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I want to check if their hybridizations are as follows:

$\mathrm{sp^3d^2}$ in $\ce{[Ti(NH3)6]^3+}$ and $\mathrm{d^2sp^3}$ in $\ce{[Ti(H2O)6]^3+}$.

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    $\begingroup$ What exactly is the difference between $d^2sp^3$ and $sp^3d^2$? $\endgroup$ – bon Dec 30 '15 at 15:23
  • $\begingroup$ In d2sp3 3d is involved while in sp3d2 4d is involved in case of Ti. $\endgroup$ – BEWARB Dec 30 '15 at 15:23
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    $\begingroup$ Do you REALLY need to use hybridization here? $\endgroup$ – permeakra Dec 30 '15 at 16:35
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Immediately stop using the hybridisation concept for transition metals. Don’t even take it a step further. It is not helpful here at all.

Both complexes have essentially identical electronic structures except that the relative orbital energies differ slightly. The following image — originally taken from Professor Klüfers’ internet scriptum for his coordination chemistry 1 course (LMU Munich; in German) — shows a typical octahedral coordination compound and only includes the ligand–metal σ interactions.

orbital scheme octahedral complex

You can see the metal’s orbitals on the left-hand side; from bottom to top they are 3d, 4s, 4p. 4d would be even higher in energy, possibly even higher up than 5s. These virtual energies can be really hard to calculate or measure accurately. On the right-hand side, you can see the six donor orbitals which transform as $\mathrm{a_{1g} + e_g + t_{1u}}$. (For some reason, the u is missing.) The ligands’ p-orbitals that can take part in π bonding would transform as $\mathrm{t_{1g} + t_{1u} + t_{2g} + t_{2u}}$, and would therefore form another bonding-antibonding interaction with the metal’s $\mathrm{t_{2g}}$ orbitals. The ligand orbitals would again be bonding and the metal’s antibonding. If you check out the link and look for the image you’re seeing, you can also see pictures of the metal orbitals, and further down the scheme that includes the π interactions (only the relevant part highlighted).

The metal orbitals that interact with ligand orbitals in a σ fashion are two 3d, 4s and three 4p. The remaining three 3d orbitals interact in a π fashion. So if you really, really, really must use hybridisation in this case (did I say it is not recommended?) then you should report titantium’s hybridisation as $\mathrm{d^2sp^3}$ and not the other way round: 4d orbitals do not play any role.

The only difference between the hexaaquacomplex and the hexaamminecomplex is that the former has lower-lying ligand orbitals than the latter so the energy difference between $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$ will be slightly different. That’s all.

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It should theoretically be the other way round. $\ce{NH3}$ is a strong field ligand as it is higher in the spectrochemical series than $\ce{H2O}$. Hence, $\ce{NH3}$ can cause force-pairing of electrons in inner shell ($\rm 3d$ orbitals) while $\ce{H2O}$ cannot, as it is a weak field ligand. This utilizes concepts from Crystal Field Theory.

Both $\ce{[Ti(H2O)6]^3+}$ and $\ce{[Ti(NH3)6]^3+}$ have $\rm d^2sp^3$ hybridization.

Thus, purely theoretically and naively, $\ce{[Ti(H2O)6]^3+}$ should involve the $\rm 4d$ subshell for accepting electrons donated from ligand molecules and have $\rm d^2sp^3$ hybridization while $\ce{[Ti(NH3)6]^3+}$ should involve the $\rm 3d$ subshell for accepting electrons donated from ligand molecules and have $\rm sp^3d^2$ hybridization.

The above is merely a theoretical and naive discussion. In fact, the $\ce{Ti^3+}$ has only 1 electron in its $\rm 3d$ subshell, and force-pairing of electrons does not occur in it. We find that in fact it has $\rm d^2sp^3$ hybridization in both cases, as it has vacant $d$ orbitals in the inner $\rm 3d$ subshell which can accept electronic pairs donated from the ligands. Here, the higher (in octahedral geometry) $\rm e_g$ subshell takes part in forming the $\rm d^2sp^3$ hybrid orbital, leaving the lower $\rm t_{2g}$ subshell as unhybridized. This happens for both $\ce{[Ti(H2O)6]^3+}$ and $\ce{[Ti(NH3)6]^3+}$

The take-home lesson here is that force-pairing of electrons only affects the overall hybridization if the penultimate $\rm d$ subshell has more than 3 or less than 8 electrons.

Source: Basic Concepts of Inorganic Chemistry.

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  • $\begingroup$ I think your answer is exactly opposite to what i wrote, and the logic also seems correct. But according to you [Ti(H2O)6]3+ must have hybridisation sp3d2 (4d). click already registered . $\endgroup$ – BEWARB Dec 30 '15 at 17:44
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    $\begingroup$ Really? 4d subshell for titanium, a 3d metal? $-1$. Would minus more if I could. $\endgroup$ – Jan Dec 30 '15 at 21:42
  • $\begingroup$ @Jan, did you bother reading the 3rd paragraph? Also, there are 3d metals from iron onwards for sure which involve the 4d orbitals in complex formation, especially zinc. $\endgroup$ – Tamoghna Chowdhury Dec 31 '15 at 2:53
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    $\begingroup$ The hybridisation concept is more than wrong in this case. I support @Jan's comment. $\endgroup$ – Martin - マーチン Dec 31 '15 at 8:00
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    $\begingroup$ @Jan well at least they don't in the SE app on Android. Well, that's your decision to keep your voting, I won't argue with you. You're welcome for your feedback, which does help expand my concept of coordination compounds (I'm no expert, just a high school student). I won't be deleting this answer, but I'll restrict myself to questions more in my field in the future. $\endgroup$ – Tamoghna Chowdhury Dec 31 '15 at 17:23

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