Question

Starting from Lewis structure, determine the hybridization types of the central atom of ${\mathrm{TeCl}}_4$ and ${{\mathrm{ICl}}_4}^{-}$.

Answer 1

Lewis dot structure:

• Lewis structures are diagrams that depict the bonding between atoms in a molecule as well as any lone pairs of electrons that may occur.
• We can determine the hybridization using the central atom's valence electron configuration and the number of electron pairs.
• By placing the total number of electrons around the central atom in the hybrid orbitals and describing the bonding.

Hybridization of ${\mathrm{TeCl}}_4$:

• ${\mathrm{TeCl}}_4$ has six valence electron on Tellurium ($\mathrm{Te}$) and Chlorine ($\mathrm{Cl}$) has seven valence electron.
• Total valence electrons in ${\mathrm{TeCl}}_4$ $=6+4\mathrm{x}7=34$
• In ${\mathrm{TeCl}}_4$, one-one electrons of four Chlorine unite with four electrons of Tellurium, leaving two Tellurium pair electrons as a lone pair. That is, it has four bonding pairs and one lone pair.
• So, the hybridization type is ${\mathrm{sp}}^3\mathrm{d}$.

Hybridization of ${{\mathrm{ICl}}_4}^{-}$:

• Iodine has seven valence electrons, and Chlorine also has seven valence electrons.
• As a result, the total number of valence electrons in ${{\mathrm{ICl}}_4}^{-}$ is $=7+4\left(7\right)+1=36$.
• In, four Chlorine electrons combine with four Iodine electrons, leaving four Iodine pair electrons as a lone pair.
• That is, it has four bonding pairs and two lone pairings.
• As a result, the hybridization will be ${\mathrm{sp}}^3{\mathrm{d}}^2$ with square planar geometry.

So, the hybridization of the central atom of ${\mathrm{TeCl}}_4$ and ${{\mathrm{ICl}}_4}^{-}$ is ${\mathrm{sp}}^3\mathrm{d}$ and ${\mathrm{sp}}^3{\mathrm{d}}^2$.