Assertion (A): The ionic radii of Na^{+} and | vedclass

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(D) The electronic configuration of $Na^{+}$ is $1s^2 2s^2 2p^6$ ($10$ electrons).The electronic configuration of $F^{-}$ is $1s^2 2s^2 2p^6$ ($10$ el

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$(A)$ is not correct but $(R)$ is correct

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(D) The electronic configuration of $Na^{+}$ is $1s^2 2s^2 2p^6$ ($10$ electrons).The electronic configuration of $F^{-}$ is $1s^2 2s^2 2p^6$ ($10$ electrons).Since both have the same number of electrons,they are isoelectronic species,so $(R)$ is correct.However,the ionic radius depends on the nuclear charge $(Z)$.For $Na^{+}$,$Z = 11$,and for $F^{-}$,$Z = 9$.As the nuclear charge increases,the attraction of the nucleus on the electrons increases,leading to a smaller ionic radius.Therefore,the ionic radius of $Na^{+}$ $(102 \ pm)$ is smaller than that of $F^{-}$ $(133 \ pm)$.Thus,$(A)$ is incorrect.

Assertion $(A)$: The ionic radii of $Na^{+}$ and $F^{-}$ are same.
Reason $(R)$: Both $Na^{+}$ and $F^{-}$ are isoelectronic species.
The correct answer is:

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Assertion $(A)$: The ionic radii of $Na^{+}$ and $F^{-}$ are same.Reason $(R)$: Both $Na^{+}$ and $F^{-}$ are isoelectronic species.The correct answer is: