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 




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. 


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