Question

Which of these statements about $\left[Co\right(CN{)}_6{]}^{3-}$ is true :-

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• A. $\left[Co\right(CN{)}_6{]}^{3-}$ has no unpaired electrons and will be in a high-spin configuration.
• B. $\left[Co\right(CN{)}_6{]}^{3-}$ has no unpaired electrons and will be in a low-spin configuration.
• C. $\left[Co\right(CN{)}_6{]}^{3-}$ has four unpaired electrons and will be in a low-spin configuration.
• D. $\left[Co\right(CN{)}_6{]}^{3-}$ has four unpaired electrons and will be in a high spin configuration.

Answer 1

The correct option is B $\left[Co\right(CN{)}_6{]}^{3-}$ has no unpaired electrons and will be in a low-spin configuration.

$\left[Co\right(CN{)}_6{]}^{3-}$
$C{o}^{3+}=3d^{6}4s^{0}4p^0$
It is a octahedral species and due to presence of strong field ligand it form low spin complex.
Its electronic configuration is $t_{2g}^6{e}_g^0$
$\because$ Due strong field ligand, pairing of electrons occurs so in this complex no unpaired electron is present and it is low spin complex.

Hence, option (d) is correct.