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Name: JEE Advanced Chemistry
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Question

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Prashant Kotian · Accepted Answer

Correct answer: 00000.27.         2 2 2 2 1 1 1 ) ( 1 R b Z ;         2 2 2 3 1 1 1 ) ( 1 R b Z 32 27 9 1 1 4 1 1          = 32 27 32 27   × (0.32 Å) = 0.27 Å

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