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The number of unpaired electrons in $\mathrm{Fe}^{3+}$ is
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5
$\mathrm{Fe}(26)=\lfloor\mathrm{Ar}\rfloor 3 d^{6} 4 s^{2}$
$\mathrm{Fe}^{3+}(23)=[\mathrm{Ar}] 3 d^{5} 4 s^{0}$
$\mathrm{Fe}^{3+}=$
$\therefore$ Number of unpaired electrons $=5$
$\mathrm{Fe}^{3+}(23)=[\mathrm{Ar}] 3 d^{5} 4 s^{0}$
$\mathrm{Fe}^{3+}=$
$\therefore$ Number of unpaired electrons $=5$
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