Question

Which of the following has maximum number of unpaired $d-$elctrons?

• A. $Zn$
• B. $F{e}^{+2}$
• C. $N{i}^{+3}$
• D. $C{u}^{+1}$

Answer 1

Answer: (B)

$F{e}^{+2}$

The electronic configuration of the given elements or ions are given below.

$Zn⟶4s^{2}3d^{10}$

$F{e}^{2+}⟶4s^{0}3d^6$

$N{i}^{3+}⟶4s^{0}3d^6$

$C{u}^{+}⟶4s^{0}3d^{10}$

$F{e}^{2+}$ has 4 unpaired electron and $N{i}^{3+}$ has $3$ unpaired $e^{-}s$. $Zn$ and $C{u}^{+1}$ have no unpaired electron.

Therefore, $F{e}^{+2}$ has the maximum no. of unpaired electrons.

Hence, the correct option is B.