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Chemical Bonding and Molecular Structure question

2020 · 8 Jan · Shift 2 · Q18
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Chemical Bonding and Molecular Structure question

2020 · 8 Jan · Shift 2 · Q18

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
Arrange the following bonds according to their average bond energies in descending order : C–Cl, C–Br, C–F, C–I
  1. A
    C–Br > C–I > C–Cl > C–F
  2. B
    C–Cl > C–Br > C–I > C–F
  3. C
    C–I > C–Br > C–Cl > C–F
  4. D
    C–F > C–Cl > C–Br > C–I
View written solutionFree

Correct answer: D

  1. Principle involved

Bond energy generally depends on:

  • Bond length: shorter bonds are usually stronger.
  • Extent of overlap between orbitals: better overlap gives stronger bonds.

For carbon–halogen bonds, as we move down the halogen group:

F<Cl<Br<I\text{F} < \text{Cl} < \text{Br} < \text{I}F<Cl<Br<I

atomic size increases, so the bond length with carbon increases:

C–F<C–Cl<C–Br<C–I\text{C–F} < \text{C–Cl} < \text{C–Br} < \text{C–I}C–F<C–Cl<C–Br<C–I

Hence, bond strength (and average bond energy) decreases in the reverse order.

  1. Order of bond energies

Therefore,

C–F>C–Cl>C–Br>C–I\text{C–F} > \text{C–Cl} > \text{C–Br} > \text{C–I}C–F>C–Cl>C–Br>C–I

  1. Check the options
  • A: C–Br>C–I>C–Cl>C–F\text{C–Br} > \text{C–I} > \text{C–Cl} > \text{C–F}C–Br>C–I>C–Cl>C–F ❌
  • B: C–Cl>C–Br>C–I>C–F\text{C–Cl} > \text{C–Br} > \text{C–I} > \text{C–F}C–Cl>C–Br>C–I>C–F ❌
  • C: C–I>C–Br>C–Cl>C–F\text{C–I} > \text{C–Br} > \text{C–Cl} > \text{C–F}C–I>C–Br>C–Cl>C–F ❌
  • D: C–F>C–Cl>C–Br>C–I\text{C–F} > \text{C–Cl} > \text{C–Br} > \text{C–I}C–F>C–Cl>C–Br>C–I ✅
  1. Final answer

The correct descending order of average bond energies is:

C–F>C–Cl>C–Br>C–I\boxed{\text{C–F} > \text{C–Cl} > \text{C–Br} > \text{C–I}}C–F>C–Cl>C–Br>C–I​

So, option D is correct.

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