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NEET CHEMISTRYMedium

Which one of the following is the correct order of decreasing bond enthalpies for the given species?

A

O22>O2>O2>N2O_2^{2-} > O_2^- > O_2 > N_2

B

N2>O2>O22>O2N_2 > O_2 > O_2^{2-} > O_2^-

C

N2>O2>O2>O22N_2 > O_2 > O_2^- > O_2^{2-}

D

O2>N2>O2>O22O_2 > N_2 > O_2^- > O_2^{2-}

Step-by-Step Solution

According to Molecular Orbital Theory (MOT), the strength of a bond is directly related to its bond order, which is defined as half the difference between the number of electrons in bonding orbitals (NbN_b) and antibonding orbitals (NaN_a) . The sources state that as the bond order increases, the bond enthalpy increases and the bond length decreases .

We can determine the bond orders for the given species as follows:

  1. N2N_2: With 14 electrons, its configuration results in a bond order of 3 . Its experimental bond enthalpy is very high, at 946.0 kJ/mol .
  2. O2O_2: With 16 electrons, its configuration results in a bond order of 2 . Its experimental bond enthalpy is 498 kJ/mol .
  3. O2O_2^- (Superoxide): This ion has 17 electrons. The additional electron enters an antibonding orbital, reducing the bond order to 1.5 .
  4. O22O_2^{2-} (Peroxide): This ion has 18 electrons. With two additional electrons in antibonding orbitals compared to O2O_2, the bond order is 1 .

Comparing these values, the order of decreasing bond order is N2(3)>O2(2)>O2(1.5)>O22(1)N_2 (3) > O_2 (2) > O_2^- (1.5) > O_2^{2-} (1). Therefore, the order of decreasing bond enthalpies follows the same sequence .

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