Question

The removal of an electron that forms an anti-bonding molecular orbital results in​

Answer 1

Answer:

Explanation:It is true that electrons in antibonding molecular orbitals have their electron density away from the internuclear region. This has the effect of increasing the repulsion between the nuclei and forcing them apart. Electrons in bonding molecular orbitals have most of their electron density between the nuclei which has the effect of holding the nuclei together. This means we must consider the net effect of the two types of orbitals.

We do this by measuring the bond order. This can be expressed as:

Bond order = (no. of bonding electrons - no. of antibonding electrons)/2

In a theoretical molecule such as He2 there are 2 electrons in the bonding orbital and 2 electrons in the antibonding orbital. This gives a bond order of zero so the molecule would fall apart.

Answer 2

The removal of an electron that forms an anti-bonding molecular orbital results in increased bond order for the molecule.

• The main reason for this is: destabilization is caused due to increased energy of the molecule due to presence of Anti-Bonding MO.

• Anti-Bonding orbital electrons decrease the bond order.

• That is why , Anti-Bonding Molecular Orbitals are placed higher in the energy diagram.

Example of H2 molecular orbital using a diagram:

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(7,7)\put(1.5,3){ \boxed{\uparrow\downarrow} }\put(2,3){\line(1,1){1.5}}\put(3.5,4.5){ \boxed{\uparrow\downarrow} }\put(4,4.5){\line(1,-1){1.5}}\put(5.5,3){ \boxed{\uparrow\downarrow} }\put(2,3){\line(1,-1){1.5}}\put(3.5,1.5){ \boxed{\uparrow\downarrow} }\put(4,1.5){\line(1,1){1.5}}\put(0.5,1){\vector(0,1){3}}\put(0.5,4.5){E}\put(3.5,5){ \sigma_{1s}^{*} }\put(3.5,1){ \sigma_{1s} }\put(2,0){Lower \: Bond \: Order}\end{picture}}$

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(7,7)\put(1.5,3){ \boxed{\uparrow} }\put(2,3){\line(1,1){1.5}}\put(3.5,4.5){ \boxed{\underline{-}} }\put(4,4.5){\line(1,-1){1.5}}\put(5.5,3){ \boxed{\uparrow} }\put(2,3){\line(1,-1){1.5}}\put(3.5,1.5){ \boxed{\uparrow\downarrow} }\put(4,1.5){\line(1,1){1.5}}\put(0.5,1){\vector(0,1){3}}\put(0.5,4.5){E}\put(3.5,5){ \sigma_{1s}^{*} }\put(3.5,1){ \sigma_{1s} }\put(2,0){Higher \:Bond\:Order}\end{picture}}$

So, here the $\sigma_{1s}^{*}$ is the Anti-Bonding Molecular Orbital , which is much higher in energy than bonding molecular orbital.

So, when electrons are removed from Anti-Bonding orbitals , the bond order increases.