Q8(a)(i). Products of four interhalogen reactions
Predict the product of the following reactions with justification:
(A) ClF + BF₃ → (B) ClF + CsF → (C) ClF₃ + BF₃ → (D) I₂ + Br₂ →
Answer in brief
(A) 2ClF + BF₃ → [Cl₂F]+[BF₄]⁻; (B) ClF + CsF → Cs+[ClF₂]⁻; (C) ClF₃ + BF₃ → [ClF₂]+[BF₄]⁻; (D) I₂ + Br₂ → 2IBr.
Key points
- (A) ClF + BF₃: BF₃ takes F⁻ to form the stable tetrahedral [BF₄]⁻. A bare Cl⁺ is far too unstable, so the cation is stabilised by a second ClF: 2pt] 2ClF + BF₃ → [Cl₂F]⁺[BF₄]⁻ [Cl₂F]⁺ is bent, Cl–Cl–F (central Cl: two bond…
- (B) ClF + CsF: CsF is an ionic fluoride donor (large Cs⁺ stabilises a large anion); ClF accepts F⁻: 2pt] ClF + CsF → Cs⁺[ClF₂]⁻ [ClF₂]⁻ is linear: five electron pairs on Cl (trigonal bipyramid), three lone pairs equatorial,…
- (C) ClF₃ + BF₃: fluoride transfer from ClF₃ to BF₃: 2pt] ClF₃ + BF₃ → [ClF₂]⁺[BF₄]⁻ [ClF₂]⁺ is bent (four electron pairs on Cl, two lone pairs), isoelectronic with SCl₂.
- (D) I₂ + Br₂: the halogens combine directly to the interhalogen iodine monobromide: 2pt] I₂ + Br₂ → 2IBr IBr is a dark solid, polar (I^(δ⁺)-Br^(δ⁻), Br more electronegative). The reaction is reversible; IBr partly dissociates…
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Principle: interhalogens are amphoteric towards fluoride. A strong fluoride acceptor (Lewis acid: BF₃, SbF₅, AsF₅) removes F⁻ and leaves a halonium cation; a fluoride donor (CsF, KF) adds F⁻ and gives a polyhalide anion.
- (A) ClF + BF₃: BF₃ takes F⁻ to form the stable tetrahedral [BF₄]⁻. A bare Cl⁺ is far too unstable, so the cation is stabilised by a second ClF: 2pt] 2ClF + BF₃ → [Cl₂F]⁺[BF₄]⁻ [Cl₂F]⁺ is bent, Cl–Cl–F (central Cl: two bond pairs, two lone pairs, sp³).
- (B) ClF + CsF: CsF is an ionic fluoride donor (large Cs⁺ stabilises a large anion); ClF accepts F⁻: 2pt] ClF + CsF → Cs⁺[ClF₂]⁻ [ClF₂]⁻ is linear: five electron pairs on Cl (trigonal bipyramid), three lone pairs equatorial, like I₃⁻ (3c–4e bond).
- (C) ClF₃ + BF₃: fluoride transfer from ClF₃ to BF₃: 2pt] ClF₃ + BF₃ → [ClF₂]⁺[BF₄]⁻ [ClF₂]⁺ is bent (four electron pairs on Cl, two lone pairs), isoelectronic with SCl₂.
- (D) I₂ + Br₂: the halogens combine directly to the interhalogen iodine monobromide: 2pt] I₂ + Br₂ → 2IBr IBr is a dark solid, polar (I^(δ⁺)-Br^(δ⁻), Br more electronegative). The reaction is reversible; IBr partly dissociates on heating.
- Interhalogens are more reactive than the halogens (except F₂) because the X–X' bond is weaker and polar. The cations and anions above are the basis of their use as non-aqueous ionising solvents (ClF₃, BrF₃).
Application. ClF₃ and BrF₃ fluorinate uranium to volatile UF₆ for isotope enrichment, and IBr and ICl (Wijs reagent) are used to measure the iodine value of fats and oils.
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