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

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Q. The reaction between $ NH_4Br $ and $Na $ metal in liquid ammonia (solvent) results in the products

The s-Block Elements

Solution:

$Na \xrightarrow{liq.NH_3 } [Na(NH_3)_x]^ + + \underset {\text{Ammoniated electron}}{[e(NH_3)_y} ]^- $
$ NH^+_4Br^- + [e(NH_3)y]^- \rightarrow NH_3 + \frac{1}{2} H_2 + _yNH_3 + Br^- $
$ [Na(NH_3)_x]^+ + Br^- \rightarrow Nabr + _xNH_3 $