Identify a molecule having highest number of lone pair of electrons in valence shell of central atom.

The number of lone pairs on the central atom is determined by applying VSEPR theory to each molecule's Lewis structure. For a central atom with valence electrons $V$ and bonding pairs $b$, lone pairs are obtained from $\frac{V - b}{2}$.

For $\mathrm{NH}_3$, nitrogen has $V=5$ and $b=3$, yielding $1$ lone pair.

For $\mathrm{SF}_4$, sulfur has $V=6$ and $b=4$, yielding $1$ lone pair.

For $\mathrm{ICl}_3$, iodine has $V=7$ and $b=3$, yielding $2$ lone pairs.

For $\mathrm{PCl}_3$, phosphorus has $V=5$ and $b=3$, yielding $1$ lone pair.

$\mathrm{ICl}_3$ has the highest number of lone pairs.

Final Answer: $\boxed{\text{C}}$