1

### AIPMT 2012 Prelims

The number of octahedral void(s) per atom present in a cubic close-packed structure is
A
1
B
3
C
2
D
4

## Explanation

Number of octahedral voids is same as number of atoms.
2

### AIPMT 2012 Mains

Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are cooupied by a monovalent metal B. The formula of the oxide is
A
ABO2
B
A2BO2
C
A2B3O4
D
AB2O2

## Explanation

Number of atoms in cubic close packing = 4 = O2-

Number of tetrahedral voids = 2 × N = 2 × 4

Number of A2+ ions = 8 $\times$ ${1 \over 4}$ = 2

Number of octahedral voids = Number of B+ ions = N = 4

Ratio of ions will be,

O2– : A2+ : B+ = 4 : 2 : 4 = 2 : 1 : 2

Formula of oxide = AB2O2
3

### AIPMT 2012 Prelims

A metal crystallises with a face-centred cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is
A
288 pm
B
408 pm
C
144 pm
D
204 pm

## Explanation

Given $a$ = 408 pm

For the face centred cubic structure

r = ${a \over {2\sqrt 2 }}$ = ${{408} \over {2\sqrt 2 }}$ = 144 pm

$\therefore$ Diameter = 2r = 2$\times$144 = 288 pm
4

### AIPMT 2011 Mains

A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y$-$) will be
A
275.1 pm
B
322.5 pm
C
241.5 pm
D
165.7 pm

## Explanation

For NaCl crystal, ${{{r_ + }} \over {{r_ - }}} = 0.414$

Given, r+ = 100 pm

$\therefore$ ${{100} \over {{r_ - }}} = 0.414$

$\Rightarrow$ ${r_ - } = {{100} \over {0.414}}$ = 241.5 pm