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Why are the bond angles in the diborane $\ce{B2H6}$ ring much less than $\ce{109.5^o}$? Below is the structure:

Diborane

This source says that the boron atoms are approximately $\ce{sp^3}$ hybridized, so I thought that the bond angles of the ring would be approximately $\ce{109.5^o}$. However, the diagram indicates that the angles of the ring are $\ce{97^o}$ and $\ce{83^o}$. Why is this so?

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    $\begingroup$ It's a 4-membered ring... $\endgroup$ – Mithoron Aug 9 '18 at 22:22
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    $\begingroup$ Diborane is better not thought of in terms of hybridization and bond angles. Think of cyclobutane: no multiple bonds, hence $sp^3$, hence should be $109.5^\circ$, but no, it is more like $90^\circ$. How so? $\endgroup$ – Ivan Neretin Aug 9 '18 at 22:28
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    $\begingroup$ Related: chemistry.stackexchange.com/questions/228/… (full disclosure, it's my question). @IvanNeretin hope you can forgive our hybridization slant to the question :) $\endgroup$ – jonsca Aug 9 '18 at 23:20
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    $\begingroup$ Possible duplicate of What makes banana bonds possible in diborane? $\endgroup$ – Abcd Aug 10 '18 at 5:42
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    $\begingroup$ It doesn’t seem to me that the proposed duplicate discusses bond angles. $\endgroup$ – a-cyclohexane-molecule Aug 10 '18 at 11:21

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