The number of S = O bonds present in sulphurous acid, peroxodisulphuric acid and pyrosulphuric acid, respectively are:

First, we recall that an S = O bond is a double bond between a sulphur atom and an oxygen atom. We must examine the accepted structures of each of the three oxo-acids and then simply count how many such double bonds are present.

We begin with sulphurous acid whose molecular formula is $$\mathrm{H_2SO_3}.$$ The usual Lewis structure is obtained by giving sulphur an expanded octet (which sulphur can do because it is a third-period element). We write one S = O double bond and attach the remaining two oxygen atoms by single bonds, converting each of those oxygens into an $$\mathrm{-OH}$$ group by adding a hydrogen. Symbolically,

$$\mathrm{HO\!-\!S(=O)-OH}.$$

There is only one S = O bond visible in this structure.

Next, we consider peroxodisulphuric acid, whose molecular formula is $$\mathrm{H_2S_2O_8}.$$ The name already suggests the presence of a peroxide linkage. Accordingly, we connect the two sulphur atoms through an $$\mathrm{-O-O-}$$ bridge and complete the octets around each sulphur by adding the necessary oxygen atoms and hydrogens. Each sulphur ends up being bonded

• by a single bond to the peroxy oxygen,
• by a single bond to an $$\mathrm{-OH}$$ group,
• and by two S = O double bonds in order to satisfy the overall count of eight oxygens.

Thus every sulphur carries $$2$$ S = O bonds. Since there are two sulphur atoms, the total number of S = O bonds is

$$2$$ bonds per S $$\times 2$$ S atoms $$=4.$$

Finally, pyrosulphuric acid—also called oleum—has the formula $$\mathrm{H_2S_2O_7}.$$ This acid is derived formally by condensing two molecules of sulphuric acid and removing one molecule of water, leaving a single $$\mathrm{-O-}$$ bridge between the sulphur atoms. The accepted structure is therefore

$$\mathrm{HO\!-\!S(=O)_2-O-S(=O)_2-OH}.$$

Each sulphur is joined

• by a single bond to the bridging oxygen,
• by a single bond to an $$\mathrm{-OH}$$ group,
• and by two S = O double bonds.

Hence, for pyrosulphuric acid as well, we count

$$2$$ bonds per S $$\times 2$$ S atoms $$=4$$

S = O bonds.

We can now summarise our findings:

$$\begin{aligned} \text{Sulphurous acid }(\mathrm{H_2SO_3}) &:&\;1\;S{=}\!O\;\text{bond},\\ \text{Peroxodisulphuric acid }(\mathrm{H_2S_2O_8}) &:&\;4\;S{=}\!O\;\text{bonds},\\ \text{Pyrosulphuric acid }(\mathrm{H_2S_2O_7}) &:&\;4\;S{=}\!O\;\text{bonds}. \end{aligned}$$

The ordered triple of counts is therefore $$1,\,4,\,4$$ which is exactly what is listed in Option B.

Hence, the correct answer is Option B.