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MO of NO- ???????????????????????????????
So I know the NO- is -N=O ...
Will there be 2 unpaired electrons? Are there any nonbonding orbitals?
I just had a midterm and want to see if i got it right so please answer if you know!!!
1 Answer
- GeorgeSiO2Lv 710 years agoFavorite Answer
Simple MO diagram no sp overlap (probably incorrect but NO- is isoelectronic with O2)
NO- = 12e- (N = 5;O= 6; charge +1) = σs(2e-) σs*(2e-) σp(2e-) πp(4e-) πp*(2e-) σs*(0)
Bond Order = ½[Σ (bonding e-) - Σ (antibonding e-)]
bo = ½[ σ1(2e-) σ3(2e-) π1(4e-) - σ2*(2e-) π2*(2e-) ] = ½[8-4] = 2
There are two e-s in the πp* (two equal energy MOs) and the e-s have their spins parallel (unpaired) (Hund's Rule). The NO- isoelectronic with O2 and will be paramagnetic.
With sp overlap σ3>π1
NO^-: 12 valence e- σ1(2e-) σ2 (2e-) π1(4e-) σ3(2e-) π2*(2e-) σ4*(0e-)
Bond Order = ½[Σ (bonding e-) - Σ (antibonding e-)]
bo = ½[Σ σ1(2e-) π1(4e-) -π2*(2e-)] = 2.0 (σ2, σ3 are considered nonbonding or their effects cancel)
Still paramagnetic with bond order of 2.0.
(I don't think NO^- exists.)
