The number of P - O bonds in P$$_4$$O$$_6$$ is:

We begin by examining the molecular formula $$\mathrm{P_4O_6}$$. This compound contains four phosphorus atoms and six oxygen atoms.

First, recall the usual valencies in such oxides. Phosphorus in its +3 oxidation state typically forms three single bonds, while oxygen in oxides forms two single bonds when it acts as a bridging atom. Hence, every phosphorus atom will try to connect itself to exactly three oxygen atoms through single $$\mathrm{P\!-\!O}$$ bonds.

With this idea in mind, picture the well-known “P4 tetrahedron” framework: the four phosphorus atoms occupy the corners of a regular tetrahedron. The six oxygen atoms sit on the six edges of that tetrahedron, each oxygen bridging the two phosphorus atoms that define that edge. Thus every edge $$\bigl(\text{there are }6\bigr)$$ carries one oxygen atom, and every such oxygen links the two adjacent phosphorus atoms.

Now let us count the actual $$\mathrm{P\!-\!O}$$ bonds one by one. Each bridging oxygen uses both of its valence positions, so it makes two single bonds—one to each of the two phosphorus atoms it bridges. We have $$6$$ such oxygen atoms, and every one of them supplies exactly $$2$$ $$\mathrm{P\!-\!O}$$ bonds. Therefore, the total number of $$\mathrm{P\!-\!O}$$ bonds is

$$\text{Number of P-O bonds}=6\times 2=12.$$

We can confirm this result by counting from the phosphorus side as well. There are $$4$$ phosphorus atoms, and each one is attached to precisely $$3$$ oxygens (valency of P in this oxide is $$3$$). Hence

$$\text{Number of P-O bonds}=4\times 3=12,$$

which is identical to the previous count, as expected.

So, $$\mathrm{P_4O_6}$$ contains exactly twelve $$\mathrm{P\!-\!O}$$ single bonds.

Hence, the correct answer is Option C.