Q. Buckminsterfullerene has the formula C60.
The bonding in buckminsterfullerene is similar to the bonding in
graphite.
Which of the following is true?
A All the bond angles in buckminsterfullerene are 120°.
B The melting temperature of buckminsterfullerene is higher than
that of graphite.
C There are delocalized electrons in buckminsterfullerene.
D On complete combustion, buckminsterfullerene forms carbon
dioxide and water.
It seems that B is the answer.
But D and A are equally right snce both will give CO2 and H2O
upon combustion and all Carbon in graphite are sp2-hydridized so
the bond angle will be 120....
Alterntively question will be "Which of the following is NOT
true"?
Please help.
The first thing to note, is that buckminsterfullerene consists of
alternating pentagons and hexagons, with 2 different bond lengths
and bond angles even in the resonance hybrid (ie. not just in the
resonance contributors).
A is wrong because if all C atoms had exactly 120 degrees bond
angles, the molecule would be planar (like the graphene layers in
graphite), but obviously buckminsterfullerene is spherical, and
therefore although the C atoms are considered (largely) sp2
hybridized, the bond angles necessarily deviate from 120 degrees in
order to achieve the spherical molecular geometry, which also
results in molecular strain due to ring strain due to angle strain.
Interestingly, although every C atom in buckminsterfullerene
belongs simultaneously to both a pentagon and a hexagon, the ring
strains are concentrated in the pentagons, which are non-contiguous
and are required for the 'wraparound closure' of the 'graphene
sheet' to form the spherical buckminsterfullerene molecule. A
mathematical model of the bond angles would be 108 degrees within
each pentagon, and 120 degrees within each hexagon. The actual
values for the 2 different types of bond angles in the resonance
hybrid of buckminsterfullerene would therefore be slightly greater
than 108 degrees (within each pentagon), and slightly lesser than
120 degrees (within each hexagon), after taking resonance into
consideration.
B is wrong because to melt buckminsterfullerene, you only need to
partially overcome the intermolecular van der Waals attractions. To
melt graphite, you need to partially overcome *both* the van der
Waals attractions between the graphene layers in graphite, as well
as the partial double covalent bonds within each graphene
layer.
C is correct, as electrons can indeed be delocalized by across the
surface of each molecule (not across molecules), due to sideways
overlapping of unhybridized p orbitals of each (largely) sp2 C
atom. However, the electron delocalization and aromaticity is only
partial, and not complete. This is related to option A and the
structure (see diagram) above. Buckminsterfullerene tends to avoid
having double bonds in the pentagonal rings (ie. in the resonance
hybrid, the pentagonal bond lengths and strengths have mostly
single bond character, while the hexagonal bond lengths and
strengths have greater partial double bond character), which makes
electron delocalization poor, and results in buckminsterfullerene
being only partially aromatic.
D is wrong because buckminsterfullerene isn't a hydrocarbon, but
just carbon (ie. an allotrope of). Hence only product of complete
combustion will be CO2 (no H2O).