# Calculation of CFSE

I need to find CFSE for these:

• $$\ce{[Ti(H2O)6]^3+}.$$ $$\ce{Ti}$$ is $$\mathrm{(4s)^2(3d)^2}$$, $$\ce{Ti^3+}$$ is $$\mathrm{(4s)^0(3d)^1}.$$ Afterwards it becomes $$\mathrm{d^4sp^2}$$ or $$\mathrm{t_{2g}^5e_g^4}$$ so CFSE is $$\frac25\Delta_0$$. And I'm given $$\bar\nu_\mathrm{max} = \pu{20300 cm-1}.$$ So, we can use $$E=hc\bar\nu.$$ And that doesn't seem to come close to answer, I think some conceptual mistake has occurred.

• $$\ce{[CoF6]^3-}.$$ It's $$\mathrm{sp^3d^2}$$ or $$\mathrm{t_{2g}^4e_g^2}.$$ Here CFSE would be $$4\frac25\Delta - 2\frac35\Delta = \frac25\Delta = \pu{4 Dq}$$. This one's correct.

• What transition do you think the max wavenumber corresponds to ? Apr 19, 2015 at 10:57
• @J.LS think?? what do you mean, please elaborate. Apr 19, 2015 at 14:14
• Well, what sort of electronic transition would that wavenumber correspond to ? Apr 19, 2015 at 14:16
• @J.LS $e_g\leftarrow t_{2g}$ Apr 19, 2015 at 14:18
• That's correct; the only other band you would see would be LMCT at a higher wavenumber. So you know the energy of the delta parameter and thus the CFSE; be careful to work in SI units if you want to convert it. Apr 19, 2015 at 14:20

$$\bar\nu_\mathrm{max} = \pu{20300 cm^-1},$$ so $$\Delta = \pu{4.03E-19 J}$$ per molecule or $$\pu{242.7 kJ mol-1}.$$ $$\mathrm{CFSE} = \pu{97.1 kJ}.$$
$$\mathrm{CFSE} = \pu{6.6E-34 J s} × \pu{3E8 m s-1} × \pu{20300 cm-1} × \pu{6.02E23 mol-1} × \pu{E-3 J kJ-1} × \frac25 = \pu{97.1 kJ mol-1}$$