When fluoride is heated with conc. ${{H}_{2}}S{{O}_{4}}$and $Mn{{O}_{2}}$ the gas evolved is_
A. HF
B. ${{F}_{2}}$
C. $S{{F}_{4}}$
D. None

Hint: The concentrated sulphuric acid protonated halide ions and thereby forming hydrogen halide. But after protonation hydrogen halides escape from the system to air. Streamy fumes are obtained if hydrogen halide is exposed to air.

Complete step by step solution:
Concentrated ${{H}_{2}}S{{O}_{4}}$ is not a strong oxidising agent to oxidise fluoride or chloride. That’s why $Mn{{O}_{2}}$ is added with concentrated sulphuric acid to facilitate the oxidation process. Even chloride and fluoride are not enough reducing agents to reduce sulphuric acid due to their strong electronegativity. Only a stream of fumes of hydrogen fluoride and hydrogen chloride is obtained. But for bromide and iodide, the same things do not happen. They (Iodide and Bromide) are enough reducing agents to reduce concentrated sulphuric acid.

Bromide reduces sulphuric acid to sulphur dioxide and oxidises itself to bromine. The overall redox reactions are shown below:
Oxidation process: $2B{{r}^{-}}\to B{{r}_{2}}+2{{e}^{-}}$
Reduction process: ${{H}_{2}}S{{O}_{4}}+2{{H}^{+}}+2{{e}^{-}}\to S{{O}_{2}}+2{{H}_{2}}O$

Iodide is a stronger reducing agent, it reduces sulphuric acid to hydrogen sulphide and oxidises itself to iodine. The whole process can be confirmed by steamy fumes of HI and a very high amount of iodine in the solution. The corresponding redox reactions are given below:
Oxidation process: $2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}$
Reduction process: ${{H}_{2}}S{{O}_{4}}+8{{H}^{+}}+8{{e}^{-}}\to {{H}_{2}}S+4{{H}_{2}}O$
Therefore, when fluorides are heated with concentrated sulphuric acid $Mn{{O}_{2}}$, a pungent smell of hydrogen fluoride (HF) evolves.
Thus, Option (A) is correct.

Note: When we move from top to bottom of a group in the periodic table, the reducing ability of halides increases because electronegativity decreases the group's ability to lose the outermost electrons increases, and also the attraction between the nucleus and outermost valence electron decreases.