Question

Which of theses statements about $\left[Co\right(CN{)}_6{]}^{3-}$ is true?

• A. $\left[Co\right(CN{)}_6{]}^{3-}$ has four unpaired electrons and will be in a high-spin configuration.
• B. $\left[Co\right(CN{)}_6{]}^{3-}$ has no unpaired electrons and will be in a high-spin configuration.
• C. $\left[Co\right(CN{)}_6{]}^{3-}$ has no unpaired electrons and will be in a low-spin configuration.
• D. $\left[Co\right(CN{)}_6{]}^{3-}$ has four unpaired electrons and will be in a low-spin configuration.

Answer 1

The correct option is C $\left[Co\right(CN{)}_6{]}^{3-}$ has no unpaired electrons and will be in a low-spin configuration.

The electronic configuration of Co is $\left[Ar\right]3d^{7}4s^2$.

The central Co atom has +3 oxidation state and $3d^6$ as outer electronic configuration.

Cyanide ion is a strong field ligand and ${\Delta }_o>P$. Hence, pairing of electrons is preferred.

Hence, the outer electronic configuration of $C{o}^{3+}$ can be written as $t_{2g}^6{e}_g^0$

Thus, $\left[Co\right(CN{)}_6{]}^{3-}$ has no unpaired electrons and will be in a low-spin configuration.