According to is M.O. theory, which is the correct increasing sequence of B.O. in `N_(2),N_(2)^(-)and N_(2)^(2-)` ?

Options

A. `N_(2)^(2-) lt N_(2)^(-) lt N_(2)`
B. `N_(2) lt N_(2)^(2-) lt N_(2)^(-)`
C. `N_(2)^(-) lt N_(2)^(2-) lt N_(2)`
D. `N_(2)^(-) lt N_(2) lt N_(2)^(2-)`

Answer

To determine the correct increasing sequence of bond order (B.O.) for the molecules \(N_2\), \(N_2^-\), and \(N_2^{2-}\) according to Molecular Orbital (M.O.) theory, we will follow these steps:

### Step 1: Determine the electronic configuration of \(N_2\)
- The total number of electrons in \(N_2\) is 14 (7 from each nitrogen atom).
- The electronic configuration is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \]
- Bonding electrons = 10 (from \(\sigma\) and \(\pi\) orbitals).
- Antibonding electrons = 4 (from \(\sigma^*\) orbitals).

### Step 2: Calculate the bond order of \(N_2\)
- Using the formula for bond order:
\[ \text{Bond Order} = \frac{(\text{Bonding Electrons} - \text{Antibonding Electrons})}{2} \]
- For \(N_2\):
\[ \text{Bond Order} = \frac{(10 - 4)}{2} = 3 \]

### Step 3: Determine the electronic configuration of \(N_2^-\)
- \(N_2^-\) has one additional electron, making the total number of electrons 15.
- The electronic configuration becomes:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^*^1 \]
- Bonding electrons = 10 (same as \(N_2\)).
- Antibonding electrons = 5 (4 from \(\sigma^*\) and 1 additional from \(\pi^*\)).

### Step 4: Calculate the bond order of \(N_2^-\)
- For \(N_2^-\):
\[ \text{Bond Order} = \frac{(10 - 5)}{2} = 2.5 \]

### Step 5: Determine the electronic configuration of \(N_2^{2-}\)
- \(N_2^{2-}\) has two additional electrons, making the total number of electrons 16.
- The electronic configuration becomes:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^*^2 \]
- Bonding electrons = 10.
- Antibonding electrons = 6 (4 from \(\sigma^*\) and 2 from \(\pi^*\)).

### Step 6: Calculate the bond order of \(N_2^{2-}\)
- For \(N_2^{2-}\):
\[ \text{Bond Order} = \frac{(10 - 6)}{2} = 2 \]

### Step 7: Compare the bond orders
- The bond orders are:
- \(N_2\): 3
- \(N_2^-\): 2.5
- \(N_2^{2-}\): 2

### Conclusion
The correct increasing sequence of bond order is:
\[ N_2^{2-} < N_2^- < N_2 \]