Species having same bond order are

Options

A. `N_(2)`
B. `N_(2)^(-)`
C. `F_(2)^(+)`
D. `O_(2)^(-)`

Explanation

To determine which species have the same bond order, we need to follow these steps:

### Step 1: Understand the Bond Order Formula
The bond order (BO) can be calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} (N_B - N_A) \]
where \( N_B \) is the number of bonding electrons and \( N_A \) is the number of antibonding electrons.

### Step 2: Identify the Species
The species we need to analyze are:
1. \( N_2 \)
2. \( N_2^- \)
3. \( F_2^+ \)
4. \( O_2^- \)

### Step 3: Calculate the Total Number of Electrons
Next, we will calculate the total number of electrons for each species:

1. **For \( N_2 \)**:
- Each nitrogen atom has 7 electrons.
- Total for \( N_2 = 2 \times 7 = 14 \) electrons.

2. **For \( N_2^- \)**:
- Total for \( N_2 = 14 \) electrons + 1 extra electron (due to the negative charge).
- Total = 15 electrons.

3. **For \( F_2^+ \)**:
- Each fluorine atom has 9 electrons.
- Total for \( F_2 = 2 \times 9 = 18 \) electrons - 1 electron (due to the positive charge).
- Total = 17 electrons.

4. **For \( O_2^- \)**:
- Each oxygen atom has 8 electrons.
- Total for \( O_2 = 2 \times 8 = 16 \) electrons + 1 extra electron (due to the negative charge).
- Total = 17 electrons.

### Step 4: Compare the Total Electrons
Now we compare the total number of electrons:
- \( N_2 \): 14 electrons
- \( N_2^- \): 15 electrons
- \( F_2^+ \): 17 electrons
- \( O_2^- \): 17 electrons

From this, we can see that \( F_2^+ \) and \( O_2^- \) both have 17 electrons.

### Step 5: Conclusion
Since \( F_2^+ \) and \( O_2^- \) have the same number of total electrons, they will have the same bond order.

Thus, the species having the same bond order are:
- **\( F_2^+ \) and \( O_2^- \)**.