JEE Advanced 2024 · Paper 2, Chemistry question 2

Charge balance in $\ce{M_XY2O4}$:

• the oxide ions carry $-8$;
• $\ce{Y2}$ carries $+6$;
• so the $X$ atoms of M must carry $+2$ in total.

Average charge of M. One third is $\ce{M^2+}$ and two thirds $\ce{M^3+}$:
$$\tfrac13(2)+\tfrac23(3)=\tfrac83.$$

$$X\cdot\frac83=2\implies X=\frac34=0.75.$$