1.

Theionic radius ofCl^(-)ion is1.81 Å . The inter-ionic distances of NaCl andNaF are2.79 Årespectively. The ionic radiusof F^(-)ionwill be

Answer»

`0.98 Å`
`0.80 Å`
`1.33 Å`
`2.29 Å`

Solution :`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`


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