Question

Which of the following has the maximum number of unpaired electrons ? $(AtomicnumberofFe=26)$

• A. Fe
• B. $F{e}^{2+}$
• C. $F{e}^{3+}$
• D. $F{e}^{4+}$

Answer 1

The correct option is C $F{e}^{3+}$

Ground state electronic configuration of $Fe=\left[Ar\right]3d^{6}4s^2$
So, Fe has 4 unpaired electrons.
$F{e}^{2+}=\left[Ar\right]3d^{6}4s^0$
So, $F{e}^{2+}$ has 4 unpaired electrons.
$F{e}^{3+}=\left[Ar\right]3d^{5}4s^0$
So, $F{e}^{3+}$ has 5 unpaired electrons.
$F{e}^{4+}=\left[Ar\right]3d^{4}4s^0$
So, $F{e}^{4+}$ has 4 unpaired electrons.