Question

please answer question 8 using question 5 and the molecular orbital diagram given.

5. Prepare a molecular orbital diagram for BN. You may consider valence orbitals only Label the atomic orbitals for B and N with the appropriate potential energy. Decide which orbitals have the correct symmetry, orientation, and potential energies to interact (see the handout provided during class). Label the valence atomic orbitals with the appropriate potential energy. Should s-p mixing be considered for BN? How will the MO diagram change if s-p mixing is not involved? Based on your diagram, is BN expected to be paramagnetic or diamagnetic? Does your prediction change depending on whether or not s-p mixing is considered? Why or why not? 6. Using your M.O. diagram, write the valence electron configuration in the general form (02)(T29)" Be sure to follow the Aufbau principle. For example, the valence electron configuration for the MOs of O2 is (02)(a2)(02)' (T2)* (T#2p) 7. Based on your M.O. diagram, what bond order is predicted for BN? 8. If placed on the same scale, how does the M.O. diagram for BN differ from the diagrams for B2 and N2? What accounts for the difference(s)?

Answer 1

UN + 7 62PX Antibonding TI2R 12P2 ma Bonding m.o AO I MO is orbital do not take part in bonding while waiting electronic configuration for each two elecpons in Gis and got is letter K is written. .. The molecular orbital elechanic configuration : IKK (62513 (6²28) 2 (T12p)? (112p) 2] Bond andered (MG of elechon in Banding Molecular 2 - Mo of eam Antibending m:o) bila! Mo unpaired efechon The molewe is Demognetic

Mo Diagrun without making s and p obftal BN - 2P li napy 123 Molecular orbital elechanic configuration [kk (623)} (6921) 3 (6287² CTT2py), (122) 2 2 unpaired elechon The molecule ir paramagnetic

B molecule combination of hos Borns alom Boron atomic Ho5. electronic anffguration 15:27; 2pl the as orbftal with twoelemes and apoobital with one eleeson of each bonn take part in bonding. erpype THY 1122 62 АО AO The molecular orbital eleebinic anfiguration [kk (Q2533,061013(714) (123) Bond order = 1 (64) two unpuired electron moleulle as para magnetic

Ha molecule DANH horhol & tt + H + 61 2P then 28 hit 23 MO. electric Conf9400 М) ekk [kk (6281} (CM2)13 (9744)? CT12/2), (622) 27 Bond oder 188.2) No unpaired election molecules demagnetic
there is difference in Bond order molecular orbital electronic configuration In MO of BN,B2 and N2