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

Options

A. `[Co(CN)_6]^(3-)` has four unpaired electrons and will be in a high-spin configuration
B. `[Co(CN)_6]^(3-)` has no unpäired electrons and will, be in a high-spin configuration
C. `[Co(CN)_6]^(3-)` has no unpaired electrons and will be in a low-spin configuration
D. `[Co(CN)_6]^(3-)` has four unpaired electrons and will be in a low-spin configuration.

Answer

C

Explanation

`[Co(CN)_6]^(3-)` has cobalt in +3 oxidation state
`Co^(3+) = 3d^6 = (##BRL_NEET_SP_CHE_XII_V02_C03_E03_019_S01.png" width="80%">
As `CN^(-)` is a strong field ligand, all electrons will be paired up and complex will be a low-spin complex.