Considering `H_2O` as a weak field ligand, the number of unpaired electrons in `[Mn(H_2O)_6]^(2+)` will be (At. no. of `Mn=25`)

Options

A. 3
B. 5
C. 2
D. 4

Answer

B

Explanation

In `[Mn(H_(2)O)_(6)]^(2+)`, Mn is present as `Mn^(2+)` or Mn (II)m so its electronic configuration
`=1s^(2),2s^(2)2p^(6),3s^(2)3p^(6),3d^(5)`

In `[Mn(H_(2)O)_(6)]^(2+)` the coordination number of Mn is six, and 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 electrons.