Long Answer TypeDiscuss the formation of N2 molecule on the basis of MO theory. Predict its:
(i) Bond order
(ii) Magnetic character.
Write the stability configuration of 
and predict their bond order, stability and magnetic character.
a) 
ion: It is formed from 
a molecule by the loss of one electron. This electron will be lost from 2p7 orbital of nitrogen molecule.
(i) Electronic configuration:
(iii) Stability: As the bond order is positive, it is quite stable.
(iv) Magnetic character: Since 2pz orbital has one unpaired electron, therefore it is paramagnetic.
(b) 
ion : It is formed by the gain of one electron by N2 molecule.
This electron will go to either 
orbital of N2 each of which is empty.
(i) Electronic configuration.
(ii) Bond order: Here Nb = 8; Na = 3
(iii) Stability. Bond order being positive, N2 is quite stable.
(iv) Magnetic character Due to the presence of an unpaired electron in 
*2px orbital, it is paramagnetic.
(c) 
ion: It is formed when molecule gains two electrons,
(i) Electronic configuration
(ii) Bond order: Here Nb = 8; Na = 4
(iii) Stability. Since bond order is positive, it is quite stable.
(iv) Magnetic character: As it has two unpaired electrons one each in 
orbitals, it is paramagnetic.
Draw the molecular orbital energy diagram for oxygen molecule (O2) and show that:
(i) It has a double bond
(ii) It has paramagnetic character.
Using a molecular orbital diagram, predict the bond order, stability and magnetic character of O2–,O2+, and 
. Also, write their electronic configurations.
Discuss the relatives stabilities, bond dissociation energies and bond lengths of O2, 
species.
Is the bond order in superoxide ion more or less than in peroxide ion? Explain on the basis of MO theory.
Short Answer TypeUsing bond order show that N2 would be expected to have a triple bond, F2 a single bond and Ne2 no bond.
Show that N2 molecule has a greater bond dissociation energy than N-2 where O2 has lower bond dissociation energy than O+2.
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