Chemistry, asked by Arifan2466, 1 year ago

CsCl crystallizes in a cubic cell that has a Cl- at each corner and Cs at the centre of the unit cell.if the radius of of Cs is 1.69Å and rcl - = 1.81Å, what is the edge length of unit cell?

Answers

Answered by disha2ooo
17
we know for simple cubic lattice, rc+ra=a 
1.69+1.81=3.5A



pknavas1997: Ith shariyanoo guyss
cool12385: 4.04 A is the answer i think
Answered by BarrettArcher
22

Answer : The edge length of unit cell is, 4.0414\AA

Solution :

From the given information, we conclude that the  CsCl has BCC (body centered cubic unit cell) structure.

The relation between the edge length and radius of unit cell.

2(r_{Cs^+}+r_{Cl^-})=\sqrt{3}\times a

where,

a = edge length of unit cell

r_{Cs^+} = radius of cesium cation = 1.69\AA

r_{Cl^-} = radius of chloride anion = 1.81\AA

Now put all the given values in the above relation, we get the edge length of the unit cell.

2(1.69\AA+1.81\AA)=\sqrt{3}\times a

a=4.0414\AA

Therefore, the edge length of unit cell is, 4.0414\AA

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