Introduction

The bond order of N₂²⁻ is a frequently asked question in competitive exams that test concepts of chemical bonding and molecular orbital theory.

Knowing how to calculate this bond order helps in predicting bond length, bond strength and magnetic behaviour of the N₂²⁻ species.

Step‑by‑Step Solution for N₂²⁻

1. Total electrons in N₂²⁻

• Neutral N₂ has 14 electrons (7 from each nitrogen atom).
• The 2− charge adds 2 more electrons, so N₂²⁻ has a total of 16 electrons.

2. Valence MO ordering for nitrogen

For molecules up to nitrogen, the energy order of the 2p molecular orbitals is:

σ(2s), σ*(2s), π(2p_x) = π(2p_y), σ(2p_z)

3. MO configuration of N₂²⁻

• Core (1s and 2s) orbitals:
  σ_1s² σ_1s*² σ_2s² σ_2s*²
  → 8 electrons (4 bonding, 4 antibonding).
• Remaining 8 electrons (in 2p orbitals) are filled as:
  π_2px² π_2py² σ_2pz² π*_2p²
  (two electrons in the π* antibonding orbitals).

4. Bonding and antibonding electrons

• Bonding electrons:
  σ_1s², σ_2s², π_2px², π_2py², σ_2pz²
  → 10 bonding electrons in total.
• Antibonding electrons:
  σ_1s*², σ_2s*², π*_2p²
  → 6 antibonding electrons in total.

5. Bond order formula and calculation

Bond order is calculated using the formula:

Bond order = (Number of bonding electrons − Number of antibonding electrons) / 2

For N₂²⁻:

Bond order = (10 − 6) / 2 = 4 / 2 = 2

Therefore, the bond order in N₂²⁻ is 2, which matches option (A).

Why the Other Options Are Incorrect

Option (B) 2.5

• Bond order 2.5 is obtained for N₂⁻ or N₂⁺, not for N₂²⁻.
• In those species, only one electron occupies a 2p antibonding orbital, giving 10 bonding and 5 antibonding electrons, so bond order = (10 − 5) / 2 = 2.5.
• N₂²⁻ has two electrons in π* antibonding orbitals, so its bond order must be lower, not 2.5.

Option (C) 3

• Neutral N₂ has bond order 3 with configuration:
  σ_1s² σ_1s*² σ_2s² σ_2s*² π_2p⁴ σ_2p².
• Here, bonding = 10 and antibonding = 4, so bond order = (10 − 4) / 2 = 3, with no 2p antibonding electrons.
• Adding two extra electrons to form N₂²⁻ fills π* antibonding orbitals and reduces the bond order to 2 instead of keeping it at 3.

Option (D) 3.5

• A bond order of 3.5 would require fewer antibonding electrons than neutral N₂, which is not possible for N₂²⁻.
• N₂²⁻ has two electrons in π* antibonding orbitals, which significantly lowers bond order, so 3.5 is inconsistent with its MO electron configuration.

Comparison of N₂ Species