CFSE of high-spin $d^{5} - M n^{2 +}$ complex is__________.

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Solution

CFSE of high-spin $d^{5} - M n^{2 +}$ complex is 0.
Octahedral complex splitting takes place in two orbitals named as
$e_{g} (d o u b l y d e g e n e r a t e)$
$t_{2 g} (t r i p l y d e g e n e r a t e)$

$d^{5} b e i n g a h i g h s p i n c o m p l e x h e n c e t h e e l e c t r o n s e n t e r s i n t o e_{g} o r b i t a l$

Hence the energy of
$e_{g} o r b i t a l i s - 0.4 Δ a n d t_{2 g} o r b i t a l i s + 0.6 Δ$
$H e n c e t h e 3 e l e c t r o n s o f t_{2 g} b e c o m e s 3 \times (- 0.4) Δ = - 1.2 Δ$
$F o r t h e 2 e l e c t r o n s o f e_{g} o r b i t a l i t b e c o m e s 2 \times (+ 0.6) = 1.2 Δ$
$H e n c e d i f f e r e n c e i n e n e r g y = 1.2 Δ - 1.2 Δ = 0$

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