The electron configuration of cobalt (Z = 27) is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}4s^2$. How many unpaired electrons are in a gaseous ${\mathrm{Co}}^{3+}\text{ ion }$ in its ground state?

The unpaired electrons are in a gaseous ${\mathrm{Co}}^{3+}\text{ ion }$ in its ground state is 4.

The electronic configuration of the given cobalt ion (${\mathrm{Co}}^{3+}\text{ ion }$) is,

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}4s^2$

${\mathrm{CO}}_{27}^{3+}\to \left[Ar\right]\left(3d^4\right)4{\mathrm{s}}^2$

From the following electronic configuration it is obtained that there are in a gaseous ${\mathrm{Co}}^{3+}\text{ ion }$ in its ground state.

Hence, the option C is correct.