CsBr has BCC structure. The length of its one side is 4 A ˚ . The minimum distance between Cs + and Br − ion will be -

r + + r − = 3 2 a = 2 1.73 × 4 = 3.46 A ˚

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CsBr has BCC structure. The length of its one side is $4 \overset{˚}{A}$ . The minimum distance between $Cs^{+}$ and $Br^{-}$ ion will be -

Answer:

Solution:

$r^{+} + r^{-} = 3 \frac{a}{2} = \frac{1.73 \times 4}{2} = 3.46 \overset{˚}{A}$

Stream:JEESubject:ChemistryTopic:Solid StateSubtopic:Cubic Crystals

⏱ 2mℹ️ Source: QB

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