I know the trend in group 1 is that ionisation energy decreases down the group due to an increase in atomic radius and more energy levels are added so more shielding, but I'm not sure why there's such a big jump between sodium and potassium.

My textbook doesn't explain it very clearly for me but says the d orbital doesn't shield the outer electrons very well explaining why K to Cs has similar values, but potassium's 3d orbital is not yet filled, so I'm not sure why it's value is so much lower.

I'll put the values from my textbook in to help explain what I mean. Li=513, Na=496, K=419, Rb=403, Cs=376. (in kJ/mol)

The gap is much more obvious when you look at a graph.

Sorry if I haven't made it very clear but any help would be appreciated. Thanks

  • 2
    $\begingroup$ Wikipedia has plotted the graph you mentioned above (in terms of eV) and it seems normal. Also check out This graph $\endgroup$ – Aniruddha Deb Jan 4 at 8:25
  • $\begingroup$ We also have Rb above the average of K and Cs. The stepwise nature of the graph seems to correlate with the subshells filled by the preceding period: Li -> s, Na and K -> s and p, Rb and Cs -> s, p and d. $\endgroup$ – Oscar Lanzi Jan 4 at 11:23
  • $\begingroup$ @aniruddhadeb the Wikipedia graph does not include lithium. If that element were included the drop from Na to K would look like a very sharp step because Li is only a little above Na. This is what the OP is asking, and there is a similar though less pronounced effect centered around Rb. $\endgroup$ – Oscar Lanzi Jan 5 at 0:53
  • $\begingroup$ For Alkali metals, the First Ionisation Energy is equal to the work function $ (\phi) $ of the metal. As metallic character increases down the group, thus $\displaystyle \phi_{Li} > \phi_{Na} > \phi_{K} > \phi_{Rb} > \phi_{Cs} $. Hence Ionisation Energies are different. $\endgroup$ – Nikola Alfredi Apr 21 at 20:21

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