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In 13th group, atomic radius increases from boron to aluminium. From aluminium to gallium, atomic radii decreases. From gallium to indium, atomic radii increases. And from indium to thallium, atomic radii decreases.

The reason for this irregular trend given is screening effect. While explaining my teacher told that due to d-orbital screening effect decreases.

I can't get why?

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The reason d-orbitals make a difference is that electrons in d-orbitals do not screen nuclear charge as effectively as those in s and p orbitals. This is because of something called penetration .

The mathematical shapes of d-orbitals prevent them from allowing electrons to penetrate very closely to the nucleus, compared with electrons in s or p-orbitals. In gallium, you have $10$ electrons in the filled $3$d-subshell, and each of these electrons is doing a slightly worse job (relatively speaking) of screening the nuclear charge than the electrons in the s and p orbitals. Therefore, the effective nuclear charge in gallium is slightly higher than that in aluminum, so the increase in the radius is a quite a bit smaller than would be expected based on the difference between boron and aluminum, or gallium and indium.
The trend goes:

  • $\pu{82 pm}$ ($\ce{B}$)

  • $\pu{118 pm}$ $\ce{(Al)}$

  • $\pu{126 pm}$ $\ce{(Ga)}$

  • $\pu{144 pm \ce{(In)}}$

    [covalent radii from www.webelements.com].

This effect is generally known as the d-block contraction. (It can be more or less pronounced depending on how you define the atomic radii.)

A similar thing happens (in principle) when you go from indium to thallium; except in this case you are now dealing with adding a filled f-subshell to the valence shell.

Electrons in f-orbitals are even worse at screening nuclear charge than those in d-orbitals, therefore again, the effective nuclear charge in thallium is a bit larger than it is in indium, so again the jump in radius is fairly small (from $144$ to $148$ pm). This effect (of the filled f-subshell) is generally known as the lanthanide contraction.

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    $\begingroup$ +1 I was always skeptical of this explanation (I thought relativity may be more important), but I found a reference in support of your answer here: en.wikipedia.org/wiki/Lanthanide_contraction. I thought of sharing it here too for others like me! $\endgroup$ – FreezingFire Oct 13 '16 at 9:57
  • $\begingroup$ The exact atomic radius values are: $$ \begin{array}\\ \text{Element}&\text{Atomic radius}\\ \ce{B}&\pu{85 pm}\\ \ce{Al}&\pu{125 pm}\\ \ce{Ga}&\pu{130 pm}\\ \ce{In}&\pu{155 pm}\\ \ce{Tl}&\pu{190 pm} \end{array} $$ The values you mention are covalent radii (webelements.com/boron/atom_sizes.html), but question is talking about atomic radii instead. If I just change the values, your answer would not make perfect sense. Can you please confirm/update? $\endgroup$ – Gaurang Tandon May 1 '18 at 4:00
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Naively, one should expect the atomic radii to increase in a proportional fashion from boron to aluminium, gallium, indium and thallium, because each of these atoms has one more populated shell.

However, the difference is not proportional because there is not a constant proton difference between the elements. From boron to aluminium 8 protons are added to the nucleus; from aluminium to gallium and from gallium to indium 18; and finally, from indium to thallium 30. Whereever there are more protons than expected, orbitals contract; thus, both gallium and thallium are smaller than one naively might assume.

Of course, one should not forget the additional electrons. However, these are added to the d orbitals (and in thallium’s case, f orbitals). These orbitals are much higher in energy than the corresponding shell’s s and p subshells (as is evident by when these orbitals are populated in the periodic table). Furthermore, these orbitals are associated with two (d) or three (f) nodal planes through the nucleus, meaning that the corresponding electrons are further away from the nucleus on average. Thus, their shielding is less effective than that of s and p orbitals.

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  • $\begingroup$ Hmmm... But the inorganic textbook I read states that nd electrons are usually much more core-like than (n+1)s electrons and then state that it is observed that the (n+1)s and (n+1)p electrons are much more involved in bonding. $\endgroup$ – Tan Yong Boon Dec 31 '17 at 4:16
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d and f orbitals experience more attraction towards nucleus due to lesser shielding effect than s and p orbitals. This is why the ionization energies of d and f block elements are high and so their electronegativities.

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