How are there #pi# bonds in #B_2# molecule without #sigma# bonds?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

5

This answer has been featured!

Featured answers represent the very best answers the Socratic community can create.

Learn more about featured answers

Nov 27, 2017

There shouldn't be, and that's why #"B"_2# is very unstable; it has two orthogonally-localized half-#pi# bonds, which is quite a weak bond.

Imagine that each #"B"# takes the place of a #"CH"# on #"HC"-="CH"#.

https://chem.libretexts.org/

Then, take away the #sigma# bond, the two #"H"# atoms, and one electron from each of the two full #pi# bonds, and you have a structure similar to #"B"_2#.

What about #"Li"_2# "molecule"? Is that stable? (Its bond length is #"267.3 pm"#, over twice the length of an average bond.)


The first chemical bond made in a molecule is preferentially a #bbsigma# bond.

#sigma# bonds are formed from a direct atomic orbital overlap. In comparison, #pi# bonds are sidelong overlaps and thus, #sigma# overlaps are made preferentially because they form the stronger bond.

#"B"_2# contains two boron atoms, which each use a basis of a #1s#, a #2s#, and three #2p# atomic orbitals.

For #"Li"_2# through #"N"_2#, there exists an orbital mixing effect that makes the #sigma# molecular orbital for a #2p_z-2p_z# overlap (#sigma_(g(2p))#) higher in energy than the #pi# molecular orbitals for a #2p_(x//y)-2p_(x//y)# overlaps (#pi_(u(2p))#).

Inorganic Chemistry, Miessler et al., Ch. 5.2.3

So, this particular orbital energy ordering takes place in the molecular orbital diagram, similar to the #"C"_2# molecule:

where #|E_(sigma_(2p_z))| > {|E_(pi_(2p_x))| = |E_(pi_(2p_y))|}#.

Since #"C"# has one more electron, #"B"_2# would have a similar MO diagram, EXCEPT for two fewer electrons. This, at first glance, seems to suggest that #"B"_2# has two half-#pi# bonds, with a molecular electron configuration of:

#color(green)((sigma_(1s))^2(sigma_(1s^"*"))^2(sigma_(2s))^2(sigma_(2s^"*"))^2(pi_(2p_x))^1(pi_(2p_y))^1)#

Since #"B"_2# is paramagnetic with two electrons, in order to make that bond, there are indeed two half-#bbpi# bonds, which form what we represent improperly in line notation as a #sigma# bond... and it isn't actually a #sigma# bond.

#:"B"-"B":#

That isn't a #sigma# bond, but two half-#pi# bonds. We would then expect #"B"_2# to be very unstable, which it is. (The lone pairs are from the #sigma_(2s)# and #sigma_(2s)^"*"#, since bonding + antibonding filled = nonbonding.)

Was this helpful? Let the contributor know!
1500
Trending questions
Impact of this question
97 views around the world
You can reuse this answer
Creative Commons License