What is the total spin value in case of ${ }_{26}{\mathrm{Fe}}^{3+}$ ion?

The electronic configration of${ }_{26}{\mathrm{Fe}}^{3+}$ ion is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3{\mathrm{d}}^5$.

The five electrons in the third shell will all be left unpaired since each orbital in the d subshell should be solely occupied before being linked.

hence, the total spin of ${ }_{26}{\mathrm{Fe}}^{3+}$

$$\begin{gathered} =\pm \frac{1}{2}\times 5 \\ =\pm 2.5 \end{gathered}$$

Thus the option C is correct.