The correct order of ionic radii is :

I^{-}

T e^{2 -}

A r^{+}

C l^{-}

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I^{-}

C l^{-}

T e^{2 -}

A r^{+}

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I^{-}

T e^{2 -}

C l^{-}

A r^{+}

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T e^{2 -}

I^{-}

C l^{-}

A r^{+}

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Solution

The correct option is D $T e^{2 -}$ > $I^{-}$ > $C l^{-}$ > $A r^{+}$
Size ∝ number of shells.
Number of shells present in $T e^{2 -}$ & $I^{-}$ (five shells) are more than that in $C l^{-}$ & $A r^{+}$ (three shells).
$T e^{2 -}$ & $I^{-}$ are isoelectronic.
Since, size ∝ negative charge.
So, $T e^{2 -}$ > $I^{-}$ and $C l^{-}$ > $A r^{+}$ .
Combining both factors, overall order is:
$T e^{2 -}$ > $I^{-}$ > $C l^{-}$ > $A r^{+}$

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