Molecular Orbital Theory JEE 2016: Why A and C Are Correct

Which options are correct in the Molecular Orbital Theory JEE 2016 question?

A and C are correct in the Molecular Orbital Theory JEE 2016 question from JEE Advanced 2016 Paper 2. This Chemistry PYQ belongs to Chemical Bonding and Molecular Structure and uses the multiple-correct format: select every claim supported by Molecular Orbital Theory.

The four claims, paraphrased, are:

The question bank tags this question medium, with a 90-second target. These are question-bank labels, not measured student performance. Test each claim through electron count, orbital occupancy and bond order.

Why is the carbon molecular dianion in option A diamagnetic?

A is true because every occupied orbital contains paired electrons. Diamagnetism means there are no unpaired electrons. Each carbon contributes four valence electrons, and the negative charge adds two.

C_{2}^{2 -} : N_{v a l e n c e} = 2 \times 4 + 2 = 10

Use the light-diatomic ordering specified in the official solution. Filling ten valence electrons gives the configuration below.

σ 2 s < σ^{*} 2 s < π 2 p_{x} = π 2 p_{y} < σ 2 p_{z}

(σ 2 s)^{2} (σ^{*} 2 s)^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2} (σ 2 p_{z})^{2}

The two equal-energy pi orbitals fill singly first, following Hund’s rule, before pairing. In the completed configuration, both pi orbitals are doubly occupied. Every other occupied orbital also contains a pair.

For bond order, count bonding and antibonding electrons separately. For second-period species, the filled core bonding and antibonding orbitals cancel, so valence-only counting is sufficient.

Bond order = \frac{bonding electrons - antibonding electrons}{2}

(σ 1 s)^{2} (σ^{*} 1 s)^{2} : \frac{2 - 2}{2} = 0

The carbon dianion has eight bonding and two antibonding valence electrons, giving bond order three. But paired occupancy, not bond order alone, establishes diamagnetism.

Why does removing two electrons shorten the oxygen bond in option B?

B is false: double ionisation removes antibonding electrons, raising bond order from two to three. For this comparison, MOT therefore predicts a shorter bond. Positive charge alone does not establish the direction of the change.

Each oxygen contributes six valence electrons. Unlike carbon and nitrogen here, oxygen places the bonding sigma orbital below the bonding pi orbitals.

O_{2} : N_{v a l e n c e} = 2 \times 6 = 12

σ 2 s < σ^{*} 2 s < σ 2 p_{z} < π 2 p_{x} = π 2 p_{y}

(σ 2 s)^{2} (σ^{*} 2 s)^{2} (σ 2 p_{z})^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2} (π^{*} 2 p_{x})^{1} (π^{*} 2 p_{y})^{1}

There are eight bonding and four antibonding electrons. The two singly occupied antibonding pi orbitals supply the electrons removed during double ionisation.

BO (O_{2}) = \frac{8 - 4}{2} = 2

Removing one electron from each leaves eight bonding and two antibonding electrons. No bonding orbital loses an electron.

BO (O_{2}^{2 +}) = \frac{8 - 2}{2} = 3

The higher bond order predicts the following bond-length relation, where each length is measured between the nuclei. This is opposite to B’s claim.

r (O_{2}^{2 +}) < r (O_{2})

Why do the two nitrogen ions in option C have equal bond orders?

C is true: both ions have bond order 2.5. Removing a bonding electron and adding an antibonding electron each reduce nitrogen’s bond order by one-half.

Each nitrogen contributes five valence electrons. The occupied valence configuration matches the carbon dianion, with eight bonding and two antibonding electrons.

N_{2} : N_{v a l e n c e} = 2 \times 5 = 10

(σ 2 s)^{2} (σ^{*} 2 s)^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2} (σ 2 p_{z})^{2}

BO (N_{2}) = \frac{8 - 2}{2} = 3

For the positive ion, remove one electron from the highest occupied bonding orbital. Its occupancy changes from two to one.

N_{2}^{+} : (σ 2 p_{z})^{2} ⟶ (σ 2 p_{z})^{1}

BO (N_{2}^{+}) = \frac{7 - 2}{2} = 2.5

For the negative ion, the added electron enters an antibonding pi orbital. The original bonding occupancy stays unchanged.

N_{2}^{-} : (π^{*} 2 p)^{1}

BO (N_{2}^{-}) = \frac{8 - 3}{2} = 2.5

The charge signs differ, but the bond orders agree. This follows from orbital occupancy, not a general rule about oppositely charged ions.

Why is the equal-energy claim for the helium ion false?

D is false under the supplied solution’s bond-order argument: the helium molecular cation has positive bond order and net bonding stabilisation. Count total electrons here, not second-shell valence electrons. Each neutral helium atom supplies two, and the positive charge removes one.

{H e}_{2}^{+} : N = 2 \times 2 - 1 = 3

(σ 1 s)^{2} (σ^{*} 1 s)^{1}

BO ({H e}_{2}^{+}) = \frac{2 - 1}{2} = 0.5

Following the official solution’s intended qualitative reasoning, this non-zero bonding stabilisation rejects D’s equal-energy claim. Bond order does not numerically calculate total energy, so no energy difference should be inferred from 0.5.

A true; B false; C true; D false. Select A and C.

What reasoning mistake would make you accept option B?

The mistake is treating fewer electrons as automatically weaker bonding. The faulty chain is: the oxygen dication has fewer electrons, so bonding must weaken, so the bond must become longer. It skips the deciding step: identifying whether the removed electrons were bonding or antibonding.

Use the initial bonding and antibonding counts below. Removing two antibonding electrons reduces only the antibonding count by two.

N_{b}, N_{a}

{BO}_{n e w} = \frac{N_{b} - (N_{a} - 2)}{2} = \frac{N_{b} - N_{a}}{2} + 1

Bond order therefore rises by one. Compare this with removing a bonding electron from nitrogen.

O_{2} \to O_{2}^{2 +} : 2 \to 3

N_{2} \to N_{2}^{+} : 3 \to 2.5

Oxygen loses antibonding electrons; nitrogen loses a bonding electron. Name the orbital before predicting the effect of ionisation.

Can you apply the same method to three related questions?

Use electron count, occupancy, bond order and unpaired-electron count in that order. These are original related practice questions within Chemical Bonding and Molecular Structure, not additional verified PYQs. Attempt all three before reading the checks.

Find the bond order and magnetic nature of the superoxide ion. $O_{2}^{-}$
Find the bond order and magnetic nature of the nitrogen dication using the same MO model. $N_{2}^{2 +}$
Compare neutral molecular helium and its singly positive ion using their configurations.

{H e}_{2}, {H e}_{2}^{+}

Answer checks

Question 1: There are 13 valence electrons: eight bonding and five antibonding. The added electron pairs in one antibonding pi orbital; one unpaired electron remains, so the ion is paramagnetic.

BO (O_{2}^{-}) = \frac{8 - 5}{2} = 1.5

Question 2: Remove both electrons from nitrogen’s highest occupied bonding sigma orbital. All remaining electrons are paired, so the dication is diamagnetic.

(σ 2 p_{z})^{2} ⟶ (σ 2 p_{z})^{0}

BO (N_{2}^{2 +}) = \frac{6 - 2}{2} = 2

Question 3: Neutral molecular helium has cancelling bonding and antibonding occupancies. Its cation has bond order 0.5 and one unpaired electron.

{H e}_{2} : (σ 1 s)^{2} (σ^{*} 1 s)^{2}, BO = 0

{H e}_{2}^{+} : (σ 1 s)^{2} (σ^{*} 1 s)^{1}, BO = 0.5

For any missed answer, rewrite the changed orbital’s occupancy. Then recalculate bond order and check for unpaired electrons.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

Which options are correct in the JEE Advanced 2016 molecular orbital theory question?

A and C are correct in this multiple-correct question from JEE Advanced 2016 Paper 2. The carbon molecular dianion has no unpaired electrons, and the singly positive and singly negative nitrogen molecular ions both have bond order 2.5.

Why is C2 2− diamagnetic?

C2 2− has ten valence electrons, which fill the molecular orbitals with all electrons paired under the light-diatomic ordering used in the solution. Its bond order is 3, but paired occupancy—not bond order alone—establishes diamagnetism.

Why is the bond in O2 2+ shorter than in O2?

Forming O2 2+ removes two electrons from antibonding pi orbitals without removing any bonding electrons. The bond order rises from 2 to 3, so molecular orbital theory predicts a shorter O–O bond.

Why do N2+ and N2− have the same bond order?

Neutral N2 has bond order 3. Forming N2+ removes one bonding electron, while forming N2− adds one antibonding electron. Each change lowers bond order by 0.5, giving both ions a bond order of 2.5.