In which of the following processes. the bond order increases and paramagnetic character changes to — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

Molecule /Ion Electronic configuration Bond order Magnetic behaviour N2 (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2 1/2(10 – 4) = 3 Diamagnetic N2 + (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)1 1/2(9 – 4) = 2.5 Paramagnetic O2 (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px1 = *2p1y) 1/2(10 – 6) = 2 Paramagnetic O2 + (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px1 = *2p0y) 1/2(10 – 5) = 2.5 Paramagnetic O2 2– (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px2 = *2p2y) 1/2(10 – 8) = 1.0 Diamagnetic NO (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px1 = *2p0y) 1/2(10 – 5) = 2.5 Paramagnetic NO+ (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) 2pz)2 1/2(10 – 4) = 3 Diamagnetic