Question

What is the magnitude of total spin value in case of ${}_{26}F{e}^{3+}$ ion?

Answer 1

Answer: 2.5

${}_{26}Fe-[Ar]3d^{6}4s^2$
$\Rightarrow F{e}^{3+}-[Ar]3d^5$
Clearly, it has $5$ unpaired electrons; hence,
Total spin $=+\frac{5}{2}$ or $-\frac{5}{2}$