Why is the C-O-H bond slightly larger than the C-S-H bond
angle?
For this question, the answer given is that O and S contains 2
lone pairs and 2 bond pairs, as O is more electronegative than S,
the bond pair of electrons are held more closely towards O, hence,
a stronger repulsion.
My question is, can i answer in terms of the size of the
molecule? O is smaller than S, and thus the C and H atoms are of
closer proximity to the molecule than when its a large molecule
like S, therefore, the lone pair - bond pair repulsion is of
greater magnitude.
Singapore JCs often use electronegativity to explain the deviation
from the basic bond angles of period 2 elements as predicted by
VSEPR theory, when discussing period 3 elements. Other than cases
involving electronegative F (eg. why NF3 bond angles deviate from
NH3 bond angles), using electronegativity to explain the deviation
is actually incorrect, or rather, only a secondary factor
(Chemistry, being a microcosm of real-life, is
multi-factorial).
The more important reason, is due to the significantly lesser
extent of orbital hybridization (which requires energy, and thus
only thermodynamically justified for less diffused valence orbitals
of period 2 elements), due to the significantly lesser extent of
electron-pair repulsions, due to the significantly more diffused
valence orbitals (which the lone pairs reside in, or the bond pairs
are formed from the overlap of) of period 3 elements.
Consequently, the atoms of period 3 elements use their mostly
unhybridized orbitals for their lone pairs to reside in, and for
overlapping head-on or end-on to form sigma bonds. In the case of
C-S-H, the S atom uses it's mostly unhybridized p orbitals to
overlap head-on or end-on to form its sigma bonds with the C and H
atoms, as well as for 1 of the lone pairs to reside in, with the
remaining lone pair residing in the mostly unhybridized s orbital.
This results in the C-S-H bond angle (with 2 orthogonally oriented
p orbitals used for head-on or end-on overlap with the C and H
atoms) being close to the unhybridized p orbital bond angle of 90
degrees.
So you're partially right that it's about the size, though
you'll need to elaborate (as I did) to secure your marks. Your
school (and Singapore JCs in general) aren't entirely wrong, it's
just that electronegativity isn't the most important contributing
factor here (unlike in cases such as NF3 vs NH3).
However, all this discussion may be moot, because as far as A
levels are concerned, Cambridge won't require the student to give
any answer other than the basic bond angles as predicted by VSEPR
theory for period 2 elements, ie. just state 104.5 degrees as if
the S atom were an O atom, and include the usual explanation about
the S atom having 2 lone pairs and 2 bond pairs, and that lone pair
- lone pair repulsions > lone pair - bond pair repulsions >
bond pair - bond pair repulsions, that would gain you full
marks.
After writing the basic answer above, if you like, if you have
the time, or if (unexpectedly, though I'd see it as a pleasant
surprise) Cambridge specifically asks you to explain why the C-S-H
bond angle deviates from the C-O-H bond angle, then you can
elaborate further to why the C-S-H bond angle is actually less than
the C-O-H bond angle, giving either my (more correct) explanation,
or your school's (less correct) explanation, or both. But again,
because asking this would be beyond the A level syllabus, it is
rather unlikely for Cambridge to ask why the C-S-H bond angle
deviates from the C-O-H bond angle for the actual A level exam (ie.
your question, which you presumably took from your school's
mid-year revision package).
