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Q.Discuss the formation of N₂ molecule using MO Theory

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Step 1. Nitrogen's atomic configuration is 1s² 2s² 2p³; each N₂ molecule combines two such sets of orbitals, giving 14 electrons total to fill.

Step 2. Filling by Aufbau/Pauli/Hund (with π2p BELOW σ2px for this lighter diatomic, same ordering as B₂/C₂): σ1s² σ1s² σ2s² σ2s² π2py² π2pz² σ2px².

Step 3. Bond order. Nb=2(N_b = 2(σ1s)+2() + 2(σ2s)+2() + 2(π2py)+2() + 2(π2pz)+2() + 2(σ2px)=10) = 10; Na=2(N_a = 2(σ1s)+2() + 2(σ2s)=4) = 4. BO=(10−4)/2=3BO = (10-4)/2 = 3.

Step 4. A bond order of 3 confirms N₂'s well-known N≡N triple bond. Every orbital in the configuration above is completely filled (no singly-occupied orbital anywhere), so N₂ has zero unpaired electrons — it is diamagnetic. …

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