Arrange the following species in the decreasing order of their bond dissociation enthalpies :
`O_(2),O_(2)^(+) and O_(2)^(-)`

Explanation

To arrange the species \( O_2 \), \( O_2^+ \), and \( O_2^- \) in decreasing order of their bond dissociation enthalpies, we need to analyze their bond orders. The bond dissociation enthalpy is directly related to the bond order; higher bond order means stronger bonds and thus higher bond dissociation enthalpy.

### Step 1: Determine the number of electrons in each species
- **For \( O_2 \)**: Oxygen has an atomic number of 8, so \( O_2 \) has \( 2 \times 8 = 16 \) electrons.
- **For \( O_2^+ \)**: This species has one less electron than \( O_2 \), so it has \( 16 - 1 = 15 \) electrons.
- **For \( O_2^- \)**: This species has one extra electron compared to \( O_2 \), so it has \( 16 + 1 = 17 \) electrons.

### Step 2: Write the molecular orbital configurations
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