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Standard XIII

Chemistry

Ionic Compounds and Radius Ratios

KCl crystalli...

Question

$K C l$ crystallises in the same type of lattice as $N a C l$ . Given that $r_{N a^{+}} / r_{C l^{-}} = 0.50$ and $r_{N a^{+}} / r_{K^{+}} = 0.70$ , calculate the ratio of the side of the unit cell of $K C l$ to that of $N a C l$ :

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
Rock salt type lattice structure means the edge length will be $a = r_{+} + r_{-}$
Hence, $\frac{a_{K C l}}{a_{N a C l}} = \frac{r_{K^{+}} + r_{C l^{-}}}{r_{N a^{+}} + r_{C l^{-}}} = \frac{\frac{r_{K^{+}}}{r_{C l^{-}}} + 1}{\frac{r_{N a^{+}}}{r_{C l^{-}}} + 1} = \frac{\frac{0.5}{0.7} + 1}{0.5 + 1} = 1.143$

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Similar questions

Q. 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.

K C l

crystallises in the same type of lattice as

N a C l

. Given that

r_{N a^{+}} / r_{C l^{-}} = 0.50

and

r_{N a^{+}} / r_{K^{+}} = 0.70

, calculate the ratio of the side of the unit cell of

K C l

to that of

N a C l

K C l

crystallises in the same type of lattice as does

N a C l

. Given that

r_{N a^{+}} / r_{C l^{-}} = 0.55

and

r_{K^{+}} / r_{C l^{-}} = 0.74

. Calculate the ratio of the side of the unit cell for

K C l

to that of

N a C l

KCl crystallizes in the same type of lattice as does NACl. Given that $\frac{r_{N a^{+}}}{r_{C l^{-}}} = 0.55$ and $\frac{r_{K^{+}}}{r_{C l^{-}}} = 0.74$ . Calculate the ratio of the side of the unit cell for KCl to that of NaCl.

K C l

crystallizes in the same type of lattice as

N a C l

. Given that

\frac{r_{N a^{+}}}{r_{C l^{-}}} = 0.50

and

\frac{r_{K^{+}}}{r_{C l -}} = 0.70

. Calculate the ratio of the side of the unit cell for

K C l

to that for

N a C l

(Give your answer upto two decimal value)

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

The correct option is A 1.143Rock salt type lattice structure means the edge length will be a=r++r− Hence, aKClaNaCl=rK++rCl−rNa++rCl−=rK+rCl−+1rNa+rCl−+1=0.50.7+10.5+1=1.143

$K C l$ crystallises in the same type of lattice as $N a C l$ . Given that $r_{N a^{+}} / r_{C l^{-}} = 0.50$ and $r_{N a^{+}} / r_{K^{+}} = 0.70$ , calculate the ratio of the side of the unit cell of $K C l$ to that of $N a C l$ :