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

14 th group elements (p-block) practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 102 questions
31 Medium
31 of 102
Analyzing +4 | -1
(A): PbI4{PbI}_4 is stable compound.
(R): Iodide stabilizes higher oxidation state.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D (Both (A) and (R) are false) is correct because PbI4PbI_4 is unstable and iodide does not generally stabilize higher oxidation states.
32 Medium
32 of 102
Remembering +4 | -1
Tetra ethyl lead is used as:
View answer and explanation
Correct option: B
Antiknock compound
Option B (Antiknock compound) is correct because tetraethyl lead (TEL) was historically used as an antiknock agent in gasoline to increase its octane rating and prevent engine knocking.
33 Medium
33 of 102
Remembering +4 | -1
Standard state of carbon is
View answer and explanation
Correct option: A
Graphite
Option A (Graphite) is correct because graphite is the most thermodynamically stable allotrope of carbon under standard conditions.
34 Medium
34 of 102
Remembering +4 | -1
Clay is an example of
View answer and explanation
Correct option: D
Sheet silicates
Option D (Sheet silicates) is correct because clay minerals are composed of layers or sheets of silicate tetrahedra.
35 Medium
35 of 102
Remembering +4 | -1
Rose metal is an alloy of.
View answer and explanation
Correct option: A
Sn + Pb + Bi
Option A (Sn + Pb + Bi) is correct because Rose metal is an alloy composed of tin, lead, and bismuth.
36 Medium
36 of 102
Remembering +4 | -1
Which of the following compounds is used in cosmetic surgery.
View answer and explanation
Correct option: C
Silicones
Option C (Silicones) is correct because silicones are used in cosmetic surgery due to their biocompatibility and flexibility.
37 Medium
37 of 102
Remembering +4 | -1
Organosilicon polymers (silicones) have repeating unit as
View answer and explanation
Correct option: A
−(R2SiO)−-\left({R}_2 {SiO}\right)-
Option A (−(R2SiO)−-\left({R}_2 {SiO}\right)-) is correct because organosilicon polymers (silicones) have a repeating unit of −(R2SiO)−-\left({R}_2 {SiO}\right)-.
38 Medium
38 of 102
Application +4 | -1
For carbon dating of a wood sample (C14C12)t=18(C14C12)t=0\left(\frac{\mathrm{C}^{14}}{\mathrm{C}^{12}}\right)_\mathrm{t}=\frac{1}{8}\left(\frac{\mathrm{C}^{14}}{\mathrm{C}^{12}}\right)_{\mathrm{t}=0}. If Half life of C14\mathrm{C}^{14} is 1580 years what is the life of wood sample (in yr)
View answer and explanation
Correct option: A
4740
Option A (4740) is correct because it is calculated using the half-life of C-14 and the given ratio of C-14 to C-12.
39 Medium
39 of 102
Analyzing +4 | -1
Assertion: The melting points of the Carbon family elements increase down the group.

Reason: The metallic bonding becomes stronger down the group due to the increased number of valence electrons.
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation for Assertion
Option A is correct because the melting points of the Carbon family elements increase down the group due to stronger metallic bonding from an increased number of valence electrons.
40 Medium
40 of 102
Remembering +4 | -1
Which gas when passed through dilute blood will impart a cherry red colour to the solution?
View answer and explanation
Correct option: D
CO\mathrm{CO}
Option D (CO) is correct because carbon monoxide (CO) reacts with hemoglobin in blood to form carboxyhemoglobin, which has a cherry-red color.
41 Medium
41 of 102
Understanding +4 | -1
Lead sulphate is soluble in :
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because lead sulfate (pbSO4pb\mathrm{S}{\mathrm{O}_4}) is generally considered insoluble in common reagents like concentrated nitric acid, potassium permanganate, or potassium dichromate.
42 Medium
42 of 102
Remembering +4 | -1
Number of 6 membered and five membered rings in C60{C}_{60} fullerene respectively are
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Correct option: A
20 and 12
Option A (20 and 12) is correct because C60 fullerene contains 20 six-membered rings and 12 five-membered rings.
43 Medium
43 of 102
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\mathrm{SiO_2}) with carbon at high temperatures.
44 Medium
44 of 102
Remembering +4 | -1
Semi water gas is mixture of :
View answer and explanation
Correct option: A
Water gas and producer gas
Option A (Water gas and producer gas) is correct because semi-water gas is a mixture of both.
45 Medium
45 of 102
Understanding +4 | -1
Mg2C3\mathrm{M}{\mathrm{g}_2}{\mathrm{C}_3} has the following characteristics:
View answer and explanation
Correct option: D
All of the above
Option D is correct because Magnesium carbide (Mg2C3\mathrm{M}{\mathrm{g}_2}{\mathrm{C}_3}) contains Mg2+  and  C34−\mathrm{M}{\mathrm{g}^{2 + }}\;and\;\mathrm{C}_3^{4 - } ions, is called magnesium allylide, and produces propyne upon hydrolysis.
46 Medium
46 of 102
Remembering +4 | -1
Tin dissolves in dilute HNO3\mathrm{HN}{\mathrm{O}_3} forming :
View answer and explanation
Correct option: C
Ammonium nitrate
Option C (Ammonium nitrate) is correct because tin reacts with dilute nitric acid to produce stannous nitrate, ammonium nitrate, and water.
47 Medium
47 of 102
Understanding +4 | -1
Assertion (A): Graphite fibres embedded in plastic material form high strength, light weight composites.

Reason (R): The composites are used in products such as tennis rackets, fishing rods, aircrafts and canoes.
View answer and explanation
Correct option: B
Both A\mathrm{A} and R\mathrm{R} are correct; R\mathrm{R} is not the correct explanation of A\mathrm{A}.
Option B is correct because both the assertion and reason are true, but the reason does not explain why graphite fibers form high-strength composites.
48 Medium
48 of 102
Remembering +4 | -1
Red lead is:
View answer and explanation
Correct option: B
Pb3O4\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}
Option B (Pb3O4\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}) is correct because red lead is the common name for lead tetroxide, which has the formula Pb3O4\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}.
49 Medium
49 of 102
Remembering +4 | -1
The variety of glass, used for the preservation of eggs is:
View answer and explanation
Correct option: C
Water glass
Option C (Water glass) is correct because water glass is a common name for sodium silicate, which is used for preserving eggs.
50 Medium
50 of 102
Remembering +4 | -1
The coal form containing maximum percentage of carbon is:
View answer and explanation
Correct option: B
Anthracite
Correct Answer + Core Reason: Option A (Anthracite) is correct because anthracite coal has the highest carbon content compared to other types of coal.
51 Medium
51 of 102
Remembering +4 | -1
Which of the following oxides is amphoteric in character?
View answer and explanation
Correct option: A
SnO2\mathrm{Sn}{\mathrm{O}_2}
Option A (SnO2\mathrm{Sn}{\mathrm{O}_2}) is correct because tin(IV) oxide (SnO2\mathrm{Sn}{\mathrm{O}_2}) exhibits amphoteric behavior, reacting with both acids and bases.
52 Medium
52 of 102
Understanding +4 | -1
The metal which does not form a polynuclear carbonyl is :
View answer and explanation
Correct option: A
Sodium
Option A (Sodium) is correct because alkali metals like sodium do not readily form polynuclear carbonyls due to their electronic structure and bonding characteristics.
53 Medium
53 of 102
Remembering +4 | -1
Stable form of ‘C’ may be represented by the formula
View answer and explanation
Correct option: B
C2{C_2}
Option B (C2{C_2}) is correct because the stable form of chlorine is diatomic, represented as Cl2 or C2.
54 Medium
54 of 102
Analyzing +4 | -1
Consider the following statement(s) regarding CO.

I. CO has one σ\sigma and two π\pi-bonds.

II. CO acts as electron donor.

III. CO has one lone pair on its carbon atom.

IV. CO reacts with all metals to produce metal carbide.

The correct statements are
View answer and explanation
Correct option: A
I, II and III
Option A (I, II and III) is correct because CO has one sigma and two pi bonds, acts as an electron donor, and has one lone pair on its carbon atom.
55 Medium
55 of 102
Remembering +4 | -1
Which of the following is a three dimensional silicate?
View answer and explanation
Correct option: C
Zeolite
Option C (Zeolite) is correct because zeolites have a three-dimensional network structure of silicate tetrahedra.
56 Medium
56 of 102
Understanding +4 | -1
Which of the following is a mixed oxide?
View answer and explanation
Correct option: C
Pb3O4\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}
Option C (Pb3O4\mathrm{P}{\mathrm{b}_3}{\mathrm{O}_4}) is correct because it can be represented as a combination of two oxides of lead, 2PbO.PbO22\mathrm{PbO}\mathrm{.Pb}{\mathrm{O}_2}.
57 Medium
57 of 102
Remembering +4 | -1
CCl4{CCl}_4 is used as a fire extinguisher because of
View answer and explanation
Correct option: D
It gives incombustible vapors
Option D (It gives incombustible vapors) is correct because carbon tetrachloride (CCl4CCl_4) is used as a fire extinguisher because it produces heavy, non-combustible vapors that displace oxygen and smother the fire.
58 Medium
58 of 102
Remembering +4 | -1
The acid used for etching the glass is:
View answer and explanation
Correct option: C
Hydrofluoric acid
Option C (Hydrofluoric acid) is correct because hydrofluoric acid (HF) reacts with silica in glass, allowing it to etch the glass surface.
59 Medium
59 of 102
Understanding +4 | -1
Monosilane on coming in contact with air burns with a luminous flame producing vortex rings. These rings are of
View answer and explanation
Correct option: A
SiO2\mathrm{Si}{\mathrm{O}_2}
The correct answer is A (SiO2\mathrm{SiO}_2) because silanes burn in air to form silicon dioxide, which creates the vortex rings.
60 Medium
60 of 102
Understanding +4 | -1
The correct statement among the following

a) Fullerene has both single and double bonds

b) C−CC-C bond lengths in fullerene are 143.5 PM and 138.3 PM

c) Fullerene has one electron at each carbon which is delocalized into molecular orbitals

d) Fullerene is aromatic in nature.
View answer and explanation
Correct option: C
a, b, c & d
Option C (a, b, c & d) is correct because fullerenes possess both single and double bonds, specific bond lengths, delocalized electrons, and exhibit aromatic character.
Revision notes and key takeaways

Review 14 TH GROUP ELEMENTS (P-BLOCK) in your textbook or class notes, then use this set to check your understanding.

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