#
VSEPR THEORY - BOND ANGLES - NSF_{3} SiF_{4} POF_{3}

# 3) The correct order of ∠FMF bond angles (where
M is the central atom) in NSF_{3}, SiF_{4} and POF_{3}
is:

1) NSF_{3} > SiF_{4} > POF_{3}

2) SiF_{4} > NSF_{3} > POF_{3}

3) NSF_{3} > POF_{3}
> SiF_{4}

4) SiF_{4} > POF_{3} > NSF_{3}

**Logic:**

* First write the Lewis dot structures for the molecules and find the number
and types of electrons pairs; magnitude of repulsions between them to arrive at
the relative bond angles.

It is known that, from VSEPR theory, the repulsion increases with
increase in the volume occupied by electron pair(s). Since the electron density
in triple bond occupies more space, it exerts more repulsion than that of double
bond and which turn exerts more repulsion than that of single bond.

**Solution:**

In NSF_{3}, there is a triple bond between N and S. Hence the
repulsions will be more on S-F bond pairs. This will reduce the bond angle more
than in other cases.

The bond angle is least affected in case of SiF_{4}, since all the Si-F
bonds are single bonds, which exert less repulsion on other bond pairs. Hence
the bond angle is maximum i.e. 109^{o}28'

In POF_{3}, there is a double bond between P and O, which also causes
more repulsion than single bond, but less than the triple bond. Hence the bond
angle will be medium in this case.

### Homework:

1) Write the complete Lewis dot structures of above molecules indicating
electrons pairs on atoms connected to central atom.

2) What will be the bond angle in PF_{3} less than in case of OPF_{3}
or greater than it? Hint: There is now lone pair on P.