Topic 9.4: Transition Elements Chemistry

Size: px

Start display at page:

Download "Topic 9.4: Transition Elements Chemistry"

Erika Reeves
5 years ago
Views:

1 Topic 9.4: Transition Elements Chemistry AJC 2009/P3/Q3c,d 1 167

2 CJC 2009/P2/Q3a-c 2 (a) Cr: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1 or [Ar] 3d 5 4s 1 Cr 3+ : 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 or [Ar] 3d 3 (b) (i) Cr S N Mols Ratio x + y + 1 = 7 x: 4 y: 2 (ii) +3 (iii) NH SCN SCN : : Cr (III).. NH 3 : NCS : NCS (c) (i) K 2 Cr 2 O HBr 2 KCrO 3 Br + H 2 O (ii) No, oxidation no. of Cr doesn t change, remains at +6 (iii) E o cell = (+1.07) = V >0 Products should be Cr 3+ (aq) and Br 2 (l) 168

3 (iv) Reaction didn t take place at standard condition. OR The reaction in step I took place under cooled conditions. (v) AgBr (vi) Cr is reduced from +6 (orange solution of Cr 2 O 7 2- ) to +3 Cr 3+ (deep green solution) DHS 2009/P3/Q4a-c (a) Explain why transition metals are denser than s-block elements. [3] They have relatively smaller atomic radius and higher relative atomic mass. Hence, they have a close-packed structure. (b) Describe the bonding in the following substances and explain why it contributes to the specified property in glass: substance (i) silicon (IV) oxide (ii) lead (IV) oxide property hardness Electrical conductivity (if any) [4] SiO 2 has a giant molecular structure with extensive covalent bonding in a giant three-dimensional structure. PbO 2 has a giant ionic structure. In the solid state, the ions can only vibrate about fixed positions. (c) The window frames enclosing the stained glass is made of aluminium due to its low density and high melting point. Account for the high melting point of aluminium. [2] High amount of energy is required to overcome the strong electrostatic forces of attraction between the cations and sea of delocalised electrons. 169

4 HCI 2009/P2/Q6a,b 4 A: [Cr(H 2 O) 6 ] 3+ B: [Cr(H 2 O) 6 ] 2+ C: Cr(OH) 3 or Cr(H 2 O) 3 (OH) 3 D: CrO 4 2 or Na 2 CrO 4 E: 3+ H 2 H 2 N N NH 2 Cr N NH 2 H 2 H 2 N Cr 3+ has a high charge density. Hence [Cr(H 2 O) 6 ] 3+ can undergo hydrolysis in water to produce H + ions, forming CO 2 with carbonate ions. Ligand exchange Cr 3+ has d 3 electronic configuration. In an octahedral ligand field, the 6 NH 3 ligands will split the five degenerate 3d orbitals into 2 groups of different energy levels. The difference in the two energy levels, E, falls within the visible region of the electromagnetic spectrum. An electron in a lower d orbital energy level can absorb radiation in the visible spectrum and be promoted into the higher d orbital energy level. This d d electron transition gives rise to the colour as the complement of the absorbed colour. 170

5 HCI 2009/P3/Q4a,b,e 5 (a) Any two properties: Fe has a higher melting/boiling point than Al. Fe has stronger metallic bonds as both the 3d and 4s electrons can be used in metallic bonding due to their proximity in energies, hence more energy is required to overcome the stronger bonds. Fe has a greater density than Al. Fe has a greater atomic mass but its atomic radius is smaller. Hence atomic volume is smaller. Since density = mass/volume, density of Fe is greater than Al. Fe has better conductivity than Al as both the 3d and 4s electrons are available in the mobile sea of electrons due to the proximity in energy of 3d and 4s orbitals. Fe is harder than Al due to stronger metallic bonding in Fe as Fe can use both its 3d and 4s electrons for metallic bonding due to the proximity in energy of 3d and 4s electrons. (b) 26Fe 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 13Al 1s 2 2s 2 2p 6 3s 2 3p 1 26Fe 3+ 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 13Al 3+ 1s 2 2s 2 2p 6 Although Fe has one more quantum shell than Al and higher nuclear charge, the increase in shielding effect due to the d electrons is proportionately less than the increase in nuclear charge, hence Fe experiences higher effective nuclear charge and is smaller. Fe 3+ has one more quantum shell than Al 3+ and the additional shielding is due to s and p electrons which are more effective in shielding compared to d electrons. (i) This is due to the presence of partially-filled 3d subshells that can accept electrons from lone pairs or form temporary bonds with reactant molecules during adsorption, thus weakening the bonds in the reactant molecules and lowering the activation energy of the reaction. (ii) E a Energy / kj N 2 + E a (catalysed) H NH 3 Reaction 171

6 IJC 2009/P2/Q2c 6 IJC 2009/P3/Q1e 7 172

7 IJC 2009/P3/Q2d 8 IJC 2009/P3/Q3b 9 JJC 2009/P2/Q2a,b 10 (a) (bi) (ii) Fe(H 2 O) 3+ 6 = Fe(H 2 O) 5 (OH) 2+ + H + 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 Colourless because Cu + does not have unfilled/ partially filled d-orbital JJC 2009/P2/Q4a (a) (i) Mn O Mass A r

8 Mole Ratio 1 2 Empirical Formula: MnO 2 [1] (ii) Identity of Z: MnO 4 [1] Type of reaction Disproportionation [1] Equation 3MnO H 2 O 2MnO 4 + MnO 2 + 4OH [1] MI 2009/P3/Q1b 12 b (i) [Al(H 2 O) 6 ] 3+ + H 2 O [Al(H 2 O) 5 (OH)] 2+ + H 3 O + b (ii) Size of anion: U 3+ (aq) < Al 3+ (aq) U 3+ (aq) has a lower charge density less able to distort the electron cloud of H 2 O less weakening of O-H bond less able to produce H + [½] [½] [½] [½] MJC 2009/P2/Q3a 13 Coordination number of Co = 6 In Co 2+ ions, the d orbitals are split into two groups due to the ability of the ligands to split them into the energy levels. d The d electrons undergoes d-d transition and is promoted to the higher d orbital. During the transition, the d electron absorbs a certain wavelength of light from the visible region of the electromagnetic spectrum and emits the remaining wavelength which appears as the colour of the complex observed. NYJC 2009/P2/Q2b 14 (bii) 2 Cu + Cu + Cu 2+ Cu + + e Cu V Cu 2+ + e Cu V E θ = (+ 0.15) = V > 0 (feasible) 174

9 PJC 2009/P3/Q5a,b

10 176

11 RI 2009/P3/Q3a,b 16 (a) The electronic configuration of V 3+ is [Ar]3d 2. Hence its 3d subshell is only partially filled. V 3+ exists as aqua complex, [V(H 2 O 6 )] 3+ in aqueous solution. H 2 O ligand splits the 3d subshell into two sets of different energy. Electron in the lower set absorbs visible light energy corresponding to the small energy gap between the 2 sets of energy and is promoted to the upper set. Since visible light is absorbed, aqueous solutions of vanadium ions are coloured, with the colour observed being the complementary colour of the light absorbed. (b)(i) SO 2 (g) + 2VO 2 + (aq) SO 4 2 (aq) + 2VO 2+ (aq) SO 2 reduces yellow VO 2 + to blue VO 2+. As the reaction proceeds, the presence of both yellow VO 2 + and blue VO 2+ causes the mixture to appear green until all VO 2 + is used up and the solution appears blue due to VO 2+. (ii) SO 2 can also react with KMnO 4. It must be boiled off completely to ensure that none of it remains so that the KMnO 4 used in titration reacts only with vanadium ions and hence the amount of vanadium ions can then be determined accurately. (iii) Amount of vanadium ions = 5 x amount of MnO 4 used = 5 x x = x 10 3 mol Mass of vanadium in sample = x 10 3 x A r of V = x 10 3 x 50.9 = g Percentage by mass of V = x 100 = 56.0% SAJC 2009/P3/Q2b 17 From data booklet, Co 3+ + e Co 2+ E 0 = V I 2 + 2e 2I - E 0 = V S 2 O e 2SO 4 2- E 0 = V 2Co I - 2Co 2+ + I 2 E 0 = = V > 0 2Co S 2 O 8 2Co SO 4 177

12 E 0 = = V >0 (ii) [Cu(H 2 O) 6 ] NH 3 [Cu(NH 3 ) 4 (H 2 O) 2 ] H 2 O [Cu(NH 3 ) 4 (H 2 O) 2 ] 2+ + edta 4- [Cuedta] H 2 O + 4 NH 3 Strength of ligands : H 2 O < NH 3 < edta 4- SRJC 2009/P3/Q1d 18 dative / coordinate/covalent bond Rh has a giant metallic structure with strong electrostatic forces of attraction between cations and sea of delocalised electrons. Both NH 3 and H 2 O have simple molecular structures with intermolecular hydrogen bonds. For the identification of hydrogen bonds, either mentioned here or in part (iv). Number / Extent of hydrogen bonds: NH 3 < H 2 O Energy required: NH 3 < H 2 O Boiling point: NH 3 < H 2 O NH 3 is soluble in water as favourable hydrogen bonds between NH 3 and water / solute-solvent interaction can be formed. SRJC 2009/P3/Q4b 19 (b) Transition metals possess variable oxidation states due to the small energy difference between the 3d and 4s electrons. Thus different number of 3d and 4s electrons may be lost to form stable ions or compounds of different oxidation states. NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) Cu 2+ (aq) + 2OH - (aq) Cu(OH) 2 (s) (1) 178

13 When NH 3 (aq) is added gradually, [OH - ] will increase Ionic product of Cu(OH) 2 > K sp of Cu(OH) 2 Pale blue ppt, Cu(OH) 2 is formed [Cu(H 2 O) 6 ] 2+ (aq) + 4NH 3 (aq) [Cu(NH 3 ) 4 (H 2 O) 2 ] 2+ (aq) + 4H 2 O(l) -----(2) deep blue When excess NH 3 is added, NH 3 ligands replaces the H 2 O ligands, forming a more stable deep blue [Cu(NH 3 ) 4 (H 2 O) 2 ] 2+ complex with Cu 2+ (aq). [Cu 2+ ] decreases as it is being used to form the complex, equilibrium position in (1) shifts left to increase [Cu 2+ ] Pale blue ppt dissolves. The d orbitals of Cu 2+ (aq) are split into two different energy level due to presence of H 2 O ligands. The d electron undergoes d-d transition and is promoted to a higher energy d orbital. During the process, red wavelength of light energy from the visible region of the electromagnetic spectrum is absorbed and blue wavelength is transmitted which appears as the colour observed. TJC 2009/P3/Q5a,b 20 (a) (i) Mn is a transition element which forms Mn 2+ in which the d subshell is partially filled with 5 electrons. Water molecule is a ligand and has a lone pair of electrons on O atom that can form a dative bond with the central metal ion Mn 2+. (ii) When the ligands approach the central metal ion, splitting of the d-orbitals occur. The energy gap, E between the non-degenerate orbitals corresponds to the wavelength of light in the visible region of the electromagnetic spectrum. When d-d transition of electrons takes place, radiation in the visible region of the electromagnetic spectrum corresponding to E is absorbed. The light energy not absorbed will be seen as the colour of the complex. (b) (i) Empirical formula of A = MnO 179

14 (ii) B = MnO 4 It is a disproportionation reaction 3 MnO H 2 O 2 MnO 4 + MnO 2 + 4OH TPJC 2009/P2/Q3b 21 (b) (i) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 (ii) Cr 2 O H + + 3Zn 2Cr H 2 O + 3Zn 3+ 2HCl + Zn ZnCl 2 + H 2 (iii) To let out the hydrogen gas produced. Or to prevent a build up of pressure due to evolution of hydrogen gas. (iv) Cr 3+ + e - Cr 2+ E θ = -0.41V Zn e - Zn E θ = -0.76V E θ cell = (-0.76) = V. Therefore, reduction of Cr 3+ to Cr 2+ is still feasible. (v) (vi) The hydrogen produced exerts greater pressure in the flask. Ability of chromium compound to vary oxidation state or to display colors. TPJC 2009/P3/Q1a-c 22 (a) X is a transition metal with a partially filled d-subshell. In the presence of ligands, the d orbitals (of the transition metal ion) split into 2 energy levels with an energy gap that falls within the visible light spectrum. Electrons are able to be promoted from a lower energy d-orbital to the higher energy one by absorbing energy from the visible spectrum. OR When an electron from the d-orbital of lower energy is promoted to one of higher energy (dd electronic transition), an amount of energy, E, in the visible region of the electromagnetic spectrum is absorbed. The light energy not absorbed will be seen as the colour of the complex. OR We observe the complementary colour / the wavelengths that are transmitted. (b) Cr: [Ar]3d 5 4s 1 180

15 (c) Step 1 Al: [Ne]3s 2 3p 1 Atomic radius of chromium is smaller than that of aluminium as it has higher nuclear charge and 3d orbitals provide poor shielding effective nuclear charge higher hence radius smaller. Cr 3+ : [Ar]3d 3 Al 3+ : [Ne]3s 0 Cr 3+ has one more quantum shell compared to Al 3+, hence the bigger ionic radius. Relevant data: Mn 3+ + e = Mn 2+ E = V MnO 4 + 8H + + 5e = Mn 2+ + H2 O E = V Overall equation for step 1: MnO 4 + 8H + + 4Mn 2+ 5Mn H2 O E = = V > 0 AND reaction is feasible. cell Step 2 Relevant data: 2CO 2 + 2e = C 2 O 4 2 Mn 3+ + e = Mn 2+ E = 0.49 V E = V Overall equation for step 2: 2Mn 3+ + C 2 O 4 2 2Mn CO2 E = ( 0.49) = V > 0 AND reaction is feasible. cell YJC 2009/P2/Q1a

16 YJC 2009/P3/Q1a,c,d

17 183

Q.1 Predict what will happen when SiCl 4 is added to water.

Q.1 Predict what will happen when SiCl 4 is added to water. Transition etals F325 1 The aqueous chemistry of cations Hydrolysis when salts dissolve in water the ions are stabilised by polar water molecules hydrolysis can occur and the resulting solution can become