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Referring to, NCERT Chemistry Part I, Textbook for Class XI[1].

ClF3

Here the text puts forward three structures of $\ce{ClF3}$.

My teacher said to me that (b) structure is unstable due to lone pair lone pair repulsion while (a) is the most stable of the given structures as there are minimal lone pair and bond pair repulsions and no lone pair lone pair repulsion.

That got me thinking what actually qualifies as a lone pair lone pair repulsions. Is it when the lone pairs are at an angle equal to or less than the 90° or is it something else?


Reference:

[1]: Chemical Bonding and Molecular Structure. NCERT Chemistry Part I, Textbook for Class XI, First Edition(October 2019 Reprint); Uppal, Shveta, Banerjee, Binoy, Eds.; National Council of Educational Research and Training: New Delhi, India, 2006; pp 113.

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When applying VSEPR, imagine that the various electron pairs occupy "orbitals", in essence volumes shaped like balloons, protruding from the central atom (in this case chlorine). The electrons ("orbitals") repel each other.

Which repulsive interaction is strongest?

One assumes that, placed in equivalent relative positions (angles), LP-LP (LP=lone pair, BP=bonding pair) repulsion is strongest, followed by LP-BP and then BP-BP. The idea is that bonding electrons are more distant from the central atom, whereas nonbonding (LP) electrons are held tightly by the central atom and therefore represent a more "concentrated" charge, and therefore repel other electrons most strongly. The repulsion is also stronger the smaller the angle between the electron pairs.

The strength of repulsion determines the order in which to prioritize the interactions, but this can be a little tricky, as your example shows, because it is a "non-linear" problem. When more than one pair of LP is involved, the geometry should minimize LP-LP repulsions followed by LP-BP repulsions, but there is a tradeoff. While (c) places both LPs on opposite ends ($180^\circ$) and would seem to minimize the repulsion, there is a bit of a tradeoff between LP-LP and LP-BP repulsions. Geometry (c) results in a lot (6) of LP-BP repulsions at right angles ($90^\circ$), whereas geometry (a) has 4 LP-BP repulsions at right angles. This leads to positioning of the LPs in equatorial (!) positions, even though LP-LP distance is not maximized.

Such arguments are not transparent, they are based on a crude method that only approximates the "real" electron distribution. The experimental geometry is in fact a distorted version of (a), with the LP-BP angles slightly greater than right angles ($>90^\circ$).

My practical advice is to simply remember that according to VSEPR lone pairs are always placed in the equatorial plane in trigonal bipyramidal electron pair geometries (those with 5 pairs). See for instance the following table:

enter image description here

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  • $\begingroup$ So like how are we comparing, what's the criteria so as to call the repulsion interaction as not much as destabilising in comparison like in (a) lp lp is at 120 degrees which is less than that of 180 degrees in (c) but still (a) is being called as more stable. $\endgroup$ – Rishi Mar 22 at 10:45
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    $\begingroup$ @Rishi They are all repulsions, but as I try to explain the interaction doesn't change linearly, which makes your question good. It is difficult to add up the total repulsion if you can't say "an LP-LP repulsion of 180 degrees is so-and-so much energy" and "an LP-LP repulsion of 120 degrees is so-and-so much energy". Maybe there is a table but I haven't seen it. It is in my opinion confusing and that is why I give my advice in the end: when you have 5 electron pairs, always place lp in equatorial positions. $\endgroup$ – Buck Thorn Mar 22 at 10:48

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