Atoms of element ‘A’ and ‘B’ form an HCP lattice such that atoms of ‘A’ are present at corners of the unit cell and atoms of ‘B’ are present at the centers of both the hexagonal faces as well as inside the unit cell. All the tetrahedral voids are occupied by atoms of element ‘C’ and all the octahedral voids are occupied by atoms of element ‘D’. Find out the formula of the crystalline solid.
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Answer: AB₂C₆D₃ is the formula of the given crystalline solid in hcp.
Explanation: 'A' is present at the corners of unit cell which is shared by 6 unit cells. So its contribution is 1/6.
No. of atoms of A=1/6×12=2
'B' is present at the face center of both hexagonal faces as well as inside the unit cell. So its contribution is 1/2 for the face center and 1 for atoms present inside the unit cell.
No. of atoms of B=1/2×2 +3×1 =4
No. of tetrahedral voids=2×(total no. of atoms)=12 1 for inside the unit cell.
Therefore, no. of atoms of C=12
No. of octahedral voids =6 (equal to the no. of atoms)
Therefore, No. of atoms of D=6
A₂B₄C₁₂D₆ which when taken in the simplest ratio gives the formula of the compound that is AB₂C₆D₃.