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KCl crystallises in the same type of lat...

KCl crystallises in the same type of lattices as does NaCl. Given that ` r_(Na^(+)) //r_(Cl^(-)) = 0.55 and r_(K^(+)) //r_(Cl^(-)) = 0.74 ` . Calculate the ratio of the side of the unit cell of KCl to that of NaCl.

A

1.123

B

0.891

C

1.414

D

0.414

Text Solution

Verified by Experts

We aim at : ` (r_(K^(+)) + r_(Cl^(-)) // ( r_(Na^(+) + r_(Cl^(-)))) `
Given ` r_(Na^(+))/(r_(Cl^(-)) )= 0.55 and r_(K^(+))/(r_(Cl^(-)) = 0.74 `
` r_(Na^(+))/(r_(Cl^(-)))+ 1=1 .55 and r_(K^(+))/(r_(Cl^(-))) + 1 = 1.74 `
i.e, ` r_(Na^(+) + r_(Cl^(-)))/(r_(Cl^(-))) = 1.55 `
` and ( r_(K^(+)) + r_(Cl^(-)))/(r_Cl^(-)) = 1.74 `
Dividing (ii) by (i), `(r_(K^(+))+r_(Cl^(-)))/(r_(NA^(+))+r_(Cl^(-)))= 1.74/1.55 = 1.123`
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