Of the following atoms, which has the largest first ionization energy?

  1. $\ce{Br}$
  2. $\ce{O}$
  3. $\ce{C}$
  4. $\ce{P}$
  5. $\ce{I}$

I got confused between $\ce{O}$ and $\ce{Br}$. The answer given is $\ce{O}$ although in the periodic table $\ce{Br}$ is closer to Group 18 (noble gases). Why is the answer $\ce{O}$ then?


3 Answers 3


I actually think this is an unfair question, unless you were given some additional information, or you were specifically instructed to memorize specific ionization energies, rather than learning the trends. The trend for ionization energies is that they increase from left to right across rows, and from bottom to top in columns. Based on only that information, I can't see how one should distinguish between $O$ and $Br$ ... yes, it happens to be true that $O$ has a higher IP than $Br$, but how are you supposed to tell? Suppose the example had used $S$ instead of $O$, and $I$ instead of $Br$? Those elements have the same relative positions in the table, but in that case, it would be $I$ that has the larger ionization energy.

So, in this case I think your confusion was quite understandable .. I would never construct a test question like this unless the student had some additional information. It's not even true that you can always use the rule about them increasing from left to right along rows .. in the first row, $N$ actually has a slightly higher IP than $O$, because there is a penalty for disrupting the half-filled p-shell in $N$, which is anomalously stable. You can see the periodic trends here.


The larger that atoms get, the outer electrons experience whats called a shielding effect.

Electrons that are closer to the nucleus have a stronger attraction than the electrons further away. Because the electrons in oxygen experience less shielding, they have a stronger affinity.

Oxygen's stronger electron affinity contributes to the observed electronegativity of the atom.

On the most basic level, electronegativity is determined by factors like the nuclear charge (the more protons an atom has, the more "pull" it will have on negative electrons) and the number/location of other electrons present in the atomic shells (the more electrons an atom has, the farther from the nucleus the valence electrons will be, and as a result the less positive charge they will experience—both because of their increased distance from the nucleus, and because the other electrons in the lower energy core orbitals will act to shield the valence electrons from the positively charged nucleus).

  • $\begingroup$ This answer is sort of right, because the trend in shielding ..usually expressed as effective nuclear charge.. is the underlying reason for the trend in ionization energies. It does also happen to be the underlying reason for the trend in electronegativities, but electronegativity actually describes a different phenomenon. The electronegativity of an atom describes its tendency to polarize electrons in covalent bonds. $\endgroup$ Jan 11, 2015 at 0:15

Ok so I thought about the question to and I thought that Bromine should have a higher first ionization energy but then if you really think about it, the answer would be oxygen. Look at it this way. Oxygen desperately needs two electrons. And Bromine needs only one more. So if you look at it, it would be harder to lose another electron when you already lost two and it would be easier to lose one more if you are Bromine. And also if you look at a basic ionization energy trend, it goes to the top right corner.

Imagine you were broke, would you be willing to pay to get something even though you know that you need money? No like Oxygen.

Think of Bromine as an investor who needs a little bit more $ for an achievement, Bromine would be more likely to sacrifice its electron or to invest in something like an investor.

Really bad analogies but hopefully you get the idea.

  • 1
    $\begingroup$ Almost everything about this answer is incorrect or misguided ... except the part about IE's increase toward the top right corner in general. $\endgroup$ Jan 11, 2015 at 0:09

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.