11
$\begingroup$

The only reason I can think of is that the lone pairs in the oxygen in water are in lower energy orbital relative to the nitrogen in ammonia due to increased effective nuclear charge in oxygen. However, as far as I am aware, ammonia is SIGNIFICANTLY more basic than water. Surely the orbital energy cannot account for this.

$\endgroup$
  • 1
    $\begingroup$ Basicity and acidity can be accounted for by the stability of the conjugate base and the conjugate acid. Try taking a look at $\ce{H3O+}$ and $\ce{NH4+}$'s $\mathrm{pK_a}$'s. $\endgroup$ – John Snow Nov 18 '14 at 18:31
  • 1
    $\begingroup$ How is that explained though? $\endgroup$ – RobChem Nov 18 '14 at 19:02
  • 1
    $\begingroup$ Hybridization changes from sp sp2 in water, sp2 to sp3 in ammonia. So in the first case there might be less relative repulsion with nucleus. However, the comparison might not be appropriate. $\endgroup$ – santimirandarp Sep 5 '18 at 18:25
6
$\begingroup$

When it comes to comparing ammonia and water, the acidity or basicity can be explained because of the central atoms electronegativity. When comparing atoms of the same periodic row, the more electronegative an atom is, the less it will want to donate an electron pair.

The electronegativity values from the Pauling Scale for $\ce{N}$ and $\ce{O}$ are

$$\ce{N}=3.04$$ $$\ce{O}=3.44$$

So based off of these values we can see that nitrogen is less electronegative, it less strongly it holds onto its electrons, and thus it is a better electron donor. This allows the addition of a $\ce{H}$ to be more stable.

The atomic reasoning behind this is that, as a general periodic trend, as you move across the periodic table, from left to right, the nucleus is increasing, which increases it's charge, producing a stronger attraction to its valence electrons.

$\endgroup$
  • 1
    $\begingroup$ Be careful with this approximation; if you continue to the left on that row, you find carbon, with less electro negativity, but it does not have a lone pair to share, so it won't behave like a base in methane (nor an acid), and if you move even more to the left, you find the weird boron, which will have an empty p orbital which, in principle, would make borane behave like a Lewis acid (I say in principle, because actually it shows an amphoteric behavior). $\endgroup$ – Altered State Nov 19 '14 at 0:37
  • 2
    $\begingroup$ Edited my answer to reflect that my statement is based off of a trend. $\endgroup$ – John Snow Nov 19 '14 at 0:44
  • 2
    $\begingroup$ Also, my statement was about electron pairs and your comment seems to not take into account molecules where carbon or boron would have lone pairs. Carbon would definitely have lone pairs if it was in a molecule like $\ce{CH3-}$, and you can sure bet that $\ce{CH3-}$ is a strong base. The same would be applied to a compound like $\ce{BH2-}$. $\endgroup$ – John Snow Nov 19 '14 at 2:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.