Which of these statements about `[Co(CN)_6]^(3-)` is true?

Options

A. `[Co(CN)_(6)]^(3-)` has no unpaired electron and will be in a low-spin configuration.
B. `[Co(CN)_(6)]^(3-)` has four unpaired electrons and will be in a low-spin configuration.
C. `[Co(CN)_(6)]^(3-)` has four unpaired electrons and will be in a high-spin configuration.
D. `[Co(CN)_(6)]^(3-)` has no unpaired electrons and will be in a high-spin configuration.

Answer

A

Explanation

`[Co(CN)_(6)]^(3-)`
`Co^(3+)=1s^(2)2s^(2)2p^(2)3s^(2)3p^(2)3d^(2)`
`CN^(-)` is a strong field ligand and as it approaches the metal ion, the electrons must pair up.
The splitting of the d-orbitals into two sets of orbitals in an octahedral `[Co(CN)_(6))^(3-)` may be represented as

Here, for `d^(6)` ions, three electrons first enter orbitals with parallel spin put the remammg may pair up in `t_(2g)` orbital giving rise to low spin complex (strong ligand) field.
`therefore[Co(CN)_(6)]^(3-)` has no unpaired electron and will be in a low spin configuration.