Use molecular orbital theory to explain why the $\mathrm{Be}_{2}$ Molecule does not exist.
$Be$ $(\mathrm{Z}=4)$, so electron configuration $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2}$.
In $\mathrm{Be}, 4$ electrons and in $\mathrm{Be}_{2}$ total eight electrons.
Electron configuration of $\mathrm{Be}_{2}:\left(\sigma_{1 s}\right)^{2}\left(\sigma_{1 s}^{*}\right)^{2}\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}$
So, $BMO$ $\left(N_{b}\right)=4$ and ABMO $\left(N_{a}\right)=4$
$BO$ $=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(4-4)=0$
Bond order of $\mathrm{Be}_{2}$ is zero, so, $\mathrm{Be}_{2}$ is unstable and not possible.
Explain Linear combination of atomic orbitals by suitable example.
The correct order of the $O-O$ bond length in $O_2$ , $H_2O_2$ and $O_3$ is
The incorrect statement is :-
When ${N_2}$ goes to $N_2^ + ,$ the $N - N$ bond distance ..... and when ${O_2}$ goes to $O_2^ + ,$ the $O - O$ bond distance .......
The correct stability order for $N_2$ and its given ions is :-