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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?

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  • $\begingroup$ @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 $\endgroup$ – orthocresol Jan 28 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. $\endgroup$ – Sir Arthur7 Jan 29 at 8:34
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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.

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  • $\begingroup$ But then the ionization energy should increase down the group due to increasing nuclear charge. Why does that not happen? $\endgroup$ – user88339 Jan 28 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. $\endgroup$ – Sujay Ghosh Jan 28 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. $\endgroup$ – Sujay Ghosh Jan 28 at 16:20
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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

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