Question

If $M{n}^{2+}$ has 'X' number of unpaired electrons and 'Y' other positive oxidation states than +2. What is the sum of X + Y here?

Answer 1

Answer: 10

$M{n}^{2+}\to \left(Ar\right)3d^5$

unpaired $e^{-}\left(X\right)=5$

Mn can show +3, +4, +5, +6 and +7 oxidation states also other than +2.

Y = 5

X + Y = 5 + 5 = 10