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Indicate the number of unpaired electrons in $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ and $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
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The correct answer is:
$1 \quad 0$
$\begin{array}{l}
\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \quad \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \\
\mathrm{Fe}=[\mathrm{Ar}] 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{2} \\
\mathrm{Fe}^{+3} \rightarrow(\mathrm{Ar}) 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0} 4 \mathrm{p}^{0} \quad \mathrm{Fe}^{+2} \rightarrow(\mathrm{Ar}) 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0} 4 \mathrm{p}^{0}
\end{array}$
As $\mathrm{CN}$ is the strong field ligand; we pair up
\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \quad \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \\
\mathrm{Fe}=[\mathrm{Ar}] 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{2} \\
\mathrm{Fe}^{+3} \rightarrow(\mathrm{Ar}) 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0} 4 \mathrm{p}^{0} \quad \mathrm{Fe}^{+2} \rightarrow(\mathrm{Ar}) 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0} 4 \mathrm{p}^{0}
\end{array}$
As $\mathrm{CN}$ is the strong field ligand; we pair up
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