I know that when hydrogen acts as a cation, it can form H-bonds with the electronegative ions or groups. But, is this also true for anionic hydrogens? Will they form similar bonds, other than the existing ionic bond, with electropositive ions, like the alkali metals and alkalikne earth metals, or maybe some other electropositive group?
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1$\begingroup$ Do the bonds in B2H6 count? $\endgroup$– Ivan NeretinApr 6, 2021 at 18:02
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1$\begingroup$ And how are you imagining this? Do you know how hydrogen bonds work, because it sounds like you talk about something else. $\endgroup$– MithoronApr 6, 2021 at 18:02
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1$\begingroup$ Particularly bridging hydride ligand that @IvanNeretin mentions and are a very different thing then hydrogen bonds. There are also agostic interactions, which are like an analogue of hydrogen bonds, but "hydridic" instead of "protic". $\endgroup$– MithoronApr 6, 2021 at 18:19
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$\begingroup$ @Mithoron.. H-bonds are formed because of the polarisation effect due to unequal sharing of electrons between the ions.. if I'm not wrong, then shouldn't any molecule having a dipole moment show coulombic attraction among its neighbours? $\endgroup$– Pratik DasApr 7, 2021 at 2:45
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1$\begingroup$ The point of hydrogen bonds is they are something more then dipole interactions. $\endgroup$– MithoronApr 7, 2021 at 13:37
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1 Answer
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On the surface, the answer is no, because that's not what we would call a hydrogen bond.
On a deeper level, the answer is... probably still no for NaH or CsH, because they are too ionic for this. Other hydrides, however, actually do have a thing which looks mighty like an "inverted" hydrogen bond (though we don't normally call it that, either).
Really, look at it this way: in $\ce{H2O}$ we have a hydrogen with partial positive charge, which gets attracted to a lone pair of another electronegative atom and forms something like a (3c, 4e) bond.
Now look at another scenario: a hydrogen in $\ce{BH3}$ (or better yet, in $\ce{AlH3}$) bears partial negative charge and gets attracted to a less electronegative atom with an empty orbital, so as to form a (3c, 2e) bond.
Whether to perceive these two scenarios as similar or as opposite is up to you.
So it goes.
answered Apr 6, 2021 at 18:33