explain hydrogen bonding in HF molecules
Answers
Answer:
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Answer:
here's the answer.
Explanation:
a) From HCl to HBr to HI, the halogen atom is getting bigger with more electrons. That increases
the size of the temporary fluctuating dipoles which cause van der Waals dispersion forces. That
makes the attractions between neighbouring molecules stronger, and so more heat is needed to
separate them.
(Note: If you are really wide-awake, you will also have considered the effect of the change of
halogen on the dipole-dipole interactions which will also be present. In fact, the permanent polarity
of these molecules is falling, because the halogens get less electronegative from Cl to Br to I, and so
there is less electronegativity difference between the halogen and the hydrogen. The permanent
dipole-dipole attractions therefore fall – but not enough to counteract the effect of the increasing
dispersion forces. Well done if you thought about that!)
b) Fluorine is a very electronegative element, and so the bonding electrons will be attracted strongly
towards the fluorine, away from the hydrogen. That leaves the hydrogen with quite a lot of positive
charge, and the fluorine quite negative.
The fluorine also has small, intense, 2-level lone pairs which will be very strongly attracted towards
anything positive.
The fairly positive hydrogen on one HF molecule will be attracted to one of these lone pairs on a
nearby HF molecule. This is a hydrogen bond.
Hydrogen bonds are attractions between a δ+ hydrogen on one molecule and a lone pair on a very
electronegative atom (N, O or F) on another molecule.
c) In HF, each molecule has one δ+ hydrogen and three active lone pairs. In the liquid as a whole
there are therefore three times as many lone pairs are there are δ+ hydrogens. On average, then,
each molecule can only form one hydrogen bond using its δ+ hydrogen and one involving one of its
lone pairs. The other lone pairs are essentially wasted.
) In HF, each molecule has one δ+ hydrogen and three active lone pairs. In the liquid as a whole
there are therefore three times as many lone pairs are there are δ+ hydrogens. On average, then,
each molecule can only form one hydrogen bond using its δ+ hydrogen and one involving one of its
lone pairs. The other lone pairs are essentially wasted.
In water, there are two δ+ hydrogens on each molecule and two lone pairs. Because the numbers
are equal, each water molecule in the liquid could in principle form four hydrogen bonds, two using
the δ+ hydrogens and two using the lone pairs.
d) This time, there are too many δ+ hydrogens and not enough active lone pairs. That means that
on average each ammonia molecule can form one hydrogen bond using its lone pair and one
involving one of its δ+ hydrogens. The other hydrogens are wasted. So both ammonia and HF can,
on average, only form two hydrogen bonds per molecule. So that isn't the reason for the difference!
The difference is that nitrogen isn't anywhere near as electronegative as fluorine. There will be less
positive charge on the hydrogens, and less negative charge on the nitrogen. The attractions will
therefore be rather weaker between the δ+ hydrogens and the lone pairs on nearby molecules.