Question

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

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

Answer 1

Answer: (A)

${\left[Co(CN{)}_6\right]}^{3-}$
$C{o}^{3+}=1s^2,2s^2,2p^6,3s^2,3p^6,3d^6$
$C{N}^{-}$is a strong field ligand and as it approaches the metal ion, the electron must pair up.
The splitting of the $d$-orbitals into two sets of orbitals in as octahedral complex ${\left[Co(CN{)}_6\right]}^{3-}$ may be represented as

Here, for $d^6$ ions, three electrons first inter orbitals with parallel spin out the remaining may pair up in $t_{2g}$ orbital giving rise to low spin complex (strong ligand) field.
$\therefore {\left[Co(CN{)}_6\right]}^{3-}$ has no unpaired electrons and will be in low spin configuration.