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CsCl crystallises in a cube cell that has a Cl- at each corner and Cs+ at the centre of the unit cell.If radius of Cs+ is 1.69 A and rCl- =1.81 A.What is the edge length of unit cell.?

CsCl crystallises in a cube cell that has a Cl- at each corner and Cs+ at the centre of the unit cell.If radius of Cs+ is 1.69 A and rCl- =1.81 A.What is the edge length of unit cell.?

Grade:12

4 Answers

Ramreddy IIT Patna
askIITians Faculty 49 Points
9 years ago
CsCl is BCC structure.
so in BCC structure atoms touch along the body diagonal, which implies that
2(rCs+ + rCl- )= \sqrt{3}\times a
you can calculate the value of a from above equation.
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Shriya Mehrotra
12 Points
9 years ago
sir the ans by dis method is coming wrong.ans is 4.04 A.
Ridham Rana
13 Points
5 years ago
No it's right,a=2*3.5/✓3=7/1.732=4.04 Å
So the formula is right
It is the same only you have to write ✓3 as 1.732 and not 1.7 to the the exact answer.
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
CsCl is BCC structure.
so in BCC structure atoms touch along the body diagonal, which implies that
2(rCs+ + rCl- ) = √3 a
2 x (1.69 + 1.81) = 1.732a
Hence, a = 2 x 3.5/1.732
a = 4.04 Å
 
Thanks and regards,
Kushagra

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