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

1 Answer
Dec 17, 2015

Answer:

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

Explanation:

#bb"NSF"#

The structure of #"NSF"# is

upload.wikimedia.org

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

cnx.org

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 °.

upload.wikimedia.org

#bb"IBr"_2^-#

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

i.ytimg.com

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

lessons.chemistnate.com

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

The bond angle is 180 °.