The correct order of decreasing ionic radius among the following anions would be:

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

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

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

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

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Solution

The correct option is C $I^{-}, S e^{2 -}, B r^{-}, O^{2 -}, F^{-}$
The correct order of decreasing ionic radius among the following anions would be $I^{-} > S e^{2 -} > B r^{-} > O^{2 -} > F^{-}$

On moving down the group, the ionic radii increases as new shells of electrons are added. The radius of the isoelectronic species decreases with increasing charge.

Hence, option C is correct.

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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^{-}

Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy ) is
(a) F > Cl > Br > I (b) F < Cl < Br < I (c) F < Cl > Br > I (d) F < Cl < Br < I

Q. The correct order of increasing radii of the ions

B r^{-}, F^{-}, O^{2 -}

and

S^{2 -}

is as follows:

Q. Among the following, the isoelectronic species is/are :
(i)

O^{2 -}, F^{-}, N a^{+}, M g^{2 +}

(ii)

N a^{+}, M g^{+}, A l^{3 +}, F^{-}

(iii)

N^{3 -}, O^{2 -}, F^{-}, N e

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