Question

$Li$ on heating with $N{H}_3$ forms $L{i}_{3}N$ whereas $Na$ does not form $N{a}_{3}N$. Explain.

Answer 1

$Li$ in heating with ${NH}_3$ from ${Li}_{3}N$ but $Na$ does not form ${Na}_{3}N$. $Na$ on reaction ${NH}_3$ forms sodamide $\left({NaNH}_2\right)$.

$6Li+2{NH}_3⟶2{Li}_{3}N+3H_2$

$2Na+2{NH}_3⟶2{NaNH}_2+H_2$