Solid State MCQ Questions & Answers in Physical Chemistry | Chemistry
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181.
In a monoclinic unit cell, the relation of sides and angles are respectively :
A
$$a = b \ne c\,\,{\text{and}}\,\,\alpha = \beta = \gamma = {90^ \circ }$$
B
$$a \ne b \ne c\,\,{\text{and}}\,\,\alpha = \beta = \gamma = {90^ \circ }$$
C
$$a \ne b \ne c\,\,{\text{and}}\,\,\beta = \gamma = {90^ \circ } \ne \alpha $$
D
$$a \ne b \ne c\,\,{\text{and}}\,\,\alpha \ne \beta \ne \gamma \ne {90^ \circ }$$
Monoclinic crystal is one of the 7 crystal system. In monoclinic system, the crystal is describe by vectors of unequal lengths,
So, we have $$a \ne b \ne c\,\,{\text{and}}\,\,\beta = \gamma = {90^ \circ } \ne \alpha $$
182.
An element with atomic mass 100 has a $$bcc$$ structure
and edge length 400 $$pm.$$ The density of element is
183.
Ionic solids conduct electricity in molten state but not in solid state because
A
in molten state, free ions are furnished which are not free to move in solid state
B
in solid state, ionic solids are hard, brittle and become soft in molten state
C
all solids conduct electricity in molten state
D
in solid state, ions are converted to atoms which are insulators.
Answer :
in molten state, free ions are furnished which are not free to move in solid state
Ionic solids conduct electricity in molten state since in molten state ionic solids dissociate to give free ions which in solid state are not free to move and are held together by strong electrostatic forces of attraction.
184.
In a face centred cubic lattice, atoms of $$A$$ form the corner points and atoms of $$B$$ form the face centred points. If two atoms of $$A$$ are missing from the corner points, the formula of the ionic compound is :
A
$$A{B_3}$$
B
$$A{B_4}$$
C
$${A_2}{B_5}$$
D
$$A{B_2}$$
Answer :
$$A{B_4}$$
$$A$$ form corner points and two atoms of $$A$$ are missing from corner
$$\therefore $$ Atoms at corner $$\left( A \right) = 6 \times \frac{1}{8} = \frac{3}{4}$$
Atoms at face centre $$\left( B \right) = 6 \times \frac{1}{2} = 3$$
$$\therefore \,\,{A_{\frac{3}{4}}}\,{B_3}\,{\text{i}}{\text{.e}}{\text{.,}}\,A{B_4}$$
185.
A crystal lattice with alternate positive and negative ions has radius ratio 0.524. Its coordination number is
A
4
B
3
C
6
D
12
Answer :
6
$$\frac{{{r^ + }}}{{{r^ - }}} = 0.524.$$ It is in between $$0.414 - 0.732.$$
Hence $$C.No. = 6.$$
186.
A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is $$'a',$$ the closest approach between two atoms in metallic crystal will be :
A
2a
B
$$2\sqrt 2 a$$
C
$$\sqrt 2 a$$
D
$$\frac{a}{{\sqrt 2 }}$$
Answer :
$$\frac{a}{{\sqrt 2 }}$$
For a fcc unit cell
$$\eqalign{
& r = \frac{{\sqrt 2 \,a}}{4} \cr
& \therefore \,\,{\text{Closest distance (2r) = }}\frac{{\sqrt 2 \,a}}{4} = \frac{a}{{\sqrt 2 }} \cr} $$
187.
$$CsBr$$ has $$bcc$$ structure with edge length 4.3. The shortest inter ionic distance in between $$C{s^ + }$$ and $$B{r^ - }$$ is :
A
3.72
B
1.86
C
7.44
D
4.3
Answer :
3.72
For $$bcc$$ structure, atomic radius,
$$r = \frac{{\sqrt 3 }}{4}a$$
$$\eqalign{
& = \frac{{\sqrt 3 }}{4} \times 4.3 \cr
& = 1.86 \cr} $$
Since, $$r$$ = half the distance between two nearest neighbouring atoms.
∴ Shortest inter ionic distance = 2 × 1.86 = 3.72
188.
Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic radius of the lithium will be:
190.
Each of the following solids show, the Frenkel defect except
A
$$ZnS$$
B
$$AgBr$$
C
$$AgI$$
D
$$KCl$$
Answer :
$$KCl$$
In $$KCl,$$ co-ordination number of cation and anion is 6 and 6 respectively. $$KCl$$ is highly ionic so Schottky defect is common.
Note : Schottky defect is common in compounds having high coordination number while Frenkel defect is common in compounds with low coordination number.