Carbon cannot be used in the reduction of ${\rm{A}}{{\rm{l}}_2}{{\rm{O}}_3}$ because

The enthalpy of formation of ${\rm{A}}{{\rm{l}}_2}{{\rm{O}}_3}$ is too high

The enthalpy of formation of ${\rm{C}}{{\rm{O}}_2}$ is more than that of ${\rm{A}}{{\rm{l}}_2}{{\rm{O}}_3}$

Pure carbon is not easily available

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{SiC}}{{\rm{l}}_2}$ undergoes hydrolysis but ${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{CC}}{{\rm{l}}_2}$ does not why?

Only 3p orbital is involved in C

${\rm{Si}} - {\rm{Cl}}$ bond is more polar than ${\rm{C}} - {\rm{Cl}}$ bond

${\rm{Si}}{{\rm{F}}_4}$ gets hydrolysed giving

${{\rm{H}}_2}{\rm{Si}}{{\rm{F}}_6}$

${\rm{Si}}{\left( {{\rm{OH}}} \right)_4}$

${\rm{Si}}{\left( {{\rm{OH}}} \right)_2}{{\rm{F}}_2}$

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NEET • CHEMISTRY • 14 TH GROUP ELEMENTS (P-BLOCK)

14 th group elements (p-block) practice · Mixed difficulty

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Showing 1-30 of 187 questions

1 Hard

1 of 187

Understanding +4 | -1

Which of the following is obtained on heating, potassium ferrocyanide with

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

View answer and explanation

Correct option: B

CO

Option B is correct because heating potassium ferrocyanide with concentrated sulfuric acid produces carbon monoxide (CO).

2 Easy

2 of 187

Understanding +4 | -1

It is because of inability of electrons of the valance shell to participate in bonding that

View answer and explanation

Correct option: D

S{n^{ + 2}}

is reducing while

P{b^{ + 4}}

is oxidising

Option D (

S{n^{ + 2}}

is reducing while

P{b^{ + 4}}

is oxidising) is correct because of the inert pair effect,

S{n^{ + 2}}

readily oxidizes to

S{n^{ + 4}}

(reducing agent), while

P{b^{ + 4}}

readily reduces to

P{b^{ + 2}}

(oxidizing agent).

3 Easy

3 of 187

Understanding +4 | -1

(A): Silicones are used in surgical cosmetic plants.

(R): Silicones are biocompatible.

View answer and explanation

Correct option: A

Both (A) and (R) are true and (R) is the correct explanation of (A)

Option A is correct because silicones are biocompatible, which is why they are used in surgical cosmetic implants.

4 Medium

4 of 187

Understanding +4 | -1

Carbon cannot be used in the reduction of

\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}

because

View answer and explanation

Correct option: D

The enthalpy of formation of

\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3}

is too high

Option D is correct because the enthalpy of formation of

Al_2O_3

is extremely high, making it thermodynamically unfavorable to reduce with carbon.

5 Easy

5 of 187

Remembering +4 | -1

Buckminster fullerene is

View answer and explanation

Correct option: B

C-60

Option B (C-60) is correct because Buckminsterfullerene is a specific allotrope of carbon containing 60 carbon atoms arranged in a spherical structure.

6 Easy

6 of 187

Remembering +4 | -1

Which form of carbon has a two-dimensional sheet-like structure?

View answer and explanation

Correct option: D

Graphite

Option D (Graphite) is correct because graphite has a layered structure consisting of two-dimensional sheets of carbon atoms.

7 Medium

7 of 187

Remembering +4 | -1

Thermodynamically, the most stable allotrope of carbon is

View answer and explanation

Correct option: D

Graphite

Option D (Graphite) is correct because graphite is the thermodynamically most stable allotrope of carbon under standard conditions.

8 Easy

8 of 187

Remembering +4 | -1

Diamond and graphite are:

View answer and explanation

Correct option: C

allotropes

Option C (allotropes) is correct because diamond and graphite are different structural forms of the same element (carbon).

9 Medium

9 of 187

Analyzing +4 | -1

Assertion (A): In silicates like

\mathrm{SiO}_2, \mathrm{Si}

, is

\mathrm{sp}^3

hybridised.

Reason (R):

\mathrm{SiO}_2

is a linear molecule.

View answer and explanation

Correct option: C

(A) is true but (R) is false

Option C ((A) is true but (R) is false) is correct because Si in SiO2 is sp3 hybridized, but SiO2 is not a linear molecule; it has a network structure.

10 Medium

10 of 187

Understanding +4 | -1

Pure CO can be obtained from:

View answer and explanation

Correct option: B

Nickel tetracarbonyl

Option B (Nickel tetracarbonyl) is correct because nickel tetracarbonyl decomposes at high temperatures to give pure CO and Ni.

11 Medium

11 of 187

Analyzing +4 | -1

Which of the following has the minimum heat of dissociation?

View answer and explanation

Correct option: D

[{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{N} \to \mathrm{B}{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}]

Option D

[{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{N} \to \mathrm{B}{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}]

has the minimum heat of dissociation because the steric hindrance is maximum in this case, weakening the bond.

12 Medium

12 of 187

Remembering +4 | -1

Carborundum is obtained when silica is heated at high temperature with

View answer and explanation

Correct option: A

Carbon

Option A (Carbon) is correct because carborundum (silicon carbide, SiC) is produced by heating silica (SiO2) with carbon at high temperatures.

13 Hard

13 of 187

Remembering +4 | -1

Lapis lazuli is

View answer and explanation

Correct option: A

Sodium alumino silicate

The correct answer is A (Sodium alumino silicate) because lapis lazuli is a complex mineral containing sodium, aluminum, silicon, and oxygen.

14 Easy

14 of 187

Understanding +4 | -1

Calorific value of producer gas is low because of

View answer and explanation

Correct option: A

High per cent of

{\mathrm{N}_2}

Option A (High per cent of

{\mathrm{N}_2}

) is correct because nitrogen is non-combustible and dilutes the combustible gases, lowering the calorific value.

15 Easy

15 of 187

Understanding +4 | -1

Which property is common in diamond and graphite?

View answer and explanation

Correct option: B

Relative atomic weight

The correct answer is B (Relative atomic weight) because both diamond and graphite are allotropes of carbon, and therefore have the same relative atomic weight.

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16 Medium

16 of 187

Remembering +4 | -1

The different layers in graphite are held together by

View answer and explanation

Correct option: D

Van\;\mathrm{der}Waals'\;forces

Option D (

Van der Waals' forces

) is correct because the layers in graphite are held together by weak Van der Waals' forces.

17 Easy

17 of 187

Application +4 | -1

What is the average atomic mass of silicon, if it occurs naturally in 3 isotopes

\mathrm{Si}^{28}, \mathrm{Si}^{29}, \mathrm{Si}^{30}

with the abundance of

92.2 \%, 4.7 \%

and

3.1 \%

respectively?

View answer and explanation

Correct option: B

28.1 \mathrm{~amu}

Option B (28.1 amu) is correct because the average atomic mass is calculated by summing the product of each isotope's mass and its abundance.

18 Easy

18 of 187

Remembering +4 | -1

The bond energy for catenation next to carbon is :

View answer and explanation

Correct option: C

The correct answer is C (Si) because silicon exhibits the strongest catenation after carbon due to its ability to form stable Si-Si bonds.

19 Easy

19 of 187

Analyzing +4 | -1

Select the incorrect statement :

a. The common oxidation states exhibited by group 14 elements are +4 and +3 .

b. Carbon and Silicon mostly show +4 oxidation state.

c. Group 14 oxides in higher oxidation states of elements are generally less acidic than those in lower oxidation state.

View answer and explanation

Correct option: D

a and c

Option D (a and c) is correct because group 14 elements commonly exhibit +4 and +2 oxidation states, not +3, and oxides in higher oxidation states are generally more acidic.

20 Easy

20 of 187

Remembering +4 | -1

Producer gas, a fuel and also a source of nitrogen is obtained by:

View answer and explanation

Correct option: B

Restricted supply of air through a bed of incandescent coke

Option B (Restricted supply of air through a bed of incandescent coke) is correct because producer gas is made by passing air over hot coke, which is primarily carbon, under controlled conditions.

21 Easy

21 of 187

Understanding +4 | -1

The tendency for catenation in the group – 14 elements is in the order

View answer and explanation

Correct option: D

\mathrm{C > > > Si > Ge} \approx \mathrm{Sn > > > Pb}

Option D (

\mathrm{C > > > Si > Ge} \approx \mathrm{Sn > > > Pb}

) is correct because the catenation tendency decreases down the group due to decreasing bond strength, with carbon having the highest tendency.

22 Medium

22 of 187

Remembering +4 | -1

The chemical formula of sindhur is

View answer and explanation

Correct option: B

\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}

Option B (

\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}

) is correct because sindhur, also known as red lead, has the chemical formula

\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}

23 Medium

23 of 187

Analyzing +4 | -1

Statement-1:

B

and

C

form their stable halides as

B X_3

and

CX_4

, while

Tl

and

Pb

produce their stable halides as

Tl

and

PbX_2

.

Statement-2: Both

PbX_4

and

TlX_3

are strong oxidizing agent.

View answer and explanation

Correct option: B

Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

Option B is correct because both statements are true, but statement 2 is not the correct explanation for statement 1. The inert pair effect explains the stability of TlX and PbX2.

24 Medium

24 of 187

Understanding +4 | -1

Tin (II) fluoride (anhydrous) can be obtained by :

View answer and explanation

Correct option: C

Dissolving SnO in HF

The correct answer is C (Dissolving SnO in HF) because tin(II) fluoride is formed by the reaction of tin(II) oxide with hydrofluoric acid.

25 Hard

25 of 187

Remembering +4 | -1

Silicones are formed by hydrolysis of

View answer and explanation

Correct option: D

Chlorosilanes

The correct answer is D (Chlorosilanes) because silicones are polymers formed from the hydrolysis and subsequent polymerization of chlorosilanes.

26 Easy

26 of 187

Remembering +4 | -1

The hybridisation of carbon in diamond, graphite and fullerenes are respectively.

View answer and explanation

Correct option: A

S{p^3},S{P^2},S{P^2}

Option A (

S{p^3},S{P^2},S{P^2}

) is correct because diamond is

sp^3

hybridized, while graphite and fullerenes are

sp^2

hybridized.

27 Medium

27 of 187

Understanding +4 | -1

Lead (IV) oxide is obtained by :

View answer and explanation

Correct option: D

Heating

\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}

with

\mathrm{conc}\mathrm{.HN}{\mathrm{O}_3}

Option D (Heating

\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}

with

\mathrm{conc}\mathrm{.HN}{\mathrm{O}_3}

) is correct because this reaction selectively dissolves PbO, leaving behind PbO2.

28 Medium

28 of 187

Understanding +4 | -1

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{SiC}{\mathrm{l}_2}

undergoes hydrolysis but

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CC}{\mathrm{l}_2}

does not why?

View answer and explanation

Correct option: A

Low lying d-orbitals present in Si but not in C

Option A is correct because silicon has low-lying d-orbitals that can accept electron pairs from water during hydrolysis, while carbon does not.

29 Medium

29 of 187

Understanding +4 | -1

Zn on heating with barium carbonate gives :

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because Zn reacts with barium carbonate upon heating to produce BaO, ZnO, and CO.

30 Easy

30 of 187

Understanding +4 | -1

\mathrm{Si}{\mathrm{F}_4}

gets hydrolysed giving

View answer and explanation

Correct option: D

{\mathrm{H}_2}\mathrm{Si}{\mathrm{F}_6}

Option D (

{\mathrm{H}_2}\mathrm{Si}{\mathrm{F}_6}

) is correct because

\mathrm{Si}{\mathrm{F}_4}

hydrolyzes to form silicic acid and hydrofluorosilicic acid (

{\mathrm{H}_2}\mathrm{Si}{\mathrm{F}_6}

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Key Takeaways

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