$\underset{\left(\mathrm{i}\right)}{{\mathrm{CH}}_2=\mathrm{CH}-{\stackrel{+}{\mathrm{NH}}}_3}\leftrightarrow \underset{\left(\mathrm{ii}\right)}{\stackrel{+}{{\mathrm{CH}}_2}-\mathrm{CH}={\mathrm{NH}}_3}$

I is more stable than II because

‘N’ does not have d - orbitals to extend its octet

All the above

N’ cannot hold more than 8 electrons

In structure I N has octet but in structure ‘II’ ‘N’ violates octet with 10 electrons