In a hypothetical ionic crystal, $B$ is arranged in cubic close packing and $A$ occupies all octahedral voids and alternative tetrahedral voids. The correct formula of the compound?

A B

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A_{2} B_{2}

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A B_{2}

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A_{2} B

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Solution

The correct option is C $A_{2} B$

No. of atoms ina ccp unit cell is $f c c = 4$

Therefore no. of $B$ atom present $= 4$

No of $A$ atoms= Alternate tetrahedral voids + octahedral voids

$= (\frac{8}{2} + 4) = 8$

The formula of the compound $= A_{8} B_{4}$ or $A_{2} B$

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In a hypothetical ionic crystal, B is arranged in cubic close packing and A occupies all octahedral voids and alternative tetrahedral voids. The correct formula of the compound?

The correct option is C A2B No. of atoms ina ccp unit cell is fcc=4 Therefore no. of B atom present =4 No of A atoms= Alternate tetrahedral voids + octahedral voids =(82+4)=8 The formula of the compound=A8B4 or A2B

In a hypothetical ionic crystal, $B$ is arranged in cubic close packing and $A$ occupies all octahedral voids and alternative tetrahedral voids. The correct formula of the compound?

The correct option is C $A_{2} B$

No. of atoms ina ccp unit cell is $f c c = 4$

Therefore no. of $B$ atom present $= 4$

No of $A$ atoms= Alternate tetrahedral voids + octahedral voids

$= (\frac{8}{2} + 4) = 8$

The formula of the compound $= A_{8} B_{4}$ or $A_{2} B$