Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic called bonding molecular orbital (BMO) and anti - bonding molecular orbital (ABMO). Energy of anti - bonding molecular orbital (BMO) and anti - bonding molecular orbital ABMO). Energy of anti - bonding orbitals 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 nitrogen increase in the order : `sigma1slt sigma^(**)1s lt sigma2sltsigma^(**)2slt(pi2p_(x)~~pi2p_(y))ltsigma2p_(z)lt(pi^(**)p2p_(x)~~pi^(**)2p_(y))ltsigma^(**)2p_(z)` and for oxygen and fluorine order of energy of molecular orbitals is given as : `sigma1s lt sigma^(**)1slt sigma2sltsigma^(**)2sltsigma2p_(z)lt(pi2p_(x)~=2pi2p_(y))lt(pi^(**)2p_(x)~=pi^(**)2p_(y))ltsigma^(**)2p_(z).` Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which 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 lateral, 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 moleculas or their ions. Bond order is one of hte most important 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 molecules has the same bond order, we will follow these steps:

### Step 1: Understand the Bond Order Formula
The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} \left( \text{Number of electrons in bonding orbitals} - \text{Number of electrons in anti-bonding orbitals} \right) \]

### Step 2: Determine the Electron Configuration for O2
Oxygen (O2) has 16 electrons. The molecular orbital configuration for O2 is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi_{2p_x}^2 = \pi_{2p_y}^2\)
- \(\pi^*_{2p_x}^1 = \pi^*_{2p_y}^1\)

Counting the electrons:
- Bonding electrons = 10 (from \(\sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y}\))
- Anti-bonding electrons = 6 (from \(\sigma^*_{1s}, \sigma^*_{2s}, \pi^*_{2p_x}, \pi^*_{2p_y}\))

Calculating bond order for O2:
\[ \text{Bond Order}_{O2} = \frac{1}{2} (10 - 6) = \frac{4}{2} = 2 \]

### Step 3: Determine the Electron Configuration for O2+
O2+ has 15 electrons (one less than O2). The configuration is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi_{2p_x}^2 = \pi_{2p_y}^2\)
- \(\pi^*_{2p_x}^1 = \pi^*_{2p_y}^0\)

Counting the electrons:
- Bonding electrons = 10
- Anti-bonding electrons = 5

Calculating bond order for O2+:
\[ \text{Bond Order}_{O2+} = \frac{1}{2} (10 - 5) = \frac{5}{2} = 2.5 \]

### Step 4: Determine the Electron Configuration for O2-
O2- has 17 electrons (one more than O2). The configuration is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi_{2p_x}^2 = \pi_{2p_y}^2\)
- \(\pi^*_{2p_x}^2 = \pi^*_{2p_y}^1\)

Counting the electrons:
- Bonding electrons = 10
- Anti-bonding electrons = 7

Calculating bond order for O2-:
\[ \text{Bond Order}_{O2-} = \frac{1}{2} (10 - 7) = \frac{3}{2} = 1.5 \]

### Step 5: Determine the Electron Configuration for N2
Nitrogen (N2) has 14 electrons. The molecular orbital configuration for N2 is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\pi_{2p_x}^2 = \pi_{2p_y}^2\)
- \(\sigma_{2p_z}^2\)

Counting the electrons:
- Bonding electrons = 10
- Anti-bonding electrons = 4

Calculating bond order for N2:
\[ \text{Bond Order}_{N2} = \frac{1}{2} (10 - 4) = \frac{6}{2} = 3 \]

### Step 6: Determine the Bond Orders for N2+ and N2-
N2+ has 13 electrons. The configuration is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\pi_{2p_x}^2 = \pi_{2p_y}^2\)
- \(\sigma_{2p_z}^1\)

Counting the electrons:
- Bonding electrons = 9
- Anti-bonding electrons = 4

Calculating bond order for N2+:
\[ \text{Bond Order}_{N2+} = \frac{1}{2} (9 - 4) = \frac{5}{2} = 2.5 \]

N2- has 15 electrons. The configuration is:
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\pi_{2p_x}^2 = \pi_{2p_y}^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi^*_{2p_x}^1 = \pi^*_{2p_y}^1\)

Counting the electrons:
- Bonding electrons = 10
- Anti-bonding electrons = 6

Calculating bond order for N2-:
\[ \text{Bond Order}_{N2-} = \frac{1}{2} (10 - 6) = \frac{4}{2} = 2 \]

### Step 7: Compare Bond Orders
- O2: 2
- O2+: 2.5
- O2-: 1.5
- N2: 3
- N2+: 2.5
- N2-: 2

### Conclusion
The pairs that have the same bond order are:
- O2+ (2.5) and N2+ (2.5)