Among the following complexes, the total number of diamagnetic species is_____…
Among the following complexes, the total number of diamagnetic species is_____
$\left[\mathrm{Mn}\left(\mathrm{NH}_3\right)_6\right]^{3+},\left[\mathrm{MnCl}_6\right]^{3-},\left[\mathrm{FeF}_6\right]^{3-},\left[\mathrm{CoF}_6\right]^{3-},\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right]^{3+}$, and $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$
[Given, atomic number: $\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27$;
$\text { en } \left.=\mathrm{H}_2 \mathrm{NCH}_2 \mathrm{CH}_2 \mathrm{NH}_2\right]$
Solution
$\mathrm{Mn}^{3+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^4$
$\mathrm{d}^4$ configuration in $\mathrm{t}_{2 \mathrm{~g}}$ and $\mathrm{eg}_{\mathrm{g}}$ orbitals will always have unpaired electrons irrespective of SFL and WFL.
$\mathrm{Fe}^{3+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^5$
$\mathrm{d}^5$ configuration will also have unpaired electron irrespective of SFL and WFL.
$\mathrm{Co}^{3+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^6$
$\mathrm{d}^6 \Rightarrow$ it can be both paramagnetic or diamagnetic based on field of ligands.
In case of $\mathrm{F}^{-} \Rightarrow$ weak field ligand, configuration will be $\mathrm{t}_{2 \mathrm{~g}}^4 \mathrm{e}_{\mathrm{g}}^2$ hence it is paramagnetic but in case of
en $\Rightarrow$ strong filed ligand, configuration will be $\mathrm{t}_{2 \mathrm{~g}}^6 \mathrm{e}_{\mathrm{g}}^0$ hence it will be diamagnetic.
So,
\(\left[\mathrm{Mn}\left(\mathrm{NH}_3\right) 6\right]^{3+}\) : Paramagnetic
\(\left[\mathrm{MnCl}_6\right]^{3-}\) : Paramagnetic
\(\left[\mathrm{FeF}_6\right]^{3-}\) : Paramagnetic
\(\left[\mathrm{CoF}_6\right]^{3-}\) : Paramagnetic
\(\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right]^{3+}\) : Paramagnetic
\(\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}\) : Diamagnetic
Only 1 complex is diamagnetic.