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

`r(K^(+)) + r(Cl^(-)) = d(K^(+) - Cl^(-)) = 3.14 Å …(1)`
`K^(+) and Cl^(-)` ions have Ar (Z=18) type configuration. The effective nuclear charge for `K^(+) and Cl^(-)` can be calculated as follows.
`{:(K^(+),=,(1s^2),(2s^2 2p^6),(3s^2 3 p^6)),("","","innershell",(n-1)^th" shell",n^(th)" shell"):}`
`Z^(**)(K^(+)) = Z-S`
`= 19 - [(0.35xx7) + (0.85 xx 8) + (1xx2)]`
`= 19 - 11.25 = 7.75`
`Z^(**)(Cl^(-)) = 17 [(0.35 xx7) + (0.85 xx8) + (1xx2)]`
`=17 - 11.25 = 5.75`
`:. (r(K^(+)))/(r(Cl^(-)))= (Z^(**)(Cl^(-)))/(Z^**(K^(+))) = 5.75/7.74 = 0.74`
`:. r(K^(+)) = 0.74r(Cl^(-))` ...(2)
Substitute (2) in (1)
`0.74 r (Cl^(-)) + r(Cl^(-)) = 3.14 Å` ...(3)
`1.74r(Cl^(-)) = 3.14Å`
`r(Cl^(-)) = (3.14Å)/(1.74) = 1.81Å`
From (2)
`r(K^(+)) = 0.74 r(Cl^(-))`
`=0.74 xx 1.81 Å`
`= 1.33Å`
`r(K^(+)) = 1.33 Å`
`r(Cl^(-)) = 1.81Å`