assign4_3350_2023 answers

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Chemistry

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Dec 6, 2023

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Assignment 4 Bonding and Ligand Field Theory 2 1. For each of the following ligands: see notes or book for this (i) Draw their structures, underline the donor atoms and give the likely modes of bonding (e.g. monodentate, bidentate, ambidentate etc.); (ii) state whether you expect the ligand to be a σ-donor, σ-donor/π-donor or σ-donor/π- acceptor; (iii) write out the full name of the ligand. (a) edta (b) bipy (c) [CN] - (d) [N 3 ] - (e) CO (f) phen (g) [ox] 2- (h) [NCS] - (i) PMe 3 (j) [Ph 2 N] - 2. Red crystalline [NiCl 2 (dppm)] (where dppm is Ph 2 PCH 2 PPh 2 ) is diamagnetic. On heating to 387 K for 2 hours, a blue-green form of the complex is obtained, which has a magnetic moment of 3.18 µB when measured at 295 K. Suggest an explanation for these observations and draw structures for the complexes. Also, comment on any isomerism that may exist. Square planar d 8 Ni 2+ complex. As long as there is no ligand addition, the compound would still be four coordinate. If we swapped the ligands so that now we had a Ni(dppm) 2 2+ square planar cation and a NiCl 4 2- tetrahedral anion. We could have a diamagnetic cation and a tetrahedral paramagnetic anion. A d 8 tetrahedral molecule is e g 4 t 2 4 and would have 2 unpaired electrons. The expected spin only moment  SO would be 2.83  B. the value of 3.18 is reasonable. As far as isomers, I can only come up with one for both the cation and anion. 3. Find x in the formulae of the following complexes by determining the oxidation state of the metal from the experimental values of µ ef : (a) [VCl x (bipy)], 1.77 µB; the moment looks like 1 unpaired electron (1.73  B), to get this we need an odd number of electrons and for cationic V this would be V(IV) d 1 , V(II) d 3 , or V(0) d 5 . Two reasonable options would be an octahedral VCl 4 (bipy) or tetrahedral VCl 2 (bipy). In a tetrahedral we would expect high spin so 3 unpaired electrons in this case. That would not work so most likely x = 4 (b) K x [V(ox) 3 ], 2.80 µB; looks like 2 unpaired electrons(2.83  B), for 6 coordinate it is likely octahedral we could have t 2g 2 or t 2g 4 (ls). The first case would be V(III) and x = 3, the second case is V(I) and x = 5. I think the first one makes the most sense to me (in fact it is a known compound, the V(II) analogue is also known)) (c) [Mn(CN) 6 ] x , 3.94 µB. looks like 3 unpaired electrons (3.87  B), for 6 coordinate it is likely octahedral we could have t 2g 3 or t 2g 5 e g 2 . The first case would be Mn(IV) and x = -2, the second case is Mn(0) and x = -6. (the first one makes the most sense to me and is known, also the Mn(III) analogue is commercially available)

4. For the p 3 electronic configuration, construct a microstate table and reduce the table to its constituent free-ion terms. Identify the lowest-energy term. Determine the possible values of J and identify the lowest energy. The nitrogen atom is an example of an element with this configuration. Microstates = (6x5x4)/3! = 120/6 = 20 We only need to fill in the blue portion of the microstate table and then we know the number of states in each cell. M L \M S 3/2 ½ -1/2 -3/2 2 1 + 1 - 0 + 1 + 1 - 0 - 1 1 + ,0 - ,0 + ; 1 - ,1, + -1 + 1 - ,0 + ,0 - ; 1,1,-1 0 1 + 0 + -1 + 1 + 0 + -1 - ; 1 - 0 + -1 + ; 1 + 0 - -1 + 1 + 0 - -1 - ; 1 - 0 + -1 - ; 1 - 0 - -1 + 1 - 0 - -1 - -1 X, X X, X -2 X X This is an alternative view of the same table with X for each microstate M L \M S 3/2 ½ -1/2 -3/2 2 X X 1 X ,X X , X 0 X X , X, X X ,X, X X -1 X , X X , X -2 X X Extract terms from this table– I have indicated green X for the microstates that would give L = 2, S = 1/2 2 D, min J = 2-1/2 = 3/2 and max J = 2+1/2 = 5/2 I have indicated red X for the microstates that would give L = 1, S = 1/2 2 P, min J = 1-1/2 = ½, max J = 1+1/2 = 3/2 I have indicated blue X for the microstates that would give L = 0, S = 3/2 4 S, min J = 0-3/2 = 3/2 and max J = 0+3/2 = 3/2

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