MO Diagram of Nitrogen

Introduction

Nitrogen gas provides a closed-shell test of the second-period MO diagram. Each N atom contributes five valence electrons, so N₂ has ten valence electrons. With the standard light-diatomic ordering, these fill bonding orbitals enough to give bond order three and no unpaired electrons.

Core explanation

Use the common B₂–N₂ valence ordering: σ2s, σ2s , the pair π2p, then σ2p, followed by antibonding 2p levels. Place ten valence electrons. σ2s² and σ2s ² use four. π2p orbitals together hold four more, two in each. The last two fill σ2p. Written compactly, the valence configuration is σ2s²σ2s ²(π2p)⁴σ2p². All occupied orbitals have paired electrons in this ground-state diagram.

Count bonding electrons: two in σ2s, four in π2p and two in σ2p, totaling eight. Antibonding valence electrons: two in σ2s , totaling two. Bond order = (8 − 2)/2 = 3. This is consistent with the strong N≡N bond in a Lewis representation. The MO model supplies electron occupancy and predicts diamagnetism because no electron remains unpaired.

The 1s core levels can be included or omitted. If included, both σ1s and σ1s are filled with two electrons each, adding equal bonding and antibonding counts that cancel. If omitted, use ten valence electrons, not fourteen total electrons. Mixing a valence-only diagram with total-electron counting is a common source of accidental extra electrons.

The light-diatomic order matters conceptually even though fully filled π2p and σ2p levels give the same total bonding count if those two levels were swapped. For B₂ and C₂, the order changes unpaired-electron predictions, and a consistent N₂ diagram prepares comparisons with its ions. If N₂ loses one electron to N₂⁺, it comes from the highest occupied σ2p bonding orbital in the standard diagram, lowering bond order to 2.5 and creating an unpaired electron. This is a model comparison for gas-phase species.

N₂'s low reactivity under ordinary conditions is often related to its strong bond, but thermodynamic bond strength alone does not determine every rate. Reactions such as nitrogen fixation need pathways that overcome activation barriers, often supplied by catalysts or high-energy conditions. The MO diagram helps explain bonding, not a complete kinetic mechanism.

Step-by-step reasoning

1. Count N₂ valence electrons: 5 + 5 = 10. 2. Choose π2p below σ2p for the standard N₂ diagram. 3. Fill σ2s²σ2s ²π2p⁴σ2p². 4. Count eight bonding and two antibonding electrons. 5. Calculate order three and note all electrons paired.

Visual explanation

Draw a vertical MO ladder with two σ2s boxes, two equal π2p boxes and one σ2p box. Put paired arrows in σ2s, σ2s , both π boxes and σ2p. Colour eight bonding arrows blue and two antibonding arrows red, then show (8−2)/2=3.

Real-world analogy

A ledger with eight contributions supporting a project and two offsetting it leaves a net six, which the bond-order scale halves to three. The analogy is arithmetic only; electrons do not literally vote for or against a bond.

Real-world example

Atmospheric nitrogen is N₂. Its strong bond and lack of unpaired electrons in the ground-state MO diagram are important features when comparing it with O₂, which has unpaired electrons despite also being a stable atmospheric gas.

Why?

Why is N₂ diamagnetic in this model? Every occupied MO contains an electron pair with opposite spins. There is no unpaired spin contribution in the ground-state configuration.

Common misconception

“N₂ has fourteen electrons, so place fourteen in the displayed valence-only 2s/2p diagram.” Ten are valence electrons; the four 1s core electrons belong in omitted σ1s and σ1s levels if an all-electron diagram is used.

Worked example

Compute N₂⁺ after N₂. Neutral N₂ has eight bonding and two antibonding valence electrons, order three. Removing one electron from highest occupied σ2p leaves seven bonding and two antibonding, so order (7 − 2)/2 = 2.5. The singly occupied σ2p gives one unpaired electron. This illustrates why identifying the removed orbital matters more than merely noting a positive charge.

Quick check

1. What is N₂'s ground-state bond order and magnetic classification in the simple MO model? Answer: Bond order 3 and diamagnetic, with no unpaired electrons.

Exam focus

State valence-electron count and π-before-σ ordering. Show occupancy and the bond-order tally, then count unpaired electrons separately. Do not equate strong bonding with a universal reaction-rate prediction.

Advanced insight

Accurate N₂ spectroscopy resolves vibrational and electronic states beyond a single MO box diagram. The introductory configuration is a qualitative ground-state model that captures strong bonding and paired-electron magnetism.

Summary

N₂'s ten valence electrons fill σ2s²σ2s ²π2p⁴σ2p² in the standard light-diatomic diagram. Eight bonding minus two antibonding electrons give order three, and paired occupancy gives diamagnetism.

Practice questions

1. How many valence electrons does N₂ have? Answer: Ten. 2. How many bonding and antibonding valence electrons are in its basic diagram? Answer: Eight bonding and two antibonding. 3. What happens to bond order on forming N₂⁺ by removing σ2p electron? Answer: It falls from 3 to 2.5. 4. Why does omitting 1s core orbitals not change bond order? Answer: Filled core bonding and antibonding levels contribute equally and cancel.