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JEE • CHEMISTRY • SOLID STATE

Solid state 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 91-98 of 98 questions
91 Medium
91 of 98
Remembering +4 | -1
The percentages of void space for simple cubic, body centered cubic and hexagonal close packed arrangement respectively are
View answer and explanation
Correct option: A
48,32,26
Option A (48,32,26) is correct because the percentages of void space for simple cubic, body-centered cubic, and hexagonal close-packed arrangements are approximately 48%, 32%, and 26%, respectively.
92 Medium
92 of 98
Understanding +4 | -1
What is the coordination number of sodium in Na2ON{a_2}O ?
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Correct option: B
4
Option B (4) is correct because in Na2ON{a_2}O, each Na+N{a^+} ion is coordinated to 4 oxide ions (O2−O^{2-}).
93 Medium
93 of 98
Understanding +4 | -1
Assertion (A): Stoichiometric compounds obey the law of constant composition.

Reason (R): Schottky and Frenkel defects are observed in stoichiometric compounds.
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Correct option: B
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Option B is correct because both the assertion and reason are true, but the reason doesn't explain the assertion. Stoichiometric compounds maintain a fixed ratio of elements, and Schottky/Frenkel defects don't change this ratio, but they are not the *reason* for the law of constant composition.
94 Medium
94 of 98
Application +4 | -1
If NaCl\mathrm{NaCl} is doped with 10−3{10^{ - 3}} mol % SrCl2\mathrm{SrC}{\mathrm{l}_2}, then the concentration of cation vacancies will be
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Correct option: A
1×10−3mol%1 \times {10^{ - 3}}mol\%
Option A (1×10−3mol%1 \times {10^{ - 3}}mol\%) is correct because each SrCl2 introduces one cation vacancy.
95 Medium
95 of 98
Analyzing +4 | -1
(Ag+Pb)alloy  →Zinc     is     addedMelt     and\left( {Ag + Pb} \right)\mathrm{alloy}\,\,\xrightarrow[{\mathrm{Zinc}\,\,\,\,\,\mathrm{is}\,\,\,\,\,\mathrm{added}}]{{\mathrm{Melt}\,\,\,\,\,\mathrm{and}}} ; (Ag+Pb+Zn)melt→Cool       →Layer    YLayer    X\left( {Ag + Pb + Zn} \right)\mathrm{melt}\xrightarrow[{}]{{\mathrm{Cool}}}\,\,\,\,\,\,\,\xrightarrow[{\mathrm{Layer}\,\,\,\,\mathrm{Y}}]{{\mathrm{Layer}\,\,\,\,\mathrm{X}}}
Select correct statement based on above scheme.
View answer and explanation
Correct option: D
All are correct answer.
Option D (All are correct answer.) is correct because zinc and lead are immiscible in the molten state, forming separate layers, and silver is soluble in both zinc and lead.
96 Medium
96 of 98
Remembering +4 | -1
The co – ordination number for the simple, FCC and BCC lattices are respectively
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Correct option: D
6, 12, 2008
Option D (6, 12, 8) is correct because the coordination numbers for simple cubic, FCC (face-centered cubic), and BCC (body-centered cubic) lattices are 6, 12, and 8, respectively.
97 Medium
97 of 98
Analyzing +4 | -1
Assertion (A): Covalent crystal have the higher melting point.

Reason (R): Covalent bonds are stronger than ionic bonds.
View answer and explanation
Correct option: C
(A) is correct but (R) is not correct
Option C is correct. Covalent crystals have high melting points, but the reason that covalent bonds are stronger than ionic bonds is incorrect. The network structure is the main reason.
98 Medium
98 of 98
Application +4 | -1
A compound CuCl\mathrm{CuCl} has face centred cubic structure. Its density is 3.4  g  cm−3\mathrm{3}\mathrm{.4}\;\mathrm{g}\;\mathrm{c}{\mathrm{m}^{ - 3}}. The length of unit cell is :
View answer and explanation
Correct option: A
5.783A∘5.783\mathop \mathrm{A}\limits^ \circ
Option A (5.783A∘5.783\mathop \mathrm{A}\limits^ \circ) is correct because it is calculated using the density formula for a face-centered cubic structure.
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