Calculate the bond order,the molecular configuration,Number of bonding electron and Nber of anti-bonding electron for the NO+

1 answer

To analyze the molecular orbital configuration and calculate the bond order for the \( \text{NO}^+ \) ion, we can follow these steps:

Step 1: Determine the total number of electrons

Nitric oxide (NO) has 11 electrons (7 from nitrogen and 8 from oxygen). The \( \text{NO}^+ \) ion has one electron removed, giving it a total of 10 electrons.

Step 2: Molecular Orbital Configuration

For diatomic molecules like NO, the molecular orbitals are filled according to the energy levels. The general order of molecular orbitals for NO and its isoelectronic species (like O2 and F2) can be represented as follows:

  1. \( \sigma_{1s} \)
  2. \( \sigma_{1s}^* \)
  3. \( \sigma_{2s} \)
  4. \( \sigma_{2s}^* \)
  5. \( \sigma_{2p_z} \)
  6. \( \pi_{2p_x} \)
  7. \( \pi_{2p_y} \)
  8. \( \pi_{2p_x}^* \)
  9. \( \pi_{2p_y}^* \)
  10. \( \sigma_{2p_z}^* \)

Filling the Molecular Orbitals

  • Fill the orbitals with 10 total electrons:
    • \( \sigma_{1s} \): 2 electrons (filled)
    • \( \sigma_{1s}^* \): 0 electrons (not filled)
    • \( \sigma_{2s} \): 2 electrons (filled)
    • \( \sigma_{2s}^* \): 0 electrons (not filled)
    • \( \sigma_{2p_z} \): 2 electrons (filled)
    • \( \pi_{2p_x} \): 2 electrons (filled)
    • \( \pi_{2p_y} \): 2 electrons (filled)
    • \( \pi_{2p_x}^* \): 0 electrons
    • \( \pi_{2p_y}^* \): 0 electrons
    • \( \sigma_{2p_z}^* \): 0 electrons

Thus, the electron configuration of \( \text{NO}^+ \) would be:

\[ ( \sigma_{1s} )^2 ( \sigma_{1s}^)^0 ( \sigma_{2s} )^2 ( \sigma_{2s}^)^0 ( \sigma_{2p_z} )^2 ( \pi_{2p_x})^2 ( \pi_{2p_y})^2 ( \pi_{2p_x}^)^0 ( \pi_{2p_y}^)^0 ( \sigma_{2p_z}^*)^0 \]

Step 3: Calculate Bond Order

The bond order (\( BO \)) is calculated using the formula:

\[ BO = \frac{(N_b - N_a)}{2} \]

where \( N_b \) is the number of bonding electrons and \( N_a \) is the number of anti-bonding electrons.

From the configuration:

  • Bonding electrons (\( N_b \)):

    • \( \sigma_{1s} = 2 \)
    • \( \sigma_{2s} = 2 \)
    • \( \sigma_{2p_z} = 2 \)
    • \( \pi_{2p_x} = 2 \)
    • \( \pi_{2p_y} = 2 \)

    Total \( N_b = 2 + 2 + 2 + 2 + 2 = 10 \)

  • Anti-bonding electrons (\( N_a \)):

    • \( N_a = 0 \) (none occupied)

So,

\[ BO = \frac{(10 - 0)}{2} = 5 \]

Summary

  • Bond Order: 5
  • Molecular Configuration: \( ( \sigma_{1s} )^2 ( \sigma_{1s}^)^0 ( \sigma_{2s} )^2 ( \sigma_{2s}^)^0 ( \sigma_{2p_z} )^2 ( \pi_{2p_x})^2 ( \pi_{2p_y})^2 ( \pi_{2p_x}^)^0 ( \pi_{2p_y}^)^0 ( \sigma_{2p_z}^*)^0 \)
  • Number of Bonding Electrons: 10
  • Number of Anti-bonding Electrons: 0

Thus, \( \text{NO}^+ \) is a strongly bonding species with a bond order of 5.