1.

The ionic radius of `Cl^(-)` ion is `1.81 Å` . The inter-ionic distances of NaCl and NaF are `2.79 Å` respectively. The ionic radiusof `F^(-)` ion will beA. `0.98 Å`B. `0.80 Å`C. `1.33 Å`D. `2.29 Å`

Answer» Correct Answer - C
`d_(NaCl)=r_(Na).^(+)+r_(Cl-)`
`2.79=r_(Na)+1.81`
or `r_(Na)+=2.79-1.81 Å =0.98 Å`
`d_(NaF)=r_(Na).^(+)+r_(F).^(-)`, i.e., `2.31=0.98+r_(F).-` or
`r_(F).- =2.31-0.98=1.33 Å`


Discussion

No Comment Found

Related InterviewSolutions