The reaction between NH4Br and Na metal in liquid ammonia (solvent) results in the products

Question

The reaction between NH4Br and Na metal in liquid ammonia (solvent) results in the products (A)  NaBr, HBr (B) NaBr, H2 (C) H2, HBr  (D)  NaNH2,H2

Tardigrade Admin · Accepted Answer

Na xrightarrowliq.NH3 [Na(NH3)x] + + underset textAmmoniated electron[e(NH3)y ]- NH+4Br- + [e(NH3)y]- arrow NH3 + (1/2) H2 + yNH3 + Br- [Na(NH3)x]+ + Br- arrow Nabr + xNH3

Q. The reaction between $N H_{4} B r$ and $N a$ metal in liquid ammonia (solvent) results in the products

$N a B r, H B r$

$N a B r, H_{2}$

$H_{2}, H B r$

$N a N H_{2}, H_{2}$

$N a l i q . N H_{3} [N a (N H_{3})_{x}]^{+} + Ammoniated electron [e (N H_{3})_{y}]^{-}$
$N H_{4}^{+} B r^{-} + [e (N H_{3}) y]^{-} \to N H_{3} + \frac{1}{2} H_{2} +_{y} N H_{3} + B r^{-}$
$[N a (N H_{3})_{x}]^{+} + B r^{-} \to N ab r +_{x} N H_{3}$

Q. The reaction between NH4​Br and Na metal in liquid ammonia (solvent) results in the products

NaBr,HBr

NaBr,H2​

H2​,HBr

NaNH2​,H2​