JEE PYQ: Solid State Question 12
Question 12 - 2020 (06 Sep Shift 2)
A crystal is made up of metal ions $\text{M}_1^{’}$ and $\text{M}_2^{’}$ and oxide ions. Oxide ions form a $ccp$ lattice structure. The cation $\text{M}_1^{’}$ occupies 50% of octahedral voids and the cation $\text{M}_2^{’}$ occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation numbers of $\text{M}_1^{’}$ and $\text{M}_2^{’}$ are, respectively:
(1) +2, +4
(2) +1, +3
(3) +3, +1
(4) +4, +2
Show Answer
Answer: (1)
Solution
$\text{M}_1$: 50% of 4 octahedral voids $= 2$. $\text{M}_2$: 12.5% of 8 tetrahedral voids $= 1$. O: 4. Charge: $2x + y = 8$. So $x = 2$, $y = 4$. Oxidation numbers: $\text{M}_1 = +2$, $\text{M}_2 = +4$.
BETA
