Question

Calculate $|C.F.S.E|$ (mod value) is term of Dp. For complex ion $|Mn{F}_6{|}^{3-}$

Answer 1

$C.F.S.E.value=-(3\times 0.4{\Delta }_0)+(1\times 0.6{\Delta }_0)=-0.6{\Delta }_0=-6Dq$

$|CFSE|=6Dq$.