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

Options

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

Answer

C

Explanation

`[Co(CN)_6]^(3-)`, oxidation no. of `Co=+3`, `Co^(3+)=3d^6`
As `CN^(-)` is a strong field ligand, so all electrons will be paired up and complex will be low spin complex.