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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
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.
