Analyzing D-Block Chemistry - Essay Example

?Mini Project – d-block chemistry Instructions There are 10 (ten) questions. Answer all questions. Question a) Complete the following table that gives the electronic configuration of the 1st row d-block elements and their corresponding ions in oxidations states (II) and (III). Atomic no. Element Symbol M(0) M(II) M(III) 21 Scandium Sc [Ar]4s23d1 [Ar] 3d1 [Ar] 3d0 22 Titanium Ti [Ar]4s23d2 [Ar] 3d2 [Ar] 3d1 23 Vanadium V [Ar]4s23d3 [Ar] 3d3 [Ar] 3d2 24 Chromium Cr [Ar]4s13d5 [Ar] 3d4 [Ar] 3d3 25 Manganese Mn [Ar]4s23d5 [Ar] 3d5 [Ar] 3d4 26 Iron Fe [Ar]4s23d6 [Ar] 3d6 [Ar] 3d5 27 Cobalt Co [Ar]4s23d7 [Ar] 3d7 [Ar] 3d6 28 Nickel Ni [Ar]4s23d8 [Ar] 3d8 [Ar]4s23d5 29 Copper Cu [Ar]4s13d10 [Ar] 3d9 [Ar]3d8 30 Zinc Zn [Ar]4s23d10 [Ar] 3d10 [Ar]3d9 b) What are the two irregularities in placing electrons in the 3d orbitals of the elements given above and why are such irregularities allowed? Answer: Cr [Ar] 4s13d5 and Cu [Ar] 4s13d10 are two irregularities present in the d-block elements. This can be explained using Hund’s rule of electrons filling all the orbitals in the subshell before they pair up with opposite spins. For chromium 4s1 3d5 will be lower in energy than 4s2 3d4 since the electron pairing energy is not needed for 4s orbital. The energy required for pairing is more than the difference in the energies of 4s and 3d orbitals. Thus, the lowest energy electron configuration will be the one which has one electron in each of the six orbitals that are available. For copper, the 3d orbital has less energy than that of the 4s because of high effective nuclear charge. Copper also has smaller size and more stable inner shell orbitals than the elements placed before it. Therefore, it is better to have the paired electrons in the d and the unpaired one in the s. Question 2. a) Give a definition of a Lewis acid and a Lewis base. Answer: Lewis acid is a substance which accepts a pair of electrons like an H+ ion. A Lewis base is any substance which donates a pair of nonbonding electrons, for example, the OH- ion. Transition metals form Lewis acids as they bond with electron-donating species. The ligands interacting with the metal ion are the Lewis bases as they donate their unshared pair of electrons to the central atom. The coordinate bond, so formed, is provided by the ligand or Lewis base. b) In a reaction of CoCl3 and NH3 a number of cobalt (III) complexes with different colour were formed. Explain this observation. When CoCl2 is dissolved in aqueous ammonia it forms several complexes in the +3 oxidation state. These complexes have different colors and different empirical formulas. Co3+ ion has a secondary valence or coordination number of six. [CoCl3(NH3)4]   green [CoCl3(NH3)5]   purple [CoCl3(NH3)6]   orange-yellow [CoCl3(NH3)5(H2O)]   red The different composition of each of these complexes gives rise to crystal fields that differ in strength. A weaker field causes the energy gap between two subsets of d-orbitals (?) to be small. It can absorb low frequency radiation to move from one level to the other. For example, [Co(NH3)4Cl2+][Cl-] absorbs red and appears green. The converse is true for [Co(NH3)5(H2O)3+][Cl-]3 which absorbs blue-green light and appears red in colour. c) Draw the structure of each of the possible complexes formed in part (b). d) Draw the structure of the complex [Cr(en)3]Cl3 Question 3. Complete the following table Metal complex Coordination number Oxidation state of the metal [Co(NH3)5(ONO)]2+ 6 2 [Cu(NH3)4]2+ 4 2 [Fe(CN)6]3 6 3 [Ni(CO)4] 4 0 Na4[Mn(ox)3] 3 2 [VO(acac)2] 2 4 Question 4. Give the chemical formula of the following metal complexes: a) Potassium hexafluorocobaltate (III) : K3[CoF6] b) Tetraamminechloronitrocobalt(III) chloride: [Co(NH3)4(NO2)]Cl2 c) Tris(ethylenediamine)nickel(II) sulphate: [Ni(en)3]SO4 d) Tris(ethylenediamine)cobalt(II) nitrate: [Co(en)3](NO3)2 e) Cobalt(II) hexanitrocobaltate(III): Co3[Co(NO2)6]2 f) Ammineaquadicarbonyldicyanoiron(III):[Fe(CN)2(NH3)(H2O)(CO)2]+ Question 5. Name the following complexes: a) [CoI(NH3)5]Cl2 : pentaammineiodocobalt(III) chloride b) [Cu(NH3)4]SO4: tetraamminecopper(II) sulfate c) [CrCl(en)2(H2O)]Cl2: aquochlorobis(ethylenediamine)chromium(III) chloride d) K3[Fe(ox)(ONO)4]: potassium tetranitritooxalatoferrate(III) e) [Sc(NC)2(NO)4]+ : diisocyanotetranitrosylscandium(II) f) [Co(NCS)3(SCN)3]3-: triisothiocyanatotrithiocyanatocobalt(III) Question 6. For each of the following complexes, draw and name the possible types of isomers that may arise. a) diaquadiiodoplatinum(II): Cis and trans forms of [Pt(OH2)2I2] b) trioxalatoiron(3-): Lambda and delta forms of [Fe(C2O4)3]3- c) dichlorotetraammineiron(+): Geometrical isomers of [Fe(NH3)4Cl2]+ given as trans and cis forms and optical isomers of the cis form of [Fe(NH3)4Cl2]+ d) triamminetrichloroiron(III): Meridional and facial forms of [Fe(NH3)3Cl3] Question 7. a) On clearly labelled three dimensional Cartesian axes give ONE clear diagram of the boundary surfaces used to represent the angular part of the eg set of 3d-orbitals (the ‘ring’of the dz 2 orbital may be omitted for clarity; this must be stated) for Mn+. In the image, the eg subset of 3d-orbitals are represented on the three axes. The dz2 orbital has been depicted without its ring so that the two lobes of the dumbbell shape are visible. b) Indicate the sites of an octahedral set of fluoro ligands in [MF6]n+. In the given image, the six fluoro ligands are depicted by the small blue dots. They will take up the positions are shown in the diagram to form an octahedral complex with the transition metal atom, depicted in red in the centre. c) Give a clear energy level diagram to show the effect that an octahedral set of fluoro ligands will have on the energies of the eg and t2g sets of 3d-orbitals relative to those in an isolated metal ion. The triple degenerate t2g orbitals consisting of dxy, dyz and dxz orbitals are pointed towards the ligands placed around the metal atom in an octahedral complex. They are likely to be affected more when there is a strong ligand in the equatorial plane. However, the dz2 and dx2-z2 orbitals have lobes directed at the axial ligand. They form higher-energy orbitals if strong ligands are placed axially in an octahedral complex. Halides are weak field ligands. Hence, they cause little splitting of the eg and t2g orbitals. In Mn+ and F6 complexes, all the positions are occupied by fluoro ligands. Hence, they do not exert a particularly greater force on any position more than the others. Question 8. Draw up a table for isolated ions (M2+) of the ten elements of the First Transition Series, using the column headings: M2+ Electronic configuration No. of unpaired electrons Magnetic moment /BM Sc2+ [Ar] 3d1 1 1.73 Ti2+ [Ar] 3d2 2 2.83 V2+ [Ar] 3d3 3 3.87 Cr2+ [Ar] 3d4 4 4.90 Mn2+ [Ar] 3d5 5 5.92 Fe2+ [Ar] 3d6 4 4.90 Co2+ [Ar] 3d7 3 3.87 Ni2+ [Ar] 3d8 2 2.83 Cu2+ [Ar] 3d9 1 1.73 Zn2+ [Ar] 3d10 0 0 Question 9. Using ‘box notation’ give the d-electron configuration of each of the following ions in (a) a weak crystal field and (b) a strong crystal field. In each case give the value of µ (spin only), and the Crystal Field Stabilisation Energy (CFSE) in terms of Dq (i.e. oct) and P (the pairing energy). (i) Cr2+ (ii) Cr3+ (iii) Fe2+ (iv) Fe3+(v) Ni3+ Question 10. a) What is the order of increasing crystal field strength of the ligands H2O, ethylenediamine and SCN- deduced from their positions in the spectrochemical series? Answer: According to the spectrochemical series of ligands, SCN- < H2O < ethylenediamine is the correct order of increasing crystal field strength. b) How can the uv-visible spectra of [Ni(Et2en)2Cl2] and [Ni(Et2en)2(SCN)2] be used to explain the difference in colour of these two complexes? Answer: When a coordination complex is formed with metal ion and ligands, the d-orbitals split into low energy and high energy orbitals. The difference in energy (?) is dependent on the metal and the ligand. If the difference is large, the electrons will need a lot of energy to jump to the high energy orbitals; instead they pair up at the lower levels- giving a low spin complex. However, if energy difference is less, the electrons can easily move into the higher energy orbitals resulting in high spin complex. Ligand with a weak field gives small energy difference, and hence, high spin configuration, and vice versa. Electrons can jump from lower energy levels to higher energy levels by absorbing photons. When certain wavelengths are absorbed, the complementary or subtractive colour becomes visible. A colour wheel can be used to predict whether a complex is high spin or low spin. The high spin complexes formed due to weak field ligands with low energy difference will absorb low frequency wavelength. So for example, if orange to yellow light is absorbed, the complex will appear blue. The same is true for the opposite. [Ni(Et2en)2Cl2] is green in colour which means it absorbs red light. It is therefore a high spin complex as is confirmed by the higher field strength of Cl- as compared to SCN- in [Ni(Et2en)2(SCN)2]. The latter compound appears red which means that it absorbs in the blue-green range. It also means that the energy absorbed is larger than that by [Ni(Et2en)2Cl2], implying a larger energy gap. Thus it is a low spin compound and must have a larger tetragonal distortion. c) The coordination geometries around the nickel atom in the nominally six-coordinate complexes [Ni(Et2en)2X2] (X = Cl-, SCN-, NO3- and I-) range from slightly axially distorted octahedral to virtually square planar. What are the likely structures for these complexes based on their magnetic properties? Answer: In a free ion state the 5 d-orbitals would be degenerate. However, according to the crystal field theory, the ions forming octahedral complexes have split d-orbitals with dx2-y2 and dz2 (egsubset) at higher energy than the dxy, dxz, dyz orbitals (the t2g subset). When the metal ion has between 4-7 electrons, the interaction between the axial ligands with the dz2 orbital determines the structure of a complex. Depending on the ligand field strength, octahedral complexes can have high spin or low spin electronic configurations. If the axial ligand effect is slightly weak, the d-orbitals will move or split slightly causing a distorted octahedral symmetry. In this case, the electrons in eg subset are placed together, which means that the dx2-y2 and dz2 are similarly placed with regard to energy. The distortion is because the orbitals are asymmetrically occupied. However, if the axial positions have very weak field ligands with respect to the ethylenediamine groups, which are bidentate ligands strongly attached to the Ni(II) ion, this distortion can be large causing the complex to have a square planar symmetry. Ligand strength of these electron donors are as from weaker to stronger: I- < SCN- < Cl-< NO3-. Read More