How many unpaired electrons are present in `N_2^+` ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To determine the number of unpaired electrons in the \( N_2^+ \) ion, we can follow these steps:

### Step 1: Determine the Total Number of Electrons
Nitrogen (N) has an atomic number of 7, meaning each nitrogen atom has 7 electrons. In \( N_2 \), there are 2 nitrogen atoms, so:
\[ \text{Total electrons in } N_2 = 7 + 7 = 14 \]
Since \( N_2^+ \) has lost one electron, the total number of electrons in \( N_2^+ \) is:
\[ \text{Total electrons in } N_2^+ = 14 - 1 = 13 \]

**Hint for Step 1:** Count the number of electrons from the individual atoms and account for the charge of the ion.

### Step 2: Apply Molecular Orbital Theory (MOT)
According to Molecular Orbital Theory, the order of filling the molecular orbitals for diatomic nitrogen is as follows:
1. \( \sigma_{1s} \)
2. \( \sigma_{1s}^* \)
3. \( \sigma_{2s} \)
4. \( \sigma_{2s}^* \)
5. \( \sigma_{2p_z} \)
6. \( \pi_{2p_x} \) and \( \pi_{2p_y} \)
7. \( \pi_{2p_x}^* \) and \( \pi_{2p_y}^* \)

### Step 3: Fill the Molecular Orbitals with Electrons
Now, we will fill the molecular orbitals with the 13 electrons:
- Fill \( \sigma_{1s} \) with 2 electrons: \( \sigma_{1s}^2 \)
- Fill \( \sigma_{1s}^* \) with 0 electrons: \( \sigma_{1s}^* \)
- Fill \( \sigma_{2s} \) with 2 electrons: \( \sigma_{2s}^2 \)
- Fill \( \sigma_{2s}^* \) with 0 electrons: \( \sigma_{2s}^* \)
- Fill \( \sigma_{2p_z} \) with 2 electrons: \( \sigma_{2p_z}^2 \)
- Fill \( \pi_{2p_x} \) and \( \pi_{2p_y} \) with 4 electrons (2 in each): \( \pi_{2p_x}^2 \) and \( \pi_{2p_y}^2 \)
- Now we have 10 electrons filled. We have 3 electrons left to fill, which will go into the next available orbitals:
- Fill \( \pi_{2p_x}^* \) and \( \pi_{2p_y}^* \) with 1 electron each, leaving 1 electron to go into \( \sigma_{2p_z} \).

The filling will look like this:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^* \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^* \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \pi_{2p_x}^* \)
- \( \pi_{2p_y}^* \)
- \( \sigma_{2p_z}^1 \)

### Step 4: Count the Unpaired Electrons
From the filling, we see that:
- The \( \sigma_{2p_z} \) has 1 unpaired electron.
- The \( \pi_{2p_x}^* \) and \( \pi_{2p_y}^* \) do not have any unpaired electrons.

Thus, the total number of unpaired electrons in \( N_2^+ \) is:
\[ \text{Total unpaired electrons} = 1 \]

### Final Answer
The number of unpaired electrons in \( N_2^+ \) is **1**.

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