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NEET CHEMISTRYMedium

The correct bond order in the following species is:

A

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

B

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

C

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

D

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

Step-by-Step Solution

According to Molecular Orbital Theory, the 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: has 16 electrons. Its 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^+: has 15 electrons (one electron removed from the antibonding π\pi^* orbital). Bond order = 1052=2.5\frac{10 - 5}{2} = 2.5.
  • O22+\text{O}_2^{2+}: has 14 electrons (two electrons removed from the antibonding π\pi^* orbitals). Bond order = 1042=3.0\frac{10 - 4}{2} = 3.0.
  • O2\text{O}_2^-: has 17 electrons (one electron added to the antibonding π\pi^* orbital). Bond order = 1072=1.5\frac{10 - 7}{2} = 1.5. Therefore, the correct decreasing sequence for bond order is O22+(3.0)>O2+(2.5)>O2(1.5)\text{O}_2^{2+} (3.0) > \text{O}_2^+ (2.5) > \text{O}_2^- (1.5).
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