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Volume occupied by single CsCl ion pair in a crystal is 7.014×10−23cm3.The smallest Cs - Cs internuclear distance is equal to length of the side of the cube corresponding to volume of one CsCl ion pair. The smallest Cs to Cs internuclear distance is nearly

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a

4.4Ao

b

4.3Ao

c

4Ao

d

4.5Ao

answer is C.

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Detailed Solution

Volume, V=7.014×10−23cm3As smallest Cs to Cs internuclear distance is equal to length of the side of the cube, i.e. a, therefore, a3=7.014×10−23cm3or ,a=7.014×10−233cm3=4.125×10−8=4.12≈4Ao

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