Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine atom from two molecular orbitals called vonding molecular orbital (BOM) and anti bonding molecular orbital (ABMO). Energy of anti bonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital is lowered than the parent atomic orbitals. Energies of various molecular orbitals for elements hydrogen to nitrogne increase in the order :
`sigma_(1s)ltsigma_(1s)^(**)ltsigma_(2s)ltsigma_(2s)^(**)lt(pi_(2py)~~pi_(2py))lt sigma_(2py)lt(pi_(2py)^(**)~~pi_(2py)^(**))ltsigma_(2pz)^(**)` and for ozygen and fluorine order of energy of molecular orbitals id given below :
`sigma_(1s)ltsigma_(1s)^(**)ltsigma_(2s)ltsigma_(2s)^(**)ltsigma_(2pz)lt(pi _(2px)^(**)~~pi_(2py)^(**))sigma_(2pz)^(**)`
Different atomic orbitals of one atom combine wiht the atomic orbitals of the second atom whihc have comparable energies and proper orientation. Further, if the overlapping is head on, the molecular orbital is called 'Sigma' `(sigma)` and if the overlap is atreal, the molecular orbital is called 'pi', `(pi).` The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions, Bond order is one the most ipmrtaint parameters to compare the strength of bonds.
Which of the following pair is expected to have the same bond order ?

Options

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

Explanation

To determine which pair of species has the same bond order, we need to calculate the bond order for each species mentioned in the options. Bond order can be calculated using the formula:

\[ \text{Bond Order} = \frac{(N_b - N_a)}{2} \]

where \( N_b \) is the number of electrons in bonding molecular orbitals and \( N_a \) is the number of electrons in antibonding molecular orbitals.

### Step-by-Step Solution:

1. **Identify the Electron Count for Each Species:**
- For \( O_2 \): 16 electrons
- For \( O_2^+ \): 15 electrons
- For \( O_2^- \): 17 electrons
- For \( N_2 \): 14 electrons
- For \( N_2^- \): 15 electrons
- For \( N_2^+ \): 13 electrons

2. **Determine the Molecular Orbital Configuration:**
- For \( O_2 \):
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- \( N_b = 10 \), \( N_a = 6 \) → Bond Order = \( \frac{(10 - 6)}{2} = 2 \)
- For \( O_2^+ \):
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \)
- \( N_b = 9 \), \( N_a = 6 \) → Bond Order = \( \frac{(9 - 6)}{2} = 1.5 \)
- For \( O_2^- \):
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^1 \)
- \( N_b = 11 \), \( N_a = 6 \) → Bond Order = \( \frac{(11 - 6)}{2} = 2.5 \)
- For \( N_2 \):
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \)
- \( N_b = 10 \), \( N_a = 0 \) → Bond Order = \( \frac{(10 - 0)}{2} = 5 \)
- For \( N_2^- \):
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^1 \)
- \( N_b = 11 \), \( N_a = 0 \) → Bond Order = \( \frac{(11 - 0)}{2} = 5.5 \)
- For \( N_2^+ \):
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \)
- \( N_b = 9 \), \( N_a = 0 \) → Bond Order = \( \frac{(9 - 0)}{2} = 4.5 \)

3. **Compare Bond Orders:**
- \( O_2 \) has a bond order of 2.
- \( O_2^+ \) has a bond order of 1.5.
- \( O_2^- \) has a bond order of 2.5.
- \( N_2 \) has a bond order of 5.
- \( N_2^- \) has a bond order of 5.5.
- \( N_2^+ \) has a bond order of 4.5.

4. **Identify the Pair with the Same Bond Order:**
- The only pair that has the same bond order is \( O_2^+ \) and \( N_2^- \), both having a bond order of 1.5.

### Final Answer:
The pair expected to have the same bond order is \( O_2^+ \) and \( N_2^- \).