Which of these statements about $\left[\mathrm{C}\mathrm{o}\right(\mathrm{C}\mathrm{N}{)}_6{]}^{3-}$ is true

• A

  $\left[\mathrm{C}\mathrm{o}\right(\mathrm{C}\mathrm{N}{)}_6{]}^{3-}$ has four unpaired electrons and will be in a low-spin configuration.
• B

  $\left[\mathrm{C}\mathrm{o}\right(\mathrm{C}\mathrm{N}{)}_6{]}^{3-}$ has four unpaired electrons and will be in a high-spin configuration.
• C

  $\left[\mathrm{C}\mathrm{o}\right(\mathrm{C}\mathrm{N}{)}_6{]}^{3-}$ has no unpaired electrons and will be in a high-spin configuration.
• Correct answer
  D

  $\left[\mathrm{C}\mathrm{o}\right(\mathrm{C}\mathrm{N}{)}_6{]}^{3-}$ has no unpaired electrons and will be in a low-spin configuration.

Solution

In $\left[\mathrm{C}\mathrm{o}\right(\mathrm{C}\mathrm{N}{)}_6{]}^{3-}$ Cobalt is in + 3 oxidation state so $\mathrm{C}{\mathrm{o}}^{+3}=\left(\mathrm{A}\mathrm{r}\right)3{\mathrm{d}}^{6}4{\mathrm{s}}^{0}4{\mathrm{p}}^0$
As in presence of strong field ligand, pairing of electrons occurs so in this complex no unpaired electron is present and it is a low-spin complex.

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