Answer_key_chapter_16-1 - Answer key Chapter 16 *16.7 The...

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Unformatted text preview: Answer key Chapter 16 *16.7 The density of mercury(Hg) is much greater than Cd because mercury has the effect of lanthanide contraction, coming from poor shielding of e‐ in f orbitals. This effect makes the 3rd group transition metal smaller than we would expect. In figure 16.4 (pp 659), you can see that atomic radii of the second and the third group metal in the same column are very close. But the third group element has greater atomic mass. From the basic equation d = m/v, you can clearly see that the greater mass the greater density (because atomic radii are similar, the volumes are quite the same). This is why mercury is much denser than Cd. However, the density of Cd is slightly greater than Zn because Cd has no lanthanide contraction effect (in fact, it is only affected by d block contraction, which is not as effective as lanthanide contraction). As a result, the densities between Cd and Zn are quite similar. *16.21 Ag (silver) and Au(gold) have more effective nuclear charge than Cu (This is because the poor shielding of the electrons in d and f orbitals as we go down the periodic table). This implies that the outermost electrons (valance e‐) receive more attractive force form the nucleus. For this reason, it is more difficult for Ag and Au to be oxidized (more difficult to lose electrons).This idea is also supported by the standard reduction potential (see the appendix) of each element. Thus, Ag and Au are less reactive than Cu. 16.29 (a) H = HN(CH2CH2NH2)2 N H2N NH2 each N has lone pair electrons, which can be used to coordinate with metal ion/atom. This molecule is a polydentate ligand, which can form 3 sigma bonds with central atom. (b) O O O C C O O O M M CO32‐ can be either mono‐ or bi‐ dentate ligand, using one lone pair of the electrons from oxygen atom to coordinate with metal atom/ion (M). (c) H2O uses one lone pair of the electrons on oxygen to coordinate with metal ion. It's a monodentate ligand. (d) O O O O M Although the oxalate ion has 4 oxygen atoms that can coordinate to metal ion because each of them has lone pair electrons, it can form only 2 bonds at a time with metal ions using two lone pairs of the electrons from 2 oxygen atoms. Thus, oxalate is bidentate ligand. 16.32 N N M N M N N M N M (a) (b) (c) A ligand that can be chelating ligand have to be polydentate ligand, which means it can form more than one bond with the same metal ion with a proper geometrical arrangement. Molecule (a) can freely rotate around the single bond and this makes it possible for two nitrogen atoms to coordinate with the same metal ion. Molecule (b) is rigid but two nitrogen atom are already in the proper position to coordinate with the same metal ion. However, molecule (c) cannot function as chelating ligand because two nitrogen don't arrange in such a way that they can coordinate the same metal ion. Instead, each of nitrogen atom can act as monodentate ligand and form a bond with different metal ion. 16.37 Key concept—Consider only within coordination sphere in the square bracket. You should—Be familiar with 3D structure and figure out how each atom orients in the space. (a) [CoCl2(NH3)4]Cl .H2O H 3 N NH3 Cl Co Cl NH3 NH3 NH3 Cl Co NH2 Cl NH3 Cl Co H3N NH3 Cl NH3 NH3 NH3 trans‐[CoCl2(NH3)4]+ NH3 NH3 cis‐[CoCl2(NH3)4]+ Important note Keep in mind that cis‐ and trans‐ isomer can be present in octahedral and square planar geometries only. Any molecule which has an empirical formula in coordination sphere as MX2Y4 for octahedral and as MX2Y2 for squar planar, where M = central atom, X and Y are two different types of ligands, always has cis‐ and trans‐ isomers, as you will have a clearer picture in (b) and (c). (b) [CoCl(NH3)5]Br NH3 H3N NH3 Cl Co H3N NH3 NH3 ≡ Co H3N NH3 Cl NH3 NH3 H 3 N H 3 N Co Cl H3N NH3 Cl H3N NH2 Cl Co Cl H3N H3N NH3 Cl Co Cl NH3 1 2 You can see that molecule 1 is equivalent to molecule 2, or precisely they are the same molecule, because if you rotate molecule 1 clockwise for 180o, you get 2. Alternatively, you see that this molecule has the empirical formula MXY5. Thus, this molecule has no cis‐ or trans‐ isomer. (c) [PtCl2(NH3)2] Cl Pt H3N NH3 Cl Pt H3N NH3 Cl Cl trans‐[PtCl2(NH3)2] H3N Pt H3N Cl Cl H3N Pt Cl Cl H3N cis‐[PtCl2(NH3)2] For square planar, it’s quite obvious to see cis‐ and trans‐ isomer. Again, this molecule has the empirical formula MX2Y2 so that it has cis‐ and trans‐ isomers. 16.45. Key concept—To illustrate electron configuration on a metal ion, you should ‐ know the oxidation of metal ion to get no. of electron in d orbital ‐ know the effect of ligands surrounding the metal ion ‐ know the geometry of the molecule (octahedral, tetrahedral or square planar) ‐ know the energy diagram of each geometry ‐ fill electron using Hund’s rule Strong field Weak field CN‐ > CO > NO2‐ > en (ethylenediamine) > NH3 H2O > ox(oxalate) > OH‐ > F‐ > SCN‐ > Cl‐ > Br‐ > I‐ Strong field → electron pairing energy (P) < Δo → low spin Weak field → electron pairing energy (P) > Δo → high spin (a) [Co(NH3)6]3+ Co3+ = d6 NH3 → strong field → low spin Coordination no. = 6 → Octahedral diamagnetic, no unpaired e‐ (b) [NiCl4]2‐ Ni2+ = d8 Cl‐ → weak field → high spin Coordination no. = 4 → Tetrahedral Important note electron configuration of tetrahedral is always high spin because ΔT < Δo so that you have to fill electron in each orbital by 1 electron first and then pair it if necessary. (c) [Fe(H2O)6]3+ Fe3+ = d5 H2O → weak field → high spin Coordination no. = 6 → Octahedral paramagnetic, 5 unpaired e‐ (d) [Fe(CN)6]3‐ Fe3+ = d5 CN‐ → strong field → low spin Coordination no. = 6 → Octahedral paramagnetic, 1 unpaired e‐ paramagnetic, 2 unpaired e‐ Note some diagrams you should be familiar with *16.50 Strong field ligands give rise to energy splitting (Δ) between atomic orbitals when replacing with weak field ligands. This energy splitting is what we call d‐d transition, which is in the region of visible light absorption. With strong field ligands, the complexes absorb light with the shorter wavelength (higher energy because of higher energy splitting) and this results to change in color of the complexes. Also, strong field ligands give low spin complexes. This decreases the no. of unpaired electrons. As a result, the complexes become less paramagnetic. 16.52 NH3 is strong field ligand (see spectrochemical series pp. 684), leading to large energy splitting. For this reason Co3+, which is d6 ion, absorbs light at shorter wavelength (more energy). The [Co(NH3)6]3+ is also low spin complex (you can write e‐ configuration using octahedral energy diagram and you will see the e‐ configuration is t2g6). In contrast, Cl‐ is a weak field ligand, causes smaller energy splitting, Co3+ then absorbs light at longer wavelength (less energy). [CoCl6]‐ is a high spin complex with e‐ configuration t2g4 eg2. 16.55 diamagnetic(no unpaired e‐) Zn2+ is d10 ion and all d orbital are filled. Then, if there were e‐ transition, there wouldn't be no available room in eg level to accept e‐ form t2g level. For this reason, there is no e‐ transition in d10. As a consequence, no light is absorbed and the complex is colorless. The d10 is also diamagnetic because there is no unpaired e‐. 16.57 key idea: ‐ using λmax to calculate E via the relation: E = hc/λ. c = 3 x 108 m/s ; h = 6.63 x 10‐34 m2 kg s‐1 1 J = 1 kg m2 s‐2 ‐ The energy obtained reflects to energy splitting between d orbitals in the complexes ‐ higher energy also reflects to stronger ligand field i.e, (a) E = 6.63 x 10‐34 m2 kg s‐1 x 3 x 108 m s‐1 x 6.02x 1023 = 1.62 x 105 J or 162 kJ 740 x 10‐9 m 1 mol (b) E = 6.63 x 10‐34 m2 kg s‐1 x 3 x 108 m s‐1 x 6.02x 1023 = 2.6 x 105 J or 260 kJ 460 x 10‐9 m 1 mol (c) E = 6.63 x 10‐34 m2 kg s‐1 x 3 x 108 m s‐1 x 6.02x 1023 = 2.08 x 105 J or 208 kJ 575 x 10‐9 m 1 mol Thus, the order of ligand field strength Cl‐ < H2O < NH3 (as in agreement to spectrochemical series) One thing you might have been noticed is that the longer wavelength absorbed, the weaker field the ligand. *16.87 (a) In [Ni(SO4)(en)2]Cl2, Cl‐ doesn't directly coordinate to Ni. It's just the counter ion to make the molecule neutral. This same idea applies to [NiCl2(en)2]SO4. In this case Cl‐ directly coordinate to Ni and SO42‐ is just the counter ion. Cl‐ can reacts with AgNO3 and form AgCl that is insoluble while SO42‐ cannot. For this reason, we can use AgNO3 to distinguish between this two isomers. Note Counter ion is the species that is out the coordination sphere (the square bracket). You can consider the molecule as the ionic compound, for example, [Ni(SO4)(en)2]Cl2 is from [Ni(SO4)(en)2]2+ cation and Cl‐ anion. The part that plays the role in d‐d transition is in square bracket. (b) The standard reduction potential (Appendix 2 pp A19) reveals that the oxidizing strength is I2 < Br2 < Cl2. For these reason [NiCI2(en)2]l2 can be oxidized by Br2 and we can get free I2 from the reaction while [NiI2(en)2]Cl2 cannot. The red highlighted parts indicate the species that would be expected to react with Br2, not the species in the coordination spheres. *16.92 Although H‐ (hydride) have negative charge like halide ions but it has no extra p orbital electrons to act as pi donor ligand like halide ions. It can coordinate with central atom via sigma bond only while halide ions can use their extra e‐ in p orbitals to act as pi donor(give e‐ to central atom). As a result, halide ions make the energy splitting (Δ)between e and t level smaller (see figure 16.38a pp 689). pi bond sigma bond sigma bond M L halides hydride 16.108 (a) ox = oxalate ion O O M L O O One oxalate ion form 1 chelate ring with central atom. There are 3 oxalate coordinate with Ru, thus the complex has 3 chelate rings. (b) trien = triethylenetetraamine M credit:‐24‐3.htm One trien can form three chelating rings, thus the complex has 3 chelating rings. (c) dien = diethylenediamine 3+ Fe credit:‐bin/paper?hb2285 One dien can form two chelating rings, because there are two dien molecules thus the complex has 4 chelating rings. ...
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