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Oxidation number Fe in $\left.\mathrm{K}_3[\mathrm{Fe(CN)}_6\right]$ is
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The correct answer is:
+3
Let, oxidation number of Fe be ' $x$ '
$\begin{aligned} \therefore \quad 3(+1)+x+6(-1) & =0 \\ 3+x-6 & =0 \\ x & =+3\end{aligned}$
$\begin{aligned} \therefore \quad 3(+1)+x+6(-1) & =0 \\ 3+x-6 & =0 \\ x & =+3\end{aligned}$
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