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If the anion (A) form hexagonal closet p...

If the anion (A) form hexagonal closet packing and cation (C ) occupy only 2/3 octahedral voids in it, then the general formula of the comound is:

A

AB

B

`A_3B_2`

C

`A_2B`

D

`AB_2`

Text Solution

Verified by Experts

The correct Answer is:
B

No of octahedral voids in hcp=6
so formula `=A_6B_4=A_3B_2`
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In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes. Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then, R (radius of tetrahedral void) = 0.225 R In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then, R (radius of octahedral void = 0.414 R). If the anions (A) form hexagonal close packing and cations (C ) occupy only 2/3rd octahedral voids in it, then the general formula of the compound is

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