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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.

 

Answers (1)

best_answer

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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prateek

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