Which of the following does not liberate `Br_(2)` form KBr ?

Options

A. `I_(2)`
B. `Cl_(2)`
C. `"Conc."H_(2)SO_(4)`
D. `F_(2)`

Answer

A

Explanation

`I_2` cannot liberate `Br_2` from KBr, since `l_2` is weaker oxidant than `Br_2 F_2 and Cl_2`, being stronger oxidants than `Br_(2)`, can liberate `Br_2` from KBr.
`F_(2)+2KBr to 2KF+Br_(2)`
`Cl_(2)+2KBr to 2KCl +Br_(2)`
Reaction of conc. `H_(2)SO_(4)` with KBr forms HBr. HBr is a strong reducing agent. It reduces `H_(2)SO_(4)` to form a mxiture of `SO_(2) and Br_(2)`
`KBr+H_(2)SO_(4) to HBr+KHSO_(4)`
`KHSO_(4)+KBr tO HBr+K_(2)SO_(4)`
HBr an formed reduces `H_(2)SO_(4)`
`H_(2)SO_(4)+2HBr to SO_(2)+2H_(2)O+Br_(2)`