Indicate the number of unpaired electrons in $\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ and $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

• Correct answer
  A

  $1 \quad 0$
• B

  $5 \quad 6$
• C

  $6 \quad 5$
• D

  $0\quad 1$

Solution

$\begin{array}{l} \mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \quad \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \\ \mathrm{Fe}=[\mathrm{Ar}] 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{2} \\ \mathrm{Fe}^{+3} \rightarrow(\mathrm{Ar}) 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{0} 4 \mathrm{p}^{0} \quad \mathrm{Fe}^{+2} \rightarrow(\mathrm{Ar}) 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0} 4 \mathrm{p}^{0} \end{array}$ As $\mathrm{CN}$ is the strong field ligand; we pair up

Hence, the correct answer is option: 'A'.