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By using paulings method calculate the ionic radii ofK^(+)andCl^(-)ions in the potassium chloride crystal . Given thatd_(K^(+)-Cl^(-))=3.14Å. |
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Answer» SOLUTION :`d_(K^(+)-Cl^(-))=3.14Å` `r_(K^(+))=`? `r_(K^(+))+r_(Cl^(-))=3.14Å`. 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"))_(Cl^(-))=Z-S` `=19-[(0.35xx7)+(0.85xx8)+(1xx2)]` `=19-11.25=7.75` `therefore(R(K^(+)))/(r(Cl^(-)))=((Z_("eff"))_(Cl^(-)))/((Z_("eff"))_(K^(+)))=(5.75)/(7.75)=0.74` `thereforer_((K^(+)))=0.74r_((Cl^(-)))` Substitute the VALUE of `r_((K^(+)))`inequation (1) `0.74r_((Cl^(-)))+r_((Cl^(-)))=3.14Å` `1.74r_((Cl^(-)))=3.14Å` `thereforer_((Cl^(-)))=(3.14Å)/(1.74)=1.81Å` From (1) `r_((K^(+)))=3.14-1.81` `=_(-)1.33Å` `thereforer_((K^(+)))=1.33Åandr_((C^(-)))=1.81Å` |
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