For an octahedral complex, which of the following d electron configuration will give maximum crystal-field stabilisation energy?

Options

A. high spin, `d^(6)`
B. low spin `d^(4)`
C. low spin `d^(5)`
D. high spin` d^(7)`

Answer

B

Explanation

`Delta_(0) =(-0.4 xx nt_(2g)) + ( 0.6 xx ne_(g))`
where, n = number of electrons in `nt_(2g)` and `ne_(g)` CFSE of given d-electrons configuration are as follows
a. For high spin, `d^(6)`
`Delta_(0)(CFSE)= - 0.4 xx 4 + 0.6 xx 2 = - 0.4 `
b . For low spin `d^(4)` ,
`Delta_(0) (CRSE ) = -0.4 xx 4 + 0.6 xx0 =1.6 `
(c ) For low spin `d^(2)`
`Delta _(0) ( CFSE) =- 0.4 xx 5 + 0.6 xx 0 = - 2.0`
(d ) For high spin `d^(7)`
`Delta (CFSE) = - -0.4 xx 5+ 0.6 xx 2 = - 0.8`
Neglecting -ve sign (as it indicates that energy is released.
Hence, maximum `Delta_(0)` (CFSE) value is shown by low spin `d^(5)` configuration.