What fraction of Fe exists as Fe(III) in  $F{e}_{0.96}O$ ? (Consider $F{e}_{0.96}O$  to be made up of Fe(II) and Fe{III) only)

$F{e}_{0.96}O$

Let Fe(II) present in $F{e}_{0.96}O$ = x

Fe (III) present = (0.96-x)

Total charge on Fe=2x+(0.96-x)3

Total charge on O = -2

2x + (0.96-x) 3 =2

2x +2.88 – 3x =2

-x = -0.88

x= 0.88

$F{e}^{2+}=0.88,F{e}^{3+}=0.08$

Fraction of $F{e}^{3+}=\frac{0.08}{0.96}=\frac{1}{12}$