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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 · 2018 · Paper I

Q1(d)(ii). P forms PCl₃ and PCl₅; N forms only NCl₃

  • 5 marks
  • Answered
Question

(ii) P forms two chlorides PCl₅ and PCl₃ whereas N forms only one NCl₃.

Answer in brief

P has vacant 3d orbitals: by promoting a 3s electron it gets five unpaired electrons and forms PCl₅ (sp³d) besides PCl₃ (sp³). N has no 2d orbitals; it cannot exceed three bonds (covalency 4 at most), so only NCl₃ forms.

Key points

  • N (2s²2p³) and P (3s²3p³) each have three unpaired electrons. sp³ hybridisation with one lone pair gives three bonds: pyramidal NCl₃ and PCl₃.
  • P has empty 3d orbitals in the valence shell. One 3s electron is promoted to 3d: 3s¹3p³3d¹, five unpaired electrons. sp³d hybrids give five P–Cl bonds: PCl₅ is trigonal bipyramidal (gas phase).
  • the second shell has only 2s and 2p; the next orbitals (3s) are far too high in energy to be used. N cannot unpair its 2s electrons, so its covalency stops at three bonds (four in NH₄⁺, using the lone pair). NCl₅ does not exist.
  • N is small, so five Cl atoms cannot fit around it. P is larger and holds five Cl (and six in [PCl₆]⁻). The extra energy of promotion is repaid by forming two more bonds.

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. PCl₃ and NCl₃: N (2s²2p³) and P (3s²3p³) each have three unpaired electrons. sp³ hybridisation with one lone pair gives three bonds: pyramidal NCl₃ and PCl₃.
  2. PCl₅: P has empty 3d orbitals in the valence shell. One 3s electron is promoted to 3d: 3s¹3p³3d¹, five unpaired electrons. sp³d hybrids give five P–Cl bonds: PCl₅ is trigonal bipyramidal (gas phase).
  3. N: the second shell has only 2s and 2p; the next orbitals (3s) are far too high in energy to be used. N cannot unpair its 2s electrons, so its covalency stops at three bonds (four in NH₄⁺, using the lone pair). NCl₅ does not exist.
  4. Size also counts: N is small, so five Cl atoms cannot fit around it. P is larger and holds five Cl (and six in [PCl₆]⁻). The extra energy of promotion is repaid by forming two more bonds.

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Model answer to Q1(d)(ii), 2018 Paper I: solution page 5
Solution page 5 of 55
Read the whole 2018 solution2018 question paper

All parts of Question 1