# How to rationalise the difference in halogen bonding?

Can anyone explain the difference in halogen bonding to me. I understand the explanation in (c). It appears to me, that accepting that would contradict the answers to parts (a), (b), and (c).
Is the reason for part (a) because the first molecule does not have resonance and for (b) the $\ce{F3C-I}$ molecule has a larger inductive effect? If so, that would not make sense for (c) to be correct.

In (b) you compare $\ce{H3C-I}$ and $\ce{F3C-I}$. Fluorine is more electronegative than hydrogen, and the statement $$\angle(\ce{H-C-H})>\angle(\ce{F-C-F})$$ holds. The carbon directs mor $\ce{p}$ orbital character into the carbon fluorine bonds than in the carbon hydrogen bonds. As a consequence the $\ce{s}$ orbital character in the carbon iodine bond (from the carbon) is higher in $\ce{F3C-I}$ compared to $\ce{H3C-I}$. That causes the bond to be slightly shorter and having more electron density at the carbon. This is also making the iodine a little more electron deficient and a better electron acceptor, i.e. a stronger acid (see also Bent's Rule). The electronic interaction will therefore also be slightly stronger.
This observed effect is also often called inductive effect, which like you correctly stated is larger in $\ce{F3C-I}$.
In (c) you are comparing the Lewis bases $\ce{NMe3}$ and $\ce{OMe2}$ acting towards $\ce{F3C-I}$ (or $\ce{HC3-I}$, the reasoning would remain the same). Oxygen is more electronegative than nitrogen, hence the lone pair orbitals of oxygen are more compact, i.e. the maximum electron density is closer to the oxygen nucleus. This causes dimethylether a weaker electron donor. Hence the the oxygen iodine interaction in $\ce{F3C-I\bond{~}OMe2}$ can be expected to be weaker than the corresponding nitrogen iodine interaction in $\ce{F3C-I\bond{~}NMe3}$. (Just like it is stated in your sheet.)
Last a word of caution. Oxygen (in neutral molecules) can in most cases not be described as $\ce{sp^{3}}$ hybridised. Symmetry considerations and it's electronegativity are usually the reason for this. It is better described as $\ce{sp}$ hybridised in terminal position (like in ketones), or $\ce{sp^{2}}$ hybridised in bridging positions (like in ethers).