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Standard XII

Chemistry

Examples of Ionic Compounds

KCl crystalli...

Question

KCl crystallizes in the same type of lattice as does NaCl. Given that $\frac{r_{N a +}}{r_{C l -}} = 0.50$ and $\frac{r_{N a +}}{r_{K +}} = 0.70$ , calculate the ratio of the side of the unit cell for KCl to that of for NaCl.

1.143

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1.224

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1.414

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0.875

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Solution

The correct option is A $1.143$
For NaCl type structure, $r^{+} + r^{-} = \frac{a}{2}$ , where a is the edge length of the unit cell.
So, $\frac{a_{K C l}}{a_{N a C l}} = \frac{r_{K}^{+} + r_{C l -}}{r_{N a}^{+} + r_{C l^{-}}} = \frac{\frac{r_{K^{+}}}{r_{C l^{-}}} + 1}{\frac{r_{N a^{+}}}{r_{C l^{-}}} + 1} = \frac{\frac{0.5}{0.7} + 1}{0.5 + 1} = 1.143$

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Similar questions

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(Give your answer upto two decimal value)

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Q. KCl crystallizes in the same type of lattice as does NaCl. Given that

\frac{r_{N a^{+}}}{r_{C l^{-}}} = 0.5

and

\frac{r_{N a^{+}}}{r_{K^{+}}} = 0.7

Find the ratio of the sides of unit cell for KCl to that for NaCl :

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KCl crystallizes in the same type of lattice as does NaCl. Given that rNa+rCl−=0.50 and rNa+rK+=0.70, calculate the ratio of the side of the unit cell for KCl to that of for NaCl.

The correct option is A 1.143For NaCl type structure, r++r−=a2, where a is the edge length of the unit cell.So, aKClaNaCl=r+K+rCl−r+Na+rCl−=rK+rCl−+1rNa+rCl−+1=0.50.7+10.5+1=1.143

KCl crystallizes in the same type of lattice as does NaCl. Given that $\frac{r_{N a +}}{r_{C l -}} = 0.50$ and $\frac{r_{N a +}}{r_{K +}} = 0.70$ , calculate the ratio of the side of the unit cell for KCl to that of for NaCl.