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KCET2014ChemistryClassification of Elements and Periodicity in Properties

Volume occupied by single ( CsCl ) ion pair in a crystal is ( 7.014 10⁻²³ ~cm ³ ). The smallest ( Cs - ) Cs inter-nuclear distance is equal to length of the side of the cube corresponding to volume of one ( CsCl ) ion pair. The smallest ( Cs - Cs ) inter-nuclear distance is nearly

Options

  1. A( 4.3 Å )
  2. B( 4.5 Å )
  3. C( 4.4 Å )
  4. DNone of the above

Correct answer

D. None of the above

Step-by-step solution

Volume of the unit cell =7.014 10⁻²³ a³=7.014 10⁻²³ (a is the smallest Cs-Cs distance) a= 7.014 10⁻²³ a=4.12 Å

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