Question

what information would you use to support the view that the molecule of He2 does not exist?​

Answer 1

Answer:

use MOT to say that

in He2 molecule no of electrons are 4

from which 2electrons enters in sigma 1s bonding orbital and 2 enters in sigma 1s Antibonding

bond order=(e- bonding + e- antibonding/2

=( 2-2) /2

=0

hence he2 does not exist

Note: if bond order is more than zero then only molecule exist

Answer 2

Answer:

Before going with explanation you should know some terms like Bonding and Anti-Bonding orbitals, and how to fill them.

A small recap of how to fill them.

1) Element having or less than 14 electrons.

$\sf \sigma\;1s\ < \ \sigma^{\star}\;1s \ < \ \sigma\;2s\ < \ \sigma^{\star}\;2s\ < \ \Bigg[\pi2p_{x} \ = \ \pi2p_{y}\Bigg]\ < \ \sigma\;2p_{z}$

For ex:- B₂ ,  C₂ , N₂ etc..

2) Elements having more than 14 electrons.

$\sf \sigma\;1s\ < \ \sigma^{\star}\;1s \ < \ \sigma\;2s\ < \ \sigma^{\star}\;2s\ < \ \sigma\;2p_{z}\ < \ \Bigg[\pi2p_{x} \ = \ \pi2p_{y}\Bigg] < \ \Bigg[\pi^{\star}2p_{x} \ = \ \pi^{\star}2p_{y}\Bigg]\ < \ \sigma^{\star}\;2p_{z}$

For ex:- F₂, O₂

Now lets get back to the problem.

As we know He₂ is in molecular form, so total no. of electrons will be equal to 4.

So, we need to use less than 14 electrons case. The condition will look like this,

$\longrightarrow \ \bigg(\sf \sigma^{\uparrow \downarrow}\;1s\ : \ \sigma^{\star \ \uparrow \downarrow}\;1s\bigg)$

From this we can find the bond order of He₂ molecule,

⇒ BO = Nᵇ - Nᵃ/2

(Here Nᵇ represents e⁻ in bonding orbital while Nᵃ represents e⁻ in anti-bonding material.)

Substituting the values,

⇒ BO = 2 - 2 / 2

⇒ BO = 0

From Molecular orbital theory (MOT) any molecule which has bond order equal to zero, doesn't exist.

Thus, He₂ molecule doesn't exist.