Which of the following pair is expected to have the same bond order? (i) O2, N2 (ii) O2^+, N2^- (iii) O2^-, N2

NCERT Exemplar · Q33

Q. Which of the following pair is expected to have the same bond order?

(i) O2, N2

(ii) O2^+, N2^-

(iii) O2^-, N2^+

(iv) O2^-, N2^-

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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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The bond order of a diatomic species is determined by the number of electrons in bonding versus antibonding molecular orbitals:

\text{Bond order} = \frac{1}{2}(\text{bonding electrons} - \text{antibonding electrons})

We count the total electrons in each species and apply the MO filling order for homonuclear diatomics (up to $\ce{N2}$ : $\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \pi_{2p} < \sigma_{2p} < \pi^*_{2p} < \sigma^*_{2p}$ ; from $\ce{O2}$ onward the $\sigma_{2p}$ and $\pi_{2p}$ order swaps).

Electron counts and bond orders:

Species	Total e⁻	Bonding e⁻	Antibonding e⁻	Bond order
$\ce{O2}$	16	10	6	2.0
$\ce{N2}$	14	10	4	3.0
$\ce{O2^+}$	15	10	5	2.5

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Bond order depends on the number of electrons in bonding and antibonding molecular orbitals. Two species have the same bond order when they are isoelectronic (same total electron count). $\mathrm{O}_2^+$ and $\mathrm{N}_2^-$ both have 15 electrons and bond order 2.5.

The bond order of a molecule tells us the net number of electron pairs holding two atoms together. It directly predicts bond strength and stability: higher bond order means a stronger, shorter bond.

For diatomic molecules of second-period elements, we calculate bond order from molecular orbital theory:

\text{Bond order} = \frac{1}{2}\left(\text{electrons in bonding MOs} - \text{electrons in antibonding MOs}\right)

The key insight is that isoelectronic species—molecules or ions with the same total number of electrons—fill their molecular orbitals identically and therefore have the same bond order. We need to count electrons carefully for each species.

Step-by-step analysis

1. Count the total electrons in each species

Start with the neutral molecules:

$\mathrm{N}_2$ : nitrogen has 7 electrons, so $\mathrm{N}_2$ has $7 + 7 = 14$ electrons
$\mathrm{O}_2$ : oxygen has 8 electrons, so $\mathrm{O}_2$ has $8 + 8 = 16$ electrons

Now adjust for charge:

$\mathrm{O}_2^+$ : one electron removed → $16 - 1 = 15$ electrons
$\mathrm{O}_2^-$ : one electron added → $16 + 1 = 17$ electrons
$\mathrm{N}_2^+$ : one electron removed → $14 - 1 = 13$ electrons
$\mathrm{N}_2^-$ : one electron added → $14 + 1 = 15$ electrons

2. Identify isoelectronic pairs

Looking at our electron counts:

$\mathrm{O}_2^+$ and $\mathrm{N}_2^-$ both have 15 electrons ✓

This is the only isoelectronic pair among the options.

3. Verify the bond orders

For 15-electron species, the MO configuration (for $\mathrm{O}_2$ and $\mathrm{N}_2$ ) is:

$\sigma_{1s}^2 \, \sigma_{1s}^*{}^2 \, \sigma_{2s}^2 \, \sigma_{2s}^*{}^2 \, \pi_{2p}^4 \, \sigma_{2p}^2 \, \pi_{2p}^*{}^1$ …

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Q.Which of the following pair is expected to have the same bond order?

Step-by-step analysis

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