Question

starting from Lewis structure,determine the hybridization types of the central atom of TeCl4 and ICl4-.

Answer 1

Tecl4

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

ICl4---see attachment

Answer 2

Answer : The hybridization $TeCl_4$ and $ICl_4^-$ are, $sp^3d$ and $sp^3d^2$

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

(1) The given molecule is, $TeCl_4$

As we know that tellurium has '6' valence electrons and chlorine has '7' valence electron.

Therefore, the total number of valence electrons in $TeCl_4$ = 6 + 4(7) = 34

In $TeCl_4$, one-one electron of 4 chlorine combine with 4 electrons of tellurium and 2 pair electrons of tellurium left as a lone pair. That means, it has 4 bonding pairs and 1 lone pairs. So, the hybridization will be, $sp^3d$

(2) The given molecule is, $ICl_4^-$

As we know that iodine has '7' valence electrons and chlorine has '7' valence electron.

Therefore, the total number of valence electrons in $ICl_4^-$ = 7 + 4(7) + 1 = 36

In $ICl_4^-$, one-one electron of 4 chlorine combine with 4 electrons of iodine and 4 pair electrons of iodine left as a lone pair. That means, it has 4 bonding pairs and 2 lone pairs. So, the hybridization will be, $sp^3d^2$

The Lewis-dot structure of $TeCl_4$ and $ICl_4^-$ are shown below.