If the interionic distance between Na^{+} and | vedclass

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

Question

(A) The interionic distance $d$ is the sum of the ionic radii: $d = r_{Na^{+}} + r_{F^{-}} = 2.31 \ \mathring{A}$.Since $F^{-}$ is an anion and $Na^{+

Vedclass · Accepted Answer

$0.95 \ \mathring{A}$ and $1.36 \ \mathring{A}$

Vedclass · Answer

(A) The interionic distance $d$ is the sum of the ionic radii: $d = r_{Na^{+}} + r_{F^{-}} = 2.31 \ \mathring{A}$.Since $F^{-}$ is an anion and $Na^{+}$ is a cation,the radius of the anion is larger than the radius of the cation $(r_{F^{-}} > r_{Na^{+}})$.Based on periodic trends and experimental values,$r_{Na^{+}} \approx 0.95 \ \mathring{A}$ and $r_{F^{-}} \approx 1.36 \ \mathring{A}$.Summing these: $0.95 \ \mathring{A} + 1.36 \ \mathring{A} = 2.31 \ \mathring{A}$.

If the interionic distance between $Na^{+}$ and $F^{-}$ ions is $2.31 \ \mathring{A}$,then the radii of $Na^{+}$ and $F^{-}$ are:

Similar Questions

The correct order of the decreasing ionic radii among the following isoelectronic species is

The ionic radii of $Na^{+}$,$F^{-}$,$O^{2-}$,and $N^{3-}$ follow the order:

The correct arrangement for the ions in the increasing order of their radii is

Identify the element having the smallest atomic radius from the following:

The ionic sizes decrease in the order :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module