[Douglas]

Suppose we had five bases. [OH]-, [R-C#C]- (triple bond), [NH₂]-, [CH₂=CH]-, and [CH₃-CH₂]-. How can I list these in order of weakest to strongest bases without considering K_b?

I believe [NH₂]- is the weakest base, but I can't decide between the pairs ([OH]- and [R-C#C]-) and ([CH₃-CH₂]- and [R-C#C]-). My feeling is [OH]- is the stongest base, but something about the hybridization of the lewis bases is bothering me.

Any ideas?
Thanks!

[RBF]

When considering the base strength of ions with central atoms within the same row of the periodic table (where size differences are not significant), the more electronegative atom can better stabilize a negative charge.  When comparing base strength of species having the same central atom but different hybridization (such as carbon), the closer the electrons are held to the nucleus, the more stable the species.

[Douglas]

Ok, so with your help, I should be able to conclude that [OH]- is the best base (given that a 1M solⁿ would have a pK_b of 14) since O is more electronegative than either N or C. But doesn't "feel" right when considering versus organic lewis bases.

When I compare the sp hybridized C versus the sp³ hybridized carbon, the sp³ electrons should be considered to be closer to the nucleus since the lobes of sp are so large. Thus [CH₃-CH₂]- should be "more stable", but I'm unclear on how stability translates to strength of the lewis base.

So I think I'm left with even more questions.

[Dan]

Ok, so with your help, I should be able to conclude that [OH]- is the best base (given that a 1M solⁿ would have a pK_b of 14) since O is more electronegative than either N or C.

Re-read RBF's post. A more electronegative atom stabilises the negative charge to a greater extent. If it is more stable it is less basic...

When I compare the sp hybridized C versus the sp³ hybridized carbon, the sp³ electrons should be considered to be closer to the nucleus since the lobes of sp are so large.

You need to review the relative radial extension of s and p orbitals (of the same principal quantum number). s orbitals are more contracted than p, so the more s character in the hybrid orbital, the smaller that orbital is.

[Douglas]

Ah, yes the misconceptions I had seem to have been my downfall here. So I can conclude several things (hopefully) correctly:

As electronegativity increases for an atom, so too does stability. This, in turn, creates a weaker base. Thus [OH]- is the weakest base in the original list, followed by [NH₂]-. All the carbon containing lewis bases are then stronger.

To compare the basicity of carboanions, you need to consider e^-_r (the electron radius) and the s character of the orbital. As e^-_r decreases, stability increases and thus a comparatively weaker base. e^-_r will decrease with increasing s character, and therefore sp will have the smallest e^-_r and be a weaker base than a comparative atom that is sp² and subsequently sp³ hybridized.

Combining these conclusions, my series from weakest to strongest base would be:

[OH]-
[NH₂]-
[R-C#C]-
[H₂C=CH]-
[CH₃-CH₂]-

Hopefully my misconceptions are remedied. Thank you for the help.