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The hydrogen atoms in a water molecule are negatively charged, so they should repulse each other completely due to the electromagnetic force, forcing the hydrogen atoms to be on opposite sides of the oxygen atom. However, this is not the case, as the H-O-H bond angle is $104.5 ^\circ$.

Why does the lone electron pairs repulse the hydrogen atoms to an angle of $104.5 ^\circ$ in $\ce{H2O}$ instead of $180 ^\circ$?

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closed as unclear what you're asking by Jannis Andreska, A.K., orthocresol, Jan, Ben Norris May 9 '16 at 20:26

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    $\begingroup$ Y'know, it gets a bit disappointing to see someone with more than 10 posts who still doesn't format their posts with Mathjax properly. $\endgroup$ – M.A.R. Sep 7 '15 at 23:21
  • $\begingroup$ @PhyCS Not sure what you're talking about... linear water in excited state? $\endgroup$ – Mithoron Sep 7 '15 at 23:55
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    $\begingroup$ Possible duplicate of Why is H₂O V shaped? $\endgroup$ – Mithoron May 9 '16 at 17:53
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According to VSEPR theory (there are other models, but they give the same qualitative results), since there are four electron pairs around oxygen they will be in a tetrahedral configuration. Assume the hydrogen atoms are at A and B as you've described them in your question; C and D (the lone pairs) are in a plane perpendicular to the one containing A and B. Below is a ball-and-stick model of water, showing the lone-pair electrons as yellow "atoms". Ball-and-stick model of water, showing the location of the lone pairs (Public domain image from Wikimedia Commons)

It sounds as if you're thinking in two dimensions with the hydrogen atoms and lone pairs in the same plane. If water existed in only two dimensions it could conceivably be linear, but it really is in three dimensions. (The molecule itself is flat because the lone pairs affect the shape but are not counted as part of it.)

All of the above applies to water under normal conditions. I'm not aware of any excited or exotic states of water in which the VSEPR prediction doesn't apply, but I can't guarantee that there aren't any.

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    $\begingroup$ Thank you so much! I have had a lot of good answers in the past, but they usually involved difficult (to me) concepts and sometimes calculus and a bunch of stuff that I did not even begin to understand. I don't know if this question was that easy, but you definitely gave me understanding to this problem and introduced to me new concepts that should serve me well. Thanks! $\endgroup$ – phi2k Sep 19 '15 at 21:07

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