The energy of σ2pz molecular orbital is greater than π2pₓ and π2py molecular orbitals in nitrogen molecule. Wr

NCERT Exemplar · Q43

Q. The energy of σ2p_z molecular orbital is greater than π2p_x and π2p_y molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species : N2, N2^+, N2^-, N2^2+

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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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Energy Levels and Comparison of N₂ Species

Concept: For N₂ (Z = 14), s–p mixing occurs, making σ2p_z higher in energy than π2p orbitals.

Energy Level Sequence (Increasing Order)

\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z

Electronic Configurations and Analysis

Species	Total e⁻	Configuration (valence)	Bond Order	Magnetic Nature
N₂²⁺	12	KK σ2s² σ*2s² π2p⁴	3	Diamagnetic
N₂⁺	13	KK σ2s² σ*2s² π2p⁴ σ2p_z¹	2.5	Paramagnetic
N₂	14	KK σ2s² σ*2s² π2p⁴ σ2p_z²	3	Diamagnetic
N₂⁻	15	KK σ2s² σ2s² π2p⁴ σ2p_z² π2p¹	2.5	Paramagnetic

Bond order = $\frac{1}{2}$ (bonding e⁻ − antibonding e⁻)

Stability order: N₂ = N₂²⁺ > N₂⁺ = N₂⁻ (higher bond order = greater stability) …

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For $\mathrm{N_2}$ ( $Z < 8$ ), the $\sigma_{2p_z}$ orbital lies above the $\pi_{2p_x,y}$ pair. Using molecular orbital theory and bond order, we find $\mathrm{N_2^{2+}}$ is least stable and paramagnetic, $\mathrm{N_2^+}$ and $\mathrm{N_2^-}$ have equal stability (both paramagnetic), and $\mathrm{N_2}$ is most stable and diamagnetic.

Why bond order governs stability

Molecular orbital theory builds molecules by combining atomic orbitals into bonding and antibonding MOs. Electrons in bonding orbitals pull nuclei together; those in antibonding orbitals push them apart. The net effect is captured by bond order:

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

where $N_b$ is the number of electrons in bonding orbitals and $N_a$ in antibonding orbitals. Higher bond order means a stronger, shorter, more stable bond. Magnetic behavior follows from unpaired electrons: any species with unpaired electrons is paramagnetic; all paired means diamagnetic.

For second-period homonuclear diatomics with $Z < 8$ (B, C, N), $s$ – $p$ mixing is significant enough that the $\sigma_{2p_z}$ orbital is pushed above the degenerate $\pi_{2p_x}$ and $\pi_{2p_y}$ pair. This is the key orbital-ordering difference from $\mathrm{O_2}$ and $\mathrm{F_2}$ .

Energy-level sequence for nitrogen

The complete MO energy sequence in increasing order for $\mathrm{N_2}$ is:

\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*

Notice the $\pi_{2p}$ bonding orbitals come before $\sigma_{2p_z}$ .

Step-by-step analysis of each species

1. Count total electrons

$\mathrm{N_2}$ : $7 + 7 = 14$ electrons
$\mathrm{N_2^+}$ : $14 - 1 = 13$ electrons
$\mathrm{N_2^-}$ : $14 + 1 = 15$ electrons
$\mathrm{N_2^{2+}}$ : $14 - 2 = 12$ electrons

2. Fill the MO diagram according to the sequence above

Species	Configuration (valence only: $2s$ and $2p$ )	$N_b$	$N_a$	Bond Order
$\mathrm{N_2}$	$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\sigma_{2p_z}^2$	8	2	$\frac{8-2}{2}=3$
$\mathrm{N_2^+}$	$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\sigma_{2p_z}^1$	7	2	$\frac{7-2}{2}=2.5$
$\mathrm{N_2^-}$	$\sigma_{2s}^2\,\sigma_{2s}^{2}\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\sigma_{2p_z}^2\,\pi_{2p_x}^{1}$	8	3	$\frac{8-3}{2}=2.5$
$\mathrm{N_2^{2+}}$	$\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\pi_{2p_x}^2\,\pi_{2p_y}^2$	6	2	$\frac{6-2}{2}=2$

3. Determine unpaired electrons

$\mathrm{N_2}$ : all orbitals fully paired → diamagnetic
$\mathrm{N_2^+}$ : one electron in $\sigma_{2p_z}$ → paramagnetic
$\mathrm{N_2^-}$ : one electron in $\pi_{2p_x}^*$ (or $\pi_{2p_y}^*$ ) → paramagnetic …

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Why bond order governs stability

Energy-level sequence for nitrogen

Step-by-step analysis of each species

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