p-Block Elements Test

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p-Block Elements Test - 3

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Weekly Quiz Competition

Question 1
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Graphite is a smooth lubricant that is extremely difficult to melt. The reason for this anomalous behaviour is that graphite

A
has carbon atoms arranged in large plates of rings of strongly bonded carbon atoms with weak interpolate forces

B
is a non-crystalline substance

C
is an allotrope of carbon

D
has molecules of variable molecular masses like polymers

Solution

Graphite is a network solid and c-atoms are arranged in layers of hexagonal rings in which each carbon is linked to three other carbon atoms.
The atoms within a layer are held by strong covalent bonds, which explains its extremely high melting point.

However, forces between the layers are weak van der Waals forces because of which the layers can slide over each other. Consequently, graphite is soft and slippery, and is used as a lubricant.
Question 2
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Directions For Questions

In the given question, two statements are provided which are labeled as Assertion (A) and Reason (R). Read both of them carefully and select the alternative that most suitably follows the statements.

...view full instructions

Assertion (A): Diamond is a bad conductor of electricity.
Reason (R): All C-C bond lengths in diamond are of 154 pm.

A
Both (A) and (R) are true and (R) is the correct reason for (A).

B
Both (A) and (R) are true and (R) is not the correct reason for (A).

C
Both (A) and (R) are false.

D
(A) is false and (R) is true

Solution

Both the given statements are correct but the reason why diamond is not a good conductor of electricity is that there are no free electrons flowing around in the structure of diamond.
Question 3
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Which of the following statements are correct?

a. Fullerenes have dangling bonds.
b. Fullerenes are cage-like molecules.
c. Graphite is thermodynamically the most stable allotrope of carbon.
d. The successive layers in graphite are held together by weak forces of attraction.

A
a and b

B
b and c

C
c and d

D
b, c and d

Solution

The statements b and c are factually correct. The statement d is the property which makes graphite smooth and slippery.
Question 4
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Aluminium vessels should not be washed with substances containing washing soda because

A
washing soda is expensive

B
washing soda is easily decomposed

C
washing soda reacts with aluminium to form insoluble oxide

D
washing soda forms soluble metal aluminate

Solution

Aluminium vessels should not be washed with substances containing washing soda because aluminium reacts with washing soda to form soluble metal aluminate.
2Al + Na₂CO₃ + 3H₂O ® 2NaAlO₂ + CO₂ + 3H₂
Question 5
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The stability of + 1 oxidation state increases in the sequence

A
In < Tl < Ga < Al

B
Tl < In < Ga < Al

C
Ga < In < Al < Tl

D
Al < Ga < In < Tl

Solution

Group 13 elements exhibit + 3 and + 1 oxidation states. Stability of the lower oxidation state increases on moving down the group due to inert pair effect. So, the order is Al < Ga < In < Tl.
Question 6
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Which of the following statements is correct?

A
BCl₃ and AlCl₃ are both Lewis acids and BCl₃ is stronger than AlCl₃.

B
BCl₃ and AlCl₃ are both Lewis acids and AlCl₃ is stronger than BCl₃.

C
BCl₃ and AlCl₃ are equally strong Lewis acids.

D
Neither BCl₃ nor AlCl₃ is a Lewis acid.

Solution

BCl₃ and AlCl₃, being electron deficient, are Lewis acids. Further, boron is smaller in size than aluminium. Hence, BCl₃ is a stronger acid than AlCl₃.
Question 7
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The tendency of BF₃, BCl₃ and BBr₃ to behave as Lewis acids decreases in the sequence

A
BCl₃ > BF₃ > BBr₃

B
BBr₃ > BCl₃ > BF₃

C
BBr₃ > BF₃ > BCl₃

D
BF₃ > BCl₃ > BBr₃

Solution

Acidic order is BI₃ > BBr₃ > BCl₃ > BF₃

In BF₃, due to back bonding, i.e. electron donation by fluorine to boron, electron density increases in boron and hence, its Lewis acid character decreases.

As we move from fluorine to chlorine, the size of orbitals increases and hence, there is not much effective back bonding in BCl₃ as compared to BF₃. Similarly, BBr₃ will have less back bonding due to much greater increase in size of overlapping orbitals and so on. So, as tendency of back bonding decreases from F to I, Lewis acid character increases from F to I.
Question 8
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Which of the following does not exist in free form?

A
BF₃

B
BCl₃

C
BBr₃

D
BH₃

Solution

In case of BF₃, BCl₃ and BBr₃, the vacant 2p-orbital of boron accepts lone pairs of electrons by overlapping with filled 2p_z orbital of F, Cl and Br. Therefore, the stability of BF₃, BCl₃ and BBr₃ is due to back bonding.

However, hydrogen atom does not have free electrons to form back bonding with boron and, therefore, BH₃ is not stable. However, to satisfy the electron deficiency of boron, it dimerizes to form B₂H₆.
Question 9
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A colourless aqueous solution, on adding water and on heating, gave a white precipitate. This precipitate, when reacted with NH₄Cl and NH₄OH in excess, resulted in dissolution of some of the precipitate and a gelatinous precipitate is obtained. What is the hydroxide formed in aqueous solution?

A
Zn(OH)₂

B
Mg(OH)₂

C
Ca(OH)₂

D
Al(OH)₃

Solution

Aluminium will form a gel-like precipitate when we add NH₄Cl and NH₄OH. Also, Al(OH)₃ forms gelatinous mass.
Question 10
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Which of the following statements is not correct?

A
PbCl₄ is less stable than SnCl₄ but PbCl₂ is more stable than SnCl₂.

B
PbO₂ is a stronger oxidant than SnO₂.

C
(SiH)₃N is a stronger base than (CH₃)₃N.

D
Sn(II) is a reducing agent but Pb(II) is not.

Solution

The nitrogen atom in trisilylamine is sp²hybridized and the lone pair is in the unhybridized p-orbital.

In trimethyl amine, the nitrogen is sp³hybridized and the lone pair is in the hybridized orbital, and hence, it is less available.

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p-Block Elements Test - 3

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p-Block Elements Test - 3

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Question 1

has carbon atoms arranged in large plates of rings of strongly bonded carbon atoms with weak interpolate forces

is a non-crystalline substance

is an allotrope of carbon

has molecules of variable molecular masses like polymers

Solution

Question 2

Both (A) and (R) are true and (R) is the correct reason for (A).

Both (A) and (R) are true and (R) is not the correct reason for (A).

Both (A) and (R) are false.

(A) is false and (R) is true

Solution

Question 3

a and b

b and c

c and d

b, c and d

Solution

Question 4

washing soda is expensive

washing soda is easily decomposed

washing soda reacts with aluminium to form insoluble oxide

washing soda forms soluble metal aluminate

Solution

Question 5

In < Tl < Ga < Al

Tl < In < Ga < Al

Ga < In < Al < Tl

Al < Ga < In < Tl

Solution

Question 6

BCl3 and AlCl3 are both Lewis acids and BCl3 is stronger than AlCl3.

BCl3 and AlCl3 are both Lewis acids and AlCl3 is stronger than BCl3.

BCl3 and AlCl3 are equally strong Lewis acids.

Neither BCl3 nor AlCl3 is a Lewis acid.

Solution

Question 7

BCl3 > BF3 > BBr3

BBr3 > BCl3 > BF3

BBr3 > BF3 > BCl3

BF3 > BCl3 > BBr3

Solution

Question 8

BF3

BCl3

BBr3

BH3

Solution

Question 9

Zn(OH)2

Mg(OH)2

Ca(OH)2

Al(OH)3

Solution

Question 10

PbCl4 is less stable than SnCl4 but PbCl2 is more stable than SnCl2.

PbO2 is a stronger oxidant than SnO2.

(SiH)3N is a stronger base than (CH3)3N.

Sn(II) is a reducing agent but Pb(II) is not.

Solution

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BCl₃ and AlCl₃ are both Lewis acids and BCl₃ is stronger than AlCl₃.

BCl₃ and AlCl₃ are both Lewis acids and AlCl₃ is stronger than BCl₃.

BCl₃ and AlCl₃ are equally strong Lewis acids.

Neither BCl₃ nor AlCl₃ is a Lewis acid.

BCl₃ > BF₃ > BBr₃

BBr₃ > BCl₃ > BF₃

BBr₃ > BF₃ > BCl₃

BF₃ > BCl₃ > BBr₃

BF₃

BCl₃

BBr₃

BH₃

Zn(OH)₂

Mg(OH)₂

Ca(OH)₂

Al(OH)₃

PbCl₄ is less stable than SnCl₄ but PbCl₂ is more stable than SnCl₂.

PbO₂ is a stronger oxidant than SnO₂.

(SiH)₃N is a stronger base than (CH₃)₃N.