Explanation

(b)
In $\left\lbrack Mn\left( H_{2}O \right)_{6} \right\rbrack^{2 +}$, Mn is present as $Mn^{2 +}$or Mn (II), so its electronic configuration
=$1s^{2},\ 2s^{2}2p^{6},\ 3s^{2}3p^{6}3d^{5}$
3$d$ 4$s$ 4$p$
[table] $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | | | | | | [/table]
$4d$
[table] | | | | [/table]
In $\left\lbrack Mn\left( H_{2}O \right)_{6} \right\rbrack^{2 +}$, the coordination number of Mn is six, but in presence of weak field ligand, there will be no pairing of electrons in $3d$. So, it will form high spin complex due to presence of five unpaired electron.
In $\left\lbrack Mn\left( H_{2}O \right)_{6} \right\rbrack^{2 +}$
[image:9_image143.emf]