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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

Which of the following is arranged in decreasing order of size?

Arrange the following in the decreasing order of radius: $S^{2-}, P^{3-}, Cl^{-}, Ca^{2+}, Ar, K^{+}$

Which of the following cases has the maximum size ratio of anion to neutral atom?

For which of the following ionic species would the radius be maximum?

Which of the following represents the correct order of atomic radius?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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