Question

Write shape, hybridization and no of lone pair of electrons for following: ${\mathrm{PCl}}_5$

Answer 1

Lone pair of ${\mathbf{PCl}}_{\mathbf{5}}$ molecule

• The atomic number of Phosphorus $\left(\mathrm{P}\right)$ is $15$.
• The electronic configuration of the Phosphorus atom is as follows:

$\mathrm{P}=1s^{2}2s^{2}2p^{6}3s^{2}3p^3$

• The number of valence electrons of Phosphorous is $5$.
• Phosphorus forms five sigma bonds with five Chlorine $\left(\mathrm{Cl}\right)$ atoms to form a Phosphorus pentachloride $\left({\mathrm{PCl}}_5\right)$ molecule.
• ${\mathrm{PCl}}_5$ molecule has no lone pair on the Phosphorous atom.

Hybridization of ${\mathbf{PCl}}_{\mathbf{5}}$ molecule

• In ${\mathrm{PCl}}_5$, Phosphorous is a center atom that forms five sigma bonds with five Chlorine atoms.
• So it requires five hybrid orbitals.
• It will form an $s{p}^{3}d$ hybridized orbital.

Shape of ${\mathbf{PCl}}_{\mathbf{5}}$ molecule

• According to VSEPR theory, the order of repulsion between electron pairs is as follows:
• lone pair-lone pair $>$ lone pair-bond pair $>$ bond pair-bond pair repulsions
• It has trigonal bipyramidal geometry and shape due to no lone pair of electrons on central atom phosphorus.
• Structure of ${\mathrm{PCl}}_5$: