1

AIPMT 2009

Lithium metal crystallises in a body-centred cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of lithium will be
A
151.8 pm
B
75.5 pm
C
300.5 pm
D
240.8 pm

Explanation

Since Li crystallises in body-centred cubic

crystal, atomic radius, r = ${{\sqrt 3 a} \over 4}$.

$\therefore$ r = ${{\sqrt 3 } \over 4} \times 351$ = 151.8 pm
2

AIPMT 2008

Percentage of free space in a body centred cubic unit cell is
A
34%
B
28%
C
30%
D
32%

Explanation

The ratio of volumes occupied by atoms in unit cell to the total volume of the unit cell is called as packing fraction or density of packing. For body centred cubic structure, packing fraction = 0.68 i.e., 68% of the unit cell is occupied by atoms and 32% is empty.
3

AIPMT 2008

Which of the following statements is not correct?
A
The number of carbon atoms in a unit cell of diamond is 4
B
The number of Bravais lattices in which a crystal can be categorized is 14
C
The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.
D
Molecular solids are generally volatile.

Explanation

Packing fraction for a cubic unit cell is given by

f = ${{Z \times {4 \over 3}\pi {r^3}} \over {{a^3}}}$

where a = edge length, r = radius of cation and anion.

Efficiency of packing in simple cubic or primitive cell = π/6 = 0.52 or 52 %

It means 52 % of unit cell is occupied by atoms and 48 % is empty.
4

AIPMT 2008

If $a$ stands for the edge length of the cubic systems: simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be respectively
A
${1 \over 2}a:{{\sqrt 3 } \over 2}a:{{\sqrt 2 } \over 2}a$
B
$1a:\sqrt 3 a:\sqrt 2 a$
C
${1 \over 2}a:{{\sqrt 3 } \over 4}a:{1 \over {2\sqrt 2 }}a$
D
${1 \over 2}a:\sqrt 3 a:{1 \over {\sqrt 2 }}a$

Explanation

For Simple cubic : r+ + r = ${a \over 2}$

For Body centred : r+ + r = ${{a\sqrt 2 } \over 4}$

For Face centered: r+ + r = ${a \over {2\sqrt 2 }}$

$\therefore$ Ratio of radii of the three will be

= ${1 \over 2}a:{{\sqrt 3 } \over 4}a:{1 \over {2\sqrt 2 }}a$