Q5 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Chemistry

The bond dissociation energy $(E)$ and bond length $(R)$ of $\mathrm{O}_2, \mathrm{~N}_2$ and $\mathrm{F}_2$ follow the order as:

  • A
    Figure
  • B
    Figure
  • Correct answer
    C
    Figure
  • D
    Figure

Solution

Greater the bond order, shorter is the bond length and thus, stronger is the bond. Since, for the given molecules, bond order follows the trend $\begin{array}{ccc}\mathrm{N}_2 & >\mathrm{O}_2 & >\mathrm{F}_2 \\ (\mathrm{~B} . \mathrm{O}=3) & (\mathrm{B} . \mathrm{O}=2) & (\mathrm{B} . \mathrm{O}=1)\end{array}$ Thus, the bond length decreases as $\mathrm{F}_2>\mathrm{O}_2>\mathrm{N}_2$ and bond dissociation energy follows the trend $\mathrm{N}_2>\mathrm{O}_2>\mathrm{F}_2$

Hence, the correct answer is option: 'C'.

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