Using the MO diagram of NO, calculate the bond order. Compare it to NO^(+)?

The MO diagram for "NO" is as follows (Miessler et al., Answer Key): (The original was this; I added the orbital depictions and symmetry labels. For further discussion on the orbital energy ordering being "N"_2-like, see here and comments.) Quick overview of what the labels correspond to what MOs: 1a_1 is the sigma_(2s) bonding MO. 2a_1 is the sigma_(2s)^"*" antibonding MO. 1b_1 is the pi_(2p_x) bonding MO. 1b_2 is the pi_(2p_y) bonding MO. 3a_1 is the sigma_(2p_z) bonding MO, but it's relatively nonbonding with respect to oxygen. 2b_1 is the pi_(2p_x)^"*" antibonding MO. 2b_2 is the pi_(2p_y)^"*" antibonding MO. 4a_1 is the sigma_(2p_z)^"*" antibonding MO. To obtain the bond order, look at the molecular orbitals formed and decide whether they are bonding or antibonding. "BO" = 1/2 ("bonding e"^(-) - "antibonding e"^(-)) = 1/2[(2+2+2+2) - (2+1)] = color(blue)(2.5) And this should make sense because "NO"^(+) is isoelectronic with "CO", which has a bond order of 3. With one additional electron in an antibonding orbital (2b_2), the bond order decreases by 1/2 relative to "NO"^(+). If paramagnetism occurs due to unpaired electrons, is "NO" paramagnetic or diamagnetic?

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