Q42 / 120 · NEET 2012 (Mains) · Chemistry

Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal $A$ and the octahedral voids are occupied by a monovalent metal $B$. The formula of the oxide is

  • A
    $\mathrm{ABO}_2$
  • B
    $\mathrm{A}_2 \mathrm{BO}_2$
  • C
    $\mathrm{A}_2 \mathrm{~B}_3 \mathrm{O}_4$
  • Correct answer
    D
    $\mathrm{AB}_2 \mathrm{O}_2$

Solution

According to ccp, Number of $\mathrm{O}^{2-}$ ions $=4$ So, tetrahedral void $=8$ and $\quad$ octahedral void $=4$ Since, $A$ ions occupied $\frac{1}{4}$ th of tetrahedral void. $\therefore$ Number of $A$ ions $=\frac{1}{4} \times 8=2$ Again, $B$ ions occupied all octahedral void. $\therefore$ Number of $B$ ions $=4$ $$ \begin{aligned} A: B: \mathrm{O} & =2: 4: 4 \\ & =1: 2: 2 \end{aligned} $$ $\therefore$ Structure of oxide $=A B_2 \mathrm{O}_2$

Hence, the correct answer is option: 'D'.

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