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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2016 · Paper I

Q1(c)(ii). Spacing of the (110) planes of a cubic cell

  • 5 marks
  • Answered
Question

What is the perpendicular distance between (110) planes of a cubic unit cell?

Answer in brief

d₁₁₀=a/√2=0.707 a, where a is the edge of the cubic cell.

Key points

  • The (110) plane cuts the x and y axes at a and runs parallel to z (intercepts a, a, ∞; reciprocals 1, 1, 0). Its family is x+y=0, a, 2a,…
  • Looking down z, each plane is a line across the square face. The perpendicular from the origin to x+y=a ends at (a/2, a/2), half-way along the face diagonal of length √2a.
  • Distance =(a/2)²+(a/2)², d₁₁₀=a/√2=0.7071 a
  • d₁₁₀=a/1²+1²+0². For a=400 pm: d₁₀₀=400, d₁₁₀=282.8, d₁₁₁=230.9 pm. Higher indices, closer planes.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

  1. The (110) plane cuts the x and y axes at a and runs parallel to z (intercepts a, a, ∞; reciprocals 1, 1, 0). Its family is x+y=0, a, 2a,…
  2. Looking down z, each plane is a line across the square face. The perpendicular from the origin to x+y=a ends at (a/2, a/2), half-way along the face diagonal of length √2a.
  3. Distance =(a/2)²+(a/2)², d₁₁₀=a/√2=0.7071 a
  4. The general formula gives the same: d₁₁₀=a/1²+1²+0². For a=400 pm: d₁₀₀=400, d₁₁₀=282.8, d₁₁₁=230.9 pm. Higher indices, closer planes.

In centred cells: in a body-centred cell the body-centre atom lies on the plane x+y=a, so a/√2 is also the real spacing. In a face-centred cell the face-centre atoms at (12,0,12) lie on an extra plane midway, so the atomic layers are a/(2√2) apart and the reflection is indexed (220).

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Model answer to Q1(c)(ii), 2016 Paper I: solution page 6
Solution page 6 of 46