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I was reading in my textbook Chemistry Part I, Textbook for Class XI by NCERT, ed. January 2021 that:

The atomic size generally decreases a period across as illustrated in Fig. 3.4 (a) for the elements in the second period. It is because within the period the outer electrons are in the same valence shell and the effective nuclear charge increases as the atomic number increases resulting in the increased attraction of the electrons to the nucleus.

I understood what they've tried to say, but when I looked at Fig. 3.4 (a) that they provided:

The rate of change for the atomic radius for Li and Be is not as drastic as the rate of change for Be and B, which in turn is more drastic than B and C and so on.

I noticed that the rate of change for the atomic radius becomes less drastic as I move across the period from left to right, albeit there's an anomaly: the slope of the graph between C and N is less steep than the slope of the graph between N and O.

What explains this?

To reiterate, I understand how the atomic radius changes across a period, and I also understand why it does so. What I don't understand, however, is why the rate of change of atomic radius changes? Why is the graph not linear?

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  • $\begingroup$ Why do you say the graph should be linear? E.N differences are not linear. $\endgroup$ Aug 31, 2022 at 5:48
  • $\begingroup$ That's exactly what I'm asking: why's it (not) linear? It doesn't help when I ask a question and I get a reply to it saying why I asked the question. $\endgroup$ Aug 31, 2022 at 7:12
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    $\begingroup$ It should help you to ask yourself why you have asked this question and not another. Like, "What users want is not always what they need.". // See also "effective nuclear charge" in context of Slater's rules. $\endgroup$
    – Poutnik
    Aug 31, 2022 at 7:17
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    $\begingroup$ @Poutnik thanks for the link. I'm new to the site and still learning the rules, so thanks once again for bearing with me. $\endgroup$ Aug 31, 2022 at 18:14
  • $\begingroup$ For linearity, especially in context of nonlinear interaction, must be very good reason. So more proper question than "Why is not it linear?" would be "Why should it be linear?" $\endgroup$
    – Poutnik
    Sep 1, 2022 at 3:59

2 Answers 2

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Within a given period, one might expect such a graph to be of roughly the same shape as the one-electron Bohr radius $r=\frac{53}Z$ pm, which is nonlinear and follows from the basic force equations of the Bohr model. Of course, the presence of additional electrons will perturb this basic structure to some extent, but there is no reason for it to become linear.

Atomic radius of hydrogen-like ions

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This question might be best answered conceptually, rather than mathematically:

Lithium has but 3 protons in its nucleus. Adding a fourth, to make it beryllium, increases the nuclear charge by 33%. But if you add yet another proton, to make it boron, you are increasing the nuclear charge by 25%. And so on, down to the change from oxygen to fluorine, an increase in the nuclear charge (really, the proton count) of only 12.5%. This accounts for the decline in rate of change that you observe.

  1. If this explanation is true, then what accounts for the jump that you would have observed if your graph had considered Period 3, and 4, and so on?

Remember that your graph only considers atoms within the same period. And now, I ask you to think about atoms in different periods.

This effect is explained by the fact that, when you move to a different period, the outermost electron (which determines the atomic radius) is forced to occupy an entirely separate principle quantum level. And, to get more mathematical in my answer, recall that the positions of the principle quantum levels varies as the square of the principle quantum number.

  1. If this explanation is true, then the rate of declination in atomic radius should decrease as you consider each period in turn.

And that is exactly what is observed.

  1. And if this explanation is true, then you should observe almost no change in atomic radius as you march across the Periodic Table, by the time you get to uranium.

Seems to be what the data shows.

Thanks for asking. I hope this helps.

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