Question

Explain the hybridization of the central atom in ${\mathrm{PCl}}_3$.

Answer 1

Hybridization:

• It is defined as the formation of a hybridized orbital by mixing two atomic orbitals.
• Electronic configuration of $\mathrm{P}$: $1{\mathrm{s}}^2,2{\mathrm{s}}^2,2{\mathrm{p}}^6,3{\mathrm{s}}^2,3{\mathrm{p}}^3$
• The valence shell electrons of $\mathrm{P}$:
• The chlorine atom gets attached to the unpaired electrons of 3s and 3p orbitals. i.e. $3{\mathrm{s}}^2,3{\mathrm{p}}_{\mathrm{x}}^1,3{\mathrm{p}}_{\mathrm{y}}^1,3{\mathrm{p}}_{\mathrm{z}}^1$.
• Thus the hybridization of $\mathrm{P}$in ${\mathrm{PCl}}_3$ comes to ${\mathrm{sp}}^3$.
• So, the shape of the molecule will be Trigonal pyramidal.
• Generally, the bond angle in ${\mathrm{sp}}^3$ hybridization is around $109°$ but in the case of ${\mathrm{PCl}}_3$ the bond angle is approx. $103°$ because of the repulsion between bond and a lone pair on $\mathrm{P}$.
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