Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
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):
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.
