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JEE Advanced
Exam
JEE Advanced
Subject
Chemistry
Chapter
Solid State

JEE Advanced · Chemistry · Solid State

Paragraph: In a cubic unit cell, atoms (A) are present at all...

Paragraph: In a cubic unit cell, atoms (A) are present at all corners, (B) are present at alternate face centre and all edge centres. Atoms (C) are present at face centres left from (B) and one at each body diagonal at distance of 1/4th 1 / 4^{\text{th }} of body diagonal from corner.Question: Formula of given solid is

Options

  1. AA3 B8C7\mathrm{A}_3 \mathrm{~B}_8 \mathrm{C}_7
  2. BAB4C6\mathrm{AB}_4 \mathrm{C}_6
  3. CA6 B4C8\mathrm{A}_6 \mathrm{~B}_4 \mathrm{C}_8
  4. DA2 B9C11\mathrm{A}_2 \mathrm{~B}_9 \mathrm{C}_{11}

Correct answer

B. AB4C6\mathrm{AB}_4 \mathrm{C}_6

Step-by-step solution

Atoms A(\mathrm{A}( at corner )=8×18=1)=8 \times \frac{1}{8}=1 Atoms B=2B=2 (alternate face centre) ×12+12( edge centre )×14=4\times \frac{1}{2}+12(\text { edge centre }) \times \frac{1}{4}=4 Atoms C=4C=4 (face centre left from B) ×12+1×4( body diagonal )=6∴\times \frac{1}{2}+1 \times 4(\text { body diagonal })=6\therefore Formula AB4C6\mathrm{AB}_4 \mathrm{C}_6
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