According to `MO` theory,

Options

A. `O_(2)^(o+)` is paramagnetic and bond order is greater than `O_(2)`
B. `O_(2)^(o+)` is paramagnetic and bond order is less than `O_(2)`
C. `O_(2)^(o+)` is diamagnetic and bond order is less than `O_(2)`
D. `O_(2)^(o+)` is diamagnetic and bond order is more than `O_(2)`

Answer

To solve the question regarding the molecular orbital theory (MO theory) for the O2 positive ion (O2⁺) and its comparison with O2, we will follow these steps:

### Step 1: Determine the Total Number of Electrons
The O2 molecule has a total of 16 electrons (8 from each oxygen atom). The O2⁺ ion has one less electron, giving it a total of 15 electrons.

### Step 2: Write the Molecular Orbital Configuration for O2
The molecular orbital configuration for O2 is:
- σ1s²
- σ*1s²
- σ2s²
- σ*2s²
- σ2p_x²
- π2p_y²
- π2p_z²
- π*2p_y¹
- π*2p_z¹

### Step 3: Count the Bonding and Antibonding Electrons for O2
- Bonding Electrons: 10 (from σ1s, σ2s, σ2p_x, π2p_y, π2p_z)
- Antibonding Electrons: 6 (from σ*1s, σ*2s, π*2p_y, π*2p_z)

### Step 4: Calculate the Bond Order for O2
Bond Order = (Number of Bonding Electrons - Number of Antibonding Electrons) / 2
Bond Order = (10 - 6) / 2 = 2

### Step 5: Write the Molecular Orbital Configuration for O2⁺
For O2⁺, we remove one electron from the highest energy orbital, which is π*2p_y. The configuration becomes:
- σ1s²
- σ*1s²
- σ2s²
- σ*2s²
- σ2p_x²
- π2p_y²
- π2p_z²
- π*2p_y⁰ (one less electron)
- π*2p_z¹

### Step 6: Count the Bonding and Antibonding Electrons for O2⁺
- Bonding Electrons: 10 (same as O2)
- Antibonding Electrons: 5 (from σ*1s, σ*2s, π*2p_y, π*2p_z)

### Step 7: Calculate the Bond Order for O2⁺
Bond Order = (Number of Bonding Electrons - Number of Antibonding Electrons) / 2
Bond Order = (10 - 5) / 2 = 2.5

### Step 8: Determine the Magnetic Property of O2⁺
Since O2⁺ has one unpaired electron in the π*2p_z orbital, it is paramagnetic.

### Step 9: Compare the Bond Orders
- Bond Order of O2 = 2
- Bond Order of O2⁺ = 2.5

Thus, the bond order of O2⁺ is greater than that of O2.

### Final Conclusion
- O2⁺ is paramagnetic due to the presence of unpaired electrons.
- The bond order of O2⁺ (2.5) is greater than that of O2 (2).