The hydrogen bonds in solid HF can be best represented as:
The correct answer is supposed to be (c), but I don't understand why this is so. Does the strength of hydrogen bonding depend on the $\ce{F-H\bond{...}F}$ or the $\ce{H-F\bond{...}H}$ bond angles?
At Wikipedia you find this structure (drawn by Benjah-bmm27 on Wikimedia Commons):
In the crystal, there is a hexagonal lattice of these chains, so the packing of the fluorine (green spheres) has an impact on the exact geometry of the hydrogen bonds.
So, the answer (c) is correct. Yomen Atassi correctly stated that in such a hydrogen bond the two electronegative partners and the hydrogen prefer a linear arrangement, as this maximizes the orbital overlap for the hydrogen bond. That the configuration (c) is preferred over (a) can basically be explained via the VSEPR theory: a fluoride ion in HF is surrounded by 3 electron pairs and 1 H–F bond: those 4 "ligands" should roughly be arranged tetrahedrally – approximately, not exactly, because electron pairs need more space than bonding electrons – around the F atom, and this leads to the zigzag chains from answer (c).
Generally hydrogen bonds $\ce{A-H\bond{...}B}$ can be taken to be approximately linear.
In fact, the highest bonding energy for the hydrogen bond is obtained when the two electronegative atoms (fluorine here) are approximately in line with the electropositive hydrogen atom in between ($\ce{O\bond{...}H-O}$ in water, $\ce{F\bond{...}H-F}$ in hydrogen fluoride). Deviations from linearity will reduce the bonding energy rapidly. It follows that the "best" hydrogen bond is "close" to linear.
The correct combination of straight and bent bond angles may be explained with the molecular orbitals description of hydrogen bonding.
In an isolated hydrogen fluoride molecule the valence orbitals combine to give an occupied bonding orbital (upper orbital in the figure below) and an empty antibonding orbital (lower orbital):
Note the positioning: the lower-energy bonding orbital is concentrated between the atoms where the bond forms and also around the more electronegative fluorine atom, while the higher-energy antibonding orbital is mostly in less stable regions outside the bond and around the hydrogen atom.
Now introduce a second hydrogen fluoride molecule. In this molecule a nonbonding electron pair on the fluorine can overlap the antibonding orbital on the first molecule, which is mostly around the hydrogen atom; this slightly weakens the bobd within the first molecule but creates a bond between the two of them. The net additional bonding energy is the net hydrogen bonding energy. Because the nonbonding orbitals on fluorine are not aligned with the bond but the antibonding orbital is so aligned, the best hydrogen-bond overlap occurs when the geometry at hydrogen is linear but the geometry at fluorine is bent:
Molecules other than hydrogen fluoride, such as water, also tend to have molecular orbitals aligned along the bonds between hydrogen atoms and electronegative atoms, so these too commonly form the strongest intermolecular hydrogen bonds if the hydrogen is linearly between the neighboring atoms.
