Chemistry - CHEMICAL BONDING Question with Solution | TestHub

ChemistryCHEMICAL BONDINGMOTHard2 minQB
ChemistryHardmultiple choice

For the following homonuclear species, include all core electrons and use the standard second-period MO ordering. Which have FEWER electrons in gerade orbitals than in ungerade orbitals?

Options:(select one or more)

Answer:
A, B, C
Solution:

The1s and2s bonding/antibonding pairs contribute4g and4u in total before p filling. C₂⁻ adds πᵤ⁴σ¹, giving 5g, 8u. N₂²⁺ adds πᵤ⁴, giving 4g, 8u. O₂²⁺ adds σ²πᵤ⁴, giving 6g, 8u. F₂²⁻ adds σ²πᵤ⁴π*⁴σ*ᵤ², giving 10g, 10u. Therefore A-C have fewer g electrons; D has equal counts.

Stream:JEESubject:ChemistryTopic:CHEMICAL BONDINGSubtopic:MOT
⏱ 2mℹ️ Source: QB

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