The number of $$\text{Cl} = \text{O}$$ bonds in perchloric acid is, $$n$$.........

Perchloric acid is written as $$\text{HClO}_{4}$$. We first recall that chlorine belongs to Group 17 and has seven valence electrons, while oxygen has six valence electrons and hydrogen has one valence electron.

To construct the Lewis structure, we begin by counting the total number of valence electrons available. We have

$$\text{Cl} : 7$$

$$4\,\text{O} : 4 \times 6 = 24$$

$$\text{H} : 1$$

So the total is

$$7 + 24 + 1 = 32 \text{ valence electrons}.$$

Chlorine is the least electronegative (among Cl and O) and thus becomes the central atom. We connect each of the four oxygen atoms to chlorine with single bonds, and we attach the hydrogen atom to one of those oxygens because hydrogen can form only one bond. This initial skeleton uses

$$5 \text{ single bonds} \times 2 \text{ electrons per bond} = 10 \text{ electrons}.$$

We subtract these 10 electrons from the total of 32, leaving

$$32 - 10 = 22 \text{ electrons}$$

to distribute as lone pairs on the outer atoms so that every oxygen obeys the octet rule.

After completing the octets on the four oxygens, we find that we need to use double bonds to satisfy chlorine’s expanded octet and remove any formal charge imbalance. We convert three of the Cl-O single bonds into Cl=O double bonds. Each conversion uses one lone pair from oxygen to make an extra shared pair, but the total electron count remains conserved.

Thus, in the final structure of $$\text{HClO}_{4}$$ we have:

• One $$\text{O-H}$$ single bond attached to an oxygen which in turn is single-bonded to chlorine.

• Three $$\text{Cl}= \text{O}$$ double bonds.

Therefore, the total number of $$\text{Cl}= \text{O}$$ bonds present in perchloric acid is

$$n = 3.$$

So, the answer is $$3$$.