I have been trying to find out the bond angle of $\ce{H2O}$, but every site I visit has a different answer.
So far, I have found the following angles listed:
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1$\begingroup$ Your second link does not say that water has a 107.5 degree bond angle. Rather, 104.5. $\endgroup$– DissenterCommented Aug 27, 2014 at 3:34
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2$\begingroup$ It actually says both... I suspect that 107.5 was a typo. $\endgroup$– daviewalesCommented Aug 27, 2014 at 4:14
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3$\begingroup$ First, there is no precise value due to Heisenberg uncertainty principle - there is only an average value. Second, under what conditions? The average bond angle will be different in liquid water at room temperature than in a vacuum at OK. $\endgroup$– Jeff HammondCommented Aug 27, 2014 at 23:07
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$\begingroup$ Hmm... I guess I was looking for the average angle at standard temperature and pressure in liquid water. $\endgroup$– daviewalesCommented Aug 28, 2014 at 10:56
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2 Answers
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In your comment, you say you are interested in "liquid water".
The values in your self-answer are for gas phase water, and the second source is citing the first source, so it is really just one source. The value is determined from gas phase rotation spectroscopy.
For liquid water, the value is not as precisely known:
105.5° (calculated) and 106° (experimental) as reported in Structural, electronic, and bonding properties of liquid water from first principles J. Chem. Phys. 111, 3572.
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While researching the above, I found some reliable sources that suggest the bond angle is 104.48º.
EDIT:
As DavePhD says, these values are for water in the gas phase. For liquid water, the bond angle is around 105.5-106º.
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$\begingroup$ Huh, I"ve never seen it written out to two decimal places. I guess this means that one of David Brin's books (which have a plot point involving 104 degrees) are technically correct and it would round down. $\endgroup$– CanageekCommented Aug 27, 2014 at 6:46
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$\begingroup$ Source 1 actually lists a theoretical value to three decimal places, but it lists the two decimal place version in the abstract. $\endgroup$ Commented Aug 28, 2014 at 23:42