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I am reading up on boron in my notes and there is a certain paragraph that I do not understand

The tendency of $\ce{B}$ to form $\ce{B-B}$ bonds is less pronounced than that of carbon, its periodic table neighbor on the right, to form $\ce{C-C}$ bond, but it is much greater than that of the nearest element in the group, $\ce{Al}$, to form $\ce{Al-Al}$ bonds

Anyone can explained to me why is $\ce{B-B}$ bonds can be formed in the first place?

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Why wouldn't they? — In order to ask a good answerable question, please give more information about what you know already, and what you don't understand. Otherwise, we don't know what helps you or not. For example, my direct answer to your question would be “why not? have you looked at the molecular orbital diagram for homonuclear diatomics?”… but I have no way to guess if that would actually help you or not. –  F'x Nov 8 '12 at 15:33
I agree with @F'x. A boron-boron bond will form because it is lower in energy than the two atoms separately (definition of a bond in all bonding theories). A better question is why the B-B bond is less common than the C-C bond, but more common than the Al-Al bond. –  Ben Norris Nov 9 '12 at 12:14

1 Answer 1

As mentioned in the comments, why not? A boron-boron bond will form as long as there is an increase in stability of the system. Forming a bond requires energy, but there is a lot of energy released once it gets formed. The feasibility of a bond depends which energy is greater.

One example of where a boron-boron bond occurs is in Aluminium diboride / Mangnesium diboride

enter image description here

Here, you have graphite-like "sheets" of boron (pink), with aluminium layers between them.

This will probably be quite stable, due to delocalization. The singly-filled aluminium $p$-orbitals will overlap with the empty $p$ orbitals of Boron, and will delocalize across the graphite-like sheet (I cannot confirm this, though). This will lead to a large increase in stability.

A few more compounds with $\ce{B-B}$ bonds are Lanthanum hexaboride and various Yttrium borides. These are probably very stable due to the crystal structure (though delocalization may be a factor as well)

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It's even possible to stabilize the B_{2} dimer with carbene ligands; this complex has an isolated B-B triple bond. See sciencemag.org/content/336/6087/1420. –  J. LS Feb 27 at 20:10

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