##### Find number of pentagonal faces for given fullerene.

In 1990, chemists synthesized the first fullerene, a molecule C60 consisting of 60 carbon atoms and nothing else. This third form of carbon (the first two being graphite and diamond) had been predicted by R. Buckminster Fuller. Less expected were C70 , C76 , C84 , C90 , and C94 , all of which had been produced by 1992. Every one of these higher fullerenes takes the shape of a convex polyhedron each of whose faces is either a pentagon or a hexagon. For any such structure, the number of pentagonal faces is exactly ____________.

We know that each carbon atom here is "sp2" hybridised which means degree of every carbon atom is "3". If we analyze the structure then every vertex(say "V") is being used thrice(or participating in formation of three faces) i.e if "p" is no. of pentagons and "h" be no. of hexagons then 5*p+6*h=3V (eqn.1).

Also from analysis each edge accounts for two different faces(i.e each edge is actually used twice) which implies 5*p+6*h=2E (eqn. 2).

Now p+h=F (total no. of faces) (eqn 3)

And Euler's formula V-E+F=2 (eqn .4)

Now C60(in case) have 60 atoms(or vertices).

each vertex results in 3 edges but each edge have been counted twice.Therefore,effectively there are 3*V/2 edges.In case of C60,edges =90

So,5p+6h=2*90=180

V-E+F=2

60-90+F=2 =>F=32

Also p+h=F=32

Now using

p+h=32 & 5p+6h=180,

we get

p=12 &h=20.

Hence pentagons = 12 and hexagons =20.

Similarly check for others.. You will get the same.