Question:
In which of the following, the species are arranged in the decreasing order of their bond dissociation enthalpies?
In which of the following, the species are arranged in the decreasing order of their bond dissociation enthalpies?
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Bond order is directly proportional to bond dissociation enthalpy in MO theory.
Updated On: Jul 18, 2026
- O\(_2^+\) > O\(_2\) > O\(_2^-\) > O\(_2^{2-}\)
- O\(_2\) > O\(_2^-\) > O\(_2^+\) > O\(_2^{2-}\)
- O\(_2^+\) > O\(_2\) > O\(_2^{2-}\) > O\(_2^-\)
- O\(_2\) > O\(_2^+\) > O\(_2^-\) > O\(_2^{2-}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understand bond dissociation enthalpy (BDE).
Bond dissociation enthalpy depends on bond order. Higher bond order means stronger bond and higher BDE. Using molecular orbital theory, we compare bond orders of oxygen species.
Step 2: Calculate bond order of O\(_2\).
For O\(_2\), bond order is: \[ \text{B.O.} = 2 \] This comes from MO configuration where antibonding electrons reduce bond strength slightly from triple bond.
Step 3: Analyze O\(_2^+\).
Removing one electron from antibonding orbital increases bond order: \[ \text{B.O.}(O_2^+) = 2.5 \] So it has the strongest bond among all given species.
Step 4: Analyze O\(_2^-\) and O\(_2^{2-}\).
Adding electrons to antibonding orbitals reduces bond order: \[ \text{B.O.}(O_2^-) = 1.5, \quad \text{B.O.}(O_2^{2-}) = 1 \] Thus, these species have weaker bonds.
Step 5: Arrange in decreasing order of bond strength.
Higher bond order → higher BDE: \[ O_2^+ \gt O_2 \gt O_2^- \gt O_2^{2-} \]
Step 6: Final conclusion.
Therefore, correct decreasing order is: \[ \boxed{O_2^+ \gt O_2 \gt O_2^- \gt O_2^{2-}} \]
Bond dissociation enthalpy depends on bond order. Higher bond order means stronger bond and higher BDE. Using molecular orbital theory, we compare bond orders of oxygen species.
Step 2: Calculate bond order of O\(_2\).
For O\(_2\), bond order is: \[ \text{B.O.} = 2 \] This comes from MO configuration where antibonding electrons reduce bond strength slightly from triple bond.
Step 3: Analyze O\(_2^+\).
Removing one electron from antibonding orbital increases bond order: \[ \text{B.O.}(O_2^+) = 2.5 \] So it has the strongest bond among all given species.
Step 4: Analyze O\(_2^-\) and O\(_2^{2-}\).
Adding electrons to antibonding orbitals reduces bond order: \[ \text{B.O.}(O_2^-) = 1.5, \quad \text{B.O.}(O_2^{2-}) = 1 \] Thus, these species have weaker bonds.
Step 5: Arrange in decreasing order of bond strength.
Higher bond order → higher BDE: \[ O_2^+ \gt O_2 \gt O_2^- \gt O_2^{2-} \]
Step 6: Final conclusion.
Therefore, correct decreasing order is: \[ \boxed{O_2^+ \gt O_2 \gt O_2^- \gt O_2^{2-}} \]
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