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

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

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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 91-120 of 197 questions
91 Hard
91 of 197
Remembering +4 | -1
Which gas is used in airated water?
View answer and explanation
Correct option: A
CO2\mathrm{C}{\mathrm{O}_2}
Option A (CO2\mathrm{C}{\mathrm{O}_2}) is correct because carbon dioxide gas is commonly used to create the fizz in aerated water.
92 Medium
92 of 197
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.
93 Easy
93 of 197
Remembering +4 | -1
Producer gas is a mixture of:
View answer and explanation
Correct option: A
CO+N2CO\mathrm{ } + {\mathrm{N}_2}
Option A (CO+N2CO\mathrm{ } + {\mathrm{N}_2}) is correct because producer gas is primarily a mixture of carbon monoxide (CO) and nitrogen (N2{\mathrm{N}_2}).
94 Medium
94 of 197
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.
95 Medium
95 of 197
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.
96 Easy
96 of 197
Remembering +4 | -1
Quartz is an example of
View answer and explanation
Correct option: D
Three dimensional network silicate
Option D (Three dimensional network silicate) is correct because quartz has a three-dimensional network structure of (SiO2)n{\left( {\mathrm{Si}{\mathrm{O}_2}} \right)_n} silicate.
97 Easy
97 of 197
Remembering +4 | -1
One recently discovered allotrope of carbon (e.g.,C60)\left( {e.g.,{C_{60}}} \right) is known as
View answer and explanation
Correct option: B
Fullerene
Option B (Fullerene) is correct because fullerenes, such as C60, are a recently discovered allotrope of carbon.
98 Medium
98 of 197
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.
99 Medium
99 of 197
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.
100 Easy
100 of 197
Remembering +4 | -1
A kettle which becomes furred-up in use has inside it a deposit composed mainly of:
View answer and explanation
Correct option: A
Calcium carbonate
Option A (Calcium carbonate) is correct because the furring in kettles is primarily due to the precipitation of calcium carbonate from hard water upon heating.
101 Medium
101 of 197
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}.
102 Medium
102 of 197
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.
103 Medium
103 of 197
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.
104 Easy
104 of 197
Remembering +4 | -1
Solid CO2\mathrm{C}{\mathrm{O}_2} is used as :
View answer and explanation
Correct option: C
Refrigerant
Option C (Refrigerant) is correct because solid CO2, also known as dry ice, sublimes at low temperatures, absorbing heat and providing a cooling effect, thus acting as a refrigerant.
105 Easy
105 of 197
Understanding +4 | -1
Which is a true acid anhydride?
View answer and explanation
Correct option: D
CO2\mathrm{C}{\mathrm{O}_2}
Option D (CO2) is correct because CO2 is the anhydride of carbonic acid (H2CO3).
106 Medium
106 of 197
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.
107 Easy
107 of 197
Understanding +4 | -1
Which is neutral to litmus?
View answer and explanation
Correct option: C
CO
Option C (CO) is correct because CO is a neutral oxide, while the others are amphoteric or acidic.
108 Medium
108 of 197
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.
109 Easy
109 of 197
Remembering +4 | -1
The fuel gas having volume composition equal to 34%CH4+48%H2+15%O2+3%  CO34\% \mathrm{C}{\mathrm{H}_4} + 48\mathrm{\% }{\mathrm{H}_2} + 15\mathrm{\% }{\mathrm{O}_2} + 3\% \;CO is:
View answer and explanation
Correct option: C
Coal gas
Option C (Coal gas) is correct because coal gas typically has a composition of around 34% CH₄, 48% H₂, 15% O₂, and 3% CO.
110 Easy
110 of 197
Remembering +4 | -1
Which is not an allotrope of carbon?
View answer and explanation
Correct option: D
Carborundum
Option D (Carborundum) is correct because carborundum (SiC) is silicon carbide, a compound, and not an allotrope of carbon.
111 Easy
111 of 197
Understanding +4 | -1
The correct order of the catenation is
View answer and explanation
Correct option: C
C>Si>Ge≈SnC > Si > Ge \approx Sn
Option C (C>Si>Ge≈SnC > Si > Ge \approx Sn) is correct because catenation ability decreases down the group due to decreasing bond strength.
112 Easy
112 of 197
Understanding +4 | -1
The hybridization of carbon in carbon monoxide is:
View answer and explanation
Correct option: C
spsp
Option C (spsp) is correct because carbon monoxide (CO) has one sigma and two pi bonds, requiring spsp hybridization for the carbon atom.
113 Easy
113 of 197
Remembering +4 | -1
Liquefied CO2C{O_2} on rapid expansion gives ______
View answer and explanation
Correct option: A
dry ice
Option A (dry ice) is correct because rapid expansion of liquefied CO2 results in the formation of dry ice, which is solid CO2.
114 Easy
114 of 197
Understanding +4 | -1
Ordinary sand (SiO2)\mathrm{(Si}{\mathrm{O}_2}) is attacked by:
View answer and explanation
Correct option: C
hot KOH
The correct answer is C (hot KOH) because silicon dioxide (sand) reacts with hot potassium hydroxide to form potassium silicate and water.
115 Easy
115 of 197
Understanding +4 | -1
The element Z=114Z = 114 has been discovered recently. It will belong to which of the following family group and electronic configuration?
View answer and explanation
Correct option: C
Carbon family, [Rn]5f146d107s27p2[Rn]5{f^{14}}6{d^{10}}7{s^2}7{p^2}
Option C (Carbon family, [Rn]5f146d107s27p2][Rn]5{f^{14}}6{d^{10}}7{s^2}7{p^2}]) is correct because element Z=114 belongs to the carbon family (group 14), and its electronic configuration reflects this placement with two electrons in the 7p orbitals.
116 Easy
116 of 197
Application +4 | -1
What is the average atomic mass of silicon, if it occurs naturally in 3 isotopes Si28,Si29,Si30\mathrm{Si}^{28}, \mathrm{Si}^{29}, \mathrm{Si}^{30} with the abundance of 92.2%,4.7%92.2 \%, 4.7 \% and 3.1%3.1 \% respectively?
View answer and explanation
Correct option: B
28.1 amu28.1 \mathrm{~amu}
Option B (28.1 amu) is correct because it is the weighted average of the isotopic masses based on their natural abundances.
117 Easy
117 of 197
Application +4 | -1
What is the change in oxidation number of carbon in the following reaction?

CH4+2O2⟶CO2+2H2O\mathrm{CH}_4+2 \mathrm{O}_2 \longrightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}
View answer and explanation
Correct option: C
-4 to +4
Option C (-4 to +4) is correct because the oxidation number of carbon changes from -4 in CH4 to +4 in CO2.
118 Medium
118 of 197
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.
119 Easy
119 of 197
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.
120 Medium
120 of 197
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.
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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