Conc. $H_{2} S O_{4}$ can not be used to prepare pure $H B r$ from $N a B r$ as it reacts slowly with $N a B r$ .

True

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False

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Solution

The correct option is B False
$H B r$ cannot be prepared from $N a B r$ by using conc. $H_{2} S O_{4}$ because $H B r$ is strong reducing agent and reduces $H_{2} S O_{4}$ to $S O_{2}$ and get itself oxidised to $B r_{2}$ .
$N a B r + H_{2} S O_{4} \to N a H S O_{4} + H B r$
$H_{2} S O_{4} + 2 H B r \to S O_{2} + B r_{2} + 2 H_{2} O$

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