1.

Volume occupied by single CsCl ion part in a crystal is 7.014xx10^(-23) cm^(3).The smallest Cs-Cs internucleardistance is equal to length of the side of the cube corresponding to volume of one CsCl ion pair. The smallest Cs to Cs intermuclear distance is nearly

Answer»

`4.4Å`
`4.3Å`
`4.0Å`
`4.5Å`

SOLUTION :Volume of `CsCl=a^(3)`
`7.014xx10^(-23)CM^(3)=a^(3)` (a=edge LENGTH)
`a=root(3)(7.014xx10^(-23))`
`a=4.12xx10^(-8)cm=4.01Å~~4Å`


Discussion

No Comment Found

Related InterviewSolutions