Q.24 Based on molecular orbital theory, the number of paramagnetic species in the following list
N2, N2+, N22-, O2, O2+, O2– and O22- is ____
(assume that there is no change in energy of the orbitals upon addition/removal of electrons
in a molecule)
Molecular orbital (MO) theory explains bonding and magnetic properties by filling electrons into orbitals. Paramagnetic species have unpaired electrons; diamagnetic ones do not. The answer is 5 paramagnetic species from the list.
Nitrogen Species Analysis
N2 has 14 electrons with MO configuration (σ2s)2 (σ*2s)2 (π2p)4 (σ2p)2—all paired, so diamagnetic.
N2+ (13 electrons) removes one from σ2p, leaving one unpaired electron—paramagnetic.
N22- (16 electrons) fills up to π*2p2 (degenerate orbitals, two unpaired by Hund’s rule)—paramagnetic.
Oxygen Species Analysis
O2 (16 electrons) configuration: KK (σ2s)2 (σ*2s)2 (σ2p)2 (π2p)4 (π*2p)2—two unpaired electrons in π*2p orbitals, paramagnetic.
O2+ (15 electrons) has one unpaired electron in π*2p—paramagnetic.
O2– (17 electrons) adds to π*2p, resulting in three electrons (two paired, one unpaired)—paramagnetic.
O22- (18 electrons) fully pairs π*2p4—no unpaired electrons, diamagnetic.
Summary Table of Properties
| Species | Total Valence Electrons | Unpaired Electrons | Magnetic Property |
|---|---|---|---|
| N2 | 14 | 0 | Diamagnetic |
| N2+ | 13 | 1 | Paramagnetic |
| N22- | 16 | 2 | Paramagnetic |
| O2 | 12 (valence) | 2 | Paramagnetic |
| O2+ | 11 | 1 | Paramagnetic |
| O2– | 13 | 1 | Paramagnetic |
| O22- | 14 | 0 | Diamagnetic |
Paramagnetic species: N2+, N22-, O2, O2+, O2– (total 5). This assumes standard MO diagrams without energy changes on ionization.
