Question

Determine the electron geometry, molecular geometry, and idealized bond angles for each of the following molecules.

1. CF4
2. NF3
3. OF2
4. H2S

In which cases do you expect deviations from the idealized bond angle?

Answer 1

Concepts and reason

The concepts used in the problem are hybridization, electron geometry, molecular geometry and bond angle. By observing the hybridization, the shape and geometry of a molecule can easily be identified.

Fundamentals

Electron geometry of a molecule is the shape of the molecule including the lone pairs. Molecular geometry of a molecule is the shape of the molecule excluding the lone pairs. A bond angle is the angle between two bonds around an atom.

When lone pairs are present in a molecule, the bond angle is generally deviated from the ideal bond angle.

Part 1

In ${\rm{C}}{{\rm{F}}_{\rm{4}}}$ , hybridization is $s{p^3}$ . So, electron geometry is tetrahedral.

In ${\rm{C}}{{\rm{F}}_{\rm{4}}}$ , hybridization is $s{p^3}$ so molecular geometry that is tetrahedral.

In ${\rm{C}}{{\rm{F}}_{\rm{4}}}$ , hybridization is $s{p^3}$ and hence bond angle is ${\rm{109}}{\rm{.}}{{\rm{8}}^{\rm{o}}}$ .

In ${\rm{C}}{{\rm{F}}_{\rm{4}}}$ , there will be no distortion as no lone pairs are present. Hence, there are no deviations from the idealized bond angle.

Part 2

In ${\rm{N}}{{\rm{F}}_{\rm{3}}}$ , hybridization is $s{p^3}$ . There are three bonds and one lone pair. So, electron geometry including the lone pair is tetrahedral.

In ${\rm{N}}{{\rm{F}}_{\rm{3}}}$ , the geometry excluding the lone pairs, three bonds are left and their molecular geometry will be pyramidal.

In ${\rm{N}}{{\rm{F}}_3}$ , hybridization is $s{p^3}$ and hence bond angle is ${\rm{109}}{\rm{.}}{{\rm{8}}^{\rm{o}}}$ .

In ${\rm{N}}{{\rm{F}}_{\rm{3}}}$ , there will be distortion as one lone pair is present. Hence, there are deviations from the idealized bond angle.

Part 3

In ${\rm{O}}{{\rm{F}}_{\rm{2}}}$ , there are two lone pairs and two bonds and hence hybridization is $s{p^3}$ . So, electron geometry is tetrahedral.

In ${\rm{O}}{{\rm{F}}_{\rm{2}}}$ , there is two bonds are present so molecular geometry is bent.

In ${\rm{O}}{{\rm{F}}_{\rm{2}}}$ , hybridization is $s{p^3}$ and hence ideal bond angle is ${\rm{109}}{\rm{.}}{{\rm{8}}^{\rm{o}}}$ .

In ${\rm{O}}{{\rm{F}}_{\rm{2}}}$ , there will be distortion as two lone pairs are present. Hence, there are deviations from the idealized bond angle.

Part 4

In ${{\rm{H}}_2}{\rm{S}}$ , hybridization is $s{p^3}$ . So, electron geometry is tetrahedral.

In ${{\rm{H}}_2}{\rm{S}}$ , two bonds and two lone pairs are present, so molecular geometry of two bonds and one central atom is bent.

In ${{\rm{H}}_2}{\rm{S}}$ , hybridization is $s{p^3}$ and hence bond angle is ${\rm{109}}{\rm{.}}{{\rm{8}}^{\rm{o}}}$ .

In ${{\rm{H}}_2}{\rm{S}}$ , there will be distortion in molecular shape as two lone pairs are present. Hence, there are deviations from the idealized bond angle.

Ans:

The electron geometry is tetrahedral

The molecular geometry is tetrahedral

The bond angle is ${\rm{109}}{\rm{.}}{{\rm{8}}^{\rm{o}}}$

There will be no deviations from ideal bond angles.

The electron geometry is tetrahedral

The molecular geometry is pyramidal

The bond angle is ${\rm{109}}{\rm{.}}{{\rm{8}}^{\rm{o}}}$

Part 2

There will be deviations from ideal bond angles.

The electron geometry is tetrahedral

The molecular geometry is bent

The ideal bond angle is ${\rm{109}}{\rm{.}}{{\rm{8}}^{\rm{o}}}$

Part 3

There will be deviations from ideal bond angles.

The electron geometry is tetrahedral

The molecular geometry is bent

The ideal bond angle is ${\rm{109}}{\rm{.}}{{\rm{8}}^{\rm{o}}}$

Part 4

There will be deviations from ideal bond angles.