Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

The ionic radius of Na$^{+}$ ion is 0.95 and the inter-nuclear distance between ion pairs in the NaCl crystal is 2.76. What is the radius of the Cl$^{-}$ ion?

Can you answer this question?

1 Answer

0 votes
Answer: 1.81
In an ionic C$^{+}$A$^{-}$ ionic crystal, d (C$^{+}$ - A$^{-}$) = r (C$^{+}$) + r(A$^{-}$)
$\Rightarrow$ r (Cl$^{-}$) = d(Na$^{+}$ - Cl$^{-}$) - r(Na$^{+}$) $= 2.76 - 0.95 = 1.81$
answered Jul 23, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App