Problem 106
Question
Bond order of nitric oxide is (a) 1 (b) \(2.5\) (c) 2 (d) \(1.5\)
Step-by-Step Solution
Verified Answer
The bond order of nitric oxide is 2.5 (option b).
1Step 1: Understand Bond Order
Bond order is the number of chemical bonds between a pair of atoms. It gives us an idea of the strength and stability of a bond. The formula to calculate bond order from molecular orbital theory is: \( \text{Bond Order} = \frac{{\text{Number of bonding electrons} - \text{Number of anti-bonding electrons}}}{2} \).
2Step 2: Write Electron Configuration of NO
Nitric oxide (NO) has a total of 15 electrons. In molecular orbital theory, the energy level diagram is represented as: \( \sigma(1s)^2, \sigma^*(1s)^2, \sigma(2s)^2, \sigma^*(2s)^2, \sigma(2p_z)^2, \pi(2p_x)^2, \pi(2p_y)^2, \pi^*(2p_x)^1 \). This accounts for all 15 electrons, with 10 electrons in bonding orbitals and 5 in anti-bonding orbitals. This configuration does not follow the sequence strictly but provides an approximation for this analysis.
3Step 3: Calculate Bond Order
Using the formula from Step 1, substitute the number of bonding and anti-bonding electrons: \( \text{Bond Order} = \frac{{10 - 5}}{2} = 2.5 \). This means NO has a bond order of 2.5.
4Step 4: Select the Correct Answer
Compare your calculated bond order to the provided options. The bond order of nitric oxide is calculated as 2.5, which corresponds to option (b).
Key Concepts
Bond OrderChemical BondsNitric Oxide
Bond Order
Understanding bond order is crucial when analyzing molecular structures. Bond order indicates the number of chemical bonds between a pair of atoms. It's a measure of bond strength: higher bond orders signify stronger and shorter bonds. To calculate bond order using Molecular Orbital Theory, use the formula:
Understanding and calculating bond order is fundamental in predicting both the reactivity and the geometric structure of molecules.
- Bond Order = \( \frac{{\text{Number of bonding electrons} - \text{Number of anti-bonding electrons}}}{2} \)
Understanding and calculating bond order is fundamental in predicting both the reactivity and the geometric structure of molecules.
Chemical Bonds
Chemical bonds are the forces that hold atoms together within molecules. These bonds form due to interactions involving the outer electrons of atoms. The main types of chemical bonds include:
The concept also introduces bonding and anti-bonding orbitals. Bonding orbitals stabilize the molecule, whereas anti-bonding orbitals destabilize it.
Understanding the balance and interaction between these two types of orbitals is key to comprehending molecular stability and predicting the properties of molecules like nitric oxide.
- Covalent bonds, where atoms share electrons.
- Ionic bonds, which occur when electrons are transferred from one atom to another.
- Metallic bonds, found in metals where electrons are free to move throughout a lattice structure.
The concept also introduces bonding and anti-bonding orbitals. Bonding orbitals stabilize the molecule, whereas anti-bonding orbitals destabilize it.
Understanding the balance and interaction between these two types of orbitals is key to comprehending molecular stability and predicting the properties of molecules like nitric oxide.
Nitric Oxide
Nitric oxide (NO) is a fascinating molecule with a total of 15 valence electrons. It plays a critical role in various biological and environmental processes. Its unique characteristics stem from having an odd number of electrons.
This results in a bond order of 2.5, as calculated by the molecular orbital method, indicating a bond that is stronger than a double bond but weaker than a triple bond.
This results in a bond order of 2.5, as calculated by the molecular orbital method, indicating a bond that is stronger than a double bond but weaker than a triple bond.
- In the molecular orbital diagram for NO, electrons are distributed across several orbitals: \( \sigma(1s)^2, \sigma^*(1s)^2, \sigma(2s)^2, \sigma^*(2s)^2, \sigma(2p_z)^2, \pi(2p_x)^2, \pi(2p_y)^2, \pi^*(2p_x)^1 \).
- This representation highlights the presence of both bonding and anti-bonding electrons.
Other exercises in this chapter
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