This O₂ Species Question Repeats Every Few Years in NEET — See the Pattern

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1. TITLE

This O₂ Species Question Repeats Every Few Years in NEET — See the Pattern

2. THE ACTUAL NEET QUESTION

NEET 2022 — Question
Which amongst the following is incorrect statement?
A) The bond orders of $O_2^+$, $O_2$, $O_2^-$ and $O_2^{2-}$ are 2.5, 2, 1.5 and 1, respectively
B) $C_2$ molecule has four electrons in its two degenerate $\pi$ molecular orbitals
C) $H_2^+$ has one electron
D) $O_2^+$ is diamagnetic

3. OFFICIAL ANSWER

NEET Answer Key: (D)

4. COMPLETE SOLUTION

Step 1: Analyze the Concept (Molecular Orbital Theory)
To solve this, we must determine the Bond Order and Magnetic Nature of oxygen species using Molecular Orbital Theory (MOT). The electronic configuration of the Oxygen molecule ($O_2$, 16 electrons) is:

$\sigma 1s^2, \sigma^1s^2, \sigma 2s^2, \sigma^2s^2, \sigma 2p_z^2, (\pi 2p_x^2 = \pi 2p_y^2), (\pi^ 2p_x^1 = \pi^ 2p_y^1)$.

Step 2: Calculate Bond Orders
Bond Order (B.O.) = $\frac{1}{2}(N_b - N_a)$.
* $O_2$ (16e⁻): $N_b=10, N_a=6$. B.O. = $\frac{10-6}{2} = 2.0$.
* $O_2^+$ (15e⁻):  Remove 1 electron from $\pi^*$. $N_b=10, N_a=5$. B.O. = $\frac{10-5}{2} = 2.5$. 
* $O_2^-$ (17e⁻):  Add 1 electron to $\pi^*$. $N_b=10, N_a=7$. B.O. = $\frac{10-7}{2} = 1.5$. 
* $O_2^{2-}$ (18e⁻): Add 2 electrons to $\pi^*$. $N_b=10, N_a=8$. B.O. = $\frac{10-8}{2} = 1.0$.
* Conclusion: Statement A is Correct*.

Step 3: Analyze Magnetic Nature
* Paramagnetic: Contains unpaired electrons.
* Diamagnetic: All electrons are paired.
* $O_2$ (16e⁻):   Has 2 unpaired electrons in $\pi^*$ orbitals. $\rightarrow$  Paramagnetic. 
* $O_2^+$ (15e⁻): Has 1 unpaired electron in $\pi^*$ orbital. $\rightarrow$   $\rightarrow$ Paramagnetic.
* Conclusion: Statement D says "$O_2^+$ is diamagnetic", which is Incorrect.

Why other options are Correct:
* Option B ($C_2$): $C_2$ (12e⁻) configuration allows for a double bond made of two $\pi$-bonds because the electrons fill the $\pi 2p$ orbitals before the $\sigma 2p_z$. Correct.
* Option C ($H_2^+$): $H_2$ has 2 electrons; $H_2^+$ has $2-1=1$ electron. Correct.

5. WHY NEET ASKED THIS

This is a classic "Concept Cluster" question. The examiner isn't just checking if you can memorize a formula; they are testing:
1. Electronic Configuration: Do you know the energy order of orbitals for $O_2$ (which differs from $N_2$)?.
2. Magnetism Rule: Can you spot unpaired electrons in Antibonding Molecular Orbitals (ABMO)?
3. The Trap: Many students assume removing an electron (forming a cation) makes a species diamagnetic or "stable" by default. For $O_2$, removing an electron from the antibonding orbital actually increases bond order but leaves one unpaired electron, keeping it paramagnetic.

6. TREND ANALYSIS

The $O_2$ family is a favorite of NTA. They rotate between asking for Stability Order, Bond Length, and Magnetism.

PATTERN ALERT:
* 2022: Asked to identify the incorrect statement about Bond Order & Magnetism.
* 2019: Asked simply "Which is paramagnetic?" ($O_2$ was the answer).
* 2015: Asked for the decreasing order of stability ($O_2^+ > O_2 > O_2^- > O_2^{2-}$).
* 2013: Asked about the change in magnetic behavior and bond energy for $O_2 \rightarrow O_2^+$.
* 2011: Asked "Which has the minimum bond length?" ($O_2^+$).

Prediction for NEET 2026:
Since they recently asked for the properties (2022) and Enthalpy (2024), expect a Sorting Question next.
* Expectation: "Arrange the following in increasing order of Bond Length."
* Remember: Bond Length is inversely proportional to Bond Order.
* Order: $O_2^+ < O_2 < O_2^- < O_2^{2-}$.

7. SIMILAR QUESTIONS TO PRACTICE

If you understood this, you can solve:
* NEET 2015: Which represents the correct bond order? ($O_2^- < O_2 < O_2^+$).
* NEET 2011: Which has the minimum bond length? ($O_2^+$).
* NEET 2013: In which process does bond energy increase and magnetic behavior change? ($NO \rightarrow NO^+$).

8. QUICK REVISION FOR THIS TYPE

For Molecular Orbital Theory (MOT) questions involving Diatomic species ($N_2, O_2$, etc.):

1. The "14-Electron" Rule (Super Trick):
Memorize that 14 electrons ($N_2$) = Bond Order 3.0.
For every electron added or removed, subtract 0.5 from the Bond Order.
* 14e ($N_2$) $\rightarrow$ B.O. 3.0
* 15e ($N_2^-, O_2^+$) $\rightarrow$ B.O. 2.5
* 16e ($O_2$) $\rightarrow$ B.O. 2.0
* 17e ($O_2^-$) $\rightarrow$ B.O. 1.5
* 18e ($F_2, O_2^{2-}$) $\rightarrow$ B.O. 1.0

2. Check Magnetism:
* Odd number of electrons? Always Paramagnetic.
* Even number? usually Diamagnetic, EXCEPT 10e ($B_2$) and 16e ($O_2$), which are Paramagnetic.

3. Relationships:
* Stability $\propto$ Bond Order.
* Bond Enthalpy $\propto$ Bond Order.
* Bond Length $\propto \frac{1}{\text{Bond Order}}$.

9. NCERT BASE

This question tests: NCERT Class 11, Chapter 4 (Chemical Bonding), Section 4.8.
* See Fig. 4.21 for the energy level diagrams and properties of homonuclear diatomic molecules.
* Paragraph on Oxygen molecule ($O_2$) explicitly discusses the unpaired electrons in $\pi^*$ orbitals.

10. CTA

All PYQs from this chapter → Chemical Bonding & Molecular Structure Guide
Practice similar questions → Mock exams