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.

Calculate the ionic radii of K^(+) and Cl^(-)ions in KCl crystal.The i

1.	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.
Answer» SOLUTION :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`