Which of the following statements is correct about `O_(2),O_(2)^(Θ),O_(2)^(2-)` species ? .

Options

A. `KO_(2)` and `K_(2)O_(2)` are diamagnetic while `O_(2)` is paramagnetic .
B. `KO_(2)` and `O_(2)` are paramagnetic while `K_(2)O_(2)` is diamagnetic .
C. Bond length increase in the order `O_(2)ltO_(2)^(Θ)ltO_(2)^(2-)`
D. Bond enthalpy increases in the order `O_(2)^(2-) ltO_(2)^(Θ) ltO_(2)` .

Explanation

To analyze the species \(O_2\), \(O_2^-\), and \(O_2^{2-}\), we will follow these steps:

### Step 1: Determine the Electronic Configuration of \(O_2\)
The electronic configuration of \(O_2\) can be written as:
\[ \sigma_{1s}^2 \, \sigma_{1s}^*^2 \, \sigma_{2s}^2 \, \sigma_{2s}^*^2 \, \sigma_{2p_x}^2 \, \pi_{2p_y}^2 \, \pi_{2p_z}^2 \, \pi_{2p_y}^*^1 \, \pi_{2p_z}^*^1 \]
This configuration shows that \(O_2\) has a total of 12 electrons.

### Step 2: Calculate the Bond Order of \(O_2\)
The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2}(N_b - N_a) \]
where \(N_b\) is the number of bonding electrons and \(N_a\) is the number of antibonding electrons.

For \(O_2\):
- Bonding electrons (\(N_b\)): 10
- Antibonding electrons (\(N_a\)): 6

Thus, the bond order is:
\[ \text{Bond Order} = \frac{1}{2}(10 - 6) = 2 \]

### Step 3: Determine the Electronic Configuration of \(O_2^-\)
For the superoxide ion \(O_2^-\), the electronic configuration is:
\[ \sigma_{1s}^2 \, \sigma_{1s}^*^2 \, \sigma_{2s}^2 \, \sigma_{2s}^*^2 \, \sigma_{2p_x}^2 \, \pi_{2p_y}^2 \, \pi_{2p_z}^2 \, \pi_{2p_y}^*^2 \, \pi_{2p_z}^*^1 \]
This configuration shows that \(O_2^-\) has a total of 13 electrons.

### Step 4: Calculate the Bond Order of \(O_2^-\)
For \(O_2^-\):
- Bonding electrons (\(N_b\)): 10
- Antibonding electrons (\(N_a\)): 7

Thus, the bond order is:
\[ \text{Bond Order} = \frac{1}{2}(10 - 7) = 1.5 \]

### Step 5: Determine the Electronic Configuration of \(O_2^{2-}\)
For the peroxide ion \(O_2^{2-}\), the electronic configuration is:
\[ \sigma_{1s}^2 \, \sigma_{1s}^*^2 \, \sigma_{2s}^2 \, \sigma_{2s}^*^2 \, \sigma_{2p_x}^2 \, \pi_{2p_y}^2 \, \pi_{2p_z}^2 \, \pi_{2p_y}^*^2 \, \pi_{2p_z}^*^2 \]
This configuration shows that \(O_2^{2-}\) has a total of 14 electrons.

### Step 6: Calculate the Bond Order of \(O_2^{2-}\)
For \(O_2^{2-}\):
- Bonding electrons (\(N_b\)): 10
- Antibonding electrons (\(N_a\)): 8

Thus, the bond order is:
\[ \text{Bond Order} = \frac{1}{2}(10 - 8) = 1 \]

### Step 7: Summary of Results
- \(O_2\): Bond order = 2, paramagnetic
- \(O_2^-\): Bond order = 1.5, paramagnetic
- \(O_2^{2-}\): Bond order = 1, diamagnetic

### Conclusion
Based on the bond order and magnetic properties:
- \(O_2\) and \(O_2^-\) are paramagnetic.
- \(O_2^{2-}\) is diamagnetic.
- The bond length increases in the order \(O_2 < O_2^- < O_2^{2-}\) (as bond length is inversely proportional to bond order).
- The bond enthalpy increases in the order \(O_2^{2-} < O_2^- < O_2\).