JEE Main · 2024 · Shift-IImediumBOND-055

The correct increasing order for bond angles among BF3, PF3 and ClF3 is:

Chemical Bonding · Class 11 · JEE Main Previous Year Question

Question

The correct increasing order for bond angles among BF3\mathrm{BF_3}, PF3\mathrm{PF_3} and ClF3\mathrm{ClF_3} is:

Options
  1. a

    BF3<PF3<ClF3\mathrm{BF_3 < PF_3 < ClF_3}

  2. b

    ClF3<PF3<BF3\mathrm{ClF_3 < PF_3 < BF_3}

  3. c

    PF3<BF3<ClF3\mathrm{PF_3 < BF_3 < ClF_3}

  4. d

    BF3=PF3<ClF3\mathrm{BF_3 = PF_3 < ClF_3}

Correct Answerb

ClF3<PF3<BF3\mathrm{ClF_3 < PF_3 < BF_3}

Detailed Solution

🧠 Increasing lone pairs on the central atom compresses bond angles. Compare these three trihalides: \ceBF3\ce{BF3} has no LP, \cePF3\ce{PF3} has 1 LP, \ceClF3\ce{ClF3} has 2 LP.

| Molecule | Lone pairs on central | Shape | Bond angle | |---|---|---|---| | \ceBF3\ce{BF3} | 0 | Trigonal planar | 120° | | \cePF3\ce{PF3} | 1 | Trigonal pyramidal | ~97.8° | | \ceClF3\ce{ClF3} | 2 | T-shaped | ~87.5° (F–Cl–F axial) |

Each additional lone pair exerts greater repulsion than bond pairs, compressing the remaining bond angles further.

Increasing bond angle: \ceClF3<\cePF3<\ceBF3\ce{ClF3} < \ce{PF3} < \ce{BF3}

⚡ In \ceClF3\ce{ClF3} the T-shape has an F–Cl–F angle of ~87.5° (well below 90°) due to 2 LP occupying equatorial positions and strongly repelling the axial F–Cl bonds.

Answer: (b) \ceClF3<PF3<BF3\boxed{\text{Answer: (b) } \ce{ClF3 < PF3 < BF3}}

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