The correct sequence of decreasing number of $$\pi$$-bonds in the structure of $$H_2SO_3$$, $$H_2SO_4$$ and $$H_2S_2O_7$$ is:

We have to arrange the three oxo-acids of sulphur $$H_2SO_3$$, $$H_2SO_4$$ and $$H_2S_2O_7$$ in the order of decreasing number of $$\pi$$-bonds. In every covalent bond, one $$\sigma$$-bond is always present, and whenever a double bond is formed it contains, in addition to that single $$\sigma$$-bond, one extra $$\pi$$-bond. Hence, to count the total number of $$\pi$$-bonds in a molecule, we simply have to count how many double bonds are present in its most accepted Lewis structure, because each double bond contributes exactly one $$\pi$$-bond.

Now we write the accepted structures of the three acids and count their double bonds one by one.

For $$H_2SO_3$$ (sulphurous acid) the canonical structure is

$$HO{-}S(=O){-}OH$$

Here the sulphur atom is doubly bonded to only one oxygen atom. The two remaining oxygens are connected through $$S{-}O{-}H$$ single bonds. So the number of double bonds, and therefore the number of $$\pi$$-bonds, is

$$n_{\pi}(H_2SO_3)=1$$.

For $$H_2SO_4$$ (sulphuric acid) the commonly accepted structure is

$$HO{-}S(=O)_2{-}OH$$

That is, the sulphur atom carries two $$S=O$$ double bonds while the other two oxygens are present as $$OH$$ groups joined by single bonds. Thus the number of double bonds equals two, giving

$$n_{\pi}(H_2SO_4)=2$$.

For $$H_2S_2O_7$$ (pyrosulphuric acid or oleum) the structure can be written as

$$HO{-}S(=O)_2{-}O{-}S(=O)_2{-}OH$$

There are two sulphur atoms, and each sulphur is doubly bonded to two separate oxygen atoms. Consequently the total number of double bonds is

$$2 \text{ (double bonds per }S) \times 2 \text{ (number of }S)=4.$$

Hence,

$$n_{\pi}(H_2S_2O_7)=4$$.

Summarising the counts we obtained:

$$H_2S_2O_7 : 4\pi\text{-bonds}, \qquad H_2SO_4 : 2\pi\text{-bonds}, \qquad H_2SO_3 : 1\pi\text{-bond}.$$

To list them in decreasing order we start from the highest and move to the lowest:

$$H_2S_2O_7 \; > \; H_2SO_4 \; > \; H_2SO_3.$$

Comparing this with the options given, we see that it matches Option B.

Hence, the correct answer is Option B.