Among the following oxoacids, the correct decreasing order of acid strength is:

We begin by recalling a very useful empirical rule for oxo-acids of the same central element:

Higher oxidation state of the central atom $$\; \Longrightarrow\;$$ stronger acid $$.$$

The above rule is explained by the greater electron-withdrawing power of a more highly oxidised central atom, which in turn weakens the $$\text{O-H}$$ bond and facilitates release of $$\text{H}^+$$.

Now all the acids given contain chlorine as the central atom, so we simply compute the oxidation number of chlorine in each case, show that calculation explicitly, and then arrange them in the required order.

For a general oxo-acid of chlorine written as $$\text{HClO}_n$$, the sum of oxidation numbers equals zero:

$$(+1)\;($$ for H $$) \;+\; x\;($$ for Cl $$) \;+\; n(-2)\;($$ for each O $$) \;=\; 0.$$

Solving for $$x$$ gives

$$x = -1 + 2n.$$

We now apply this to each acid one by one.

1. Hypochlorous acid, $$\text{HOCl}$$ (i.e., $$\text{HClO}$$, so $$n = 1$$):

$$x = -1 + 2(1) = +1.$$

Hence the oxidation state of Cl is $$+1$$.

2. Chlorous acid, $$\text{HClO}_2$$ (so $$n = 2$$):

$$x = -1 + 2(2) = -1 + 4 = +3.$$

Hence the oxidation state of Cl is $$+3$$.

3. Chloric acid, $$\text{HClO}_3$$ (so $$n = 3$$):

$$x = -1 + 2(3) = -1 + 6 = +5.$$

Hence the oxidation state of Cl is $$+5$$.

4. Perchloric acid, $$\text{HClO}_4$$ (so $$n = 4$$):

$$x = -1 + 2(4) = -1 + 8 = +7.$$

Hence the oxidation state of Cl is $$+7$$.

We have therefore obtained the following oxidation states:

$$\begin{aligned} \text{HOCl} &:& +1,\\[4pt] \text{HClO}_2 &:& +3,\\[4pt] \text{HClO}_3 &:& +5,\\[4pt] \text{HClO}_4 &:& +7. \end{aligned}$$

Using our stated rule, the acid strength increases with oxidation state, so the decreasing order (i.e., strongest first and weakest last) is

HClO $$_4 \; > \;$$ HClO $$_3 \; > \;$$ HClO $$_2 \; > \;$$ HOCl $$.$$

Comparing this order with the options supplied, we find that it exactly matches Option C.

Hence, the correct answer is Option C.