Question

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

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

Answer 1

The correct option is A

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

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



Here, for $d^8$ ions, three electrons first enter orbitals with parallel spin put the remaining may pair up in $t_{2g}$ orbital giving rise to low spin complex (strong ligand) field.
$\therefore \left[Co\right(CN{)}_6{]}^{3-}$ has no unpaired electron and will be in low spin configuration.