When the following five anions are arranged in order of decreasing ionic radius, the correct sequence is.

S e^{2 -}, I^{-}, B r^{-}, O^{2 -}, F^{-}

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I^{-}, S e^{2 -}, B r^{-}, O^{2 -}, F^{-}

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I^{-}, S e^{2 -}, O^{2 -}, B r^{-}, F^{-}

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S e^{2 -}, I^{-}, B r^{-}, F^{-}, O^{2 -}

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Solution

The correct option is B $I^{-}, S e^{2 -}, B r^{-}, O^{2 -}, F^{-}$
When the following five anions are arranged in order of decreasing ionic radius, the correct
sequence is $I^{-}, S e^{2 -}, B r^{-}, O^{2 -}, F^{-}$ .
Thus, the option (B) is the correct answer.
Note: As we move down the group, the atomic/ionic radii increases as additional shells of electrons are added. In a period, the ionic radii of isoelectronic species decreases with increase in the atomic number. Hence, the ionic radius of $O^{2 -}$ is greater than the ionic radius of $F^{-}$ . The same reason is true for the ionic radii of $S e^{2 -}, and B r^{-}$

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Similar questions

Among the following species, how many have their ionic size greater than $O^{2 -}$ ?___

$S e^{2 -}, F^{-}, N^{3 -}, P^{3 -}$

Q. The correct order of polarizability is

I^{-}, B r^{-}, C l^{-}, F^{-}

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and

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is as follows:

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