Hybridization of TeCl4?
Could you please explain the process for determining the hybridization of TeCl4? I would appreciate it if you could provide a clear answer so I can understand what I'm working towards. Thank you!
1 Answers
Tellurium is in group 6, so it has six valence electrons. Chlorine is in group 7 so it has 7 valence electrons. That means in TeCl4 there are 6 + 7*4 = 34 valence electrons. Drawing the electron dot diagram for the molecule, we lose 2*4 electrons for the bonds to tellurium, so there are 34-8 = 26 electrons left to distribute. Each of four chlorine atoms needs 8 valence electrons in its outer shell, but we already accounted for 2 in the bond pairs. So 24 electrons are distributed to the chlorine atoms, leaving 26-24 = 2 electrons. Since TeCl4 has four bond pairs and one unbounded pair, its geometry is based off of the trigonal bipyramidal structure. But since there are only four bond pairs, the molecule takes a see-saw shape and the unbonded electrons take the place of a bonded element. For trigonal bipyramidal structures, the hybridization is sp^3 d. Please if you are serious see the link for an image Fig. 9.23 and additional descriptions.
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