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In a bcc lattice having the edge length of $200 \mathrm{pm}$, the cation has the radius of $70 \mathrm{pm}$.
The radius ratio of $\frac{r^{+}}{r^{-}}$is
(Given, $\sqrt{2}=1.4, \sqrt{3}=1.7$ and $\sqrt{6}=2.4$ )
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The radius ratio of $\frac{r^{+}}{r^{-}}$is
(Given, $\sqrt{2}=1.4, \sqrt{3}=1.7$ and $\sqrt{6}=2.4$ )
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Verified Answer
The correct answer is:
0.7
$\because$ For bcc lattice, radius ratio for $r^{+} / r^{-}$is 0.7 , i.e. radius ratio of cation by anion ranges from 0.732 to 1.0. Also, for bcc $r^{-}$ions are at corners, Thus,
$4 \times r^{-}=\sqrt{3} a \quad(\because \sqrt{3} a=1.7 \times 200)$
(where, $a=$ edge length)
and $\quad 4 \times r^{-}=1.7 \times 200$
$r^{-}=\frac{1.7 \times 200}{4}=85 \mathrm{pm}$
$\therefore \quad \frac{r^{+}}{r^{-}}=\frac{70}{85}=0.82$, which is close to 0.7.
$4 \times r^{-}=\sqrt{3} a \quad(\because \sqrt{3} a=1.7 \times 200)$
(where, $a=$ edge length)
and $\quad 4 \times r^{-}=1.7 \times 200$
$r^{-}=\frac{1.7 \times 200}{4}=85 \mathrm{pm}$
$\therefore \quad \frac{r^{+}}{r^{-}}=\frac{70}{85}=0.82$, which is close to 0.7.
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