Chemical Bonding NEET 2008: Bond Order and Option D

Q: What is the correct answer to the Chemical Bonding NEET 2008 bond order question?

Option D is correct: He₂⁺ < O₂⁻ < NO < C₂²⁻. Their bond orders are 0.5, 1.5, 2.5 and 3, respectively, calculated using bond order = (bonding electrons − antibonding electrons)/2.

Q: How do you calculate the bond order of He₂⁺?

He₂⁺ has three electrons: two in the bonding sigma 1s orbital and one in the antibonding sigma-star 1s orbital. Its bond order is (2 − 1)/2 = 0.5. The 1s electrons must be included because their bonding and antibonding contributions do not cancel.

Q: Why is the bond order of O₂⁻ 1.5?

Neutral O₂ has bond order 2, and the extra electron in O₂⁻ enters a pi-star antibonding orbital. Adding one antibonding electron lowers bond order by 0.5, giving O₂⁻ a bond order of 1.5.

Q: Why does NO have a bond order of 2.5?

NO has 11 valence electrons, distributed as eight bonding and three antibonding electrons. Its bond order is therefore (8 − 3)/2 = 2.5. The fractional value follows directly from molecular orbital occupancy.

Q: What is the bond order of C₂²⁻?

C₂²⁻ has ten valence electrons, like N₂: eight occupy bonding orbitals and two occupy antibonding orbitals. Its bond order is (8 − 2)/2 = 3. The 2− charge adds two electrons to the whole species, not two to each carbon atom.

What is the correct order in the Chemical Bonding NEET 2008 question?

Option D is correct for the Chemical Bonding NEET 2008 question comparing He₂⁺, O₂⁻, NO and C₂²⁻. Their bond orders increase as follows:

H e_{2}^{+} < O_{2}^{-} < N O < C_{2}^{2 -}

$0.5 < 1.5 < 2.5 < 3$ Use Molecular Orbital Theory to choose the sequence from smallest to largest bond order. The original option mappings are:

A) O_{2}^{-} < N O < C_{2}^{2 -} < H e_{2}^{+}

B) N O < C_{2}^{2 -} < O_{2}^{-} < H e_{2}^{+}

C) C_{2}^{2 -} < H e_{2}^{+} < N O < O_{2}^{-}

D) H e_{2}^{+} < O_{2}^{-} < N O < C_{2}^{2 -}

The question bank tags this as hard and sets an expected solving time of 90 seconds. These are question-bank labels, not an official exam difficulty rating or a measured student average.

How do you calculate the bond order of He₂⁺?

He₂⁺ has bond order 0.5 because it contains two bonding electrons and one antibonding electron. A positive charge means one electron has been removed from the neutral species: $N_{e} = 2 + 2 - 1 = 3$

Use the bond-order formula:

B O = \frac{N_{b} - N_{a}}{2}

$N_{b} = number of electrons in bonding molecular orbitals$ $N_{a} = number of electrons in antibonding molecular orbitals$ An asterisk marks an antibonding orbital. Place the three electrons, then count each population:

(σ_{1 s})^{2} (σ_{1 s}^{*})^{1}

N_{b} = 2, N_{a} = 1

B O (H e_{2}^{+}) = \frac{2 - 1}{2} = 0.5

Bond order has no unit. Retain the 1s electrons here: their unequal bonding and antibonding populations supply the entire result.

Why does adding an electron to O₂ give O₂⁻ a bond order of 1.5?

Neutral O₂ has bond order 2. The extra electron in O₂⁻ enters a pi-star antibonding orbital, decreasing bond order by half a unit, not by one whole unit.

The negative charge adds one electron to the valence count:

N_{v a l e n c e} = 6 + 6 + 1 = 13

Adding one antibonding electron changes the formula to:

B O_{n e w} = \frac{N_{b} - (N_{a} + 1)}{2} = B O_{o l d} - \frac{1}{2}

B O (O_{2}^{-}) = 2 - 0.5 = 1.5

A direct valence-occupancy check gives the same answer:

(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}^{*})^{3}

The final term represents three electrons across the two pi-star orbitals:

N_{b} = 8, N_{a} = 5, B O = \frac{8 - 5}{2} = 1.5

For these second-period species, filled 1s bonding and antibonding core orbitals contribute equally and cancel. Omitting both leaves bond order unchanged. That cancellation does not apply to He₂⁺.

Why does NO have a fractional bond order of 2.5?

NO has eight valence bonding electrons and three valence antibonding electrons. Their difference is five, so dividing by two gives 2.5. The fractional result follows from orbital occupancy, not a special rule for NO.

Nitrogen contributes five valence electrons and oxygen contributes six: $N_{v a l e n c e} = 5 + 6 = 11$

Every electron belongs in one of these occupied valence orbitals:

Bonding: the sigma 2s orbital contains two electrons, the sigma 2p orbital contains two, and the two pi 2p orbitals contain four altogether.
Antibonding: the sigma-star 2s orbital contains two electrons, while the pi-star 2p set contains one.

In orbital notation:

Bonding: (σ_{2 s})^{2} (σ_{2 p_{z}})^{2} (π_{2 p_{x}})^{2} (π_{2 p_{y}})^{2}

Antibonding: (σ_{2 s}^{*})^{2} (π_{2 p}^{*})^{1}

N_{b} = 2 + 2 + 4 = 8, N_{a} = 2 + 1 = 3

B O (N O) = \frac{8 - 3}{2} = 2.5

Total electron count is not bond order. Eleven tells you how many electrons to place; their bonding and antibonding distribution supplies the answer.

How does the N₂ comparison give C₂²⁻ the highest bond order?

C₂²⁻ has ten valence electrons, like N₂, and the supplied solution uses this comparison to obtain bond order 3. The doubly negative charge adds two electrons to the whole species, not two to each carbon atom.

N_{v a l e n c e} = 4 + 4 + 2 = 10

Its valence occupancy is:

(σ_{2 s})^{2} (σ_{2 s}^{*})^{2} (π_{2 p_{x}})^{2} (π_{2 p_{y}})^{2} (σ_{2 p_{z}})^{2}

N_{b} = 8, N_{a} = 2

B O (C_{2}^{2 -}) = \frac{8 - 2}{2} = 3

Counting convention: use all electrons for He₂⁺ and valence electrons for the other three. Their cancelling core contributions are omitted.

He₂⁺: 2 bonding, 1 antibonding; bond order 0.5.
O₂⁻: 8 bonding, 5 antibonding; bond order 1.5.
NO: 8 bonding, 3 antibonding; bond order 2.5.
C₂²⁻: 8 bonding, 2 antibonding; bond order 3.

The increasing order is:

H e_{2}^{+} < O_{2}^{-} < N O < C_{2}^{2 -}

Option D is correct.

How can counting only antibonding electrons lead to option A?

Sorting only by antibonding population produces option A if you assume fewer antibonding electrons always means greater bond order. This is an illustrative route to a wrong option, not a claim about how often students make this mistake.

The displayed antibonding counts are:

N_{a} (O_{2}^{-}) = 5, N_{a} (N O) = 3, N_{a} (C_{2}^{2 -}) = 2, N_{a} (H e_{2}^{+}) = 1

Treating the largest antibonding count as the smallest bond order gives:

O_{2}^{-} < N O < C_{2}^{2 -} < H e_{2}^{+} [incorrect: option A]

The missing condition is equal bonding populations under a consistent counting convention. The three second-period species each have eight valence bonding electrons, so comparing their antibonding counts works. He₂⁺ has only two bonding electrons and cannot join that shortcut.

Calculate the bonding-minus-antibonding difference for every species before sorting:

N_{b} - N_{a} : 1, 3, 5, 6

These differences correspond to He₂⁺, O₂⁻, NO and C₂²⁻, respectively. Dividing each by two recovers option D.

Which three practice questions check the same method?

These original related practice questions are not additional verified PYQs. They test electron addition, electron removal and unequal 1s occupancy. Attempt each before reading its worked answer.

Arrange O₂⁺, O₂ and O₂⁻ by increasing bond order.

Adding an antibonding electron lowers bond order by half. Removing one raises it by half.

O_{2}^{-} < O_{2} < O_{2}^{+}

$1.5 < 2 < 2.5$ Calculate the bond order of N₂⁺ when its electron is removed from the highest occupied bonding MO.

Removing a bonding electron lowers the parent bond order by half. The antibonding population stays unchanged.

B O (N_{2}^{+}) = 3 - 0.5 = \frac{7 - 2}{2} = 2.5

Compare He₂ and He₂⁺ using their MO occupancies.

Neutral He₂ has equal bonding and antibonding populations. Removing one antibonding electron gives He₂⁺ a positive bond order.

H e_{2} : (σ_{1 s})^{2} (σ_{1 s}^{*})^{2}, B O = \frac{2 - 2}{2} = 0

H e_{2}^{+} : (σ_{1 s})^{2} (σ_{1 s}^{*})^{1}, B O = \frac{2 - 1}{2} = 0.5

For your next question: count electrons, assign bonding and antibonding occupancy, calculate the difference, divide by two, then order the numerical results.

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For a worked example of the same idea, see Laws of Motion NEET 2004: Accelerating Wedge Solution.

Frequently asked questions

What is the correct answer to the Chemical Bonding NEET 2008 bond order question?

Option D is correct: He₂⁺ < O₂⁻ < NO < C₂²⁻. Their bond orders are 0.5, 1.5, 2.5 and 3, respectively, calculated using bond order = (bonding electrons − antibonding electrons)/2.

How do you calculate the bond order of He₂⁺?

He₂⁺ has three electrons: two in the bonding sigma 1s orbital and one in the antibonding sigma-star 1s orbital. Its bond order is (2 − 1)/2 = 0.5. The 1s electrons must be included because their bonding and antibonding contributions do not cancel.

Why is the bond order of O₂⁻ 1.5?

Neutral O₂ has bond order 2, and the extra electron in O₂⁻ enters a pi-star antibonding orbital. Adding one antibonding electron lowers bond order by 0.5, giving O₂⁻ a bond order of 1.5.

Why does NO have a bond order of 2.5?

NO has 11 valence electrons, distributed as eight bonding and three antibonding electrons. Its bond order is therefore (8 − 3)/2 = 2.5. The fractional value follows directly from molecular orbital occupancy.

What is the bond order of C₂²⁻?

C₂²⁻ has ten valence electrons, like N₂: eight occupy bonding orbitals and two occupy antibonding orbitals. Its bond order is (8 − 2)/2 = 3. The 2− charge adds two electrons to the whole species, not two to each carbon atom.

Chemical Bonding NEET 2008: Bond Order Explained

What is the correct order in the Chemical Bonding NEET 2008 question?

How do you calculate the bond order of He₂⁺?

Why does adding an electron to O₂ give O₂⁻ a bond order of 1.5?

Why does NO have a fractional bond order of 2.5?

How does the N₂ comparison give C₂²⁻ the highest bond order?

How can counting only antibonding electrons lead to option A?

Which three practice questions check the same method?

Frequently asked questions

What is the correct answer to the Chemical Bonding NEET 2008 bond order question?

How do you calculate the bond order of He₂⁺?

Why is the bond order of O₂⁻ 1.5?

Why does NO have a bond order of 2.5?

What is the bond order of C₂²⁻?

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