Question

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

• A. It has no unpaired electron and will be in a low spin configuration.
• B. It has four unpaired electron and will be in low spin configuration
• C. It has four unpaired electron and will be in a high-spin configuration
• D. It has no unparied electrons and will be in a high-spin configuration.

Answer 1

The correct option is A It has no unpaired electron 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 electron must pair up. the splitting of the d-orbitals in two sets octahedral $\left[Co\right(CN{)}_6{]}^{3-}$ may be represnted as
here for $d^6$ions there electrons first orbitals with parallel spin put the remaining may pair up in $t_{2g}$ orbital giving rise to low spin complex.
therefore [$Co(CN{)}_6{]}^{3-}$ has no unpaired electron and will be in a low spin configuration.