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My teacher told me that lone pair and lone pair are closer together hence the repulsion is greater and they take up more space, but I do not understand? can anyone explain it to me?

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  • $\begingroup$ en.wikipedia.org/wiki/VSEPR_theory#Degree_of_repulsion $\endgroup$ – Wildcat May 30 '15 at 14:22
  • $\begingroup$ The key point: the bonding pair shared in a bond lies further from the atom than a nonbonding pair of that atom, which is held close to its positively charged nucleus. $\endgroup$ – Wildcat May 30 '15 at 14:26
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So, from Wikipedia article on VSEPR theory we read:

The overall geometry is further refined by distinguishing between bonding and nonbonding electron pairs. The bonding electron pair shared in a sigma bond with an adjacent atom lies further from the central atom than a nonbonding (lone) pair of that atom, which is held close to its positively charged nucleus. VSEPR theory therefore views repulsion by the lone pair to be greater than the repulsion by a bonding pair. As such, when a molecule has 2 interactions with different degrees of repulsion, VSEPR theory predicts the structure where lone pairs occupy positions that allow them to experience less repulsion. Lone pair-lone pair (lp-lp) repulsions are considered stronger than lone pair-bonding pair (lp-bp) repulsions, which in turn are considered stronger than bonding pair-bonding pair (bp-bp) repulsions, distinctions that then guide decisions about overall geometry when 2 or more non-equivalent positions are possible.

As I said in my comment, the key point is given in the second sentence: the bonding pair shared in a bond lies further from the atom than a nonbonding pair of that atom which is held close to its positively charged nucleus. Consequently, lone pairs are closer to each other than any other combination of pairs (lone pair-bonding pair and bonding pair-bonding pair), and thus, repell each other more strongly.

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  • $\begingroup$ so the stronger repulsion will make them stay further apart and occupy more space too? $\endgroup$ – XQ. Matsci May 30 '15 at 23:25
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    $\begingroup$ @XQ.Matsci, yes. Stronger repulsion -> bigger spatial separation. $\endgroup$ – Wildcat May 30 '15 at 23:31
  • $\begingroup$ Is it like - a lone pair consists of $2 \ e^-$ and hence they repel each other. Consequently occupying more space. On the other hand, the shared pair also repels, but there is attraction from $2$ nuclei due to which they are localized around the centre of the bond. $\endgroup$ – Kaushik Sep 27 at 17:00
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Simply put: a lone-pair orbital is "fatter" than a bonding orbital as I have illustrated here. This can be rationalized as follows: in a bonding orbital the electron pair is mainly attracted to 2 different nuclei which helps localize the electron pair. In a lone-pair orbital the electron pair is mainly attracted to a single nucleus and can therefore spread out more without reducing this attraction.

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  • $\begingroup$ Why does being attracted 2 different nuclei help localize the electron pair? $\endgroup$ – XQ. Matsci May 31 '15 at 8:44
  • $\begingroup$ For example in $\ce{H2}$ the point that is closest to both nuclei is the bond midpoint, i.e. the middle of the line connecting the two nuclei. This point will have the lowest electron-nuclear attraction energy and if you move away from that line this energy will increase. Thus, the electron pair will tend to localize between the atoms. $\endgroup$ – Jan Jensen May 31 '15 at 8:50
  • $\begingroup$ This VSEPR theory actually confused me for days. $\endgroup$ – XQ. Matsci May 31 '15 at 14:40
  • $\begingroup$ is coulombs law related to the repulsion? $\endgroup$ – XQ. Matsci May 31 '15 at 15:57
  • $\begingroup$ @JanJensen, thanks for this answer. It's simple and makes a lot of sense. $\endgroup$ – Hal May 31 '15 at 22:40

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