# What does it mean that an electron pair “spends more time” nearer an atom?

In a book I am reading, it is said

In the $$\ce{HCl}$$ molecule, the shared pair of electrons spend more time nearer the chlorine atom. In the $$\ce{HF}$$ molecule, the shared electrons spend more time nearer the fluorine atom.

What does "spend more time" mean here? Has this been experimentally validated? If yes, how?

When you come to the quantum level (looking at one electron) things don't behave anymore like we normally think. To make it anyway understandable those expressions were invented such as "spending more time". Electrons are less described as a dot but as a smearing (electron density). Chlorine and fluorine have higher electronegativities than hydrogen. The electron density therefore is not equally distributed between the two bonding partners but drawn towards fluorine/chlorine. To illustrate this, you can say "the electron spends more time at F/Cl".

This thinking was established as molecular orbital theory (s orbitals, p orbitals, ...). There have been other models before, e.g. you might know the shell model (k shell, l shell, ...). For experimental evidence you might want to take at look at this -- but most of all the citations therein: http://en.wikipedia.org/wiki/User:Chem507f10grp4/sandbox#Experimental_Evidence_Supporting_Molecular_Orbital_Theory Basically, the shell model ran into a series of properties it did not calculate correctly or precisely. Molecular orbital theory was found to predict many properties and came to well calculated values.

• There is now a technique that can visualize chemical bonds vis a vis their electron density and how that electron density interacts with a carbon monoxide molecule on the tip of an atomic force microscope. rsc.org/chemistryworld/2013/09/… – Ben Norris Jan 23 '15 at 11:15
• It would also be notable to say that what the textbook said was to clarify the meaning of electronegativity, along with MO basics. – M.A.R. Jan 23 '15 at 11:16
• This is terrible terminology. Especially in MO theory, the approximations are based on the time-independent Schrödinger equation. It very much is expressions like these, which produce inaccuracies and misunderstandings. And to be fair here to the students as well, when they learn about MO theory, they probably already know something about probability; at which point it becomes just stupid to dumb it down again. – Martin - マーチン Feb 18 at 14:28

On the question: 'What does "spend more time" mean here?', from a theoretical vantage, why explore approximations that are based on the time-independent Schrödinger equation?

Perhaps instead, this could be viewed as an application of the nonrelativistic version of the time-dependent Schrödinger equation for the wave function in position space of a single particle (a shared electron) subject to a potential (from an electric field from the respective cases of the chlorine and fluorine atoms).

Conceivably (for those more of an expert in this field), the product could be a theoretically based graphic depiction of the expected position over time. Visually, does the shared pair of electrons spend more time nearer the chlorine atom with HCl? In the case of the HF molecule, does the shared electrons spend more time nearer the fluorine atom?

As an example of one of the more readable discussions on the time-dependent Schrödinger equation in a position basis, complete with graphs, please see, for example,this source. Here is a good educational video describing the probabilistic nature of the quantum realm at here and also this. A related video is found here.

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Mithoron Feb 19 at 16:08
• To spend more time in a nonrelativistic sense means literally in a probabilistic answer, one is more likely to be in one place (near an atom, for example) than away. This what precisely what the time-dependent Schrödinger equation for the wave function in position space addresses for a single particle (like an electron). Please review all my sources, and explain to me why this is not correct. It is actually, in my opinion, the most appropriate, albeit advanced and erudite, explanation presented. I also have an advanced degree in Statistics and Applied Mathematics (ask advanced questions). – AJKOER Feb 19 at 18:20