than those of Ge²⁺ and Ge⁴⁺

Diamond is unaffected by concentrated acids, but graphite reacts with hot concentrated HNO₃, forming mellitic acid C₆(COOH)₆.

COCl₂ is phosphine gas.

C₃O₂ (carbon suboxide) is a foul-smelling gas.