CM2021-Inorganic and Bioinorganic Chemistry Exam paper solution AY , sem 1
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1 CM2021-Inorganic and Bioinorganic Chemistry Exam paper solution AY , sem 1 Do Thanh Nhut School of Physical and Mathematical Sciences Nanyang Technological University Singapore 06/11/2015
2 1 Question 1 a) Indole. Point group: C s. b) 2,8,14-trimethyl-5,11,17-triazatrinaphthylene. Point group: C 3v. c) Perchlorate ion. Point group: T d. d) stagger-c 2 H 6. Point group: D 3d. 1
3 2 Question 2 a) The dissolving process contains of 2 steps: breaking down of the lattice (endothermic process, H = + H lattice ) and solvation of the dissociated ions (exothermic process, H = ΣE solvation ). If the second step is highly exothermic enough, the whole process is totally exothermic and thermodynamically favoured. In the case of Ba(OH) 2 and Ca(OH) 2. Because the lattice energy of Ba(OH) 2 is lower than Ca(OH) 2 s (the radius of Ba 2+ and OH are not as similar as in case of Ca 2+ and OH ) and the solvation energy of the Ba 2+ is higher than Ca 2+ (larger cation radius, lower electrostatic repulsion between surrounding ligands). Hence, the Ba(OH) 2 is more soluble than Ca(OH) 2. However, in BaSO 4, the radii of Ba 2+ and SO 2 4 are similar There is much more efficient orbital overlapping and this leads to very high lattice energy. Therefore, the BaSO 4 is less soluble than CaSO 4. b) Melting point of the metal is depends on how strong the metal bonds in metal lattice are. The stronger metal bonds, the higher melting point of that metal. The metal bonds are form by the valence electrons coming from all metal atoms in the lattice. All metal atoms release their valence electrons to share together and form an electron sea covering all cation and this forms metal bonds in the lattice. From Na to Al, the number of valence electron is increase and as the results, the metal bonds in Al lattice is much stronger than in Na lattice (higher electron density). Therefore, the melting point increases along the direction Na < Mg < Al. c) In isolated ions form (Al 3+ and Cl ), the cation Al 3+ has very small radius and very high positive charge. The charge density is so high It is not so stable. In AlCl 3 form, the Al atom is not satisfy the octet rule. Then, it has tendency to dimerize into Al 2 Cl 6 and in this dimer, around 2 Al atoms is 8e (stable). d) Because at lower valence group, the metal has lesser valence electrons and it has tendency to give away that electrons to obtain the octet configuration. However, when the number of valence electrons increases, the valence orbitals are fulfilled gradually, the metal has another choice to receive more lone-pair to obtain octet configuration. Hence, it becomes less basic. In addition, in metal oxides, the higher valence, the higher positive charge in metal ions in the lattice The O 2 is stronger polarized with the negative terminus of the dipole pointing towards the metal ion. Hence, its proton accepting ability decreases. Therefore, Al 2 O 3 is less basic than NaO. 2
4 3 Question 3 a) Oxidation state of Co center: +2. b) Magnetic moment: µ S = n(n + 2). In that, n is the number of unpaired electrons. At 200 K, µ S = Hence, n = 1 Low spin. At 350 K, µ S = Hence, n = 2 High spin. c) Term symbol: 2S+1 L. In that, 2S + 1 is the multiplicity and L is the total angular momentum symbol. At 200 K, term symbol: 4 F. At 350 K, term symbol: 2 H. d) At 350 K, the complex is in low spin. Hence, there are 4 possible transitions: 2 E 2 T 1, 2 E 2 T 2, 2 E 2 A 1 and 2 E 2 A 2. e) In both case, the Co center is sp 3 d 2 hybridized. However, at 200 K, all 3d orbitals are occupied, so Co atom uses 2 4d orbitals for the hybridization. Meanwhile, at 350 K, low spin state, 1 3d orbital is vacant, hence, the Co center uses 1 3d and 1 4d orbital for the hybridization. f) The shorter bonds are origin from Jahn-Teller distortion. At 200 K, low spin state, the Jahn-Teller distortion occurs at e g set of orbitals. Because e g are the non-bonding orbitals, hence the distortion here is strongly affect the bondlength. At 350 K, high spin state, the Jahn-Teller distortion only occurs at t 2g set orbitals. These are non-bonding orbital. Hence, the distortion in bondlength is smaller. g) The reaction: [Co(tppz) 2 ](tcm) 2 + 6CH 3 NH 3 2(tppz) + [Co(CH 3 NH 3 ) 6 ](tcm) 2 The reaction cannot occur due to: Decreasing of entropy Thermodynamically unfavoured. The tppz is a tridentate ligand, which can be partially dissociated and then swing back and re-association can occur. It s really difficult to be dissociated and substituted by a monodentate ligand. 3
5 4 Question 4 The bonds between S or N with Pd in the complexes are not a normal σ bond. It s a double bond with some π bond contribution as shown below: Electron density flows from filled p orbital of S or N through empty d orbital of Pd into empty d or P-C σ MO of the phosphine. i) The complex is bidentate from S, because S has larger p orbital. Hence, the π overlapping is more efficient as compared to N. ii) The complex is bidentate from N, because of the steric effect created from bulky diphosphine ligand. The Pd-S-C angle is nearly 110 o while the Pd-N-C angle is 180 o. Hence, the N-terminal ligand is less spatial hindrance. Therefore, with bulkier ligand, the N coordination is more favoured. iii) In the third complex, the bidentate ligand is N,P-terminii. Due to strong σ ability of N atom, the trans position of N terminus must be an atom have electron accepting ability. That s the reason why the thiocyante ligand trans with N terminus is S coordinate. S atom can accept electron from Pd through π back bonding. 4
6 5 Question 5 a) From the left to the right: 3 F, 3 D, 3 P, 3 S, 3 P, 3 D, 3 F, 3 P, 3 S, 3 P. b) From the MO diagram of 2 cases, the F induces the decreasing of o, meanwhile the CO raise the o. Hence, F is a weak field ligand and CO is a strong field ligand. 5
7 c) [Cr(NH 2 )(CH 3 NH 2 )(NH 2 CH 2 CH 2 NH 2 ) 2 ]OH amidobisethylenediaminemethylaminechromium(ii) hydroxide. 6
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