Question

What Is ${\mathrm{C}}_2{\mathrm{H}}_2$ Hybridization?

Answer 1

The structure ${\mathrm{C}}_2{\mathrm{H}}_2$ is determined as$\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}$. Its IUPAC name is Ethyne and its common name is Acetylene.

The hybridization of ${\mathrm{C}}_2{\mathrm{H}}_2$can be explained by the following steps-

Step 1- Formation of sigma bond by the first Carbon atom

1. The electronic configuration of $\mathrm{C}$ is .$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^2$in ground state.
2. When it is excited one electron from $2\mathrm{s}$ the orbital jump to the $2{\mathrm{p}}_{\mathrm{z}}$ orbital and then the electronic configuration becomes$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{{\mathrm{p}}_{\mathrm{x}}}^{1}2{{\mathrm{p}}_{\mathrm{y}}}^{1}2{{\mathrm{p}}_{\mathrm{z}}}^1$.
3. The CH molecule has only one hydrogen atom therefore $2{\mathrm{s}}^1$ and $2{{\mathrm{p}}_{\mathrm{z}}}^1$ combines to form $\mathrm{sp}$ a hybridized atom.
4. Thus, there will be the formation of two sigma bonds between Carbon-Hydrogen $\mathrm{C}-\mathrm{H}$ and Carbon-Carbon$\mathrm{C}-\mathrm{C}$.

Step 2- Formation of sigma bond by the second Carbon atom

1. Similarly, the second Carbon also forms a sigma bond between $\mathrm{C}-\mathrm{H}$and $\mathrm{C}-\mathrm{C}$by $\mathrm{sp}$ hybridization.
2. Thus, four $\mathrm{sp}$hybrid orbitals are formed during the formation of${\mathrm{C}}_2{\mathrm{H}}_2$.

Step 3- $\mathrm{\pi }$ Bonding in hybrid orbitals.

1. In this, the Carbon atom will have two half-filled $2\mathrm{p}$ orbitals. These two pairs of $\mathrm{p}$ orbitals do not participate in the hybridization and instead form two $\mathrm{\pi }$ bonds resulting in the formation of a triple bond.

Thus the hybridization $\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}$ is $\mathrm{sp}$ hybridization.