Question

KCL crystallises same type of lattice as does Nacl calcutale the ratio of the side of unit cell of KCl to NaCl

Answer 1

it seems your question is incomplete. A constant question is ------> KCl crystallizes in the same type of lattice as does NaCl.Given that,$\frac{r_{Na^+}}{_r{Cl^-}}$=0.5,$\frac{r_{Na^+}}{r_{K^+}}$=0.7.Calculate the ratio of the side of the unit cell for KCl to that for NaCl.

solution : we know, NaCl crystallizes in the face centered cubic lattice.

so, $r_{Na^+}+r_{Cl^-}=\frac{a}{2}$

where a is edge length of NaCl crystal lattice.

given, $\frac{r_{Na^+}}{_r{Cl^-}}$=0.5

$\frac{r_{Na^+}+r_{Cl^-}}{r_{Cl^-}}=1.5$

or, $r_{Na^+}+r_{Cl^-}=1.5r_{Cl^-}$

again, $\frac{r_{Na^+}}{r_{K^+}}=0.7$

or, $\frac{\frac{r_{Na^+}}{r_{Cl^-}}}{\frac{r_{K^+}}{r_{Cl^-}}}=0.7$

or,$\frac{0.5}{\frac{r_{K^+}}{r_{Cl^-}}}=0.7$

or,$\frac{r_{K^+}}{r_{Cl^-}}=\frac{0.5}{0.7}$

or, $r_{K^+}+r_{Cl^-}=\frac{1.2}{0.7}r_{Cl^-}$

now, $\frac{a_{KCl}}{a_{NaCl}}=\frac{r_{K^+}+r_{Cl^-}}{r_{Na^+}+r_{Cl^-}}$

= $\frac{1.2}{0.7}\times\frac{1}{1.5}$

= 1.142