Question

1. For each of the following examples, (a) Draw a molecular orbital diagram. Include both the atomic orbitals that combine and the molecular orbitals that are formed. Clearly show the relative energies of all orbitals. Label all orbitals and show the appropriate number of electrons in the molecular orbitals. Give both labeling conventions (e.g. sigma2s and 1sigmag) for each of the molecular orbitals. You do not have to sketch the actual orbitals. (b) Give the bond order. (c) Indicate the effect of a 1-electron oxidation on bond strength. (d) Indicate the effect of a 1-electron reduction on bond length. (e) Indicate whether the example is diamagnetic or paramagnetic. (f) Give the electron configuration.
   1. B₂
   2. F₂
   3. C₂^2-

Answer 1

@] Draw the molec Jar orbitad diagoam of: ① B2 @ F2 © Ba molecule :- * M.o diagram: - *2P2 B . B 'n spy Th82927- . 67% 1 1 - - - 22x2py 282, 2PxzP4 283 Tiap! 1 Tap 12 28 . ... 6*25 1 - - .... 12 23 625 6 Bond order = II No. of ee in Bonding -No.of eo in Anti- -t mole wiar orbital bonding mol-orbita = £[ 4-2] ..Bond order = & 1 © effect of lee oxidation on bond length on leo oreidation B2 becomes Bat. eo loses from B.M.O om :. Bond order = £[3-2] = 0.5- Bond length increases. (d effect of leo reductim: - Bond order = I [52] =. 3/2-2.5 .. Bond length decreases. @ :: It contains unpaided eeIt is paragnetic. 6 Electronic configuration : KK 625 62 2Py 12P2

F2 2 Fa molecule. 5*28n F AL 14 '1*11 PP AL1L n. 2px 2py 212 1 1 ALi eße apy 212 1L 12 - 6*28 AV 1L 282 .-.1L.. 628 6 Bonded order = I No.of ee in Bro-Horof e o in] ABRO - } [8-6) = 1. © effect of, leo oreidation of bond length eo coth i t loses from ABMO. . B.0 558-5) = 1.5 :. Bond length decreases. effect of lee reduction on bond length e added into AB MO Bond order = 1 [8-7] = feco.s Bond length increases. .. ce . It does not contain hence molecule is unpaired eo paramagnetic 2 2 6 Electronic configuration= kk 625 res 62pu slipy T12P2² 2 Tik apy *

2 - c- *s* 2px Tropy na 1 i 122 12 1 1 1 1 1 1 62px 12py 6212 . Sapn bape seph парч л П AL ---- 1L 68 02. . 10 AL 673 Bond order I No.of ein B.MO - Noicfe Qin ABMO] = {[8-2] - 3 on length © Effect of leg oxidation eo goes from BMO :. Bond order = t [8-2] . Bond order = 2.5 d effect of lee reduction on length. Bond order = [8-3] | = | x" F25 Bond length increases. @ :. It does not molecule contain unpaired eo is diamagnetic. & Electronic configuration= kk 628 ott lapy T 212 6293