Question

Based on the bond angles in CH4, NH3, and H2O. rank the magnitude of these repulsions. Rank from strongest to weakest repulsion. To rank items as equivalent, overlap them.

Answer 1

Concepts and reason

The question is based on the magnitude of repulsion between the bond pairs and lone pairs present in the given molecules.

All the given molecules exhibit same hybridization. But, it is not necessary for all the compounds having same hybridization to have same bond angles. Bond angles are dependent on the repulsions between the bond pairs and lone pairs.

Fundamentals

Hybridization is defined as mixing of atomic orbitals having comparable energy to form hybrid orbital having lower energy than the former.

Hybridization of the central atom of any compound can be determined using VSEPR theory.

Number of electron pairs are the total number of bond pairs and lone pairs in a given compound. According to the VSEPR theory, number of electron pairs give the hybridization as follows:

In a molecule, three types of repulsions take place depending on the presence of number of bond pairs and lone pairs. They are:

• Bond pair – Bond pair: It occurs between two bonds.

• Bond pair – Lone pair: It occurs between a lone pair and a bond.

• Lone pair – Lone pair: It occurs between two lone pairs.

All the given molecules, , , and are hybridized according to VSEPR theory. Therefore, their expected geometry should be tetrahedral having bond angles as .

But, the observed bond angles and their shapes are as follows:

In molecule, only bond pair -bond pair repulsion will take place.

However, in two repulsive forces will act that is bond pair-bond pair and lone pair- bond pair.

In case of , three repulsions – bond pair – bond pair, bond pair – lone pair and lone pair – lone pair, will occur due to the presence of two lone pairs.

Ans:

The magnitude of repulsion from strongest to weakest is as follows: