Give electron configuration, bond order and magnetic property, energy diagram in $\mathrm{MO}$ for Helium $\left( {{\rm{H}}{{\rm{e}}_2}} \right)$ molecule.
Give electron configuration, bond order and magnetic property, energy diagram in $\mathrm{MO}$ for Helium $\left( {{\rm{H}}{{\rm{e}}_2}} \right)$ molecule.
He $(Z=2)$, So, Total electron in $\mathrm{He}_{2}=4$
Electron configuration in $\mathrm{MO} \mathrm{He}_{2}:\left(\sigma_{1 s}\right)^{2}\left(\sigma_{1 s}^{*}\right)^{2}$
All electron are paired in $\mathrm{He}_{2}$, so it is diamagnetic.
Bond order $=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(2-2)=0$
Bond order in $\mathrm{He}_{2}$ is zero. So it is unstable and does not exist.
$\mathrm{MO}$ energy diagram of $\mathrm{He}_{2}$ is as under.
Similar Questions
Which is not paramagnetic
Which is not paramagnetic
The correct stability order for $N_2$ and its given ions is :-
The correct stability order for $N_2$ and its given ions is :-
The bond order is maximum in
The bond order is maximum in
- [AIIMS 1985]
Match $List-I$ with $List-II$.
$List-I$
$List-II$
$(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$
$(I)$ Dipole moment
$(B)$ $\mu=Q \times I$
$(II)$ Bonding molecular orbital
$(C)$ $\frac{N_{b}-N_{a}}{2}$
$(III)$ Anti-bonding molecualr orbital
$(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$
$(IV)$ Bond order
Match $List-I$ with $List-II$.
| $List-I$ | $List-II$ |
| $(A)$ $\Psi_{ MO }=\Psi_{ A }-\Psi_{ B }$ | $(I)$ Dipole moment |
| $(B)$ $\mu=Q \times I$ | $(II)$ Bonding molecular orbital |
| $(C)$ $\frac{N_{b}-N_{a}}{2}$ | $(III)$ Anti-bonding molecualr orbital |
| $(D)$ $\Psi_{ MO }=\Psi_{ A }+\Psi_{ B }$ | $(IV)$ Bond order |
- [JEE MAIN 2022]
During the change of $O _{2}$ to $O _{2}^{-}$, the incoming electron goes to the orbital :
During the change of $O _{2}$ to $O _{2}^{-}$, the incoming electron goes to the orbital :
- [AIIMS 2019]