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Help me answer: If edge length and radius of anion in anti fluorite structure are 320 pm and 120 pm respectively then radius of cation is

If edge length and radius of anion in anti fluorite structure are 320 pm and 120 pm respectively then radius of cation is

  • Option 1)

    36.8 pm

  • Option 2)

    18.6 pm

  • Option 3)

    60 pm

  • Option 4)

    116 pm

 
Answers (1)
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S subam

Edge\; length=\frac{4}{\sqrt{3}}\left ( r_{cation} +r_{anion}\right )

r_{cation}=\left ( \frac{320\times \sqrt{3}}{4} \right )-r_{anion}=138.6-120

=18.6pm

 

Antifluoride Structure -

O^{2-} arrange in CCP

Na^{+} found in tetrahedral void

Edge length = \frac{4}{\sqrt{3}} (r_{Na^{+}} + r_{O^{2-}})

Coordination number = 4 : 8

 

 

 

- wherein

 Number of Na^{+} = 8

Number of O^{2-} = 4

Number of Na_{2}O molecule per unit cell = 4

Ex :  K_{2}O , Rb_{2}O , Na_{2}S etc

 

 


Option 1)

36.8 pm

This is incorrect

Option 2)

18.6 pm

This is correct

Option 3)

60 pm

This is incorrect

Option 4)

116 pm

This is incorrect

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