NaCl on heating with conc. `H_(2)SO_(4)` gives HCl where as NaBr and NaI give `Br_(2)` and `I_(2)` respetively, why ?

Explanation

HBr and Hl are reducing agent where as `H_(2)SO_(4)` is oxidising agent and thus `Br_(2) " and" l_(2)` are formed.
`NaCl+H_(2)SO_(4) to NaHSO_(4)+HCl`
`NaBr // Nal +H_(2)SO_(4) to NaHSO_(4)+HBr//Hl`
`HBr//Hl +H_(2)SO_(4) to Br_(2)//l_(2)+SO_(2)+2H_(2)O`