Question

Using VSEPR theory, draw the shape of ${\mathrm{PCl}}_5$ and ${\mathrm{BrF}}_5$ molecules. Write the state of hybridization of the central atoms.

Answer 1

VSEPR theory

• During bond formation, an atom's atomic orbitals are mixed to make equivalent orbitals.
• Hybridization refers to the mixing of orbitals.
• According to the VSEPR theory, the form of the molecule is determined by the arrangement of these hybrid orbitals.

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

• The atomic number of phosphorus is $15$.
• The electronic configuration of phosphorus is as follows:

$\mathrm{P}:\left[\mathrm{Ne}\right]3s^{2}3p^3$

• One electron is excited to the $3d$ orbital.
• The excited-state configuration is as follows:

$\mathrm{P}:\left[\mathrm{Ne}\right]3s^{1}3p^{3}3d^1$

• It now has five hybrid orbitals for bonding and no additional electrons.
• According to the VSEPR theory, three bonds are arranged in an equatorial configuration, with the remaining two bonds arranged axially perpendicular to the equatorial bonds.
• ${\mathrm{PCl}}_5$ has a trigonal bipyramidal shape.

Shape of ${\mathbf{BrF}}_{\mathbf{5}}$

• Bromine belongs to group $17$.
• The electronic configuration of bromine is as follows:

$\mathrm{Br}:\left[\mathrm{Ar}\right]4s^{2}3d^{10}4p^5$

• Around the central bromine atom are six electron pairs, five of which are bond pairs and one of which is a lone pair.
• According to VSEPR theory, the ${\mathrm{BrF}}_5$ molecule is octahedral in geometry but square pyramidal in shape.

Hybridization

• In ${\mathrm{PCl}}_5$, the central atom phosphorus is ${\mathrm{sp}}^3\mathrm{d}$ hybridized.
• In ${\mathrm{BrF}}_5$, the central atom bromine is ${\mathrm{sp}}^3{\mathrm{d}}^2$ hybridized.