I found here that the ionization energy of hydrogen is $\pu{1312kJ/mol}$ and for fluorine, it is $\pu{1681kJ/mol}$. Now clearly, from the data, we can see that hydrogen has a lower ionization energy when compared to fluorine.

I did not expect this because I saw here that fluorine is twice as big as hydrogen (I am referring to the empirical values of the atomic radii). This means that it would be significantly easier to remove an electron from a fluorine atom than from a hydrogen atom. That's why I expected hydrogen to have a higher ionization energy. But I found that it's the other way round. Why is it so? What is wrong with my reasoning? Is this an exception?

  • $\begingroup$ @SirArthur7 btw, you don't need to typeset everything in MathJax, so you can save yourself some effort; 1681 kJ/mol is typographically just as correct as $\pu{1681 kJ/mol}$, and H<sub>2</sub>O (try it on a post) is just as correct as $\ce{H2O} $\ce{H2O}$. The most important thing is to be consistent. In this case, it arguably looks better without the Jax, because then it fits into the surrounding text better. See: chemistry.meta.stackexchange.com/a/2935/16683 $\endgroup$ Jan 28, 2020 at 22:34
  • $\begingroup$ @orthocresol You know what? Your linked post was for long an ideal doubt of mine, Thanks a lot for clearing out this, this'd surely save myself from needlessly editing units, but if it's really not a problem could I use \ce for $\ce{H2O}$, is it the buffering time that you want me to avoid? If so then H<sub>2</sub>O could be fine. $\endgroup$ Jan 29, 2020 at 8:34

3 Answers 3


Fluorine is bigger, but there are $9$ protons in the nucleus, compared to $1$ in case of hydrogen. So, the outermost electron in fluorine experiences more attraction than what hydrogen's single electron experiences, in spite of larger size of fluorine.

However, the outermost electron in fluorine doesn't experience attraction from all the $9$ protons (nuclear charge). Due to improper shielding, the effective nuclear charge experienced by one $2p$ electron of fluorine becomes - $\mathrm{ [9 - {(6 \times 0.35) + (2 \times0.85)}]= 5.2}$.

In comparison, the single electron in hydrogen experiences effective nuclear charge of $1.0$

Note: For hydrogen, there is no other electron to shield nuclear charge. So, effective nuclear charge = actual nuclear charge.

Source: To know how to calculate effective nuclear charge visit this Wikihow link.

  • $\begingroup$ But then the ionization energy should increase down the group due to increasing nuclear charge. Why does that not happen? $\endgroup$
    – user88339
    Jan 28, 2020 at 16:05
  • $\begingroup$ According to Coulomb's law, attractive force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. So, it is basically a competition between the two as Ionization Energy estimation involves calculation of attractive force between the nucleus & the outermost electron. $\endgroup$ Jan 28, 2020 at 16:20
  • $\begingroup$ In the same group, the size factor dominates the charge factor. And that is pretty regular in most cases. But the comparison is tricky when you compare between different groups & periods. (Note: If you calculate the Effective Nuclear Charge for each members of a group, you will get a clearer idea about why the size factor is getting dominated.) Li has lesser ionization energy than H. But, along a period Ionization energy increases gradually, with some exceptions, and becomes the maximum at the Noble Gas element. $\endgroup$ Jan 28, 2020 at 16:20

There is confirmation of the ionization values being for Flourine $1680$ $\pu{kJ/mol}$ and Hydrogen $1312$ $\pu{KJ/mol}$ from the Theoretical Chemistry Textbook (online link). Theoretically, if you have a chemical like $\ce{HCl}$ it is very understandable that it forms an ionic bond, because the the out-most electron is held with less force for Hydrogen than with Chlorine which takes the electron from Hydrogen in an ionic bond. And Florine is in the same column in the periodic table as Chlorine, so its behavior is similar.

At ValenceElectrons.com the Aufbau electron configuration of Florine is listed as $1S^2$ $2S^2$ $2P^5$:

Aufbau configuration for Florine

It is then important to understand the Aufbau principle further as it represents hybrid energy levels because of a combination of Theoretical Hydrogen charge distributions. The $S$ $orbitals$ and $P$ $orbitals$ are somewhat combined in atoms aside from Hydrogen, and their energies can be shown in the picture as hybrid $P$ $Bonds$:

Aufbau principle energy levels for Florine and other atoms

In the referenced diagram, just above, due to the Aufbau principle and hybrid orbitals, the energy of the $2$ $S$ $Orbitals$ are lowered, making the energy-well deeper because of the hybrid bonds from the $2$ $S$ and $2$ $P$ levels. That means that going across a row in the periodic table (with $S$ and $P$ levels to it) that generally it becomes more difficult from left-most (easiest) to right-most (most-difficult) element to remove the outer-most electron, as a general trend across the row.

In the far-field, the $2$ $S$ electrons are not fully shielded and are attracted with a partial $+2e^+$ charge, resulting on a tighter hold of these electrons and a greater bonding energy then normally expected for Hydrogen which in the far-field has no-shielding and only a $+1e^+$ to hold the electron in its shell.

The Theoretical Chemistry Textbook further explains:

The Effects of Electron Shells on Ionization Energy

Electron orbitals are separated into various shells which have strong impacts on the ionization energies of the various electrons. For instance, let us look at aluminum. Aluminum is the first element of its period with electrons in the 3p shell. This makes the first ionization energy comparably low to the other elements in the same period, because it only has to get rid of one electron to make a stable 3s shell, the new valence electron shell. However, once you've moved past the first ionization energy into the second ionization energy, there is a large jump in the amount of energy required to expel another electron. This is because you now are trying to take an electron from a fairly stable and full 3s electron shell. Electron shells are also responsible for the shielding that was explained above.

For Florine, before ionization, it has seven outer shell electrons. Before ionization, the outer shell structure can be described as s s p p p p p (since the inner $1$ $S$ shell stays basically the same in before and after ionization). The shell has more of a s characteristic then after ionization.

Before ionization, the outer shell structure of Florine can be described as s s p p p p p. After ionization, the outer shell structure can be described as s s p p p p which has a greater s influence.

The general trend versus row position can be viewed at the below diagram:

Ionization energies versus row position

I have calculated this effect of the ionization energy of Helium being greater than that of Hydrogen, with some accuracy. The calculations for Helium are referenced my the free online-book here. The same concept of incomplete shielding for Florine is what is dominating its higher biding energy than that for Hydrogen also.


Your intuition is spot on : the issue is with the definition of atomic radius, which according to that same Wikipedia page "is not a well-defined physical entity, for which there are various non-equivalent definitions". This ambiguity finds its roots in the quantum mechanical description of the atom.

If you use the column with the "calculated" value, you'll see that the radius for hydrogen is 53 and for fluorine 42, i.e. hydrogen is bigger than fluorine. Correspondingly, the ionization energy is smaller, roughly by a factor $\frac{42}{53}$, in line with your reasoning.

Mills* classical model of physics allows us to calculate these radii and ionization energies fairly precisely, as it describes electrons as having fixed distances from the nucleus. The classical Bohr radius (52.91pm) for hydrogen, and 0.7807 times the Bohr radius (41.39pm) for fluorine. The approximate ionization energy, in Rydberg atomic units, is then simply given by $\frac{Z}{r}$ i.e. 1 for hydrogen and $\frac{Z-8}{r}$, or 1.281 for fluorine, with each electron shielding exactly one proton. (Multiply by 1313 to convert to kJ/mol)

*Randell L. Mills, The Grand Unified Theory of Classical Physics, 2018 edition (ISBN: 978-0-9635171-5-9) see equations 10.182 and 10.183


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.