Question

Find the number of unpaired electrons calculated in [Co(NH${}_3$)${}_6$]${}^{3+}$ and [CoF${}_6$]${}^{3-}$.

• A. 4 unpaired electrons in both complexes
• B. 4 unpaired electrons in [Co(NH${}_3$)${}_6$]${}^{3+}$ and no unpaired electrons in [CoF${}_6$]${}^{3-}$
• C. 4 unpaired electrons in [CoF${}_6$]${}^{3-}$ and no unpaired electrons in [Co(NH${}_3$)${}_6$]${}^{3+}$
• D. no unpaired electrons in both complexes

Answer 1

The correct option is C 4 unpaired electrons in [CoF${}_6$]${}^{3-}$ and no unpaired electrons in [Co(NH${}_3$)${}_6$]${}^{3+}$

In both the complexes, [Co(NH${}_3$)${}_6$]${}^{3+}$ and [CoF${}_6$]${}^{3-}$, the central Co atom is in +3 oxidation state. Ammonia is a strong field ligand. In [Co(NH${}_3$)${}_6$]${}^{3+}$, ammonia cause spin pairing. Cobalt has $3d^6$ outer electronic configuration which can also be written as $t_{2g}^6{e}_g^0$.
Thus, all the electrons are paired.
In the complex [CoF${}_6$]${}^{3-}$, the weak field ligand fluoride ion do not cause spin pairing.
Cobalt has $3d^6$ outer electronic configuration which can also be written as $t_{2g}^4{e}_g^2$.
The complex contain 4 unpaired electrons.