Experimentally it was found that a metal oxide has formula $\mathrm{M}_{0.98} \mathrm{O}$. Metal M , is present as $\mathrm{M}^{2+}$ and $\mathrm{M}^{3+}$ in its oxide. Fraction of the metal which exists as $\mathrm{M}^{3+}$ would be:

JEE Main · 2013 · 07 Apr 2013 · Offline · Solid State

Experimentally it was found that a metal oxide has formula

M_{0.98} O

. Metal M , is present as

M^{2 +}

and

M^{3 +}

in its oxide. Fraction of the metal which exists as

M^{3 +}

would be:

Answer: (A)

Overall the oxide is electrically neutral, so the positive charge from the metal must balance the

2 -

on oxygen. Let a fraction

x

of the

0.98

metal ions be

M X^{3 +}

and the rest

M X^{2 +}

3 x + 2 (0.98 - x) = 2

, which gives

x = 0.04

. The share of metal present as

M X^{3 +}

\frac{0.04}{0.98} \times 100 \approx 4.08%

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option