Number of Cl = O bonds in chlorous acid, chloric acid and perchloric acid respectively are:

We first recall that an oxo-acid of chlorine has the general formula $$\mathrm{HClO_n}$$ where $$n$$ goes from $$1$$ to $$4$$. For each acid we must draw the most accepted Lewis structure, keeping in mind two guiding principles: (i) chlorine is in period $$3$$, therefore it can expand its octet, and (ii) the total number of valence electrons must be conserved. A Cl = O bond is a double bond between chlorine and oxygen, so after drawing the structure we simply count how many such double bonds appear.

We begin with chlorous acid, $$\mathrm{HClO_2}$$ (also written $$\mathrm{HOClO}$$). Chlorine contributes $$7$$ valence electrons, each oxygen contributes $$6$$, and hydrogen contributes $$1$$, so the electron count is

$$N_e = 7 + 2\times 6 + 1 = 20$$

We connect chlorine to two oxygens and attach the hydrogen to one of those oxygens so that $$\mathrm{O\!-\!H}$$ appears, which satisfies the acidic character. After distributing lone pairs so that every atom reaches an octet where possible, we are left with one double bond required to reduce the electron count to $$20$$. Thus only one of the two Cl-O links becomes a Cl = O double bond, whereas the other remains a single bond (bearing the hydrogen). Consequently,

$$\text{Number of Cl\!=\!O bonds in } \mathrm{HClO_2} = 1$$

Now we turn to chloric acid, $$\mathrm{HClO_3}$$. The valence-electron total is

$$N_e = 7 + 3\times 6 + 1 = 26$$

Chlorine is joined to three oxygens, with one of those oxygens still bonded to hydrogen. If we try an initial arrangement with all single bonds, the electron count is too high, so we convert single bonds to double bonds until the total matches $$26$$. Every conversion of a single bond to a double bond removes two electrons from the lone-pair pool. After algebraic checking we find that two of the three Cl-O links must be double bonds, while the O bearing hydrogen remains in a single bond. Therefore,

$$\text{Number of Cl\!=\!O bonds in } \mathrm{HClO_3} = 2$$

Finally, we analyse perchloric acid, $$\mathrm{HClO_4}$$. The electron count is

$$N_e = 7 + 4\times 6 + 1 = 32$$

Chlorine is attached to four oxygens, one of which carries the hydrogen. With four single Cl-O bonds the electron total would be too large, so we systematically introduce double bonds until we reach $$32$$ electrons. Algebra shows that three of the four Cl-O links must be double bonds; the remaining bond is the single bond to the $$\mathrm{OH}$$ group. Consequently,

$$\text{Number of Cl\!=\!O bonds in } \mathrm{HClO_4} = 3$$

Collecting the three counts we have:

$$\mathrm{HClO_2}: 1,\qquad \mathrm{HClO_3}: 2,\qquad \mathrm{HClO_4}: 3$$

These numbers correspond exactly to Option D, which lists “1, 2 and 3”.

Hence, the correct answer is Option D.