The electronic configuration (outermost) of Mn+2 ion.(atomic number of Mn = 25 ) in its ground state is

The electronic configuration (outermost) of ${Mn}^{+ 2} ion.$
(atomic number of $Mn = 25$ ) in its ground state is

A
$3 d^{5} 4 s^{0}$
B
$3 d^{4} 4 s^{1}$
C
$3 d^{3} 4 s^{2}$
D
$3 d^{2} 4 s^{2} 4 p^{2}$

Solution:

$_{25} Mn = [Ar] 3 d^{5} 4 s^{2} ⇄_{+ 2 electrons}^{- 2 electrons} {Mn}^{+ 2} = [Ar] 3 d^{5} 4 s^{0}]$

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