Which of the following complex ions has three unpaired electrons ?

Options

A. `[Cr(en)_(3)]^(3+)`
B. `[Mn(CN)_(6)]^(3-)`
C. `[Co(H_(2)O)_(6)]^(2+)`
D. Both (1) and (3)

Answer

D

Explanation

All three complexes are octahedral , so the crystal field orbital energy level diagram will show three lower energy and two higher energy d orbitals. F or `d^(1)-d^(3)` and `d^(8)-d^(10)` complexes, the electrons occupy the orbitals in accord with Hund's rule so as to give the maximun number of unpaired electrons. But for `d^(4)-d^(7)` complex, the orbital occupancy and number of unpaired electrons depend on the position of the ligand in the spectrochemical series. `Cr^(3+) ([Ar]3d^(3))` has three unpaired electrons. In the complex, they occupy the lower energy set of d orbitals:

`Mn^(3+) ([Ar]3d^(4))` can have a high spin or low spin configuration. Because `CN^(-)` is a strong field ligand, all four 3d electrons go into the lower energy `t_(2g)` orbitals. Consequently, the complex ion is low spin with two unpaired electrons.

`Co^(2+)([Ar]3d^(7)` has a high spin configuration `t_(2g)^(5) e_(g)^(2)` with three unpaired electrons because `H_(2)O` is a weak field ligand.