Use the molecular orbital energy level diagram to show that N2 would be expected to have a triple bond, F2, a

NCERT Exemplar · Q65

Q. Use the molecular orbital energy level diagram to show that N2 would be expected to have a triple bond, F2, a single bond and Ne2, no bond.

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Concept understanding — Bond Order Strength

Bond Order Strength: From Intuition to Precision

Imagine two people holding hands. If they just touch fingertips, a gentle breeze can separate them. If they clasp firmly, it takes more effort to pull them apart. If they lock arms, you need real force. That's the core idea behind bond order — it tells you how strongly two atoms are connected in a molecule.

The Intuition

In a chemical bond, atoms share electrons. The more electron pairs they share, the tighter the grip. A single bond (one shared pair) is like a handshake — it works, but it's easy to break. A double bond (two shared pairs) is like a firm clasp — stronger, shorter, harder to pull apart. A triple bond (three shared pairs) is like a wrestler's lock — very strong and very short.

This directly translates to real molecules:

C–C single bond: bond energy ≈ 350 kJ/mol, bond length ≈ 154 pm
C=C double bond: bond energy ≈ 610 kJ/mol, bond length ≈ 134 pm
C≡C triple bond: bond energy ≈ 835 kJ/mol, bond length ≈ 120 pm

More shared electrons → stronger bond → shorter bond. That's the pattern.

The Precise Definition

Bond order is the number of chemical bonds between a pair of atoms. For simple molecules, it's just the number of shared electron pairs:

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

This formula matters most when you move beyond simple Lewis structures — for molecules with resonance or molecular orbital theory.

How Bond Order Determines Strength

Bond order and bond strength are directly proportional. Here's why:

More electron density between nuclei: Higher bond order means more electrons are concentrated in the region between the two nuclei. These electrons simultaneously attract both nuclei, pulling them together.
Greater electrostatic attraction: The shared electrons act like "glue." More glue means stronger adhesion.
Shorter bond length: Stronger attraction pulls the nuclei closer. Shorter bonds are harder to stretch or break.

Watch out

Don't confuse bond order with bond energy. Bond order tells you the number of bonds; bond energy tells you the energy required to break them. They're proportional, but not identical — a C=C bond isn't exactly twice as strong as a C–C bond (it's about 1.7 times stronger).

Real Examples

Molecule	Bond	Bond Order	Bond Energy (kJ/mol)	Bond Length (pm)
H₂	H–H	1	436	74
O₂	O=O	2	498	121
N₂	N≡N	3	945	110
F₂	F–F	1	159	142

Notice how N₂ with a triple bond is the strongest diatomic molecule — it takes 945 kJ/mol to break that bond. That's why nitrogen gas is so unreactive.

When Bond Order Gets Tricky

Some molecules don't have simple whole-number bond orders. Consider ozone (O₃): …

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Bond order $= \tfrac{1}{2}(N_b - N_a)$ . Filling the MO diagram gives N₂ a bond order of 3 (triple bond), F₂ a bond order of 1 (single bond), and Ne₂ a bond order of 0 (no bond).

In molecular orbital theory the atomic orbitals of the two atoms combine to form bonding molecular orbitals (lower energy) and antibonding molecular orbitals (higher energy). Electrons fill these from lowest energy upward. The bond order decides whether a bond exists and how strong it is:

\text{Bond order} = \frac{1}{2}\left(N_b - N_a\right)

A bond order of 0 means no bond; 1, 2, 3 mean single, double and triple bonds respectively.

For second-period diatomics the MO ordering differs on either side of N₂: up to and including N₂ the $\pi_{2p}$ pair lies below $\sigma_{2p_z}$ ; for O₂, F₂ and Ne₂ the order reverses so $\sigma_{2p_z}$ lies below $\pi_{2p}$ . The bond order is unaffected by this swap.

1. N₂ (14 electrons)

Configuration:

\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 valence contributions: $N_b = 2(\sigma_{2s}) + 4(\pi_{2p}) + 2(\sigma_{2p_z}) = 8$ and $N_a = 2(\sigma_{2s}^*)$ .

\text{Bond order} = \frac{1}{2}(8-2) = 3

A triple bond — one $\sigma$ and two $\pi$ bonds — consistent with N₂'s high stability and short bond length ( $109.8$ pm).

2. F₂ (18 electrons)

Configuration (F₂ ordering, $\sigma_{2p_z}$ below $\pi_{2p}$ ):

\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}^{*2}

The $\sigma_{2p_z}^*$ orbital stays empty. Valence counts: $N_b = 2(\sigma_{2s}) + 2(\sigma_{2p_z}) + 4(\pi_{2p}) = 8$ and $N_a = 2(\sigma_{2s}^*) + 4(\pi_{2p}^*) = 6$ .

$\text{Bond order} = \frac{1}{2}(8-6) = 1$ …

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Q.Use the molecular orbital energy level diagram to show that N2 would be expected to have a triple bond, F2, a single bond and Ne2, no bond.

1. N₂ (14 electrons)

2. F₂ (18 electrons)

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