Question

Which types of atomic orbitals of the central atom mix to form hybrid orbitals in (a) \(\mathrm{Cl}_{2} \mathrm{O} ;\) (b) \(\mathrm{BrCl}_{3} ;\) (c) \(\mathrm{PF}_{5}\); (d) \(\mathrm{SO}_{3}^{2-} ?\)

Short answer

(a) sp^3 (b) sp^3d (c) sp^3d (d) sp^3

Step-by-step solution

Determine the central atom and its steric number for \(\text{Cl}_{2}\text{O}\)

The central atom in \(\text{Cl}_{2}\text{O}\) is oxygen. Oxygen has two lone pairs and forms two bonds with chlorine atoms. The steric number is the total number of bonds and lone pairs, which is 4, indicating sp^3 hybridization.

Determine the central atom and its steric number for \(\text{BrCl}_{3}\)

The central atom in \(\text{BrCl}_{3}\) is bromine. Bromine forms three bonds with chlorine atoms and has two lone pairs. The steric number is 5, indicating sp^3d hybridization.

Determine the central atom and its steric number for \(\text{PF}_{5}\)

The central atom in \(\text{PF}_{5}\) is phosphorus. Phosphorus forms five bonds with fluorine atoms with no lone pairs. The steric number is 5, indicating sp^3d hybridization.

Determine the central atom and its steric number for \(\text{SO}_{3}^{2-}\)

The central atom in \(\text{SO}_{3}^{2-}\) is sulfur. Sulfur in this ion typically forms three bonds with oxygen atoms and has one lone pair. The steric number is 4, indicating sp^3 hybridization.

Key concepts

central atom determination

To solve problems related to hybridization, you first need to identify the central atom of the molecule. The central atom is typically the least electronegative element that connects to multiple other atoms. This determination plays a crucial role in understanding the molecule's shape and its hybridization. For example:

• In \(\text{Cl}_{2}\text{O}\), oxygen is the central atom because it forms bonds with both chlorine atoms.
• In \(\text{BrCl}_{3}\), bromine is the central atom forming bonds with three chlorine atoms.
• In \(\text{PF}_{5}\), phosphorus is the central atom bonded to five fluorine atoms.
• In \(\text{SO}_{3}^{2-}\), sulfur is the central atom forming bonds with three oxygen atoms.

Identifying the central atom simplifies the steps that follow, especially steric number calculation and deciding the type of hybridization.

steric number calculation

The steric number is a critical value used to determine the hybridization of the central atom. It is the sum of the number of atoms bonded to the central atom and the number of lone pairs of electrons on the central atom. For calculations:

• In \(\text{Cl}_{2}\text{O}\), the central atom is oxygen, which has 2 bonds and 2 lone pairs, making the steric number 4.
• In \(\text{BrCl}_{3}\), the central atom is bromine, with 3 bonds and 2 lone pairs, giving a steric number of 5.
• In \(\text{PF}_{5}\), phosphorus is the central atom with 5 bonds and 0 lone pairs, resulting in a steric number of 5.
• In \(\text{SO}_{3}^{2-}\), sulfur is the central atom having 3 bonds and 1 lone pair, leading to a steric number of 4.

The steric number helps in deciding the type of hybridization and ultimately the shape of the molecule.

sp3 hybridization

When the steric number is 4, the central atom undergoes sp^3 hybridization. This involves the mixing of one s orbital and three p orbitals to form four equivalent sp^3 hybrid orbitals. These orbitals arrange in a tetrahedral geometry to minimize repulsion:
In \(\text{Cl}_{2}\text{O}\), the central atom (oxygen) has a steric number of 4, leading to sp^3 hybridization. This results in a bent molecular shape due to the lone pairs.
Similarly, in \(\text{SO}_{3}^{2-}\), sulfur has a steric number of 4, suggesting sp^3 hybridization. Here, the sulfate ion has a trigonal pyramidal shape due to lone pairs.

sp3d hybridization

For molecules with a steric number of 5, the central atom undergoes sp^3d hybridization. This results from the combination of one s orbital, three p orbitals, and one d orbital to create five sp^3d hybrid orbitals. These hybrid orbitals adopt a trigonal bipyramidal geometry:
In \(\text{BrCl}_{3}\), the central atom bromine has a steric number of 5, indicating sp^3d hybridization. Due to the lone pairs, the geometry is T-shaped.
Additionally, in \(\text{PF}_{5}\), phosphorus has a steric number of 5, which also leads to sp^3d hybridization. Here, the molecular geometry remains trigonal bipyramidal with five bonding pairs.