# How do atomic radius and nuclear charge influence ionisation energies?

This is actually a multiple choice question.

Why is the first ionisation energy of neon higher than that of fluorine?

• A: Fluorine is more electronegative than neon
• B: Neon has a complete octet, but fluorine does not
• C: The atomic radius of fluorine is less than that of neon
• D: The nuclear charge in neon is greater than that in fluorine

I understand why A is incorrect and why B is correct. But I am confused about C and D. I did try to search online but my questions just increased.

Firstly, why is the atomic radius of fluorine less than that of neon? Both fluorine and neon are in the same period which means that they have the same amount of shells. Since the nuclear charge of neon is more, shouldn't the nucleus attract the electrons in the outer shells closer? Which would result in neon having a smaller atomic radius than fluorine. Secondly, according to a source, the largest factor is that the 2p sub shell in neon is full whereas in the fluorine it is not. They said that a full sub shell means that the atom is more stable which is the way neon has a higher first I.E. but since the 2p sub shell is full, wouldn't there be mutual repulsion?

I'm sorry if this is a lot, I'm trying to get a head start for my next education level and I cannot find any textbooks for that level's chemistry. I've been reading a summarized guide book and trying to do assessments whilst trying to find guides online as well.

• I don’t fully understand the bit about mutual repulsion, sorry. – Jan Jul 10 '17 at 8:18

## 2 Answers

‘The atomic radius’ is a somewhat wavy concept. In quantum mechanics, there is no such thing as an atomic radius which would imply an ‘end of the atom’. Instead, electrons are said to occupy orbitals and these orbitals technically extend into infinity albeit at very low values (i.e. they converge towards zero pretty rapidly on macroscopic scales). Therefore, as soon as you are talking about quantum mechanics or calculations, you need to define an arbitrary cutoff value as an ‘end point’ or ‘radius’ of your atom. These cutoff values are typically chosen so that the probability of locating an electron inside the border are $95~\%$ or some other similar percentage.

For the same reason, it is nontrivial to determine experimental atomic radii. If you were to just experimentally measure the electron cloud, all you would see is fog getting less dense outwards from the centre. Hence why experimental values rely on measuring compounds and attributing the distance between nuclei (whose positions can be accurately determined) in part to one atom’s radius and in part to the other’s. Neon has, to the best of my knowledge, not been persuaded to form compounds with other elements yet, so there is no way to experimentally measure its radius.

Varius calculation cutoffs do indeed state that neon’s atomic radius is smaller than fluorine’s — at least within the theory employed. Other definitions such as the van der Waals radius give a different picture because they rely on different means of calculation. For a comparison of different definitions and values, check out the Wikipedia data page on atomic radii. Note that especially the van der Waals radii do not follow the periodic trend of decreasing across a period as the other definitions largely do.

With that out of the way we can dismiss C as factually probably incorrect.

The larger nuclear charge means the electrons in the shell experience a greater potential. Ionisation means removing one of those electrons from the atom against the potential that is pulling it back. Just by means of a simple thought experiment, we can see that removing an electron from a nucleus with $10+$ charge is harder than removing one from a $9+$ nucleus. (In addition, the greater nuclear charge also leads to the orbitals being less diffuse i.e. the radii being smaller regardless of the cutoff used. This is indeed as you said.)

When we are considering atomic radius:
The atomic radius of noble gases are considered the highest in a period since we only consider their Van der Waals radius (distance between two similar atoms in their closest approach). Although technically they should have lower atomic radius due to increased $Z_\text{eff}$.

As for ionisation energy:
The ionisation energy of noble gases are way higher than the other elements of the same period due to stable configuration