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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

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

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

$=18.6pm$

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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