I saw a test question today that basically boiled down to comparing the ionization energies of $\ce{C}$ and $\ce{Cl}$. I know that in the periodic table, ionization energy generally increases as we move right across a row (because nuclear charge increases), and decreases as we move down a column (because atomic radius increases, so valence electrons feel less attraction to the nucleus). Note that $\ce{Cl}$ is in the row below $\ce{C}$, but is also to the right of $\ce{C}$, so it is not obvious how to compare the two. So, without looking up the ionization energies and using only the periodic table, how can I answer this question?


1 Answer 1


It's the other way around,

  • moving left to right across a row increases the IP due to increasing nuclear charge, and
  • moving down a column decreases the IP, due to increasing shielding of the nuclear charge by core electrons.

Chlorine is only one row below carbon, but it is three columns to the right. As the following chart illustrates, moving to the right by one position generally has a similar or larger effect than moving down one position.

enter image description here (image source)

Therefore, in this case moving further to the right by 3 positions (and increasing the IP) strongly outweighs moving down by only one position (and decreasing the IP). Hence, in this case the IP of chlorine would be predicted to be greater than the IP of carbon. Indeed, the IP of chlorine is reported to be 13.0 eV, while that of carbon is only 11.3 eV.

  • $\begingroup$ Sorry I knew it was the other way around I made a mistake when typing. But I don't think you have answered my question because how do you know that moving further to the right 3 times wins out over moving down one time. What if moving to the right affects the ionization energy less than moving down? This is not solid reasoning. $\endgroup$
    – math_lover
    Apr 26, 2015 at 19:12
  • $\begingroup$ There are no what if's, this is how it is. The reasoning is provided in the chart, it evidently shows the reasoning behind this concept. ron clearly explains this. $\endgroup$
    – Asker123
    Apr 26, 2015 at 19:51
  • $\begingroup$ @JoshuaBenabou I've edited my answer in an attempt to address your comment. $\endgroup$
    – ron
    Apr 26, 2015 at 19:57
  • $\begingroup$ @JoshuaBenabou I think you or your lecturer would have needed to analyse this graph in some detail, prior to the test, to be able to answer this question. I don't see how a qualitative argument from principal quantum number, effective nuclear charge or another periodic trend would provide a clear answer in this case, as the effects are opposite. This is the sort of test question I do not appreciate; when multiple qualitative arguments produce opposing conclusions, quantifying which one "wins out" often quickly becomes untenable in Chemistry, and it becomes a toss-up whether you were right. $\endgroup$ Apr 26, 2015 at 21:15
  • 1
    $\begingroup$ I agree I don't like this kind of question because deciding qualitatively which factor wins out always comes down to sloppy reasoning. Thanks ron, I understand now. $\endgroup$
    – math_lover
    Apr 26, 2015 at 22:08

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