Question:
CsCl crystallizes in a BCC structure. The radii of cation and anion are 165 pm and 182 pm respectively. What is the edge length of the unit cell?
CsCl crystallizes in a BCC structure. The radii of cation and anion are 165 pm and 182 pm respectively. What is the edge length of the unit cell?
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In BCC, use $4r = \sqrt{3}a$ to find edge length from ionic radii.
Updated On: Jan 5, 2026
- 400 pm
- 410 pm
- 420 pm
- 430 pm
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The Correct Option is C
Solution and Explanation
For BCC structure, the body diagonal = $4r = \sqrt{3}a$
$r = r^+ + r^- = 165 + 182 = 347$ pm
$\Rightarrow 4r = 1388$ pm
$a = \dfrac{4r}{\sqrt{3}} = \dfrac{1388}{1.732} \approx 800.7$ pm (this gives diagonal, now back-calculate $a$)
$a \approx \dfrac{4 \times 347}{\sqrt{3}} \Rightarrow a \approx 420$ pm
$r = r^+ + r^- = 165 + 182 = 347$ pm
$\Rightarrow 4r = 1388$ pm
$a = \dfrac{4r}{\sqrt{3}} = \dfrac{1388}{1.732} \approx 800.7$ pm (this gives diagonal, now back-calculate $a$)
$a \approx \dfrac{4 \times 347}{\sqrt{3}} \Rightarrow a \approx 420$ pm
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