In which pair is the stronger bond found in the first species ?

Options

A. `O_(2)^(-),O_(2)`
B. `N_(2),N_(2)^(+)`
C. `NO^(-),NO^(+)`
D. `CO,O_(2)^(2+)`

Explanation

To determine which pair has the stronger bond in the first species, we need to analyze the bond orders of the given molecules. The bond order can be calculated using the formula:

\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]

Let's evaluate the bond orders for the pairs mentioned in the question.

### Step 1: Calculate the bond order for O2
- O2 has 10 bonding electrons and 6 antibonding electrons.
- Bond Order = \(\frac{10 - 6}{2} = \frac{4}{2} = 2\)

### Step 2: Calculate the bond order for O2^2+
- O2^2+ has 10 bonding electrons and 4 antibonding electrons.
- Bond Order = \(\frac{10 - 4}{2} = \frac{6}{2} = 3\)

### Step 3: Calculate the bond order for N2
- N2 has 10 bonding electrons and 4 antibonding electrons.
- Bond Order = \(\frac{10 - 4}{2} = \frac{6}{2} = 3\)

### Step 4: Calculate the bond order for N2^+
- N2^+ has 9 bonding electrons and 4 antibonding electrons.
- Bond Order = \(\frac{9 - 4}{2} = \frac{5}{2} = 2.5\)

### Step 5: Calculate the bond order for NO^+
- NO^+ has 10 bonding electrons and 5 antibonding electrons.
- Bond Order = \(\frac{10 - 5}{2} = \frac{5}{2} = 2.5\)

### Step 6: Calculate the bond order for NO^−
- NO^− has 10 bonding electrons and 6 antibonding electrons.
- Bond Order = \(\frac{10 - 6}{2} = \frac{4}{2} = 2\)

### Step 7: Calculate the bond order for CO
- CO has 10 bonding electrons and 4 antibonding electrons.
- Bond Order = \(\frac{10 - 4}{2} = \frac{6}{2} = 3\)

### Step 8: Compare bond orders
Now we can summarize the bond orders:
- O2: 2
- O2^2+: 3
- N2: 3
- N2^+: 2.5
- NO^+: 2.5
- NO^−: 2
- CO: 3

### Conclusion
From the calculations, the species with the highest bond order is N2 (3). Therefore, the stronger bond is found in the first species, which is N2.

Thus, the correct option is **B**.