Q32.What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : N2, N +2 , O2, O+2 ? (1) 0, 1, 2, 1 (2) 2, 1, 2, 1 (3) 0, 1, 0, 1 (4) 2, 1, 0, 1
What This Question Tests
This question tests the application of Molecular Orbital Theory (MOT) to determine the electronic configuration of diatomic species and count the number of unpaired electrons in their highest occupied molecular orbital (HOMO).
Concepts Tested
📚 NCERT Sections This Tests
5.15 — Discuss The Nature Of Bonding In The Following Coordination Entities On The
Chemistry Class 11 · Chapter 5
5.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i) [Fe(CN)6] 4– (ii) [FeF6] 3– (iii) [Co(C2O4)3]3– (iv) [CoF6] 3–
5.16 — Draw Figure To Show The Splitting Of D Orbitals In An Octahedral Crystal Field.
Chemistry Class 11 · Chapter 5
5.16 Draw figure to show the splitting of d orbitals in an octahedral crystal field.
5.19 — [Cr(Nh3)6] 3+ Is Paramagnetic While [Ni(Cn)4] 2– Is Diamagnetic. Explain Why?
Chemistry Class 11 · Chapter 5
5.19 [Cr(NH3)6] 3+ is paramagnetic while [Ni(CN)4] 2– is diamagnetic. Explain why? 5.20 A solution of [Ni(H2O)6] 2+ is green but a solution of [Ni(CN)4] 2– is colourless. Explain. 5.21 [Fe(CN)6] 4– and [Fe(H2O)6] 2+ are of different colours in dilute solutions. Why?
📋 Question Details
- Chapter
- Chemical Bonding
- Topic
- Molecular Orbital Theory
- Year
- 2023
- Shift
- 24 Jan Shift 2
- Q Number
- Q32
- Type
- MCQ
- NCERT Ref
- Class 11 Chemistry Ch 4: Chemical Bonding and Molecular Structure
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