Nitric oxide (NO) is a heteronuclear diatomic molecule with 7 electrons from N and 8 electrons from O, totalling 15 electrons. The molecular orbital energy sequence for NO follows the same pattern as second-period homonuclear diatomics: \sigma(1s)^2 \sigma^(1s)^2 \sigma(2s)^2 \sigma^(2s)^2 \pi(2p_x)^2 \pi(2p_y)^2 \sigma(2p_z)^2 \pi^(2p_x)^1 \pi^(2p_y)^0. The 15th electron enters \pi^(2p_x). Bond order = (bonding electrons - antibonding electrons)/2 = (10 - 5)/2 = 5/2 = 2.5. Since there is one unpaired electron in the \pi^(2p_x) orbital, NO is paramagnetic. Bond order of 2.5 is intermediate between a double bond and a triple bond. Option 'Diamagnetic, bond order 2.0' is incorrect because NO has an odd number of electrons (15), making it inherently paramagnetic with at least one unpaired electron; a bond order of 2.0 would require equal numbers of bonding and antibonding electrons that are all paired. Option 'Paramagnetic, bond order 2.0' gives the correct magnetic character but miscalculates bond order by putting the extra electron in a bonding orbital. Option 'Diamagnetic, bond order 2.5' correctly identifies bond order but wrongly predicts diamagnetism despite the unpaired \pi^* electron. This is a standard JEE Advanced MOT question. Plausibility: NO is experimentally known to be paramagnetic (ESR active) and can easily lose one electron to form NO^+ (which has bond order 3, isoelectronic with N_2), supporting the assignment of the 15th electron to an antibonding orbital.