Question

CFSE of d⁶ high spin in Octahedral complex is?​

Answer 1

As we know that -

• $\mathsf{CFSE = 0.6e_{g} - 0.4t_{2g}}$

Here, we see that

In d⁶ electron pairing doesn't occur because it has high spin.

Then,

$\mathsf{e_{g} \: has \: 2 {e}^{ - } \: and \: t_{2g} \: has \: 4 {e}^{ - } }$

So,

CFSE = 0.6 × 2 - 0.4 × 4

= 1.2 - 1.6

= - 0.4

Hence,

CFSE of d⁶ high spin in Octahedral complex is $\mathsf{ - 0.4 \triangle_{o}}$