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This question already has an answer here:

In Concise Inorganic Chemistry by J.D.Lee (Adapted by Sudarsan Guha, Fourth Edition), on page 83, under the topic "Bridge Bonding" it is given for diborane $(\ce{B2H6}):$

The $d_\ce{B-H}(\text{terminal bonds}) < d_\ce{B-H}(\text{bridge bonds})$

The energy required to replace hydrogen atoms from the bridged position is more than that needed for the terminal position. This is supported by the fact that on reaction with methyl chloride, terminal hydrogen atoms are replaced by methyl group in preference to the bridged hydrogen atoms.

The reason for this could be the fact that to break the bridge bond, the overlapping zone should be broken from both sides and the overall energy required is more as compared to do so for terminal bonds.

Structure of Diborane:

Diborane Structure Wikipedia Image Credit: Wikipedia

Based on the second paragraph it can be concluded that terminal bonds are weaker than bridge bonds. But some sources on the internet including Wikipedia state the bridge bonds are weaker than the terminal bonds without any satisfactory reason. I wish to know the reason for this.

Further, kindly explain why the terminal H-B-H bond angle is larger than bridging H-B-H bond angle as evident from the molecular structure as given above. I have searched my book as well as google to my best, but I didn't get any reason for this.

Kindly note, I read the unaccepted answer to this question. But still my doubt persists and makes me type this question by adding relevant details from my book and internet.

Kindly clarify this doubt.

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marked as duplicate by Mithoron, Mathew Mahindaratne, Nilay Ghosh, Todd Minehardt, M.A.R. Aug 28 at 11:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Well, thanks for linking, I'd need to look for it otherwise. In future do not ask about things that are already asked and answered and perhaps get a better book... $\endgroup$ – Mithoron Aug 25 at 19:47
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    $\begingroup$ Even though I agree that this question and sub-questions have been more or less covered previously, this shouldn't discourage you from asking new questions. From what I noticed, so far your questions were nicely formulated and well referenced (especially for a new user), and if you don't manage to immediately find the answer by searching the Chemistry.SE (BTW see this Meta post), it's better to ask rather than keep staying in the dark. $\endgroup$ – andselisk Aug 25 at 22:04
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    $\begingroup$ @Mithoron, I read the answer and the comments following it. One of my doubts is clarified. I have highlighted the other doubt and added some details about a related problem. Now, I feel this question is different from the one mentioned as a possible duplicate. To be honest, the book I am currently reading is considered to be the best Inorganic Chemistry book for JEE Advanced. But I came to know (after asking questions on Chem.SE) many concepts are outdated. But we are supposed to learn these! $\endgroup$ – Intellex Aug 26 at 9:44
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    $\begingroup$ @Mithoron, The answer to that question is good, but I feel the statement in my book is equally good proving bridge bonds are stronger contrary to what most reliable sources say. I need to know why this kind of reasoning is incorrect. That's why I posted it. $\endgroup$ – Intellex Aug 26 at 10:01
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    $\begingroup$ @andselisk, Thank you for your comment :). I read that Meta post. First of all, I thought of editing that post and starting a bounty for that. I felt that would be compressing another question into the existing question, and hence started off with a new one. $\endgroup$ – Intellex Aug 26 at 10:29
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It depends on what you define as a bond. A three-center two electron bond is in general stronger than an ordinary two-center bond because delocalizing the bond over three atoms instead of two makes the bonding MO more stable. But, the bridge bond is also shared between two linkages so each individual linkage has less bonding than a single two-center bond linkage.

The question of bond as angles is a bit more complex because it hides a couple other issues. Superficially, the bond angles in the bridge region are forcibly reduced by forming a four-membered ring. But that begs the following questions:

  • Why is the dimer formed instead of a trimer or tetramer, which would enable a larger ring and more natural bond anges?

  • Why is the bond angle through boron actually greater than the bond angle through hydrogen? Since hydrogen forms smaller atoms steric factors would favor the reverse.

An answer to both questions may lie in the tendency of 3c-2e bonds to form closed structures. The bonding orbital described above gets an extra bonding linkage by bringing the end atoms of a chain together. Such a tendency is seen in the formation of a triangular trihydrogen cation in interstellar space, and in the intermediate stage of a carbocation rearrangement.

In the case of diborane this tendency shows up as boron-boron bonding. This has in fact been proposed here. The dimer has a favorable geometry for forming such a boron-boron bond, whereas an oligomer with larger bond angles in the center would tend to hold the boron atoms too far apart.

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  • $\begingroup$ Thank you for your answer. I have made some corrections to the question. I learnt from other sources the bridge bond - a 3c-2e bond is intact weaker than ordinary bonds. I agree with that fact since even Wikipedia says the bridge bonds are weaker. But I feel the statement (2nd paragraph in the quoted text) is also equally true. I am confused. Kindly help me clarify my doubts. Sorry for irritating you. $\endgroup$ – Intellex Aug 26 at 9:49
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    $\begingroup$ Slight correction. The 3c-2e bond is weaker in each of its individual linkages, but adding up all the linkages gives a stronger bond overall. Otherwise the 3c-2e bond would just break up to make an ordinary bond. $\endgroup$ – Oscar Lanzi Aug 26 at 11:14

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