1

### NEET 2016 Phase 1

The ionic radii of A+ and B$-$ ions are 0.98 $\times$ 10$-$10 m and 1.81 $\times$ 10$-$10 m. The coordination number of each ion in AB is
A
8
B
2
C
6
D
4

## Explanation

From radius ratio, ${{{r_ + }} \over {{r_ - }}} = {{0.98 \times {{10}^{ - 10}}} \over {1.81 \times {{10}^{ - 10}}}}$ = 0.541

It lies in the range of 0.414 to 0.732 hence coordination number of each ion will be 6 as the compound will have NaCl type structure i.e., octahedral arrangement.
2

### NEET 2016 Phase 2

In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca2+) and fluoride ion (F$-$) are
A
4 and 2
B
6 and 6
C
8 and 4
D
4 and 8

## Explanation

In fluorite (CaF2) structure,

C.N. of Ca+2 = 8

C.N. of F = 4.
3

### NEET 2016 Phase 1

Lithium has a bcc structure. Its density is 530 kg m$-$3 and its atomic mass is 6.94 g mol$-$. Calculate the edge length of a unit cell of lithium metal. (NA = 6.02 $\times$ 1023 mol$-$1)
A
527 pm
B
264 pm
C
154 pm
D
352 pm

## Explanation

For bcc structure we have, Z = 2

Given, $\rho$ = 530 kg m–3

atomic mass of Li = 6.94 mol–1

NA= 6.02 × 1023 mol–1

$\rho$ = 530 kg m–3 = ${{530 \times 1000} \over {{{\left( {100} \right)}^3}}} =$ 0.53 g cm–3

Also, $\rho$ = ${{z \times At.mass} \over {{N_A} \times {a^3}}}$

$\Rightarrow$ ${a^3} = {{Z \times At.mass} \over {{N_A} \times \rho }}$

= ${{2 \times 6.94} \over {6.02 \times {{10}^{23}} \times 0.53}}$

= 43.5 $\times$10-10 cm3

$\Rightarrow$ $a$ = 352 × 10–10 cm = 352 pm
4

### AIPMT 2015 Cancelled Paper

A given metal crystallises out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom?
A
80 pm
B
108 pm
C
40 pm
D
127 pm

## Explanation

As Z = 4, so, structure is fcc.

Hence, r = ${a \over {2\sqrt 2 }} = {{361} \over {2\sqrt 2 }}$ = 127.65 pm = 127 pm