JEE Advanced 2024 Paper 2, Chemistry Section 1 Q2: Oxidation and Reduction

JEE Advanced2024Paper 2Chemistry Section 1

In a metal deficient oxide sample, $M_{X} Y_{2} O_{4}$ ( $M$ and $Y$ are metals), $M$ is present in both $+ 2$ and $+ 3$ oxidation states and $Y$ is in $+ 3$ oxidation state. If the fraction of $M^{2 +}$ ions present in $M$ is $\frac{1}{3}$ , the value of $X$ is _____.

A
$0.25$
B
$0.33$
C
$0.67$
D
$0.75$

Solution

Out of $X$ total $M$ ions, $\frac{1}{3}$ are in the $+ 2$ state and $\frac{2}{3}$ are in the $+ 3$ state. So the numbers of each are

$n (M^{2 +}) = \frac{X}{3}, n (M^{3 +}) = \frac{2 X}{3} .$

Applying charge neutrality to the formula $M_{X} Y_{2} O_{4}$ with $Y$ in $+ 3$ and $O$ in $- 2$ :

$(+ 2) \cdot \frac{X}{3} + (+ 3) \cdot \frac{2 X}{3} + 2 \cdot (+ 3) + 4 \cdot (- 2) = 0.$

$\frac{2 X}{3} + 2 X + 6 - 8 = 0 \Rightarrow \frac{8 X}{3} = 2 \Rightarrow X = \frac{3}{4} = 0.75.$

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JEE Advanced 2024 Paper 2, Chemistry Section 1 Q2: Oxidation and Reduction

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