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Question

For a compound, the anion (X) crystallizes in hexagonal close packing and cation (Y) occupies only 13rd of octahedral voids in it, what will be the general formula of the compound?

A
YX
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B
Y2X
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C
YX2
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D
YX3
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Solution

The correct option is D YX3
In hexagonal closed packing, there are 6 octahedral voids completely inside the unit cell.
No. of atoms in hexagonal close packing =6
Since, anions (X) form hexagonal lattice
No. of X atoms =6
Cation 'Y' occupies 13 of the octahedral voids in the unit cell.
No. of Y atoms =6×13=2

Emperical formula of the compound =Y2X6 or YX3

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