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The order of decreasing basicity in the four halide ions is:
- $\ce{I- > Br- > Cl- > F-}$
- $\ce{Cl- > Br- > I- > F-}$
- $\ce{F- > Cl- > Br- > I-}$
- $\ce{Cl- > F- > Br- > I-}$
The answer is option 3) but i'm getting as option 1).
My reasoning is that these elements belong to the same group, and the atomic radius increases down the group. Bases are electron donors, and iodide is the largest, so it can donate electrons the easiest, hence it should be most basic. Why is this incorrect?
Define basicity. For Brønsted-Lowry, it's all about $\ce{H+}$ and the conjugate bases. In Brønsted theory, acceptance of a hydrogen ion is what a base does. If you didn't know before, you should know that the hydrogen halides acid strength is $\ce{HF < HCl < HBr < HI}$. So for the conjugate bases, you just flip the sign $\ce{F- > Cl- > Br- > I-}$.
You way overthought the question. You should also think about where you got this idea about "electron donation". (Lewis acid-base theory deals with electron pair donation/acceptance but there's a huge difference between electron donation and electron pair donation.) You claim for some species $\ce{X-}$, that if $\ce{X- -> X^0 + e-}$ easily, then $\ce{X-}$ is a stronger base (than some $\ce{Y- -> Y^0 + e-}$ which is less easy). (By 'easily', assume I mean in the thermodynamic (energetic) sense)
Well, take a look at that and ask yourself whether it is relevant to Brønsted acid-base theory. I say no, it's not. Whether it is true or not, it is not relevant to the base strength as defined by Brønsted-Lowry (at least it has no clear, direct relevance).
