Question

The number of unpaired electrons calculated in $[Co(N{H}_3{)}_6{]}^{3+}$ 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] $3{\text{d}}^6,4{\text{s}}^0$