Question

Explain hybridisation of $HCN$ and $H{N}_3$.

Answer 1

$HCN$ : In $HCN$ molecule $C-$atom is $sp-$hybridised orbital. One of the $sp$hybrid orbital overlap with the $1s$ orbital of $H-$atom, while the other mixes with one of the $3$ atomic $p-$orbitals of $N-$atom.

$H-C\equiv N:$ It is a linear molecule.

$C\left[He\right]{2s}^2{2p}^2$

$H{1s}^1$

$N\left[He\right]{2s}^2{2p}^3$

${HN}_3$ :

Octet rule states that most stable structure assure $8$ electrons around every atom.

$\therefore$ stable structure of ${HN}_3$ is

$\stackrel{\cdot \cdot }{\underset{\cdot \cdot }{N}}=N=\stackrel{\cdot \cdot }{N}-H$

Here the central $N-$atom is $sp-$hybridised where as outer $N-$atoms play ${sp}^2$ hybridisation.