Q5 / 160 · TS EAMCET 2024 (09 May Shift 2) · Chemistry

Identify the pair in which difference in bond order value is maximum

  • A
    $\mathrm{O}_2^{-}, \mathrm{O}_2^{+}$
  • Correct answer
    B
    $\mathrm{O}_2^{2-}, \mathrm{O}_2^{2+}$
  • C
    $\mathrm{O}_2^2, \mathrm{O}_2{ }^{2+}$
  • D
    $\mathrm{O}_2^{+}, \mathrm{O}_2^{2+}$

Solution

Bond order $=\frac{1}{2}$ [Number of bonding electrons - Number of anti Bonding electrons) Bond order of $\mathrm{O}_2=\frac{1}{2}[10-6]=2$ The bond order of $\mathrm{O}_2^{-}=\frac{1}{2}[10-7]=1.5$ The bond order of $\mathrm{O}_2^{+}=\frac{1}{2}[10-5]=2.5$ The bond order of $\mathrm{O}_2^{2-}=\frac{1}{2}[10-8]=1$ The bond order of $\mathrm{O}_2^{2+}=\frac{1}{2}[10-4]=3$ The difference Between $\mathrm{O}_2{ }^{2-}$ and $\mathrm{O}_2{ }^{2+}$ is $(3-1)=2$ is maximum.

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

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