Medium MCQ +4 / -1 PYQ · JEE Mains 2020

A crystal is made up of metal ions 'M1' and 'M2' and oxide ions. Oxide ions form a ccp lattice structure. The cation 'M1' occupies 50% of octahedral voids and the cation 'M2' occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation numbers of 'M1' and 'M2' are, respectively :

  1. A + 3, + 1
  2. B + 4, + 2
  3. C + 1, + 3
  4. D + 2, + 4 Correct answer

Solution

O<sup>–2</sup> ions form ccp $\Rightarrow$ O<sub>4</sub>. <br><br>M<sub>1</sub> = 50% octahedral void = ${{50} \over {100}} \times 4$ = 2 <br><br>M<sub>2</sub> = 12.5% tetrahedral void = ${{12.5} \over {100}} \times 8$ = 1 <br><br>So formula is : (M<sub>1</sub>)<sub>2</sub>(M<sub>2</sub>)<sub>1</sub>O<sub>4</sub> <br><br>Let charge on M<sub>1</sub> and M<sub>2</sub> are +x and +y respectively. And O<sub>4</sub> has -8 charge. <br><br>As crystal is neutral. So metals must have +8 charge in total. <br><br>$\therefore$ +2x + y = 8 ....(1) <br><br>By checking options we found eq (1) satisfy when <br>x = +2 <br>y = +4

About this question

Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws

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