Question

Which of the following has the maximum number of unpaired electrons?
( Given Atomic number: $Sc=21,Fe=26,V=23,Ni=28)$

• A. $S{c}^{2+}$
• B. $F{e}^{3+}$
• C. $V^{4+}$
• D. $N{i}^{2+}$

Answer 1

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

The electronic configuration of:
$Sc$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^1$

$S{c}^{2+}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^1$

$Fe$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^6$

$F{e}^{3+}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$

$V$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^3$

$V^{4+}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^1$

$Ni$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^8$

$N{i}^{2+}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^8$

So, the maximum number of unpaired electrons is in $F{e}^{3+}$.