IMO Practice Test: The Solid State | Class 12 Chemistry

Question 1 of 13

An element crystallises in fcc with edge $a$. The relation between the atomic radius $r$ and $a$ is:

$r=\frac{a}{2}$

$r=\frac{\sqrt{3}}{4}a$

$r=\frac{a}{2\sqrt{2}}$

$r=\frac{a}{4}$

Explanation: fcc: $\sqrt{2}\,a=4r\Rightarrow r=\frac{a}{2\sqrt{2}}$.

Question 2 of 13

A solid AB has B atoms in ccp and A atoms in all the octahedral voids. The formula and coordination of A is:

AB, A is 6-coordinate

$\text{A}_2\text{B}$, A is 4-coordinate

$\text{AB}_2$, A is 8-coordinate

$\text{A}_2\text{B}_3$, A is 6-coordinate

Explanation: Octahedral voids $=N$ = number of B atoms, so A:B $=1:1$ (AB) and each A sits in an octahedral hole (6-coordinate).

Question 3 of 13

A bcc metal has $a=3.0\times10^{-8}\ \text{cm}$ and density $5.0\ \text{g cm}^{-3}$. Its molar mass (approx) is:

$20\ \text{g/mol}$

$41\ \text{g/mol}$

$81\ \text{g/mol}$

$162\ \text{g/mol}$

Explanation: $M=\frac{d a^3 N_A}{z}=\frac{5.0\times(2.7\times10^{-23})\times6.022\times10^{23}}{2}\approx41\ \text{g/mol}$.

Question 4 of 13

If the radius of an octahedral void is $r$ and that of the close-packed sphere is $R$, then the limiting value of $r/R$ is:

0.225

0.155

0.732

0.414

Explanation: The geometry of six spheres around a void gives the limiting ratio $r/R=0.414$.

Question 5 of 13

In a ccp lattice of N atoms, the total number of voids (tetrahedral + octahedral) is:

$2N$

$3N$

$4N$

$N$

Explanation: Tetrahedral $=2N$ and octahedral $=N$, giving $3N$ in total.

Question 6 of 13

A compound has the formula $\text{M}_x\text{N}$ where N forms ccp and M occupies all tetrahedral voids. Then $x$ equals:

1

2

3

4

Explanation: Tetrahedral voids $=2N$ atoms, so M:N $=2:1$ and $x=2$ (formula $\text{M}_2\text{N}$).

Question 7 of 13

On heating, the conductivity of an intrinsic semiconductor:

decreases

remains constant

becomes zero

increases

Explanation: More electrons gain enough energy to cross the small band gap, so conductivity increases with temperature.

Question 8 of 13

The ratio of packing efficiencies of fcc to bcc is closest to:

0.92

1.09

1.41

0.71

Explanation: $\frac{74}{68}\approx1.09$.

Question 9 of 13

AgBr can show:

only Schottky defect

only Frenkel defect

both Schottky and Frenkel defects

neither defect

Explanation: AgBr is a classic example that exhibits both Schottky and Frenkel defects.

Question 10 of 13

The number of nearest neighbours (coordination number) of an atom in a bcc lattice is:

4

6

8

12

Explanation: The body-centre atom touches the 8 corner atoms, so the coordination number is 8.

Question 11 of 13

A metal deficiency defect such as $\text{Fe}_{0.95}\text{O}$ is balanced by:

extra trapped electrons

some Fe present as $\text{Fe}^{3+}$

interstitial oxygen

anion vacancies

Explanation: Missing $\text{Fe}^{2+}$ cations are charge-balanced by converting some $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$.

Question 12 of 13

A face-centred cubic metal has edge $a=361\ \text{pm}$. Its atomic radius is approximately:

$128\ \text{pm}$

$156\ \text{pm}$

$181\ \text{pm}$

$90\ \text{pm}$

Explanation: $r=\frac{a}{2\sqrt{2}}=\frac{361}{2.828}\approx128\ \text{pm}$.

Question 13 of 13

Which arrangement gives the lowest packing efficiency?

simple cubic

body-centred cubic

face-centred cubic

hexagonal close packing

Explanation: Simple cubic occupies only 52.4% of space, the lowest among these.

IMO Practice Test — The Solid State