The bond orders for `O_(2)^(+) and C_(2)` respectively are

Options

A. 2.5, 2
B. 3, 2
C. 2, 2.5
D. 2, 3

Answer

To determine the bond orders for the molecules \(O_2^+\) and \(C_2\), we will follow these steps:

### Step 1: Determine the total number of electrons
- For \(O_2^+\): Each oxygen atom has 8 electrons, so two oxygen atoms have a total of \(16\) electrons. Since \(O_2^+\) has a positive charge, it means one electron has been removed. Therefore, the total number of electrons in \(O_2^+\) is \(16 - 1 = 15\).
- For \(C_2\): Each carbon atom has 6 electrons, so two carbon atoms have a total of \(12\) electrons.

### Step 2: Write the molecular orbital configuration
- For \(O_2^+\) (15 electrons):
- The molecular orbital configuration is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \]
- For \(C_2\) (12 electrons):
- The molecular orbital configuration is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \]

### Step 3: Count the electrons in bonding and anti-bonding orbitals
- For \(O_2^+\):
- Bonding electrons: \(2 (\sigma_{1s}) + 2 (\sigma_{2s}) + 2 (\sigma_{2p_z}) + 2 (\pi_{2p_x}) + 2 (\pi_{2p_y}) = 10\)
- Anti-bonding electrons: \(2 (\sigma_{1s}^*) + 2 (\sigma_{2s}^*) + 1 (\pi_{2p_x}^*) = 5\)

- For \(C_2\):
- Bonding electrons: \(2 (\sigma_{1s}) + 2 (\sigma_{2s}) + 2 (\pi_{2p_x}) + 2 (\pi_{2p_y}) = 8\)
- Anti-bonding electrons: \(2 (\sigma_{1s}^*) + 2 (\sigma_{2s}^*) = 4\)

### Step 4: Calculate the bond order
- The formula for bond order is:
\[ \text{Bond Order} = \frac{\text{Number of electrons in bonding MOs} - \text{Number of electrons in anti-bonding MOs}}{2} \]

- For \(O_2^+\):
\[ \text{Bond Order} = \frac{10 - 5}{2} = \frac{5}{2} = 2.5 \]

- For \(C_2\):
\[ \text{Bond Order} = \frac{8 - 4}{2} = \frac{4}{2} = 2 \]

### Final Answer
The bond orders for \(O_2^+\) and \(C_2\) are \(2.5\) and \(2\) respectively.

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