The correct decreasing order of ionic radius is:

C^{4 -} > N^{3 -} > O^{2 -} > F^{-}

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F^{-} > O^{2 -} > N^{3 -} > C^{4 -}

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O^{2 -} > F^{-} > C^{4 -} > N^{3 -}

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N^{3 -} > O^{2 -} > F^{-} > C^{4 -}

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Solution

The correct option is A $C^{4 -} > N^{3 -} > O^{2 -} > F^{-}$
All the specified elements has the same number of electrons in the ionic state mentioned, i.e., they are isoelectronic. For isoelectronic species the ionic size decreases with increasing atomic number.

Hence correct order is: $C^{4 -} > N^{3 -} > O^{2 -} > F^{-}$

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Similar questions

C^{4 -}, N^{3 -}, O^{2 -}

F^{-}

N^{3 -}

N a^{+}

F^{-}

M g^{2 +}

O^{2 -}

The correct order of ionic radii of the species $N^{3 -}, O^{2 -}, N a^{+}$ and $F^{-}$ is:

The ionic radii (in Å) of N3–, O2–and F–are respectively:

The ionic radii (in $A^{\circ}$ of $N^{3}, O^{2 -} a n d F^{-}$ ) respectively are

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