Which one of the following complexes will have $\Delta_0=0$ and $\mu=5.96$ B.M.?

• A

  $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^4$
• B

  $\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
• C

  $\left[\mathrm{FeF}_6\right]^4$
• Correct answer
  D

  $\left[\mathrm{Mn}(\mathrm{SCN})_6\right]^4$

Solution

(1) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \quad \mathrm{Co}^{3+} \Rightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^0$

$=[-0.4 \times 6+0.6 \times(0)] \Delta_0=-2.4 \Delta_0$
(2)
$\begin{aligned}<br />& {\left[\mathrm{Mn}(\mathrm{SCN})_6\right]^4} \\<br>& \mathrm{Mn}^{2+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0<br />\end{aligned}$
$\begin{aligned}<br />& \mu=\sqrt{35} \text { B.M. }=5.96 \text { B.M. } \\<br>& \mathrm{CFSE}=(-0.4 \times 3+0.6 \times 2) \Delta_0<br />\end{aligned}$
So $\Delta_0=0$
(3)
$\begin{aligned}<br />& {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{-4} \quad \mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^0}<br />\end{aligned}$
$\mathrm{CFSE}=-2.4 \Delta_0$
(4) $\left[\mathrm{FeF}_6\right]^{4-}$
$\mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6 4 \mathrm{~s}^0$
$\begin{aligned}<br />& \mu=\sqrt{24} \text { B.M. }=4.89 \text { B.M. } \\<br>& \text { CFSE }=(-0.4 \times 4+0.6 \times 2) \Delta_0=-1.2 \Delta_0<br />\end{aligned}$

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