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Which of these statement about ${[Co (CN)_{6}]}^{3 -}$ is true

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${[Co (CN)_{6}]}^{3 -}$ has four unpaired electrons and will be in a low-spin configuration.

${[Co (CN)_{6}]}^{3 -}$ has no unpaired electrons and will be in a low-spin configuration.

${[Co (CN)_{6}]}^{3 -}$ has no unpaired electrons and will be in a high-spin configuration.

${[Co (CN)_{6}]}^{3 -}$ has four unpaired electrons and will be in a high-spin configuration.

answer is C.

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Detailed Solution

${[Co (CN)_{6}]}^{3 -}$ $oxidation no. of Co = + 3$

${Co}^{3 +} = 3 d^{6}$

As $C N^{-}$ is a strong field ligand, so all electrons will be paired up and complex will be low spin complex.

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