Question

The ratio of $\mathrm{Fe}{}^{3+}$ and $\mathrm{Fe}{}^{2+}$ in $\mathrm{Fe}{}_{0.9}\mathrm{S}{}_{1.0}$ is

• A. 0.28
• B. 0.5
• C. 2.0
• D. 4.0

Answer 1

Answer: (A)

$F{e}^{3+}=x$
$F{e}^{2+}=0.93-x$
Using charge balance
$F{e}_{0.9}{S}_{1.0}=$ Charge on neutral compound $=0$
$\underbrace{3\times x2\times (0.93-x)}+2\times 1=0$
Charge on $F{e}_{0.9}$
$\left(F{e}^{2+}\&F{e}^{3+}\right)$
$\Rightarrow 3x+2(0.9-x)=2\times 2$
$3x+1.8-2x=2$
$x=2-1.8$
$x=0.2$
Thus the no. of $F{e}^{+2}$ ion $=0.93-x$ $=0.93-0.2$ $=0.73$
Ratio $\frac{F{e}^{3+}}{F{e}^{2+}}=\frac{0.2}{0.73}$ $=0.28$