Q. Calculate the number of unpaired electrons in $\left[\right.Mn\left(\right.H_{2}O\left(\left.\right)_{6}\left(\left]\right.\right)^{2 +}$ , Considering $H_{2}O$ as a weak field ligand $\left(\right.At.No.of \, Mn=25$

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A

Two

B

Three

C

Four

D

Five

Solution:

$Mn^{25} \rightarrow 3d^{5}4s^{2}$
$Mn^{2 +} \rightarrow 3d^{5}$
In presence of weak ligand field, there will be no pairing of electrons. So, it will form a high spin complex. i.e. the number of unpaired electrons = 5.

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