JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Solid State

A crystal is made up of metal ions '

M_{1}

' and '

M_{2}

' and oxide ions. Oxide ions form a ccp lattice structure. The cation '

M_{1}

' occupies

50%

of octahedral voids and the cation '

M_{2}

' occupies

12.5%

of tetrahedral voids of oxide lattice. The oxidation numbers of '

M_{1}

' and '

M_{2}

' are respectively:

Answer: (A)

Solution

A ccp array of oxide ions holds

4

oxide ions per cell, with

4

octahedral and

8

tetrahedral voids.

M X_{1}

fills half the octahedral voids,

4 \times 0.5 = 2

ions, and

M X_{2}

fills an eighth of the tetrahedral voids,

8 \times 0.125 = 1

ion, so the formula is

(M X_{1}) X_{2} (M X_{2}) O X_{4}

. Charge balance against the four

O X^{2 -}

needs

2 x + y = 8

, and the pair

x = + 2

for

M X_{1}

with

y = + 4

for

M X_{2}

satisfies this.

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option