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The composition of a sample of wiistite is $Fe_{0.93} O_{1.00-}$. What percentage of the ion is present in the form of $Fe(III)?$

$\begin{array}{1 1} 15.05 \\ 1.143 \\ 8.4 \\ 9.5 \end{array} $

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Answer :$15.05$
The compound $Fe_{0.93}O_{1.00} $ is a non- stoichiometric where electrical neutrality is achieved by converting appropriate $Fe^{2+}$ ions into $Fe^{3+}$ ions.
There are $7FE^{2+}$ ions missing out of the expected $100 \;Fe^{2+}$ ions.
The missing $2 \times 7$ positive charge is compensated by the presence of $Fe^{3+}$ ions.
Replacement of one $Fe^{2+}$ ions by $Fe^{3+}$ ion increasing one positive charge.
Thus , 14 positive charges is compensated by the presence of $14 \;Fe^{3+}$ ions out of a total of 93 ions.
Hence per cent of $Fe^{3+} $ ions present $=\large\frac{14}{93} $$ \times 100 =15.05$
answered Aug 11, 2014 by meena.p

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