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Step 1:
I'll provide detailed solutions for each part of the practice problems.

Since some of the questions involve conceptual explanations, I will format the LaTeX syntax where applicable.

Step 2:
(a) The individual bonds in $\mathrm{PCl}_{5}$ are polar due to the difference in electronegativity between phosphorus (P) and chlorine (Cl).

Experimentally, the $\mathrm{P-Cl}$ bond angle in $\mathrm{PCl}_{3}$ is about $100.3\degree$, which is a relatively small deviation from the ideal angle.
However, the molecule as a whole is nonpolar because the polar bonds cancel each other out. The vector sum of all the bond dipoles is zero, resulting in a nonpolar molecule.

Step 3:
(ii) Using VSEPR, we can determine the molecular geometry of the following molecules:

d) $\mathrm{IBr}_{3}$: sp$^3$d hybridization, T-shaped geometry (with two lone pairs)
(iii) Using Valence Bond Theory, we can determine the hybridization of the central atom and the geometry of the following molecules:

Step 4:
(a) The electron-domain geometry of $\mathrm{SO}_{3}{ }^{2 -}$ is trigonal planar, and the central sulfur atom is sp$^2$ hybridized.

(b) The electron-domain geometry of $\mathrm{SF}_{6}$ is octahedral, and the central sulfur atom is sp$^3$d$^2$ hybridized.

Step 5:
(a) $\mathrm{SF}_{6}$ has a larger bond angle compared to $\mathrm{H}_{2} \mathrm{~S}$.

However, $\mathrm{NH}_{3}$ has a smaller deviation from the ideal bond angle compared to $\mathrm{H}_{2} \mathrm{O}$.
However, due to lone pairs on the sulfur atom, the actual bond angle in $\mathrm{H}_{2} \mathrm{~S}$ is reduced to about $92\degree$. In both cases, the bond angles are reduced due to the presence of lone pairs on the central atom.

Final Answer

However, $\mathrm{NH}_{3}$ has a smaller deviation from the ideal bond angle compared to $\mathrm{H}_{2} \mathrm{O}$.