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

Options

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

Answer

A

Explanation

`[Co (CN)_(6)]^(3-) , Co ^(3+ )=1s^(2) , 2S^(2),2p^(6) ,3S^(2),3P^(6) ,3d^(6)`
`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 octachedral complex `,[Co (CN)_(6)]^(3-)` may be represented as

Here , for `d^(6) ` ions , three electrons first enter orbitals with parallel spin and the remaining electron pair up in `t_(29)` orbital giving rise to low spin complex
`therefore [Co(CN)_(6)]^(3-)` has no unpaired electron and will exist as a low spin complex .