I was reading about the p-block elements and found that the inert pair effect is mentioned everywhere in this topic. However, the book does not explain it very well. So, what is the inert pair effect? Please give a full explanation (and an example would be great!).
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8$\begingroup$ The inert pair effect is an example of a relativistic effect. The simplified explanation for it is given here. $\endgroup$– Nicolau Saker NetoCommented Feb 25, 2014 at 22:03
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3 Answers
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The inert pair effect describes the preference of late p-block elements (elements of the 3rd to 6th main group, starting from the 4th period but getting really important for elements from the 6th period onward) to form ions whose oxidation state is 2 less than the group valency.
So much for the phenomenological part. But what's the reason for this preference? The 1s electrons of heavier elements have such high momenta that they move at speeds close to the speed of light which means relativistic corrections become important. This leads to an increase of the electron mass. Since it's known from the quantum mechanical calculations of the hydrogen atom that the electron mass is inversely proportional to the orbital radius, this results in a contraction of the 1s orbital. Now, this contraction of the 1s orbital leads to a decreased degree of shielding for the outer s electrons (the reason for this is a decreased "core repulsion" whose origin is explained in this answer of mine, see the part titled "Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?") which in turn leads to a cascade of contractions of those outer s orbitals. The result of this relativistic contraction of the s orbitals is that the valence s electrons behave less like valence electrons and more like core electrons, i.e. they are less likely to take part in chemical reactions and they are harder to remove via ionization, because the s orbitals' decreased size lessens the orbital overlap with potential reaction partners' orbitals and leads to a lower energy. So, while lighter p-block elements (like $\ce{Al}$) usually "give away" their s and p electrons when they form chemical compounds, heavier p-block elements (like $\ce{Tl}$) tend to "give away" their p electrons but keep their s electrons. That's the reason why for example $\ce{Al(III)}$ is preferred over $\ce{Al(I)}$ but $\ce{Tl(I)}$ is preferred over $\ce{Tl(III)}$.
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$\begingroup$ Please don't refer to relativistic mass. Mass is invariant, the energies associated with it are not. $\endgroup$ Commented Feb 16, 2016 at 19:46
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2$\begingroup$ @gsurfer04 Could you elaborate on what you mean by your second sentence, because the part "the energies associated with it are not" needs more explanation. If you are talking about rest mass, which indeed would be a constant, then your argument is correct. I know that the whole argument that mass increases with speed is not the most correct way of expressing what happens as the term mass used here is not identical with the rest mass but I didn't want to dive into that discussion in my answer. Thanks for raising awareness of this issue. $\endgroup$– PhilippCommented Feb 16, 2016 at 20:29
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1$\begingroup$ For reference: physics.stackexchange.com/q/1686/7768 $\endgroup$– PhilippCommented Feb 16, 2016 at 20:29
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$\begingroup$ @Philipp I am not an expert to judge the correctness as gsurfero4 did but your answer is absolutely amazing. Such good explanation is nowhere available! $\endgroup$– SohamCommented May 28, 2018 at 18:43
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$\begingroup$ @Philipp But in wave mechanics electron is not supposed to be object so why are we assuming electron to behave as a planet and spin? $\endgroup$ Commented Feb 28, 2019 at 10:02
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Inert pair effect is mostly shown by the 15-17th group elements.
That is, the oxidation state reduces by 2 for elements below ($\ce{As}$, $\ce{Sb}$), which is more stable than the other oxidation states.
The reason for this is the inertness of the inner $s$ electrons due to poor shielding.
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$\begingroup$ However, the s and p orbitals are thought to be the most effective in terms of shielding, whereas the d and f orbitals have poor shielding. Why is it the other way around in this situation? $\endgroup$– user100905Commented Dec 25, 2021 at 18:38
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As we move down the group, the intervening d and f electrons shield the ns2 electrons poorly and hence the ns2 electrons are pushed inside towards the nucleus This results in the reluctance of ns electrons to participate in bond formation and this phenomenon is called Inert pair effect.
