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NEET CHEMISTRYMedium

Decreasing order of stability of O2\text{O}_2, O2\text{O}_2^-, O2+\text{O}_2^+ and O22\text{O}_2^{2-} is:

A

O2+>O2>O2>O22\text{O}_2^+ > \text{O}_2 > \text{O}_2^- > \text{O}_2^{2-}

B

O22>O2>O2>O2+\text{O}_2^{2-} > \text{O}_2^- > \text{O}_2 > \text{O}_2^+

C

O2>O2+>O22>O2\text{O}_2 > \text{O}_2^+ > \text{O}_2^{2-} > \text{O}_2^-

D

O2>O22>O2+>O2\text{O}_2^- > \text{O}_2^{2-} > \text{O}_2^+ > \text{O}_2

Step-by-Step Solution

According to Molecular Orbital Theory, the stability of a species is directly proportional to its bond order. Bond order is calculated as 12(NbNa)\frac{1}{2}(N_b - N_a), where NbN_b is the number of bonding electrons and NaN_a is the number of antibonding electrons.

  • O2\text{O}_2 (16 electrons): The configuration is σ1s2σ1s2σ2s2σ2s2σ2pz2(π2px2=π2py2)(π2px1=π2py1)\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 (\pi 2p_x^2 = \pi 2p_y^2) (\pi^* 2p_x^1 = \pi^* 2p_y^1). Bond order = 1062=2.0\frac{10 - 6}{2} = 2.0.
  • O2+\text{O}_2^+ (15 electrons): Formed by removing one electron from the antibonding π\pi^* orbital. Bond order = 1052=2.5\frac{10 - 5}{2} = 2.5.
  • O2\text{O}_2^- (17 electrons): Formed by adding one electron to the antibonding π\pi^* orbital. Bond order = 1072=1.5\frac{10 - 7}{2} = 1.5.
  • O22\text{O}_2^{2-} (18 electrons): Formed by adding two electrons to the antibonding π\pi^* orbitals. Bond order = 1082=1.0\frac{10 - 8}{2} = 1.0. Since stability increases with an increase in bond order, the decreasing order of stability is O2+ (2.5)>O2 (2.0)>O2 (1.5)>O22 (1.0)\text{O}_2^+ \text{ (2.5)} > \text{O}_2 \text{ (2.0)} > \text{O}_2^- \text{ (1.5)} > \text{O}_2^{2-} \text{ (1.0)}.
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