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The no. of particle present in octahedral void in ZnS (Blend) structure is:
    Correct answer is '0'. Can you explain this answer?
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    The no. of particle present in octahedral void in ZnS (Blend) structur...
    Explanation:

    Zinc sulfide (ZnS) has a blend structure where both zinc and sulfur atoms occupy the positions of face-centered cubic (fcc) lattice, and half of the tetrahedral voids are occupied by sulfur atoms.

    Octahedral voids are present in the fcc lattice between the layers of zinc atoms.

    However, the size of the octahedral voids in ZnS is much smaller than the radius of the sulfur atom. Hence, no sulfur atom can fit into the octahedral voids.

    Therefore, the number of particles present in the octahedral voids in ZnS is '0'.

    Conclusion:

    In ZnS (Blend) structure, the number of particles present in the octahedral voids is '0' as the size of the octahedral voids is smaller than the radius of sulfur atoms, and hence, no sulfur atom can fit into the octahedral voids.
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    The no. of particle present in octahedral void in ZnS (Blend) structur...
    Yes because all are present in tetrahedral void
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    The no. of particle present in octahedral void in ZnS (Blend) structure is:Correct answer is '0'. Can you explain this answer?
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