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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?

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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
 

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