Question

Starting from Lewis structure, determine the hybridization type of the central atom of Tecl2 and ICI4-

Answer 1

Hybridization type of Tecl2 = sp3d - since it has 6 valence electrons.


Hybridization type of ICI4- = sp3d2 - since  it has 8 valence electrons.

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.