Four species are listed below: i. $$HCO_3^{-}$$ ii. $$H_3O^{+}$$ iii. $$HSO_4^{-}$$ iv. $$HSO_3F$$. Which one of the following is the correct sequence of their acid strength?

For Brønsted-Lowry acids, stronger acidity means a greater tendency to donate $$H^+$$. Numerically, the smaller (more negative) the $$pK_a$$, the stronger the acid. Hence, arrange the four species in order of increasing acidity by comparing their $$pK_a$$ values or by using standard qualitative arguments (conjugate-base stability, electronegativity, resonance, inductive effects).

Case 1: $$HCO_3^-$$ (bicarbonate)
Bicarbonate can lose one more proton to give $$CO_3^{2-}$$. The $$pK_a$$ of this step is about $$10.3$$. Thus it is a weak acid.

Case 2: $$HSO_4^-$$ (bisulfate)
Bisulfate is the conjugate base of the first, very strong, dissociation of $$H_2SO_4$$. For its second dissociation $$HSO_4^- \rightarrow SO_4^{2-}+H^+$$, $$pK_a \approx 1.99$$. Therefore it is significantly stronger than $$HCO_3^-$$ but still weaker than the hydronium ion.

Case 3: $$H_3O^+$$ (hydronium)
In aqueous solution, $$H_3O^+$$ is the strongest acid that can exist to an appreciable extent, with $$pK_a \approx -1.7$$ (sometimes quoted as $$-1.74$$). Consequently it is stronger than both bicarbonate and bisulfate.

Case 4: $$HSO_3F$$ (fluorosulfuric acid)
Fluorosulfuric acid is a “superacid.” The highly electronegative $$F$$ atom withdraws electron density, and the conjugate base $$SO_3F^-$$ is strongly resonance-stabilised. Reported $$pK_a$$ values lie near $$-15$$, making it far stronger than even $$H_3O^+$$.

Putting the four in ascending order of acid strength (largest $$pK_a$$ to most negative $$pK_a$$):
$$HCO_3^- \; (pK_a \approx 10.3) \; \lt \; HSO_4^- \; (pK_a \approx 2.0) \; \lt \; H_3O^+ \; (pK_a \approx -1.7) \; \lt \; HSO_3F \; (pK_a \approx -15)$$.

Therefore the correct sequence is:
$$\text{i} \; \lt \; \text{iii} \; \lt \; \text{ii} \; \lt \; \text{iv}$$.

Option C which is: i < iii < ii < iv