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NEET CHEMISTRYMedium

Increasing order of bond length is:

A

NO<NO<NO+<O2NO^{-} < NO < NO^{+} < O_{2}^{-}

B

O2<NO<NO<NO+O_{2}^{-} < NO < NO^{-} < NO^{+}

C

O2<NO<NO<NO+O_{2}^{-} < NO^{-} < NO < NO^{+}

D

NO+<NO<NO<O2NO^{+} < NO < NO^{-} < O_{2}^{-}

Step-by-Step Solution

According to Molecular Orbital Theory, bond length is inversely proportional to bond order; as the bond order increases, the bond length decreases . To find the increasing order of bond length, we must determine the bond order (B.O.) for each species:

  1. NO+NO^{+}: This species has 14 electrons and is isoelectronic with N2N_{2}. Its bond order is 3.0 .
  2. NONO: This molecule has 15 electrons. Compared to N2N_{2}, the additional electron enters an antibonding π\pi^{*} orbital, reducing the bond order to 2.5 .
  3. NONO^{-}: This ion has 16 electrons and is isoelectronic with O2O_{2}. Following the molecular orbital configuration of O2O_{2}, it has a bond order of 2.0 .
  4. O2O_{2}^{-} (superoxide ion): This species has 17 electrons. The extra electron further occupies antibonding π\pi^{*} orbitals, resulting in a bond order of 1.5 .

Comparing the bond orders: NO+(3.0)>NO(2.5)>NO(2.0)>O2(1.5)NO^{+} (3.0) > NO (2.5) > NO^{-} (2.0) > O_{2}^{-} (1.5). Therefore, the increasing order of bond length (the reverse of bond order) is NO+<NO<NO<O2NO^{+} < NO < NO^{-} < O_{2}^{-}.

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