To understand this question properly, we must know what an acid is and what a base is. Asn acid is a substance that is a proton donor (H+). So if you look at the halides in terms of acid strength, their ability to donate hydrogen ions upon dissociation in water, the fluorine ion would have the strongest electronegative pull. So HF would be the weakest acid, because the fluorine strongly holds on to its hydrogen ion. The iodide ion would be the strongest acid (HI) because it has a larger electron shell structure, is less electronegative, and will give up its hydrogen ion more willingly. Bases are the exact opposite of acids, ranked in their ability to donate an electron pair. So if fluorine were the most electronegative in terms of acidity, it would be the opposite in terms of base classification. The F would be the strongest base, followed by the Cl, the Br, and lastly, the I. My suggestion would be to understand either the way the acids work and are classified, or the bases, and then it would be vice-versa for the other one.
to undertand the basicity we take a look at the acidity of the halides.
HF, HCl, HBr, HI. All of them act as acid so
HX(x=for any halide F, Cl, Br, I)
HX + H2O = H3O+ + X-
among the four HI is the strongest and it follows the series
HI> HBr> HCl> HF
their conjugate base are of the different, so F- becomes the strongest
so we have:
F->> Cl- > Br-> I-
hope this helps :)