Among the following complexes, the one which shows zero CFSE is
(Atomic Nos : `Mn=25`, `Fe=26`, `Co=27`)

Options

A. `[Mn(H_2O)_6]^(3+)`
B. `[Fe(H_2O)_6]^(3+)`
C. `[Co(H_2O)_6]^(2+)`
D. `[Co(H_2O)_6]^(3+)`

Answer

B

Explanation

`H_2O` is a weak field ligand. Hence, `Detla_0lt ` pairign energy. `CFSE = (-0.4 x + 0.6y) Delta_0`
where x and y are number of electrons occupying `t_(2g) and e_(g)` orbitals repectively.
For `[Fe(H_2O)_6]^(3+)" complex ion, " Fe^(3+) (3d^5) = t_(2g)^3 e_(g)^2 " ":. CFSE = - 0.4 xx 3 + 0.6 xx 2 = 0 Dq`.