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
- A( 4.3 Å )
- B( 4.5 Å )
- C( 4.4 Å )
- 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