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This question concerns the dihalo addition reactions. I've learned that a halogen molecule (which, of course, consists of two atoms of the respective halogen) is more easily polarised by the double bond present in an alkene if it has a larger atomic number.

My question is this: Is this because the electrons of the halogens of higher atomic number are, on average, further away from its nucleus, so that they are easier to dislocate because the force that the protons exert on them is weaker?

A follow-up question would be: Is it, for the same reason, easier (in terms of energy requirement) to ionise elements of higher atomic number?

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    $\begingroup$ Welcome to Chemistry SE! In an attempt to increase the chances of a response, I would recommend adding a reputable reference. $\endgroup$
    – z1273
    Jan 31 at 23:18
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    $\begingroup$ The ionisation energy goes as $(Z_\mathrm{eff}/n)^2$, where $Z_\mathrm{eff}$ is effective nuclear charge and $n$ is principal quantum number. What you're describing is $Z_\mathrm{eff}$, but you can see here both play a role: smaller $Z_\mathrm{eff}$ and larger $n$ means smaller IE. For polarisability, your explanation makes sense conceptually, although I'm not sure if there's a similar formula. $\endgroup$
    – orthocresol
    Jan 31 at 23:19
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Short answer: yes.

The Wikipedia article on Polarizability has a nice section on this, which I would like to supplement with further quotations from its cited source.$^1$

pg 24

Since electrons are charged particles they respond to electric fields. ... The electron cloud is said to become polarized in response to the electric field. The ability of the electron cloud to distort in response to an external field is known as its polarizability.

pg 25

We define the polarizability of a molecule as the magnitude of the dipole induced by one unit of field gradient, which works out to be in units of volume. Often, the larger the volume occupied by the electrons, the more polarizable those electrons.

My understanding is that it's less that the electrons are farther away from the nucleus in higher elements, but rather that they occupy a larger volume and are therefore more greatly affected by the electric field.


There is probably a nice analytical expression for the relationship of the polarizability to the terms we're used to seeing in the Schrodinger equation, but I'm not aware of it. In my experience, it is typically calculated numerically using perturbation theory. I did find an example where they worked through the perturbation solution for a particle in a box being subjected to an electric field here, section 4.1.1.


$^1$ Anslyn, Eric V. and Dougherty, Dennis A. Modern Physical Organic Chemistry. University Science Books, California, 2006.

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  • $\begingroup$ Well, a larger electric field does, of course, mean, that the electrons are more subjected to the shielding effect and less by the charge of the nucleus, because they are also more spaced out. I'm very grateful for the reference describing the relevant parts of perturbation theory, even though I'm as yet unfamiliar with Schrödinger's equation, whence I'll get back to it once I study the latter. $\endgroup$ Feb 8 at 10:01

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