Question

The number of unpaired electrons calculated in $[Co(NH_3)_6]^{3+}$ and $[Co(F_6)]^{3-}]$ are

• A. 4 and 4
• B. 0 and 2
• C. 2 and 4
• D. 0 and 4

Answer 1

Answer: (D)

In both $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{CoF}_6\right]^{3+}$, Co is present as $\text{C}\text{o}^{3 +}.$

Thus, the electronic configuration of Co is

Co=[Ar] $3\text{d}^{7},4\text{s}^{2}$

Co³⁺=[Ar] $3\text{d}^{6}, \, 4\text{s}^{0}$

In case of $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is a strong field ligand, so pairing of electrons in 3d-orbital takes place.

Co³⁺=[Ar] $3\text{d}^{6}, \, 4\text{s}^{0}$





In $\left[\text{C} \text{o} \text{F}_{6}\right]^{3 +}, \, \text{F}$ is a weak field ligand, thus doesn't cause pairing. Hence,

Co³⁺=[Ar] $3\text{d}^{6}, \, 4\text{s}^{0}$