Question

What is the Hybridization of ${\mathrm{HgCl}}_2$?

Answer 1

Hybridization:-

• The process of combining the atomic orbitals to produce new orbitals (hybrid orbitals) of varying energies and shapes than the initial orbitals, is called hybridization.

VSEPR theory:-

• VSEPR stands for, Valence Shell Electron Pair Repulsion theory.
• In 1940, Sidgwick and Powell proposed this theory on the basis of the repulsion interaction of the electron pairs in the valence shell of the atoms.
• Basically from this theory we can calculate the number of electrons pairs present in a molecule with the help of this formula : $\frac{1}{2}\times [V+N-C+A]$
• Here, V= the number of valence electrons present in the central atom, N= total number of monovalent atoms, C= cationic charge, and A= anionic charge.

Hybridisation of Mercury (II) chloride using VSEPR theory:-

• In Mercury (II) chloride ${\mathrm{HgCl}}_2$, there are 2 Chlorine atoms that are linked with 1 Mercury atom.
• Electronic configuration of Mercury is$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^{6}5{\mathrm{s}}^{2}4{\mathrm{d}}^{10}5{\mathrm{p}}^{6}6{\mathrm{s}}^{2}4{\mathrm{f}}^{14}5{\mathrm{d}}^{10}$, hence it has 2 valence electrons in its outermost shell.
• Here Chlorine is the monovalent atom as its valency is 1.
• So according to the formula, for calculating electron pair, the values of V= 2 ($\because$Mercury has 2 valence electrons), N= 2($\because$as there are 2 Chlorine monovalent atom is present), and C= 0 and A=0 ($\because$No cations or anions are present in Mercury (II) chloride, hence charge is 0).
• Putting the values in the formula we get the number of electron pair

$$\begin{gathered} =\frac{1}{2}\times [V+N-C+A] \\ =\frac{1}{2}\times [2+2-0+0] \\ =\frac{1}{2}\times 4 \\ =2 \end{gathered}$$

• Here total of 2 electron pairs are present, and both of them are bond pairs $(\mathrm{Hg}-\mathrm{Cl})$
• Now since there are 2 electron pairs present, the hybridization will be $\mathrm{sp}$, so its shape will be linear:

Therefore, the hybridisation of ${\mathrm{HgCl}}_2$ is $\mathrm{sp}$.