# Predict the order of ionic radius

We have arrange the followig sequence for decreasing order of ionic radius .

$Se^{-2},I^-,Br^{-},O^{-2},F^{-}$

In my book the answer given as $I^{-},Se^{-2},Br^{-},O^{-2},F^{-}$

But according to me $O^{-2}$ should be greater than F$^-$

As oxygen has two more electrons , so its zeff should decrease more in comparism to fluorine

• If you're showing it in decreasing order, as you've said, then oxygen and fluorine are in the correct order. Oxygen, being larger, would appear on the list in the order you've shown. Feb 2 '17 at 5:56
• Could you clarify your question? You expect the oxygen ion to be greater in size than the fluorine ion, which is also what your book is saying. I don't see your question other than confusion in what is meant by "decreasing order"? Feb 2 '17 at 5:59

You're right about the expected order. From largest to smallest the expected order would be:

$\ce{_{53}I^-}\quad$ = [Kr] $4d^{10}\text{ }5s^2\text{ }5p^6$ = [Xe]

$\ce{_{34}Se^{2-}}$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr]

$\ce{_{35}Br^-}\text{ }$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr]

$\ce{_8O^{2-}}\text{ }\text{ }$ = [He] $2s^2\text{ }2p^6$ = [Ne]

$\ce{_9F^-}\quad$ = [He] $2s^2\text{ }2p^6$ = [Ne]

You expressed the idea poorly though. $\ce{_8O^{2-}}$ and $\ce{_9F^-}$ have the same number of electrons. However $\ce{_9F^-}$ has an extra proton which should draw the electrons in closer.

There are a lot of ways to measure atomic/ionic size. No idea why the book would have $\ce{_8O^{2-}}$ and $\ce{_9F^-}$ flipped.

• I don't see how his book has it switched. It appears your ordering is identical to what he's saying his book shows. However, he says he expects the oxygen ion to be greater in size than the fluorine ion, which is also what his book is saying. I don't see a question other than confusion in what is meant by "decreasing order"? Feb 2 '17 at 5:57