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  • Need explanation for: CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct ?

CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct ?

  • Option 1)

    r_{Cs^{+}}+r_{CI^{-}}=3a

  • Option 2)

    r_{Cs^{+}}+r_{CI^{-}}=\frac{3a}{2}

  • Option 3)

    r_{Cs^{+}}+r_{CI^{-}}=\frac{\sqrt{3}}{2}a\;

  • Option 4)

    r_{Cs^{+}}+r_{CI^{-}}=\sqrt{3}a

 

Answers (2)

best_answer

As we have learned

CsCl type structure -

Cl^{-} located at all corners

Cs^{+} located at body centre

Edge length = \frac{2}{\sqrt{3}} (r_{Cs^{+}} + r_{Cl^{-}})

Coordination number = 8:8

- wherein

Number of Cl^{-} = 1

Number of Cs^{+} = 1

Number of CsCl molecule per unit cell = 1

 

 

In  C_5 Cl structure , C_5^+  ion is in contact with Cl ^- ion at the nearest distance which is equal to \frac{\sqrt 3 a}{2}

 


Option 1)

r_{Cs^{+}}+r_{CI^{-}}=3a

Option 2)

r_{Cs^{+}}+r_{CI^{-}}=\frac{3a}{2}

Option 3)

r_{Cs^{+}}+r_{CI^{-}}=\frac{\sqrt{3}}{2}a\;

Option 4)

r_{Cs^{+}}+r_{CI^{-}}=\sqrt{3}a

Posted by

SudhirSol

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