In the conversion of `N_(2)` into `N_(2)^(+)` the electron will be lost from which of the following molecular orbital ?

Options

A. `sigma^(*)2p_(z)`
B. `sigma2p_(z)`
C. `pi 2p_(x)`
D. `pi^(**)2p_(x)`

Explanation

To determine from which molecular orbital an electron is lost when converting \( N_2 \) into \( N_2^+ \), we can follow these steps:

### Step 1: Determine the Total Number of Electrons in \( N_2 \)
- Each nitrogen atom has 7 electrons, so for \( N_2 \):
\[ \text{Total electrons} = 7 \times 2 = 14 \]

### Step 2: Write the Molecular Orbital Electronic Configuration for \( N_2 \)
- The molecular orbital configuration for \( N_2 \) with 14 electrons is:
\[ \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2, \sigma 2p_z^2 \]
- The configuration can be summarized as:
\[ \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2, \sigma 2p_z^2 \]

### Step 3: Identify the Highest Energy Molecular Orbital
- The order of energy levels for the molecular orbitals in \( N_2 \) is:
\[ \sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z \]
- The highest energy orbital filled in \( N_2 \) is \( \sigma 2p_z \).

### Step 4: Remove One Electron to Form \( N_2^+ \)
- When converting \( N_2 \) to \( N_2^+ \), we remove one electron from the highest energy molecular orbital, which is \( \sigma 2p_z \).

### Conclusion
- The electron will be lost from the \( \sigma 2p_z \) molecular orbital when converting \( N_2 \) into \( N_2^+ \).

### Final Answer
The electron will be lost from the \( \sigma 2p_z \) molecular orbital.

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