Q

# Tell me? - Coordination compounds: - NEET-2

Which of these statements about $\left [ Co(CN)_{6} \right ]^{3-}$ is true?

• Option 1)

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

• Option 2)

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

• Option 3)

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

• Option 4)

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

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As we learnt in

Magnetic Moments (spin only) -

$\sqrt{n\left ( n+2 \right )}$ where n= number of unpaired electron.

- wherein

Number of unpaired $e^{-} s$ and corresponding magnetic moments

$1\rightarrow 1.7\\2\rightarrow 2.8\\3\rightarrow 3.9\\4\rightarrow 4.9\\5\rightarrow 5.9$

Electronic config of $[Co(CN)_{6}]^{3-}$ for $Co^{3+}$   =

$w^{-}s$ will be paired up because $CN^{-}$ is an SFL

$\therefore [Co(CN)_{6}]^{3-}$ has no unpaired $e^{-}s$ and is in a low spin config.

Option 1)

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

This solution is incorrect.

Option 2)

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

This solution is incorrect.

Option 3)

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

This solution is incorrect.

Option 4)

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

This solution is correct.

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