JEE Advanced · Chemistry · 9. Redox Reactions
In a metal deficient oxide sample, \(\mathbf{M}_{\mathbf{x}} \mathbf{Y}_2 \mathbf{O}_4\) ( \(\mathbf{M}\) and \(\mathbf{Y}\) are metals), \(\mathbf{M}\) is present in both +2 and +3 oxidation states and \(\mathbf{Y}\) is in +3 oxidation state. If the fraction of \(\mathbf{M}^{2+}\) ions present in \(\mathbf{M}\) is \(\frac{1}{3}\), the value of \(\mathbf{X}\) is ________.
- A 0.25
- B 0.33
- C 0.67
- D 0.75
Answer & Solution
Correct Answer
(D) 0.75
Step-by-step Solution
Detailed explanation
Average oxidation state of \(\mathrm{M}=\frac{1}{3} \times 2+\frac{2}{3} \times 3=+\frac{8}{3}\)
\(\begin{aligned} & \therefore \text { For } \mathrm{M}_{\mathrm{X}} \mathrm{M}_{2} \mathrm{Y}_4 \\ & \frac{8}{3} \times \mathrm{x}+3 \times 2+4(-2)=0 \\ & \frac{8}{3} \times \mathrm{x}=2 \\ & \mathrm{x}=\frac{3}{4}=0.75\end{aligned}\)
\(\begin{aligned} & \therefore \text { For } \mathrm{M}_{\mathrm{X}} \mathrm{M}_{2} \mathrm{Y}_4 \\ & \frac{8}{3} \times \mathrm{x}+3 \times 2+4(-2)=0 \\ & \frac{8}{3} \times \mathrm{x}=2 \\ & \mathrm{x}=\frac{3}{4}=0.75\end{aligned}\)
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