Decreasing order of stability of O 2 , O 2 - , O 2 + and O 2 2 - is:

Decreasing order of stability of O2, O2-, O2+ and O22- is:

  1. O2+>O2> O2-> O22-
  2. O22-> O2->O2> O2+
  3. O2> O2+> O22-> O2-
  4. O2-> O22->O2+>O2

Solution

$B.O=\frac{N_b-N_a}{2}$ $O_2^- \Rightarrow \sigma_1s^2 < \sigma^*_1s^2 < \sigma_2s^2 < \sigma^*_2s^2 < \sigma_{p_z}^2 < \pi_2p_x^2 = \pi_2p_y^2 < \pi^*_2p_x^2 = \pi^*_2p_y^1$ $B.O$ for $O_2^- = 1.5 \Rightarrow \frac{10-7}{2} = 1.5$ $O_2 \Rightarrow \sigma_1s^2 < \sigma^*_1s^2 < \sigma_2s^2 < \sigma^*_2s^2 < \sigma_{p_z}^2 < \pi_2p_x^2 = \pi_2p_y^2 < \pi^*_2p_x^1 = \pi^*_2p_y^1$ $B.O$ for $O_2 = 2.0 \Rightarrow \frac{10-6}{2}= 2$ $O_2^+ \Rightarrow \sigma_1s^2 < \sigma^*_1s^2 < \sigma_2s^2 < \sigma^*_2s^2 < \sigma_{p_z}^2 < \pi_2p_x^2 < \pi_2p_y^2 < \pi^*_2p_x^1 < \pi^*_2p_y^0$ $B.O$ for $O_2^+ = 2.5 \Rightarrow \frac{10-5}{2} = 2.5$ $O_2^{2-} \Rightarrow \sigma_1s^2 < \sigma^*_1s^2 < \sigma_2s^2 < \sigma^*_2s^2 < \sigma_{p_z}^2 < \pi_2p_x^2 = \pi_2p_y^2 < \pi^*_2p_x^2 = \pi^*_2p_y^2$ $O_2^{2+} \Rightarrow B.O = \frac{10-4}{2} = 3.0$ $O_2^- < O_2^+ < O_2^{2+}$ $O_2 \rightarrow 16e^-$ $O_2^- \rightarrow 17e^-$ $O_2^+ \rightarrow 15e^-$ $O_2^{2-} \rightarrow 18e^-$ Stability α Bond order
Stability order O2+>O2>O2->O22-

Asked in: NEET 2015 (Phase 2)

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