Calculate the ionic radii of `K^(+) and Cl^(-)` ions in KCl crystal. The internuclear distance between `K^(+) " an " Cl^(-)` ions are found to be 3.14Å.

Explanation

Given `d_(K^(+)-Cl^(-))=3.144oversetoA`
`r_(K^(+))=?` `r_(Cl^(-))=?`
i.e. `r_(K^(+))+r_(Cl^(-))=3.14oversetoA` .........(i)
We know that,
`(r_(K^(+)))/(r_(Cl^(-)))=(Z_(eff))_(Cl^(-))/((Z_(eff))_(K^(+)))`
`(Z_(eff))_(Cl^(-))=Z-S`
=`17-[(0.35xx7)+(0.85xx8)+(1xx2)]`
=17-11.25=5.75
`(Z_(eff))_(K^(+))=Z-S`
=`19-[(0.35xx7)+(0.85xx8)+(1xx2)]`
=19-11.25=7.75
therefore `(r_((K^(+))))/(r_((Cl^(-))))=((Z_(eff))_(Cl^(-)))/((Z_(eff))_(Cl^(-)))=(5.75)/(7.75)=0.74`
therefore`r_((K^(+)))=0.74r_((Cl^(-)))`
`0.74r_((Cl^(-)))+r_((Cl^(-)))=3.14oversetoA`
`1.74r_((Cl^(-)))=3.14oversetoA`
therefore `r_((Cl^(-)))=(3.14oversetoA)/(1.74)=1.81oversetA`
From(1),
`r_(K^(+))=3.14-1.81`
=`1.33oversetoA`
therefore`r_(K^(+))=1.33oversetoA` and `r_(Cl^(-))=1.81 oversetoA`