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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 · 2017 · Paper I

Q4(a)(ii). Complete 1s wavefunction and its spherical shape

  • 10 marks
  • Answered
Question

Write the complete wave function for 1s-orbital and deduce its shape.

Answer in brief

ψ₁ₛ=(1)/(π)((Z)/(a₀))^(3/2)e^(-Zr/a₀). It depends on r only, not on θ or φ, so the orbital is spherically symmetric: a sphere with no node.

Key points

  • Complete wavefunction = radial part × angular part, ψₙₗₘ=Rₙₗ(r) Yₗₘ(θ,φ). For 1s (n=1, l=0, mₗ=0): R₁₀=2((Z)/(a₀))^(3/2)e^(-Zr/a₀), Y₀₀=(1)/(√(4π)) ψ₁ₛ=(1)/(√(π))((Z)/(a₀))^(3/2)e^(-Zr/a₀) (a₀=52.9 pm; Z=1 for H)
  • at a given r it has the same value in every direction. Every surface of constant ψ is a sphere centred on the nucleus: the 1s orbital is spherically symmetric.
  • Nodes. Radial nodes =n-l-1=0; angular nodes =l=0. ψ never changes sign; it is largest at the nucleus and falls exponentially.
  • Radial distribution. The probability in a shell of radius r and thickness dr is P(r) dr=4π r²ψ²dr. dP/dr=0 gives the most probable radius r=a₀/Z (52.9 pm for H).

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. Complete wavefunction = radial part × angular part, ψₙₗₘ=Rₙₗ(r) Yₗₘ(θ,φ). For 1s (n=1, l=0, mₗ=0): R₁₀=2((Z)/(a₀))^(3/2)e^(-Zr/a₀), Y₀₀=(1)/(√(4π)) ψ₁ₛ=(1)/(√(π))((Z)/(a₀))^(3/2)e^(-Zr/a₀) (a₀=52.9 pm; Z=1 for H)
  2. Shape. ψ₁ₛ contains no θ or φ: at a given r it has the same value in every direction. Every surface of constant ψ is a sphere centred on the nucleus: the 1s orbital is spherically symmetric.
  3. Nodes. Radial nodes =n-l-1=0; angular nodes =l=0. ψ never changes sign; it is largest at the nucleus and falls exponentially.
  4. Radial distribution. The probability in a shell of radius r and thickness dr is P(r) dr=4π r²ψ²dr. dP/dr=0 gives the most probable radius r=a₀/Z (52.9 pm for H).
  5. Boundary surface. 32.3 % of the electron density lies within r=a₀, 76.2 % within 2a₀, and 90 % within a sphere of radius 2.66a₀: this sphere is drawn as the shape of the 1s orbital.

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Model answer to Q4(a)(ii), 2017 Paper I: solution page 16
Solution page 16 of 38

All parts of Question 4