Answer to Question #88664 in General Chemistry for Edidiong Michael inyang

Question

What are the hybridization, bond angle and dipole moment of the following compounds.
1. IF5
2. XeF2
3.XeF4
4. SF4.

2.) In one sentence account for the differences in bond angle of the following isoelectronic species.
1 CH4 - 109.5
2. NH3 - 107
3. H2O - 105

Answer 1

1. IF₅

Hybridization: sp³d²

In IF₅, there are 5 bond pairs and 1 lone pairs giving a total of 6 electron pairs on central atom. This suggests for square bipyramidal geometry. However the presence of lone pair give the molecule a different shape. The lone pair is situated on axial position of the geometry and hence making the shape as square pyramidal. In this shape, all the bond angles are somewhat lower than 90 degrees due to the distortion caused by lone pair present on iodine.

IF₅ is polar

Dipole moment: yes

2. XeF₂

XeF₂ is a linear compound with F-Xe-F angle being 180 degree. Xe atom has sp³d hybridisation, while that of F is sp³.

XeF₂ is nonpolar

Dipole moment: no

3. XeF₄

Hybridization: sp³d²

XeF₄ is nonpolar

Dipole moment: no

4. SF₄

Hybridization: sp³d²

SF₄ bond angles are around 102 degrees in the equatorial plane and around 173 degrees between the axial and equatorial positions.

SF₄ is polar

Dipole moment: yes

As H₂O has two lone pairs so It repels the bond pairs much more and makes bond angle shorter of 104.5 degrees and as NH₃ has one lone pair that repels the three bond pair but not much effectively and strongly as two lone pairs of water repel the bond pair so the bond angle between hydrogen atom of ammonia is 107.5 greater than that of water. Similarly an other molecule have sp³ hybridization that is methane CH₄ molecule have no lone pair and each bond pair repels each other with equal force and bond angle between two adjacent hydrogen atoms becomes 109.5 degrees.