Fill in the blanks with appropriate choice.
Bond ordr of `N_(2)^(+) is ul(" "P)`while that of `N_(2) is ul(" "Q)`.
Bond order of `O_(2)^(+) is ul(" "R)` while that of `O_(2) is ul(" "S)`.
N - N bond distance `ul(" "T)` when `N_(2)` changes to `N_(2)^(+) ` and when `O_(2)` changes to `O_(2)^(+)` , the O - O bond distance `ul(" "U)`.

Options

A. `{:(P,Q,R,S," "T," "U),(2,2.5,2.5,1,"increases","decreases"):}`
B. `{:(P,Q,R,S," "T," "U),(2.5,3,2,1.5,"decreases","increases"):}`
C. `{:(P,Q,R,S," "T," "U),(3,2,1.5,1,"increases","decreases"):}`
D. `{:(P,Q,R,S," "T," "U),(2.5,3,2.5,2,"increases","decreases"):}`

Explanation

To solve the question step by step, we need to determine the bond orders and bond distances for the given molecular species.

### Step 1: Determine the bond order of \( N_2 \) and \( N_2^+ \)

1. **Calculate the bond order of \( N_2 \)**:
- \( N_2 \) has 14 electrons.
- The electronic configuration is:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^1 \]
- Bond order formula:
\[ \text{Bond Order} = \frac{n_b - n_a}{2} \]
where \( n_b \) is the number of bonding electrons and \( n_a \) is the number of antibonding electrons.
- For \( N_2 \):
- Bonding electrons (\( n_b \)) = 10 (from \( \sigma \) and \( \pi \) orbitals)
- Antibonding electrons (\( n_a \)) = 4 (from \( \sigma^* \) orbitals)
- Thus, bond order \( Q \) for \( N_2 \):
\[ Q = \frac{10 - 4}{2} = 3 \]

2. **Calculate the bond order of \( N_2^+ \)**:
- \( N_2^+ \) has 13 electrons (one electron is removed).
- The electronic configuration becomes:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^0 \]
- For \( N_2^+ \):
- Bonding electrons (\( n_b \)) = 9
- Antibonding electrons (\( n_a \)) = 4
- Thus, bond order \( P \) for \( N_2^+ \):
\[ P = \frac{9 - 4}{2} = 2.5 \]

### Step 2: Determine the bond order of \( O_2 \) and \( O_2^+ \)

1. **Calculate the bond order of \( O_2 \)**:
- \( O_2 \) has 16 electrons.
- The electronic configuration is:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \pi^*_{2p_y}^1 \]
- For \( O_2 \):
- Bonding electrons (\( n_b \)) = 10
- Antibonding electrons (\( n_a \)) = 6
- Thus, bond order \( S \) for \( O_2 \):
\[ S = \frac{10 - 6}{2} = 2 \]

2. **Calculate the bond order of \( O_2^+ \)**:
- \( O_2^+ \) has 15 electrons (one electron is removed).
- The electronic configuration becomes:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \]
- For \( O_2^+ \):
- Bonding electrons (\( n_b \)) = 10
- Antibonding electrons (\( n_a \)) = 5
- Thus, bond order \( R \) for \( O_2^+ \):
\[ R = \frac{10 - 5}{2} = 2.5 \]

### Step 3: Determine bond distances \( T \) and \( U \)

1. **Bond distance \( T \) when \( N_2 \) changes to \( N_2^+ \)**:
- As the bond order decreases from 3 to 2.5, the bond distance increases.
- Therefore, \( T \) is "increases".

2. **Bond distance \( U \) when \( O_2 \) changes to \( O_2^+ \)**:
- As the bond order increases from 2 to 2.5, the bond distance decreases.
- Therefore, \( U \) is "decreases".

### Final Answers:
- \( P = 2.5 \)
- \( Q = 3 \)
- \( R = 2.5 \)
- \( S = 2 \)
- \( T = \text{increases} \)
- \( U = \text{decreases} \)