Answer:
No. Because bond order becomes zero,
e.g., in case of \[H{{e}_{2}},\,B{{e}_{2}},\,N{{e}_{2}},\] etc. Note that in \[{{H}_{2}},\,{{\sigma
}_{1s}}M.O.\]is full but \[\sigma _{1s}^{*}\]is empty.
You need to login to perform this action.
You will be redirected in
3 sec