Valence-bond theory is one of the two quantum mechanical approaches that explains bonding in molecules In some cases, valence bond theroy cannot cannot satisfactorily account for observed properties of molecules
Consider the following molecular geometries
`{:(NO,,CO,,O_(2)):}`
Select correct statement`(s)` about these (More than one correct) .

Options

A. `NO` is paramagnetic `CO` and `O_(2)` are diamagnetic
B. `NO` and `O_(2)` are paramagnetic `CO` is diamagentic
C. Bond order is in order `O_(2) lt NO lt CO`
D. Number of unpaired electrons are
`{:(NO,,1,,),(CO,,0,,),(O_(2),,0,,):}`

Explanation

To solve the question regarding the molecular geometries of NO, CO, and O2, we will analyze the properties of each molecule based on their electron configurations, unpaired electrons, and bond orders.

### Step-by-Step Solution:

1. **Determine the Total Number of Electrons:**
- For **NO**: Nitrogen (N) has 7 electrons, and Oxygen (O) has 8 electrons. Therefore, total electrons = 7 + 8 = **15 electrons**.
- For **CO**: Carbon (C) has 6 electrons, and Oxygen (O) has 8 electrons. Therefore, total electrons = 6 + 8 = **14 electrons**.
- For **O2**: Each Oxygen (O) has 8 electrons, so for two Oxygens, total electrons = 8 + 8 = **16 electrons**.

2. **Molecular Orbital Configuration:**
- For **CO (14 electrons)**:
- Molecular orbital configuration:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_z² (bonding) π2p_x² = π2p_y² (bonding)
- Total bonding electrons = 10, anti-bonding = 4.
- Bond order = (10 - 4) / 2 = **3**.
- Unpaired electrons = **0** (Diamagnetic).

- For **NO (15 electrons)**:
- Molecular orbital configuration:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_z² (bonding) π2p_x² = π2p_y² (bonding) π*2p_x¹ (anti-bonding)
- Total bonding electrons = 10, anti-bonding = 5.
- Bond order = (10 - 5) / 2 = **2.5**.
- Unpaired electrons = **1** (Paramagnetic).

- For **O2 (16 electrons)**:
- Molecular orbital configuration:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_z² (bonding) π2p_x² = π2p_y² (bonding) π*2p_x¹ = π*2p_y¹ (anti-bonding)
- Total bonding electrons = 10, anti-bonding = 6.
- Bond order = (10 - 6) / 2 = **2**.
- Unpaired electrons = **2** (Paramagnetic).

3. **Summarize the Properties:**
- **NO**:
- Paramagnetic (1 unpaired electron)
- Bond order = 2.5
- **CO**:
- Diamagnetic (0 unpaired electrons)
- Bond order = 3
- **O2**:
- Paramagnetic (2 unpaired electrons)
- Bond order = 2

4. **Select the Correct Statements:**
- **NO is paramagnetic, CO is diamagnetic, O2 is paramagnetic.**
- **Bond order of CO is higher than that of NO and O2.**
- **Number of unpaired electrons: NO has 1, CO has 0, O2 has 2.**

### Final Answer:
The correct statements are:
- NO is paramagnetic, CO is diamagnetic, O2 is paramagnetic.
- Bond order: CO > NO > O2.
- Number of unpaired electrons: NO has 1, CO has 0, O2 has 2.