# Is there a direct proportionality between base strength and ionic radius for anions in aprotic solvents?

##### 1 Answer
Jun 4, 2016

Let's look at a fairly straightforward example. Consider the hydrohalogenic acids, $\text{HI}$, $\text{HBr}$, $\text{HCl}$, and $\text{HF}$.

RADIUS AND BOND LENGTH VS. ACID/BASE STRENGTH

The ionic radii of the halides go as follows:

What this leads to is the following trend in $\text{H"-"X}$ bond length $r$, where $\text{X" = "F", "Cl", "Br} ,$ or $\text{I}$:

${r}_{\text{H"-"I") > r_("H"-"Br") > r_("H"-"Cl") > r_("H"-"F}}$

• The longer the $\text{H"-"X}$ bond length, the weaker the $\text{H"-"X}$ bond, and the stronger the acid.
• The stronger the acid, the weaker the conjugate base.

So, the strongest acid here is $\text{HI}$, which means ${\text{I}}^{-}$ is the weakest base.

CHARGE DENSITY VS. BASE STRENGTH

This correlates with a decreasing charge density as well.

Since ${\text{I}}^{-}$ has the largest ionic radius in this series, its charge density is the least concentrated (it is the most polarizable), and iodide is classified as a "soft base". In comparison, fluoride is a "hard base".

Iodide's higher polarizability means it is harder to localize (to concentrate into one spot) the electron density in between $\text{H}$ and $\text{I}$ than it is to do so between $\text{H}$ and $\text{F}$, making the bond weaker in $\text{H"-"I}$.

(Another interpretation is that the orbital overlap becomes less stabilizing because not all the electron density is in the right spot for the resultant molecular orbital to be as stable as possible.)

Hence, $\text{HI}$ is the strongest hydrohalogenic acid and ${\text{I}}^{-}$ is the weakest conjugate base in this series.

So, in this sense, there is a direct correlation where the increasing charge density leads to an increasing basicity.

(Note that this only holds true for nucleophilicity in aprotic solvent.)