The distance between adjacent, oppositely charged ions is rubidium chloride is $3.285 \overset{°}{A}$ ; in potassium chloride is $3.139 \overset{°}{A}$ ; in sodium bromide is $2.981 \overset{°}{A}$ and in potassium bromide is $3.293 \overset{°}{A}$ . The distance between adjacent oppositely charged ions in rubidium bromide is :

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3.393 $\overset{°}{A}$

3.439 $\overset{°}{A}$

3.385 $\overset{°}{A}$

3.147 $\overset{°}{A}$

answer is D.

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Detailed Solution

For Rubidium Bromide

Let distance be 'd'

${Rb}^{+} + {Br}^{-}$

We get,

${Rb}^{+} + {Br}^{-}$ $= (3.285 + 3.293 - 3.139) \overset{°}{A}$

$d = 3.439 \overset{°}{A}$

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