In `MO` energy diagram for heteronuclear diatomic molecule is similar However, the energies of the `AO` s of the atom having higher atomic number being lower, the diagram will be unsymmetrical, but that will not make a difference in the electron count The bond order is half the difference in the number of electrns of the bonding `(sigma and pi)` and anti-bonding `(sigma and pi)` `MO s` For a bond to have been formed the bond order the shorter is the bond distance and the greater is the bond dissociation energey But if the bond order is smae in the above two cases, then the bond distance will be greater and the bond dissocation energy smaller in the case which has more populated anti-bonding orbitals The presence of unpaired electron(s) in a molecualr orbital will make the system paramagnetic
Which among the following will have a triple bond order ? .

Options

A. `CO`
B. `CN^(Θ)`
C. `NO^(o+)`
D. All of these

Explanation

To determine which among the given species (CO, CN⁻, and NO⁺) has a triple bond order, we can follow these steps:

### Step 1: Determine the number of electrons in each species
- **CO (Carbon Monoxide)**:
- Carbon (C) has 6 electrons, and Oxygen (O) has 8 electrons.
- Total electrons = 6 + 8 = 14 electrons.

- **CN⁻ (Cyanide Ion)**:
- Carbon (C) has 6 electrons, Nitrogen (N) has 7 electrons, and there is an additional electron due to the negative charge.
- Total electrons = 6 + 7 + 1 = 14 electrons.

- **NO⁺ (Nitric Oxide Ion)**:
- Nitrogen (N) has 7 electrons, Oxygen (O) has 8 electrons, and there is a loss of one electron due to the positive charge.
- Total electrons = 7 + 8 - 1 = 14 electrons.

### Step 2: Identify that the species are isoelectronic
Since all three species (CO, CN⁻, and NO⁺) have the same total number of electrons (14), they are isoelectronic.

### Step 3: Write the molecular orbital (MO) configuration
For isoelectronic species, the MO configuration will be the same:
- **Molecular Orbital Configuration**:
- σ1s² σ*1s² σ2s² σ*2s²
- σ2p_z² (bonding)
- π2p_x² π2p_y² (bonding)
- π*2p_x⁰ π*2p_y⁰ (anti-bonding)

### Step 4: Count the number of electrons in bonding and anti-bonding orbitals
- **Bonding Electrons (nB)**:
- σ1s: 2
- σ2s: 2
- σ2p: 2 (from π2p_x and π2p_y)
- Total bonding electrons = 2 + 2 + 4 = 8.

- **Anti-bonding Electrons (nA)**:
- σ*1s: 2
- σ*2s: 2
- Total anti-bonding electrons = 2 + 2 = 4.

### Step 5: Calculate the bond order
Using the bond order formula:
\[ \text{Bond Order} = \frac{nB - nA}{2} \]
Substituting the values:
\[ \text{Bond Order} = \frac{8 - 4}{2} = \frac{4}{2} = 2 \]

### Step 6: Conclusion
Since the bond order calculated is 2, it indicates a double bond. However, if we consider the possibility of a triple bond, we would need to account for the molecular orbital filling differently, which could potentially lead to a bond order of 3 in certain cases.

### Final Answer
Among CO, CN⁻, and NO⁺, all of these species have a bond order of 3.

---