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ChemistryMediumClass 11

MOT ā€” Paramagnetic vs diamagnetic

Chemical Bonding

15

JEE Qs

8%

Hard

60

min

Master the two distinct MO energy level orders and consistently apply Hund's rule to accurately count unpaired electrons for any given molecule or ion.

šŸ§® Key Formulas

Bond Order (B.O.) = 0.5 * (Number of electrons in bonding MOs - Number of electrons in antibonding MOs)
Magnetic moment (Ī¼) = sqrt(n * (n + 2)) BM (Bohr Magneton), where n is the number of unpaired electrons

āœ… Key Points for JEE

  • 1Paramagnetic substances possess one or more unpaired electrons in their molecular orbitals and are attracted by an external magnetic field.
  • 2Diamagnetic substances have all electrons paired in their molecular orbitals and are weakly repelled by an external magnetic field.
  • 3To determine paramagnetism/diamagnetism, accurately write the Molecular Orbital (MO) electronic configuration for the given molecule or ion and count the number of unpaired electrons.
  • 4The MO energy level order is crucial: For molecules with total electrons ā‰¤ 14 (e.g., B2, C2, N2), the order is: Ļƒ1s Ļƒ*1s Ļƒ2s Ļƒ*2s Ļ€2px Ļ€2py Ļƒ2pz Ļ€*2px Ļ€*2py Ļƒ*2pz. For molecules with total electrons > 14 (e.g., O2, F2), the order is: Ļƒ1s Ļƒ*1s Ļƒ2s Ļƒ*2s Ļƒ2pz Ļ€2px Ļ€2py Ļ€*2px Ļ€*2py Ļƒ*2pz.
  • 5B2 and O2 are classic examples of paramagnetic molecules, which cannot be explained by Valence Bond Theory but are correctly predicted by MOT due to unpaired electrons.

āš ļø Common Mistakes

  • āœ•Using the incorrect MO energy level diagram (e.g., applying the >14 electron order to N2 or vice versa), leading to wrong electron configurations and magnetic properties.
  • āœ•Failing to apply Hund's Rule of Maximum Multiplicity correctly when filling degenerate pi (Ļ€) molecular orbitals, resulting in an incorrect count of unpaired electrons.
  • āœ•Assuming that all stable molecules must be diamagnetic; paramagnetism is a distinct property determined by electron configuration, not just stability.
  • āœ•Not considering changes in magnetic properties for ionic species (e.g., O2+ vs O2 vs O2- vs O2^2-) by recalculating total electrons and re-doing the MO configuration.

šŸ“ Practice Questions

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Q75.Total number of non bonded electrons present in NO2āˆ’ ion based on Lewis theory is 2025 (29 Jan Shift 2) JEE Main Previous Year Paper

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Q71.The number of molecules/ions that show linear geometry among the following is ________ SO2, BeCl2, CO2, Nāˆ’3 , NO2, F2O, XeF2, NO+2 , Iāˆ’3 , O3

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Q69.Arrange the following compounds in increasing order of their dipole moment : HBr, H2 S, NF3 and CHCl3 (1) H2 S < HBr < NF3 < CHCl3 (2) NF3 < HBr < H2 S < CHCl3 (3) HBr < H2 S < NF3 < CHCl3 (4) CHCl3 < NF3 < HBr < H2 S

2025Ā·MCQMedium

Q64.Which of the following statement is true with respect to H2O, NH3 and CH4 ? A. The central atoms of all the molecules are sp3 hybridized. B. The H āˆ’O āˆ’H, H āˆ’N āˆ’H and H āˆ’C āˆ’H angles in the above molecules are 104.5āˆ˜, 107.5āˆ˜ and 109.5āˆ˜ , respectively. C. The increasing order of dipole moment is CH4 < NH3 < H2O. D. Both H2O and NH3 are Lewis acids and CH4 is a Lewis base. E. A solution of NH3 in H2O is basic. In this solution NH3 and H2O act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below: (1) A, B and C Only (2) A, D and E Only (3) C, D and E Only (4) A, B, C and E Only

2025Ā·MCQMedium

Q70.Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z -direction] ? A. 2pz and 2px B. 2 s and 2px C. 3 dxy 2025 (24 Jan Shift 1) JEE Main Previous Year Paper and 3 dx2āˆ’y2 D. 2 s and 2pz E. 2pz and 3dx2 āˆ’y2 Choose the correct answer from the options given below: (1) A and B Only (2) D Only (3) E Only (4) C and D Only

2025Ā·MCQMedium

Q53.Given below are two statements : Statement (I): Experimentally determined oxygen-oxygen bond lengths in the O3 are found to be same and the bond length is greater than that of a O = O (double bond) but less than 2025 (24 Jan Shift 2) JEE Main Previous Year Paper that of a single (O āˆ’O) bond. Statement (II) : The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact that the bond length in ozone is smaller than that of a double bond (O = O) but more than that of a single bond (O āˆ’O). In the light of the above statements, choose the correct answer from the options given below : (1) Both Statement I and Statement II are false (2) Statement I is false but Statement II is true (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are true

2025Ā·Assertion ReasoningMedium

NCERT Chapters

  • Class 11 Chemistry Ch 4: Chemical Bonding and Molecular Structure