Question

What are the bond angles in the central atom of the following: `"NSF"`, `"OF"_2`, and `"IBr"_2^−"`?

Answer 1

The bond angles are slightly less than 120°, slightly less than 109.5 °, and 180 °, respectively.

Explanation:

`bb"NSF"`

The structure of `"NSF"` is

There is a lone pair on the `"N"` atom, a lone pair on the `"S"` atom, and there are three lone pairs on the `"F"` atom.

This is an `"AX"_2"E"` molecule, so the electron geometry is trigonal planar and the molecular shape is bent.

The theoretical bond angle is 120 °, but repulsion by the lone pairs decreases the bond angle to about 117 °.

`bb"OF"_2`

The Lewis structure of `"OF"_2` is

This is an `"AX"_2"E"_2` molecule, so the electron geometry is tetrahedral and the molecular shape is bent.

The theoretical bond angle is 109.5 °, but repulsions by the lone pairs decrease the bond angle to about 103 °.

`bb"IBr"_2^-`

The Lewis structure of `"IBr"_2^-` is

This is an `"AX"_2"E"_3` structure, so the electron geometry is trigonal bipyramidal.

The lone pairs occupy the equatorial positions, with the `"Br"` atoms in the axial locations.

The bond angle is 180 °.