Get Answers to all your Questions

header-bg qa

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)

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

Posted by

subam

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE