Molecular Orbital Theory CBT Mock Test | MOT PYQs Solved

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Molecular Orbital Theory CBT Mock Test & PYQs

Master Molecular Orbital Theory (MOT) with our online Computer Based Test (CBT). Practice authentic previous years' questions on LCAO principles, sp-mixing phenomena, bond order calculation shortcuts, HOMO-LUMO transitions, and magnetic properties for homonuclear and heteronuclear diatomics (CO, NO, O2, N2).

⚡ Start Live Full-Width CBT Test
💡 Related Chemical Bonding & Quantum Notes: Strengthen your bonding foundations with our Chemical Bonding Notes PDF, test molecular geometry with the VSEPR Theory PYQ CBT Test, and review quantum derivations with our Quantum Chemistry Notes by Sahendra Sir.
📑 Table of Contents
  1. 1. Core Postulates of MOT & LCAO Linear Combination Rules
  2. 2. Energy Level Diagrams: ≤ 14e− vs > 14e− (The sp-Mixing Effect)
  3. 3. Start Live MOT CBT Mock Test (Full Width Unboxed)
  4. 4. The 14-Electron Rapid Bond Order & Magnetism Matrix
  5. 5. Heteronuclear Diatomics (CO, NO, HF) & Frontier Orbitals (HOMO/LUMO)
  6. 6. Step-by-Step Solved Representative PYQs
  7. 7. Frequently Asked Questions (FAQs)

⚛️ 1. Core Postulates of MOT & LCAO Principles

Developed by F. Hund and R.S. Mulliken, Molecular Orbital Theory (MOT) views electrons in a molecule as delocalized over all atomic nuclei. Atomic orbitals combine through the Linear Combination of Atomic Orbitals (LCAO) approximation:

• Bonding Molecular Orbital (BMO): ψBMO = ψA + ψB  →  Constructive interference (lower energy, higher electron density between nuclei)
• Antibonding Molecular Orbital (ABMO): ψABMO = ψA − ψB  →  Destructive interference (higher energy, nodal plane between nuclei)

Three Mandatory LCAO Conditions:
1. Combining atomic orbitals must possess comparable energies.
2. Combining atomic orbitals must have the same symmetry relative to the molecular axis (typically z-axis).
3. Combining atomic orbitals must have maximum spatial overlap.

📈 2. Orbital Energy Sequences: ≤ 14e− vs > 14e− (sp-Mixing)

The energy sequence of second-row homonuclear diatomic molecules depends fundamentally on whether 2s–2p mixing (sp-mixing) takes place:

Electron RegimeMoleculessp-Mixing StatusIncreasing Energy Sequence
≤ 14 ElectronsLi2, Be2, B2, C2, N2Yes (Small ΔE between 2s & 2p)σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz
> 14 ElectronsO2, F2, Ne2No (Large ΔE prevents mixing)σ1s < σ*1s < σ2s < σ*2s < σ2pz < (π2px = π2py) < (π*2px = π*2py) < σ*2pz

Key Takeaway: In B2, C2, and N2, π2p orbitals are lower in energy than the σ2pz orbital due to repulsions between σ2s and σ2pz orbitals of identical symmetry.

💻 Live Test: Molecular Orbital Theory PYQ CBT Mock Test

Interactive Online Test

Attempt the mock test below to test your understanding of bond orders, magnetic properties, and frontier molecular orbitals under real examination conditions:

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Molecular Orbital Theory CBT Mock Test
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📖 Test Questions & Syllabus Outline (Google Search Index & Study Reference) Total: 28 Questions

This online mock test covers the following chemistry practice questions with bilingual explanations, negative marking schemes, and interactive CBT interface. Students searching for these topics can practice the complete test online with real-time timers:

  1. Match the following items of Column I with the appropriate items in Column II:Column I (Molecular species)Column II (Bond order and magnetic property)(P) O2−I. 2.5 paramagnetic(Q) O2II. 2.0 diamagnetic(R) O2+III. 1.5 diamagnetic(S) O22−IV. 1.0 diamagneticV. 2.0 paramagneticVI. 1.5 paramagnetic GATE 2005 | MOT of Diatomic Molecules | Match Bond Order and Magnetic Property of Dioxygen Species
    कॉलम I की मदों का कॉलम II की उपयुक्त मदों से मिलान कीजिए:कॉलम I (आण्विक स्पीशीज)कॉलम II (आबंध कोटि एवं चुंबकीय गुण)(P) O2−I. 2.5 अनुचुंबकीय(Q) O2II. 2.0 प्रतिचुंबकीय(R) O2+III. 1.5 प्रतिचुंबकीय(S) O22−IV. 1.0 प्रतिचुंबकीयV. 2.0 अनुचुंबकीयVI. 1.5 अनुचुंबकीय
  2. For a homonuclear diatomic molecule, the bonding molecular orbital is: GATE 2007 | MOT of Diatomic Molecules | Bonding MO Parity in Homonuclear Diatomics
    एक समनाभिकीय द्विपरमाणुक अणु के लिए, आबंधन आण्विक कक्षक (bonding molecular orbital) कौन सा है?
  3. The highest occupied molecular orbital of HF is: GATE 2008 | MOT of Heteronuclear Diatomics | HOMO of HF Molecule
    हाइड्रोजन फ्लोराइड ($ ext{HF}$) अणु का उच्चतम अधिग्रहीत आण्विक कक्षक (HOMO) कैसा होता है?
  4. The bond order of C2 molecule is: GATE 2010 | MOT of Diatomic Molecules | Bond Order of C2 Molecule
    $C_2$ अणु की आबंध कोटि (bond order) है:
  5. The number of antibonding electrons in NO and CO according to MO theory are respectively: CSIR NET Dec 2011 | MOT of Diatomic Molecules | Antibonding Electrons in NO & CO
    आण्विक कक्षक (MO) सिद्धांत के अनुसार $\text{NO}$ और $\text{CO}$ में संयोजी प्रति-आबंधन (antibonding $\pi^*/\sigma^*$) इलेक्ट्रॉनों की संख्या क्रमशः है:
  6. If a homonuclear diatomic molecule is oriented along the Z-axis, the molecular orbital formed by linear combination of $p_x$ orbitals of the two atoms is: CSIR NET Dec 2013 | MOT of Diatomic Molecules | Linear Combination of px Orbitals
    यदि कोई समनाभिकीय द्विपरमाणुक अणु $Z$-अक्ष के अनुदिश अभिविन्यस्त (oriented) है, तो दोनों परमाणुओं के $p_x$ कक्षकों के रैखिक संयोजन (LCAO) से बनने वाला आण्विक कक्षक है:
  7. The molecule in which the bond order increases upon addition of an electron is: CSIR NET Dec 2013 | MOT of Diatomic Molecules | Bond Order & Electron Addition
    वह अणु जिसमें एक इलेक्ट्रॉन जोड़ने पर आबंध कोटि (bond order) बढ़ जाती है, वह है:
  8. The reason for the chemical inertness of gaseous nitrogen at room temperature is best given by its: CSIR NET Dec 2013 | MOT of Diatomic Molecules | Chemical Inertness of N2
    कमरे के तापमान पर गैसीय नाइट्रोजन की रासायनिक अक्रियता (chemical inertness) का सबसे उपयुक्त कारण है:
  9. The correct statement among the following is: CSIR NET Dec 2015 | MOT of Diatomic Molecules | Bond Order & Dissociation Energy of N2 vs N2+
    निम्नलिखित में से सही कथन का चयन कीजिए:
  10. Correct combination for $\pi$ and $\pi^*$ orbitals in $B_2$ molecule is: CSIR NET June 2016 | MOT of Diatomic Molecules | Parity of pi and pi* in B2
    $B_2$ अणु में $\pi$ और $\pi^*$ कक्षकों के लिए समता (parity - gerade/ungerade) का सही संयोजन है:
  11. The first excited state of hydrogen molecule is: CSIR NET June 2016 | MOT of Diatomic Molecules | First Excited State Term Symbol of H2
    हाइड्रोजन अणु ($H_2$) की प्रथम उत्तेजित अवस्था का आण्विक पद प्रतीक (molecular term symbol) है:
  12. The HOMO (highest occupied molecular orbital) to LUMO (lowest unoccupied molecular orbital) electronic transition responsible for the observed colours of halogen molecules(g) is: CSIR NET Dec 2016 | MOT of Diatomic Molecules | Electronic Transition in Halogens
    हैलोजन अणुओं ($X_{2(g)}$) के प्रेक्षित रंगों के लिए उत्तरदायी HOMO से LUMO इलेक्ट्रॉनिक संक्रमण है:
  13. Removal of an electron from NO molecule results in: A. an increase in the v(NO) in the IR spectrum B. an EPR active species C. electrons in HOMOs being closer to the oxygen than to nitrogen 2p orbitals. The correct answer is: CSIR NET June 2018 | MOT of Diatomic Molecules | Ionization of NO Molecule
    $\text{NO}$ अणु से एक इलेक्ट्रॉन निकालने (अर्थात $\text{NO}^+$ बनने) के परिणामस्वरूप: A. IR स्पेक्ट्रम में $\nu(\text{NO})$ तनन आवृत्ति में वृद्धि होती है B. एक EPR सक्रिय स्पीशीज प्राप्त होती है C. HOMO में उपस्थित इलेक्ट्रॉन नाइट्रोजन 2p कक्षकों की तुलना में ऑक्सीजन के अधिक निकट होते हैं। सही उत्तर है:
  14. The term symbol for the ground state of $B_2$ is: CSIR NET June 2018 | MOT of Diatomic Molecules | Ground State Term Symbol of B2
    $B_2$ अणु की मूल अवस्था (ground state) का पद प्रतीक (term symbol) है:
  15. The characters of LUMO of CN– and O2 respectively, are: CSIR NET Dec 2018 | MOT of Diatomic Molecules | LUMO Symmetry of CN- and O2
    $CN^-$ और $O_2$ के निम्नतम अनधिग्रहीत आण्विक कक्षक (LUMO) की सममितीय विशेषताएं (characters) क्रमशः हैं:
  16. The ion having the highest bond order is: CSIR NET Dec 2019 | MOT of Diatomic Molecules | Highest Bond Order Cation
    निम्नलिखित में से उच्चतम आबंध कोटि (highest bond order) वाला आयन है:
  17. Choose the correct order of energy of 2σg and 1Πu molecular orbitals for B2, C2 and O2: CSIR NET 26 Nov 2020 | MOT of Diatomic Molecules | Energy Order of 2sigma_g and 1pi_u
    $B_2, C_2$ और $O_2$ के लिए $2\sigma_g$ (या $\sigma_{2p_z}$) और $1\Pi_u$ (या $\pi_{2p}$) आण्विक कक्षकों की ऊर्जा का सही क्रम चुनिए:
  18. The correct statement(s) for the oxygen from the following: A. The ground state for the O atom is 3P1 B. Both the atoms and the diatomic molecule are paramagnetic with two unpaired electrons. C. The most readily accessible singlet excited state for dioxygen has an empty * orbital Choose the correct answer from the options given below: CSIR NET 30 Nov 2020 | MOT of Diatomic Molecules | Electronic States of Oxygen
    ऑक्सीजन के लिए निम्नलिखित कथनों में से सही कथन चुनिए: A. O परमाणु की मूल अवस्था $^3P_1$ है B. परमाणु और द्विपरमाणुक अणु दोनों दो अयुग्मित इलेक्ट्रॉनों के साथ अनुचुंबकीय हैं। C. डाइऑक्सीजन की सबसे आसानी से सुलभ सिंगलेट उत्तेजित अवस्था में एक रिक्त $\pi^*$ कक्षक होता है। नीचे दिए गए विकल्पों में से सही उत्तर चुनिए:
  19. The highest occupied MO in N2 and O2+ respectively are (take x-axis as internuclear axis): CSIR NET June 2011 | MOT of Diatomic Molecules | HOMO in N2 and O2+
    $N_2$ और $O_2^+$ में उच्चतम अधिग्रहीत आण्विक कक्षक (HOMO) क्रमशः हैं ($x$-अक्ष को अंतरनाभिकीय अक्ष मानिए):
  20. A molecular orbital of a diatomic molecule changes sign when it is rotated by 180° around the molecular axis. This orbital is: CSIR NET June 2015 | MOT of Diatomic Molecules | Rotational Symmetry of Molecular Orbitals
    द्विपरमाणुक अणु का एक आण्विक कक्षक जब आण्विक अक्ष के चारों ओर $180^\circ$ ($C_2$) घुमाया जाता है, तो अपना चिह्न बदल लेता है। यह कक्षक है:
  21. The ground state electronic configuration of C2 using all electrons is: CSIR NET Dec 2015 | MOT of Diatomic Molecules | Ground State Configuration of C2
    सभी इलेक्ट्रॉनों का उपयोग करते हुए $C_2$ का मूल अवस्था इलेक्ट्रॉनिक विन्यास है:
  22. In simple molecular orbital theory of hydrogen molecule, bonding σg and antibonding σu molecular orbitals are constructed as linear combinations of atomic orbitals of two hydrogen atoms. The spatial part of a purely covalent singlet wavefunction is obtained by: CSIR NET June 2016 | MOT of Diatomic Molecules | Purely Covalent Wavefunction of H2
    हाइड्रोजन अणु के सरल आण्विक कक्षक सिद्धांत में, दो हाइड्रोजन परमाणुओं के परमाण्विक कक्षकों के रैखिक संयोजन द्वारा आबंधन $\sigma_g$ और प्रति-आबंधन $\sigma_u$ कक्षक निर्मित होते हैं। विशुद्ध रूप से सहसंयोजक सिंगलेट तरंगफलन (purely covalent singlet wavefunction) का स्थानिक भाग किसके द्वारा प्राप्त होता है?
  23. The spatial part of an excited state 3Σu+ of hydrogen molecule is proportional to [1σg(1) 1σu(2) − 1σg(2) 1σu(1)]. Using LCAO-MO expansion of 1σg and 1σu in terms of 1s-atomic orbitals, one can infer that this wavefunction has: CSIR NET Dec 2016 | MOT of Diatomic Molecules | Excited State Triplet Wavefunction of H2
    हाइड्रोजन अणु की उत्तेजित अवस्था $^3\Sigma_u^+$ का स्थानिक भाग $[1\sigma_g(1) 1\sigma_u(2) - 1\sigma_g(2) 1\sigma_u(1)]$ के समानुपाती है। 1s-परमाण्विक कक्षकों के पदों में $1\sigma_g$ और $1\sigma_u$ के LCAO-MO प्रसार का उपयोग करते हुए, यह निष्कर्ष निकाला जा सकता है कि इस तरंगफलन में उपस्थित है:
  24. The highest molecular orbitals for an excited electronic configuration of the oxygen molecule are [1πg][3σu]. A possible molecular term symbol for oxygen with this electronic configuration is: CSIR NET Dec 2016 | MOT of Diatomic Molecules | Excited State Term Symbol of Oxygen
    ऑक्सीजन अणु के एक उत्तेजित इलेक्ट्रॉनिक विन्यास के लिए उच्चतम आण्विक कक्षक $[1\pi_g][3\sigma_u]$ हैं। इस इलेक्ट्रॉनिक विन्यास वाले ऑक्सीजन के लिए एक संभावित आण्विक पद प्रतीक है:
  25. Arrange the following molecules in order of increasing fundamental vibrational frequencies: CSIR NET Dec 2018 | MOT of Diatomic Molecules | Fundamental Vibrational Frequencies of O2 Species
    निम्नलिखित स्पीशीज को उनकी मूल कंपन आवृत्तियों (fundamental vibrational frequencies) के बढ़ते क्रम में व्यवस्थित कीजिए: $O_2^{2-}, O_2^-, O_2, O_2^+$
  26. The spatial part of the dominant resonance structure of the LiH molecule is (only valence part of the wavefunction is shown): CSIR NET Dec 2019 | MOT of Diatomic Molecules | Dominant Resonance Structure of LiH
    $\text{LiH}$ अणु की प्रमुख अनुनादी संरचना का स्थानिक भाग (spatial part) है (तरंगफलन का केवल संयोजी भाग दर्शाया गया है):
  27. The term symbol for the ground state of dinitrogen cation radical N2+ is: CSIR NET 26 Nov 2020 | MOT of Diatomic Molecules | Ground State Term Symbol of N2+
    डाइनाइट्रोजन धनायन मूलक ($N_2^+$) की मूल अवस्था का आण्विक पद प्रतीक (molecular term symbol) है:
  28. For CO, the contour plot of the molecular orbital which best represents the HOMO is:(a) C O(b) C O(c) C O(d) C O CSIR NET 30 Nov 2020 | MOT of Diatomic Molecules | Contour Plot of CO HOMO
    $\text{CO}$ अणु के लिए, उच्चतम अधिग्रहीत आण्विक कक्षक (HOMO) को सबसे सटीक रूप से प्रदर्शित करने वाला समोच्च आरेख (contour plot) कौन सा है?(a) C O(b) C O(c) C O(d) C O

⚡ 3. The 14-Electron Rapid Bond Order & Magnetism Matrix

Bond Order Formula: Bond Order = ½ [ Nb − Na ] (where Nb = bonding electrons, Na = antibonding electrons).
🎯 Shortcut Trick: 14 electrons (N2) corresponds to maximum Bond Order = 3.0. For every electron added or removed between 10 and 18, the bond order changes by exactly 0.5:

Total ElectronsBond Order (BO)Magnetic PropertyClassic Species Examples
101.0Paramagnetic (2 unpaired e− in π2p)B2
111.5Paramagnetic (1 unpaired e−)B2−, C2+
122.0Diamagnetic (all paired e−)C2
132.5Paramagnetic (1 unpaired e−)C2−, N2+
14 (Reference Point)3.0DiamagneticN2, CO, CN−, NO+
152.5Paramagnetic (1 unpaired e− in π*2p)NO, O2+
162.0Paramagnetic (2 unpaired e− in π*2p)O2
171.5Paramagnetic (1 unpaired e− in π*2p)O2− (Superoxide)
181.0DiamagneticO22− (Peroxide), F2
Important Corollaries:
• Bond Order ∝ Bond Dissociation Energy (BDE) ∝ Force Constant (k)
• Bond Order ∝ 1 / Bond Length
• Oxygen Series Bond Length Order: O2+ (112 pm) < O2 (121 pm) < O2− (128 pm) < O22− (149 pm)

🔬 4. Heteronuclear Diatomics: CO, NO, and Frontier Orbitals

In heteronuclear diatomics (A−B), atomic orbitals of the more electronegative atom lie lower in energy:

Carbon Monoxide (CO - 14 Valence Electrons):
• Oxygen 2s and 2p orbitals are much lower in energy than Carbon 2s and 2p.
• HOMO of CO (3σ): Predominantly a non-bonding sp-hybridized lone pair localized on the Carbon atom. This explains why CO acts as a σ-donor through carbon to transition metals in metal carbonyls!
• LUMO of CO (2π*): Empty antibonding π* orbitals polarized toward the Carbon atom, making CO a potent π-acceptor ligand.

Nitric Oxide (NO - 15 Electrons):
• Contains 1 unpaired electron in the antibonding π*2p orbital (HOMO).
• Oxidation to NO+ (Nitrosonium ion, 14e−) removes this antibonding electron, increasing the bond order from 2.5 to 3.0 and shifting the IR stretching frequency from 1876 cm−1 up to 2200–2300 cm−1!

📝 5. Step-by-Step Solved Representative PYQs

CSIR NET & GATE Chemical Sciences

Q1: Arrange the following oxygen species in increasing order of their bond lengths: O2, O2+, O2−, O22−.

Correct Answer: O2+ < O2 < O2− < O22−
Step-by-Step Derivation:
1. Count total valence/molecular electrons and apply the 14-electron rule:
  • O2+ (15e−): Bond Order = 2.5
  • O2 (16e−): Bond Order = 2.0
  • O2− (17e−): Bond Order = 1.5
  • O22− (18e−): Bond Order = 1.0
2. Because Bond Length is inversely proportional to Bond Order, higher bond order yields shorter bond length: O2+ < O2 < O2− < O22−.
NEET & JEE Advanced PYQ

Q2: When N2 is converted into N2+, the bond distance increases. However, when O2 is converted into O2+, the bond distance decreases. Explain.

Correct Answer: Removal of electron from BMO in N2 vs. ABMO in O2
Explanation:
• In N2 (14e−), ionization removes an electron from the bonding molecular orbital (σ2pz). The bond order decreases from 3.0 to 2.5, weakening the bond and increasing the bond distance.
• In O2 (16e−), ionization removes an electron from an antibonding molecular orbital (π*2p). The bond order increases from 2.0 to 2.5, strengthening the bond and shortening the bond distance.
CSIR NET Chemical Sciences

Q3: According to Molecular Orbital Theory, the double bond in the diatomic carbon molecule (C2) consists of:

Correct Answer: Two π-bonds (no σ-bond)
Explanation:
C2 has 12 electrons (≤ 14e−, sp-mixing active). Its MO configuration is:
σ1s2 σ*1s2 σ2s2 σ*2s2 π2px2 π2py2.
The 4 valence electrons occupy the degenerate π2px and π2py bonding orbitals. Therefore, both bonds in C2 are strictly π-bonds.

❓ 6. Frequently Asked Questions (FAQs)

Q1: Why is O2 paramagnetic according to Molecular Orbital Theory?

In the MO electronic configuration of O2 (16 electrons), the last two electrons occupy the doubly degenerate π*2px and π*2py antibonding molecular orbitals with parallel spins according to Hund's rule, imparting paramagnetism (2 unpaired electrons).

Q2: Why does sp-mixing occur in N2 and C2, but not in O2 and F2?

Across the second period, the effective nuclear charge increases, which widens the energy gap (ΔE) between 2s and 2p atomic orbitals. In Li, Be, B, C, and N, ΔE is sufficiently small to allow mixing between 2s and 2pz orbitals of identical σ-symmetry. In O and F, ΔE is too large, preventing significant mixing.

Q3: What does a Bond Order of zero signify in MOT?

A bond order of zero (e.g., in He2 or Be2) signifies that the number of bonding electrons equals the number of antibonding electrons, meaning there is no net stabilization and the diatomic molecule does not exist under ordinary conditions.

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