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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

Q6(a)(i). Interhalogen compounds; preparation and structure of IF₇

  • 15 marks
  • Answered
Question

What are the main reasons for the formation of interhalogen compounds? Give one example of each type. Give the method of formation of IF₇ and draw its structure.

Answer in brief

Halogens differ in size and electronegativity, and the larger, less electronegative halogen can expand its octet; so XX'ₙ compounds (n=1,3,5,7) form. Types: ClF, ClF₃, BrF₅, IF₇. IF₇: IF₅ + F₂ heated (250–300°C); pentagonal bipyramid (sp³d³, D₅ₕ).

Key points

  • the X–X' bond is polar, with partial ionic character, so a mixed bond can be as strong as or stronger than the mean of X–X and X'–X'; all the fluorine interhalogens are exergonic (Δ_fG°<0).
  • the larger halogen (Cl, Br, I) is the central atom and can hold several small F atoms around it. The number of F atoms rises with the size ratio: Cl and Br go up to five, only iodine reaches seven (IF₇).
  • Cl, Br and I have vacant nd orbitals of suitable energy and can use 3, 5 or 7 bonding pairs (sp³d, sp³d², sp³d³).
  • the F–F bond is weak and F is the most electronegative element, so it oxidises other halogens to high oxidation states (+3, +5, +7).

The full model answer

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

Why interhalogens form

  1. Difference in electronegativity: the X–X' bond is polar, with partial ionic character, so a mixed bond can be as strong as or stronger than the mean of X–X and X'–X'; all the fluorine interhalogens are exergonic (Δ_fG°<0).
  2. Difference in size: the larger halogen (Cl, Br, I) is the central atom and can hold several small F atoms around it. The number of F atoms rises with the size ratio: Cl and Br go up to five, only iodine reaches seven (IF₇).
  3. Expanded octet: Cl, Br and I have vacant nd orbitals of suitable energy and can use 3, 5 or 7 bonding pairs (sp³d, sp³d², sp³d³).
  4. Fluorine: the F–F bond is weak and F is the most electronegative element, so it oxidises other halogens to high oxidation states (+3, +5, +7).
  5. n is always odd: X has seven valence electrons, each X' uses one, and the total must be even (all electrons paired), so 1+n is even.

The four types

TypeExamplePairs on XHybridShape
AXClF (also BrCl, ICl)1 bp, 3 lp—linear
AX₃ClF₃ (BrF₃)3 bp, 2 lpsp³dT-shaped
AX₅BrF₅ (ClF₅, IF₅)5 bp, 1 lpsp³d²square pyramidal
AX₇IF₇ (only)7 bp, 0 lpsp³d³pentagonal bipyramidal

IF₇: preparation and structure

  1. Preparation: fluorination of IF₅ with F₂ at 250–300°C: IF₅ + F₂ → IF₇ (or iodine directly with excess fluorine: I₂ + 7F₂ → 2IF₇). IF₇ is a colourless solid that sublimes at about 5°C.
  2. Structure: I (+7) uses all seven valence electrons in seven I–F bonds; no lone pair. Seven bond pairs take a pentagonal bipyramid: five F in an equatorial pentagon (F–I–F 72°), two axial F at 90° to the pentagon; sp³d³, D₅ₕ.
  3. Only iodine is big enough to hold seven F atoms; ClF₇ and BrF₇ are unknown.

Application. interhalogens are strong fluorinating agents: ClF₃ and BrF₃ fluorinate uranium to volatile UF₆ for isotope enrichment; IF₅ is a mild fluorinating agent that can be handled in glass; liquid BrF₃ is an ionising, aprotic solvent (2BrF₃ ⇌ BrF₂⁺ + BrF₄⁻).

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Model answer to Q6(a)(i), 2016 Paper I: solution page 31
Solution page 31 of 46
Read the whole 2016 solution2016 question paper

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