Question

Why is the first ionization energy of oxygen lower than the first I.E of fluorine? (3 marks)

Answer 1

The 1st ionisation energy of oxygen is less than that of fluorine because the outer electrons experience a smaller effective nuclear charge.

Explanation:

The energy of an outer electron decreases with increasing nuclear charge `Z`.

As the nuclear charge and the number of electrons in an atom increases, account must be taken of the repulsive effect of the extra electrons as well as that of the increased nuclear charge.

To do this we replace the nuclear charge `Z` with the effective nuclear charge `Z_(eff)` which arises from the shielding effect of the other electrons.

So:

`Z_(eff)=Z-sigma`

Where `sigma` is the sum of all the shielding contributions of all the other electrons.

We can work this out for `O` first:

The electron structure is: `color(red)(1s^(2)color(blue)(2s^(2)2p^(4)`

It has been calculated that for the n=2 electrons in blue each electron shields by 0.35 units.

Each electron in the n=1 shell in red shields by 0.85 units.

So for an outer 2p electron:

`sigma=(5xx0.35)+(2xx0.85)=1.75+1.7=3.45`

So `Z_(eff)=8-3.45=4.55`

For `F` the electron structure is:

`color(red)(1s^(2)color(blue)(2s^(2)2p^(5)`

`sigma=(6xx0.35)+(2xx0.85)=2.1+1.7=3.8`

So `Z_(eff)=9-3.8=5.2`

This shows that the effective charge that an outer electron experiences in fluorine is greater than that of oxygen so it will take more energy to remove it.

For more details on this you can look up Slater's Rules.