C$$_{60}$$, an allotrope of carbon contains:

We are asked to find how many pentagonal and hexagonal rings are present in the molecule C60, also called buckminsterfullerene. Let us treat the molecule as a closed polyhedron whose vertices are the 60 carbon atoms and whose faces are only pentagons or hexagons.

First, we recall a well-known fact about the geometry of a truncated icosahedron (the shape of a soccer ball): every vertex is the meeting point of exactly three polygons, namely one pentagon and two hexagons. We shall now translate this verbal information into algebra.

Let $$P$$ be the number of pentagons and $$H$$ the number of hexagons on the surface. The total number of faces is then $$F = P + H.$$

Because each pentagon has 5 vertices and each vertex belongs to exactly one pentagon, we have

$$5P = 60 \; \Longrightarrow \; P = \frac{60}{5} = 12.$$

Next, we need a second relation that links $$H$$ with the known quantities. For any convex polyhedron, we can use Euler’s famous formula

$$V - E + F = 2,$$

where $$V$$ is the number of vertices, $$E$$ the number of edges, and $$F$$ the number of faces. We already know $$V = 60$$ and have just obtained $$P = 12$$, so $$F = P + H = 12 + H.$$

To express $$E$$ in terms of $$P$$ and $$H$$, observe that each pentagon contributes 5 edges and each hexagon contributes 6 edges, but every edge is shared by exactly two faces. Therefore

$$E = \frac{5P + 6H}{2}.$$

We now substitute all these expressions into Euler’s relation:

$$V - E + F = 2$$

$$\Longrightarrow 60 - \frac{5P + 6H}{2} + (P + H) = 2.$$

Substituting $$P = 12$$ gives

$$60 - \frac{5 \times 12 + 6H}{2} + (12 + H) = 2.$$

First compute the numerator in the fraction:

$$5 \times 12 = 60,$$

so $$5 \times 12 + 6H = 60 + 6H.$$

Hence

$$\frac{60 + 6H}{2} = 30 + 3H.$$

Substituting back, we have

$$60 - (30 + 3H) + (12 + H) = 2.$$

Now perform the subtractions and additions step by step:

$$60 - 30 - 3H + 12 + H = 2,$$

$$42 - 2H = 2,$$

$$-2H = 2 - 42 = -40,$$

$$H = \frac{-40}{-2} = 20.$$

We have finally obtained

$$P = 12, \qquad H = 20.$$

That is, C60 contains exactly 12 pentagons and 20 hexagons.

Hence, the correct answer is Option D.