Does basicity always vary inversely with ionic atomic radius???

1 Answer
Jun 22, 2016

Not necessarily... It's only a guideline, and one that easily gets confusing. So it's not a good guideline to try to make sense of.

We need to determine what "basicity" means. You did not specify either Lewis or Bronsted basicity, which both make a difference.

Take #"I"^(-)# and #"Cl"^(-)#, for example...


Bronsted basicity generally depends on the ability to polarize a proton's bond with another atom, weakening the bond and promoting the Bronsted base's acceptance of that proton.

When you place #"I"^(-)# and #"Cl"^(-)# in water... #"Cl"^(-)# is the stronger Bronsted base, because:

  1. It is smaller, making its electron density, well, denser, making it less polarizable by another species. Thus, it is a harder base.
  2. It is more electronegative, making it more polarizing towards the proton on water than #"I"^(-)# is on water.

That allows a stronger hydrogen-bonding interaction than #"I"^(-)# would have with water, which in this case makes it a stronger Bronsted base (a stronger proton acceptor).

(Indeed, we say that #"HI"# is the stronger acid, and so its conjugate base is generally stated to be a weak base.)


Lewis basicity may depend on the following factors:

  • electronegativity
  • solvent effects (hydrogen-bonding, for example)
  • strength of the conjugate acid's #"H"-"A"# bond(s), and the trade-off you get upon trying to make that bond.
  • and probably other stuff.

In terms of Lewis basicity, since #"I"^(-)# is less electronegative, its valence electrons are held less tightly to the nucleus (it's a softer base).

That increases its effectiveness as a Lewis base., pg. 16

However, #"I"^(-)# is larger than #"Cl"^(-)#, so the bonds it makes are naturally weaker, which is not a great tradeoff for donating its electrons, lowering its Lewis basicity.

It's unclear how this all balances out without more context. So, keep in mind... context is extremely important, and defining your words is as well.