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I just normally calculated the effective nuclear charge (Zeff) for thallium and Indium from Slater's law, and I found it same for both! That is 5.(If you want calculation for answering or correcting any of my mistake please inform me)

So further moving on, My teacher told that thallium and Indium has approximately equal atomic size due to 4f contraction.

(Note : I am in 11th standard , Inter, so I just started studying basics and inorganic chemistry for the first time , so pardon on any foolish mistake)

So I can't get "The valence electron is basically at somewhat equal distance from nucleus due to same atom size, plus those valence electron experience same effective nuclear charge (Zeff) , than according to me Ionisation potential should be same as they will expirience same protons' charge(sorry for that word) in equal distace right? But its not! "

Please Answer it , I am extremely confused.

Some people explain it as in Thallium we are having full filled d-subshell along withy fully filled f-shell, both poor shielder and 32 proton are being added on in Thallium comparison to Indium,

But I dont find it relevant as "even after poorest of poor shielding the Overall charge an valence electron experience is same!(5) For both, so no point of saying more proton=more pull and thus higher ionisation energy.

One such answer is

enter image description here enter image description here

I don't know whether I am correct or not, as I just started learning. Please Can you explain in easy way and actual reason according to my calculation.

Where I am wrong with concept? too

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First, your Zeff calculations based on Slater's rules are correct, the values of the Zeff are 5 for both Indium's 5p and Thalium's 6p outermost electrons, as their respective shielding values are 44 and 76.

Your explnation about the lanthanide contraction makes sense, but let me take another approach. When we are talking about very heavy elements, we have to consider the so called "relativistic effects", which usually appear when the atomic number is greater than 70 aprox. So basically with these atoms that have large masses, the electrons are forced to move at higher speeds, so they are closer to the nucleus (smaller radius) and as a consequence they are more difficult to remove (higher IE).

https://www.nature.com/articles/s41467-018-06745-6 https://en.wikipedia.org/wiki/Ionization_energy#:~:text=Relativistic%20effects%3A%20Heavier%20elements%20 https://www.hindawi.com/journals/isrn/2013/689040/

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  • $\begingroup$ Sorry but 1.) what is 'relativistic effects' ? As i told , i am a beginner. Please sir can you explain that effect? 2.) And in last para you told they are forced to move at higher speed thus lower radius but isnt it denying the fact that both thallium and Indium has approximately equal size or radii $\endgroup$ Aug 13 at 8:24
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So,

1/ Relativistic effects are based on the theory of relativity. So with these heavy atoms is important to consider it because in these elements electrons do attain sufficient speeds for the elements to have properties that differ from what non-relativistic chemistry predicts. In other words, relativistic effects are those discrepancies between values calculated by models that consider relativity and those that do not. They can be considered to be perturbations, or small corrections, to the non-relativistic theory of chemistry, which is developed from the solutions of the Schrödinger equation. These corrections affect the electrons differently depending on the electron speed compared to the speed of light.

https://books.google.es/books?id=0xcAM5BzS-wC&pg=PA4&redir_esc=y#v=onepage&q&f=false

2/Yes, despite the fact that they have a very similar atomic radius, thalium has a few more electrons (in the same volume), so there's a much higher "electron density" (note that this term is not very accurate because is used with other mening).

"In quantum chemistry, electron density or electronic density is the measure of the probability of an electron being present at an infinitesimal element of space surrounding any given point." (From wikipedia).


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    $\begingroup$ If OP comments for further clarification, answerer should edit the answer, or comment, not make a new answer. $\endgroup$
    – Mithoron
    Aug 13 at 23:06
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    $\begingroup$ It wasn't for further clarification, it was because @Shinchan Nohara had more questions. But it seems that he hasn't seen the answer... :( $\endgroup$ Aug 15 at 16:04
  • $\begingroup$ Than he should ask a new, follow-up question, this isn't a forum thread. $\endgroup$
    – Mithoron
    Aug 15 at 18:19
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    $\begingroup$ @jorge Bonifaz , sorry , I come to this site at long interval, but thank you for the answer , may be I am just a little immature to understand these heavy terminology until I study them completely further more . But i know this answer is well explained for people who know all this, and help them. Thanks again $\endgroup$ Aug 17 at 16:54
  • $\begingroup$ Nono, don't worry, I was just worried cause I thought you weren't answering me, I'm new at this, and I'm glad to be able to help you even a little, so if you have any other question just ask. $\endgroup$ Aug 17 at 20:36

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