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My textbook states:

In the staggered form of ethane, the electron clouds of C-H bonds are as far apart as possible, thus there are minimum repulsive forces, minimum energy and maximum stability. On the other hand, when the staggered form changes into the eclipsed form, the C-H bond clouds come closer to each other resulting in increased electron cloud repulsion. To check the increased repulsive forces, molecule will have to possess more energy. And therefore it is less stable

My question pertains to the statement in bold text. Am I correct in thinking that in the eclipsed conformer the electron clouds repel more, this repulsion would (sort of) want to break up the molecule so to fight this breaking tendency, the molecule has to possess greater energy to control these new repulsive forces?

If I am wrong, what does this statement mean?

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  • $\begingroup$ I think you have the right idea. To move from the staggered to the eclipsed conformation, you have to put in additional energy because otherwise the repulsive forces are enough to keep the molecule in the staggered form. $\endgroup$
    – Tyberius ♦
    Commented Sep 29, 2017 at 15:35

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Just take two electrons. If you want to push them closer together you need to input some energy and the system overall will have higher energy. There is a preference for the two electrons to be farther apart.

This situation is not any different if we look at pairs of electrons in bonds instead of individual electrons.

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  • $\begingroup$ The way I think of the two electron system is that the potential energy content is higher once we push them together is that that is the energy needed to set up the closer configuration on bringing the charges from infinite separation to the closer separation, which is the classical picture of potential energy of a system. While your answer does give insight, do you essentially mean that the molecule tries to push the electrons together so has greater energy? Which is what I asked, my question remains, am I correct in thinking the way I did? $\endgroup$
    – Inkjet
    Commented Sep 29, 2017 at 13:31
  • $\begingroup$ It's a little hard to parse what you mean. The classical picture doesn't matter because electron-electron repulsion is still defined in the same way. The electrons are overall closer in the eclipsed versus the staggered conformation. Not sure why you need to invoke breaking up the molecule. Increased repulsion increases energy. $\endgroup$
    – Zhe
    Commented Sep 29, 2017 at 14:20

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