The correct order of bond order among following species isI. NO–3NO3–II.  NO+2NO2+III. He2  IV. O2–2O22–

1. Break Down the Problem

To determine the correct order of bond order among the given species (NO₃⁻, NO₂⁺, He₂, O₂²⁻, and O₂), we first need to calculate the bond order for each species individually.

2. Relevant Concepts

The bond order can be calculated using the formula:

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

3. Analysis and Detail

1. For NO₃⁻:

   • Nitrogen has five valence electrons, and each oxygen has six.
   • The total number of valence electrons = $5 + 3 \times 6 + 1 = 24$ (adding one for the negative charge).
   • This results in resonance structures that primarily contribute to a bond order of 1.

   $\text{Bond Order (NO₃⁻)} = 1$

2. For NO₂⁺:

   • Nitrogen has five valence electrons and each oxygen has six.
   • The total number of valence electrons = $5 + 2 \times 6 - 1 = 16$ (subtracting one for the positive charge).
   • The bond order for NO₂⁺ is calculated as follows:

   $\text{Bond Order (NO₂⁺)} = 2$

3. For He₂:

   • Helium has two electrons.
   • Total electrons = 2 (bonding) in the σ(1s) orbital.
   • There are no antibonding electrons in He₂.

   $\text{Bond Order (He₂)} = \frac{(2 - 0)}{2} = 1$

4. For O₂²⁻:

   • Oxygen has six electrons; thus the total for O₂ is 12. For the O₂²⁻, we add two more due to the charges.
   • Total = $12 + 2 = 14$ electrons.
   • The bond order is calculated:

   $\text{Bond Order (O₂²⁻)} = \frac{(10 - 4)}{2} = 3$

5. For O₂:

   • O₂ has a total of 12 electrons.
   • The bond order is:

   $\text{Bond Order (O₂)} = \frac{(10 - 2)}{2} = 4$

4. Verify and Summarize

After calculating, we have:

• NO₃⁻: Bond Order = 1
• NO₂⁺: Bond Order = 2
• He₂: Bond Order = 1
• O₂²⁻: Bond Order = 3
• O₂: Bond Order = 4

Final Answer

The correct order of bond order among the species is: NO₃⁻ < He₂ = NO₂⁺ < O₂²⁻ < O₂.