Question

Explain the nature of the covalent bond using the bond formation in ${\mathrm{CH}}_3\mathrm{Cl}$

Answer 1

Nature of the covalent bond in ${\mathbf{CH}}_{\mathbf{3}}\mathbf{Cl}$:

1. In ${\mathrm{CH}}_3\mathrm{Cl}$, each Hydrogen atom has one valence electron and it requires one electron to complete its duplet ( since $\mathrm{H}$ is a $\mathrm{s}$-block element it need maximum of two electrons in the outermost shell to attain stability ).
2. Chlorine also has seven valence electrons and it requires one electron to complete the octet ( the tendency of atoms to have eight electrons in the valence shell to attain stability ).
3. Carbon forms four single covalent bonds by sharing one electron with three hydrogen atoms( that is, three $\mathrm{C}-\mathrm{H}$ bonds) and one electron with a Chlorine atom ( that is, one$\mathrm{C}-\mathrm{Cl}$ bond ).
4. Chlorine ($\mathrm{Cl}$) has comparatively huger size than carbon ($\mathrm{C}$) and needs just one electron to fill its valence shell and be stable, so $\mathrm{Cl}$ is more elctronegative than $\mathrm{C}$.
5. As the difference in electronegativity between chlorine and carbon is huge (since chlorine is more electronegative than carbon), the $\mathrm{C}-\mathrm{Cl}$ bond is polar in nature.

Therefore in ${\mathrm{CH}}_3\mathrm{Cl}$, Carbon forms four covalent bonds by sharing each of the four electrons with each of the three hydrogen atoms and one chloride atom. Since chlorine is more electronegative than carbon the $\mathrm{C}-\mathrm{Cl}$ bond is polar in nature.