The total number of unpaired


The total number of unpaired electrons present in $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ is


${\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2} }$

$\mathrm{Co}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^{7} 4 \mathrm{~s}^{0} 4 \mathrm{p}^{0}$

For this complex $\Delta_{0}<$ P.E., so pairing of electron does not take place.

$\mathrm{sp}^{3} \mathrm{~d}^{2}$ hybridisation

Total 3 unpaired electrons are present.

$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$

$\begin{aligned} \mathrm{Co}^{3+} &:[\mathrm{Ar}] 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{0} 4 \mathrm{p}^{0} \\ & \mathrm{~d}^{2} \mathrm{sp}^{3} \text { hybridisation } \end{aligned}$

$\mathrm{NH}_{3}$ acts as SFL because $\Delta_{0}>$ P.E.

So here all electrons becomes paired.

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