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For which pair of species is the difference in radii the greatest?

(A) $\ce{Li}$ and $\ce{F}$

(B) $\ce{Li+}$ and $\ce{F^-}$

(C) $\ce{Li+}$ and $\ce{O^2-}$

(D) $\ce{O^2-}$ and $\ce{F^-}$

I am pretty sure it is not (D) as $\ce{O}$ and $\ce{F}$ are quite close to each other on the periodic table and should have atomic radii pretty much the same.

I also know that gaining an electron increases the atomic radius whereas losing one decreases atomic radius. Also, as you move to the right of the periodic table atomic radius decreases.

I think that the answer could be (A) because $\ce{Li}$ is larger than $\ce{F}$, and the other choices result in $\ce{Li}$ becoming smaller and $\ce{F}$ becoming larger which decreases the difference in radii, but this is not the correct answer.

The answer given was C. How can this be explained?

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Your argument that $r_\ce{Li} > r_\ce{F}$ is valid only for the neutral atoms.Consider the lithium ion. Its electronic configuration is exactly similar to helium but the nuclear charge is greater by one unit, and hence is even smaller than the helium atom. On gaining an electron, fluorine now has the electronic configuration of neon with lesser nuclear charge and hence is larger than a neon atom. Similar, oxide ion, with even lesser nuclear charge is the largest. The decrease in radius of lithium on losing one electron (reduction by one entire shell), more than compensates for the difference in the radii of neutral lithium and fluorine. Finally, the order appears to be:-

$$\ce{Li+ < He < F < Li < F- < O^{2-} }$$

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