The least number of oxyacids are formed by:

We first recall that an oxyacid is an acid which contains the element hydrogen bonded to oxygen, and the oxygen in turn is bonded to some other central element. Therefore the ability of an element to form oxyacids depends on two factors: the element should be able to bond simultaneously with oxygen and also have available orbitals (generally d-orbitals) or lower electronegativity so that it can accommodate multiple bonding.

Now we examine, one by one, the four elements given in the options and count how many oxyacids each of them is known to form.

We have chlorine. Chlorine belongs to the same group as fluorine but it possesses empty 3d-orbitals and a moderate electronegativity. It is able to expand its octet and show variable oxidation states $$+1, +3, +5, +7$$. Hence chlorine forms the four well-known oxyacids: hypochlorous acid $$HOCl$$, chlorous acid $$HClO_2$$, chloric acid $$HClO_3$$ and perchloric acid $$HClO_4$$. So chlorine contributes a count of $$4$$ distinct oxyacids.

Next we consider fluorine. Fluorine is the most electronegative element ($$\chi_{F}=4.0$$ on the Pauling scale) and is restricted to the small $$2p$$ valence shell; it has no available d-orbitals. Being so electronegative, fluorine strongly withdraws electron density from oxygen, making the $$O-F$$ bond extremely unstable when hydrogen is also present. Consequently fluorine is unable to stabilise any compound that can be written in the general oxyacid form $$H-O-F$$ or $$F-O-(something)$$. In other words fluorine forms no oxyacids at all. The only common acid of fluorine is hydrofluoric acid $$HF$$, and that does not contain oxygen, so it is not an oxyacid. Therefore the tally for fluorine is $$0$$.

Now we look at sulphur. Sulphur lies in Period $$3$$, possesses empty $$3d$$ orbitals and shows a wide variety of oxidation states such as $$+2, +4, +6$$. Because of this versatility sulphur forms many oxyacids, for example sulphurous acid $$H_2SO_3$$ and sulphuric acid $$H_2SO_4$$ in the simple series, as well as peroxymonosulphuric acid $$H_2SO_5$$ (Caro’s acid), peroxydisulphuric acid $$H_2S_2O_8$$ (Marshall’s acid), thiosulphuric acid $$H_2S_2O_3$$ and several others. Even the most conservative list gives at least $$5$$ distinct sulphur oxyacids, obviously far more than chlorine and certainly more than fluorine.

Lastly we have nitrogen. Nitrogen has no d-orbitals but it still manages two very common oxyacids because of its moderate electronegativity and ability to form $$\pi$$-bonds with oxygen. These are nitrous acid $$HNO_2$$ with nitrogen in oxidation state $$+3$$ and nitric acid $$HNO_3$$ with nitrogen in oxidation state $$+5$$. Besides those, hyponitrous acid $$H_2N_2O_2$$ and peroxynitric acid $$HNO_4$$ are also known, so nitrogen’s count is at least $$3$$ or $$4$$, again larger than fluorine’s.

Collecting all these counts we have:

$$\begin{aligned} \text{Chlorine} & : 4 \text{ oxyacids} \\ \text{Fluorine} & : 0 \text{ oxyacids} \\ \text{Sulphur} & : \ge 5 \text{ oxyacids} \\ \text{Nitrogen} & : \ge 3 \text{ oxyacids} \end{aligned}$$

Clearly the smallest number is $$0$$, which belongs to fluorine. Thus, among the elements listed, fluorine forms the least number of oxyacids.

Hence, the correct answer is Option B.