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In general, we would expect IE to decrease down a group. This is because atomic radius increases $\implies$ valence electrons are further from nucleus $\implies$ less effective nuclear attraction on valence electrons, so they are easier to remove. Also, shielding effect due to more subshells also contributes to this. In group 13, for example, going down from B to Al, we do observe this trend:

  • Boron IE = 801 kJ/mol

  • Aluminium IE = 577 kJ/mol (numbers from my textbook)

Now when we go from Al to Ga, the IE increases:

  • Gallium IE = 579 kJ/mol

Question 1) Shouldn't there be a decrease in IE from Al to Ga?

When we go from Ga to In, IE decreases, as it should:

  • Indium IE = 558 kJ/mol

From In to Tl it increases again:

  • Thallium IE = 589 kJ/mol

Why?

My teacher said that:

IE increases from Al to Ga because in Ga there are now electrons in the $3d$ subshell, which are poor shielders of nuclear charge.

I don't understand how the poor shielders can INCREASE ionization energy. I get that they do not shield as much as $s$ or $p$ subshells, but after all when we go from Al to Ga, Ga not only has the $4s$ electrons for shielding effect, but now there are even MORE electrons that will shield nuclear charge (from the $3d$ subshell). Thus effective nuclear charge on valence electrons should decrease, and the IE trend should still follow. (According to me, even though $3d$ electrons don't shield as well as others, they should not "unshield" the valence electrons just because of being poor shielders.) There are 12 non-valence electrons in Al and 30 in Ga, so how can shielding effect decrease overall just from 10 of those 30 being poor shielders? Can someone explain this to me?

(I am in 11th grade so please try to explain in simple terms, thanks.)

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Here is one textbook explanation of the trends:

[Averill] In contrast to groups 1 and 2, the group 13 elements show no consistent trends in ionization energies, electron affinities, and reduction potentials, whereas electronegativities actually increase from aluminum to thallium. Some of these anomalies, especially for the series Ga, In, Tl, can be explained by the increase in the effective nuclear charge (Zeff) that results from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Consequently, although the actual nuclear charge increases by 32 as we go from indium to thallium, screening by the filled 5d and 4f subshells is so poor that Zeff increases significantly from indium to thallium. Thus the first ionization energy of thallium is actually greater than that of indium.

This explanation uses the same argument as the one the OP cited. However, the OP was not convinced:

[OP] There are 12 non-valence electrons in Al and 30 in Ga, so how can shielding effect decrease overall just from 10 of those 30 being poor shielders?

You have to consider that the nuclear charge goes up from 13 (Al) to 31 (Ga). Most of that charge is shielded, but the effective nuclear charge is a small difference between two large numbers (the nuclear charge and the "effective number of electrons shielding").

In the end, the ionization energies are very similar for the non-boron elements, and the explanation for the exact order will be more subtle than a neat set of rules that works well for other parts of the periodic table.

You can also take a look at similar questions (some with answers) posted in the comment by Nilay Gosh:

Ionization enthalpy for group 13 elements

... Why is ionization energy of indium less than gallium?

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  • $\begingroup$ Ah, I completely missed the fact that the absolute charge from nucleus is increasing anyway, and the $Z_{eff}$ only decreases if there is no reduced shielding effect. $\endgroup$
    – AVS
    Aug 26, 2022 at 15:03
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the d-subshell is not effective at shielding, increasing the effective nuclear charge and ionization energy. According to Wikipedia, The 3d10 electrons do not shield the outer electrons very well from the nucleus and hence the first ionisation energy of gallium is greater than that of aluminium.

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