Given below are two statements : Statement I: Both $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{CoF}_6\right]^{3-}$ complexes are octahedral but differ in their magnetic behaviour. Statement II: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is diamagnetic whereas $\left[\mathrm{CoF}_6\right]^{3-}$ is paramagnetic. In the light of the above statements, choose the correct answer from the options given below:

• A

  Both Statement I and Statement II are false
• B

  Statement I is true but Statement II is false
• C

  Statement I is false but Statement II is true
• Correct answer
  D

  Both Statement I and Statement II are true

Solution

In $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}, \mathrm{Co}^{3+}$ ion is having $3 \mathrm{~d}^6$ configuration. Electronic configuration of $\mathrm{Co}^{3+}$ :
In presence of $\mathrm{NH}_3$ ligand, pairing of electrons takes place and it becomes diamagnetic complex ion. In presence of $\mathrm{NH}_3$ ligand :
$\therefore \quad\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is octahedral with $\mathrm{d}^2 \mathrm{sp}^3$ hybridisation and it is diamagnetic in nature. In case of $\left[\mathrm{CoF}_6\right]^{3-}$, Co is in +3 oxidation state and it is having $3 \mathrm{~d}^6$ configuration. In presence of weak field $\mathrm{F}^{-}$ligand, pairing does not take place. In presence of $\mathrm{F}^{-}$ligands :
$\therefore$ In $\left[\mathrm{CoF}_6\right]^{3-}, \mathrm{Co}^{3+}$ is $\mathrm{sp}^3 \mathrm{~d}^2$ hybridised with four unpaired electrons, so it is paramagnetic in nature.

Hence, the correct answer is option: 'D'.