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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.

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  • $\begingroup$ What transition do you think the max wavenumber corresponds to ? $\endgroup$
    – J. LS
    Apr 19, 2015 at 10:57
  • $\begingroup$ @J.LS think?? what do you mean, please elaborate. $\endgroup$
    – RE60K
    Apr 19, 2015 at 14:14
  • $\begingroup$ Well, what sort of electronic transition would that wavenumber correspond to ? $\endgroup$
    – J. LS
    Apr 19, 2015 at 14:16
  • $\begingroup$ @J.LS $e_g\leftarrow t_{2g}$ $\endgroup$
    – RE60K
    Apr 19, 2015 at 14:18
  • $\begingroup$ 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. $\endgroup$
    – J. LS
    Apr 19, 2015 at 14:20

1 Answer 1

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Calculation for the first part:

$\bar\nu_\mathrm{max} = \pu{20300 cm^-1},$ so $\Delta = \pu{4.03E-19 J}$ per molecule or $\pu{242.7 kJ mol-1}.$ $\mathrm{CFSE} = \pu{97.1 kJ}.$

$$\mathrm{CFSE} = \pu{6.6E-34 J s} × \pu{3E8 m s-1} × \pu{20300 cm-1} × \pu{6.02E23 mol-1} × \pu{E-3 J kJ-1} × \frac25 = \pu{97.1 kJ mol-1}$$

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