1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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