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In a cubic close packed structure(ccp) of mixed oxide, it is found that lattice has O2- ions and one half of the octahedral voids are occupied by trivalent cations (X3+) and one eighth of the tetrahedral voids are occupied by divalent cations (Y2+). What is the formula of the mixed oxide? 

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)


We have:

For ccp, we know:

Number of  Oxide ions = 4

We know that in any closed packed structure consisting of n atoms, the number of Octahedral voids is n while the number of Tetrahedral voids is equal to 2n

\therefore Number of trivalent ions in octahedral voids =\frac{1}{2}\times4=2

Similarly, number of divalent ions in tetrahedral voids =\frac{1}{8}\times8=1

Thus the formula is  X_2YO_4

Therefore, ?Option(4) is correct 

Posted by

Suraj Bhandari

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