Problem 14: Lead iodide Lead nitrate (Pb(NO3)2) and potassium iodide (KI) react in aqueous solution to form a yellow precipitate of lead iodide (PbI2). In one series of experiments the masses of the two reactants were varied, but the total mass of the two was held constant at 5.000 g. The lead iodide formed was filtered from solution, washed and dried. Data for a series of reactions are given below, together with a blank graph. Experiment Mass of lead nitrate (g) Mass of lead iodide (g) 1 0.500 0.692 2 1.000 1.388 3 1.500 2.093 4 3.000 2.778 5 4.000 1.391 Mass of PbI2(s) in g
_ _ _ _ _
Mass of Pb(NO3) 2(aq) in g 1. Complete the graph; that is, plot the data and draw the approximate curve(s) connecting the data points. Determine graphically what the maximum mass of precipitate is that can be obtained? 2. Write the balanced equation for the reaction and use it to calculate the maximum mass of PbI2 and the corresponding amount of Pb(NO3)2. Problem 15: Octahedral complexes The doubly charged metal ions of the first row transition elements with the configuration d1, d2, … d9 form mainly octahedral complexes of the general formula ML62+ (L = neutral monodentate ligand). From the magnetic behavior point of view these complexes are of two types. Those with the same number of unpaired electrons with the M(g)2+, the so called “high spin” complexes, and those in which this number is smaller or even zero, the “low spin” ones. If the separation between the energy levels t2g and eg is Δ and the pairing energy (the energy necessary for two electrons to be coupled into a singlet) is P, predict the ground state electron configuration for these
complexes. Hint: The Aufbau principle and the Pauli exclusion principle must be obeyed. Problem 16: Isomerism in Inorganic Chemistry Isomerism is traditionally introduced in Organic chemistry courses. However, Inorganic Chemistry also offers many different types of isomerism, especially structural. The modern study of coordination compounds began with Alfred Werner1 and Sophus Mads Jorgensen, more than one hundred years ago. It was their experiments that established the octahedron as the premier structural entity for coordination number six (6) rather than the planar hexagon or the trigonal prism, for example in the complex ion Co(NH3)63+. Thus, there are two isomers corresponding to the chemical formula CoCl3·4NH3 , the praseo and violeo complexes. There are also two structural isomers for the complex Rh(py)3Cl3 (py: pyridine – only its N-atom need be considered). The far infrared spectra in the 450–200 cm-1 region of these two complexes are shown in the figure below. Rotation of plane polarized light is not limited to organic molecules with a chiral carbon atom. The general condition for optical activity is that the molecule or ion should not possess an improper axis of rotation, and because S2 ≡ i and S1≡ σ the molecule may not possess a center or plane of symmetry. Thus, although there is only one cis–[Co(NH3)4Cl2]Cl complex, there are two cis–[Co(en)2Cl2]Cl complexes (en: ethylene diamine or 1,2 diamino ethane, a symmetric bidentate ligand) which are enantiomers. 1 Nobel prize in Chemistry 1913 Questions: 1. What is the electronic configuration of Co in Co(NH3)63+? 3+ 2. What is the type of bonding of NH3 in Co(NH3)6 , which is a low spin complex, and of fluoride ion in CoF63–, which is a high spin complex? Place electrons from the ligands in appropriate metal orbitals according to Valence Bond Theory. 3. Write the stereochemical formulae and structures for the praseo and violeo complexes. 4. Write the stereochemical formulae and structures for the two Rh(py)3Cl3 complexes. Assign the spectra to each of the two complexes. 5. How many isomers does the complex with the formula [Co(en)3]I3 exibit ? Problem 17: Tetrahedral and square complexes Four coordination is much less common than six coordination in first row d-block elements, mainly due to energy stabilization by the ligands. The most symmetric spatial location of ligands about a central ion is the tetrahedron, but square planar complexes are also well known. In tetrahedral geometry only one MA2B2 complex is possible, whereas two isomers are possible in square planar geometry. The hybridization of metal atomic orbitals for tetrahedral geometry is sp3 or sd3 and for square planar geometry dsp2. Ni2+ is [Ar]3d8. With chloride ion a paramagnetic NiCl42– complex ion is formed, whereas with cyanide (CN–) ion a diamagnetic Ni(CN)42– complex ion is formed. 8
Show the distribution of metal ion valence electrons and ligand electrons in the 3d, 4s and 4p subshells for both complexes. Problem 18: Copper enzyme Life on our planet is based on the conversion of solar to chemical energy during photosynthesis. Plastocyanin (PC), is a Cu-containing protein that mediates electron transfer in the photosynthetic electron transfer chain. The active site of PC consists of a Cu ion coordinated by the N atoms of the side chains of two histidines and the sulfur atoms of one cysteine and one methionine residue. Electron transfer by PC involves interconversion of the Cu ion between the Cu(I) and Cu(II) oxidation states. The three-dimensional structure of PC can be found at http://www.rcsb.org/, Protein Data Bank file for poplar PC: 1PNC.
1. What is the full electron configuration of Cu, Cu(I), Cu(II) ? 2. Which oxidation state of PC is EPR active ? 3. Based on the amino-acid sequence of PC (the “primary structure” of the protein), the molar mass of PC is 10500. An aqueous solution of oxidized PC has an intense blue color (PC belongs to the family of “Blue Copper Proteins”), since it absorbs strongly at 597 nm (molar absorption coefficient 4500 mol?1dm3 cm?1). A sample of oxidized PC in a 1cm UV-Vis cell has an absorption of 0,700. Calculate the amount (mg) of PC and the number of Cu(II) ions that are present in 5 cm3 of this PC solution. 4. The Cu ions of certain Blue Copper proteins can be removed to give the corresponding metal-free proteins, known as apo-proteins. These forms can bind various metal ions such as Co(II), Ni(II), Zn(II) and Cd(II), and the metal sites adopt similar structures with the “native” Cu protein. Which reconstituted Blue Copper Proteins are expected to be redox inactive? Problem 19: Palladium nanoclusters Nanoclusters -near monodispersed metal particles that are generally less than 10 nm (100?) in diameter- have attracted intense interest over the past decade. One reason for this is the belief that nanoclusters will have unique properties, derived in part from the fact that these particles and their properties lie somewhere between those of the bulk and single-particle species. These strange “morsels of matter” have fascinating potential uses; nanoclusters have significant potential especially for catalysis as new types of higher activity and selectivity catalysts. There are four general synthetic methods for transition metal nanoclusters. These four methods are: (i) transition metal salts reduction (ii) thermal decomposition and photochemical methods (iii) ligand reduction and displacement from organometallics, and (iv) metal vapor synthesis. Furthermore, nanoclusters must be stabilized against aggregation into larger particles. Stabilization can be achieved by electrostatic (charge or “inorganic”) stabilization, steric (“organic”) stabilization or a combination of both. 9
Electrostatic stabilization occurs by the adsorption of ions to the often electrophilic metal surface. This adsorption creates an electrical double (real multi-) layer, which results in a Coulombic repulsion force between individual particles (Fig. 1a). Steric stabilization is achieved by surrounding the metal center by layers of material that are sterically bulky, such as polymers or surfactants. These large adsorbates provide a steric barrier, which prevents close contact of the metal particles centers (Fig. 1b). - - - + - ++ + - + + M + - + ++ + - - + a
Figure 1. A schematic illustration for (a) an electrostatically stabilized metal (M) particle and (b) a sterically stabilized metal particle. Metal clusters are constructed by successively packing layers- or shells- of metal atoms around a single metal atom. Metal clusters that have a complete, regular outer geometry are designated full-shell or “magic number” clusters. The total number of metal atoms, y, per nth shell is given by the equation y = 10n2+2 (n = 1, 2, 3, …) (Fig. 2.) Full-Shell or “Magic Number” Clusters 1 2 3 4 5 Number of Shells Figure 2. Idealized representation of hexagonal close-packed full-shell “magic number” clusters. Each metal atom has the maximum number of nearest neighbors, which impart some degree of extra stability to full-shell clusters. The most widely used technique for characterizing nanoclusters is transmission electron microscopy (TEM) or high resolution TEM (HR-TEM), techniques which provide direct visual information on the size, shape, dispersity, structure and morphology of nanoclusters.
0 1,0 1,5 2,0 2,5 3,0 3,5 4,0 4,5 5,0
Nanocluster diameters (nm)
Figure 3. (a) Transmission electron microscopy image of Pd(0) nanoclusters stabilized by a polymer. (b) Histogram of the Pd(0) nanocluster diameters. 1 By direct reaction of Pd(II)-polymer complex (1 mM in water solution) with gas H2, Pd(0) nanoclusters are prepared, as following: nPd(II) + nH2 → Pd(0)n + 2nH+ A transmission electron micrograph of the isolated Pd(0)n nanoclusters shows spherical Pd(0) nanoclusters, protected by the polymer, with an average diameter of 2.05 nm. Calculate the number (N) of Pd atoms per cluster. Are these nanoclusters full-shell nanoclusters? Calculate the number of shells (n) in the above Pd(0) nanoclusters. Density of Pd, ρ = 12.02 g/cm3. 2 The catalytic activity of the polymer stabilized Pd(0)n nanoclusters is detected by a catalytic olefin hydrogenation reaction, such as the cyclohexene plus H2 reaction:
Pd(0)n nanocluster catalysts
In a 400 cm-3 high pressure reactor, an amount of the above polymer-protected Pd(0)n nanoclusters containing a total of 50 ?mol of Pd(0) was dissolved in 50 cm-3 of acetone, followed by the addition of 5 cm-3 of cyclohexene. The reactor was then sealed, purged several times with prepurified H2 (dry and O2 free) and the H2 pressure was set to the desired value, approximately 4 atm. The solution was stirred continuously during the reaction and the temperature was kept constant at 30oC. H2 pressure vs. time until the end of the reaction is presented in Fig. 4.
4,6 4,4 4,2 4,0
H2 pressure (atm)
3,8 3,6 3,4 3,2 3,0 2,8 2,6 2,4 2,2 2,0 0 25 50 75 100 125 150 175 200
Figure 4. Hydrogen uptake curve. Temperature 30oC, 0.5 ?mol of Pd(0), 5 cm-3 of cyclohexene. (i) Calculate the % conversion of the cyclohexene. (ii) Taking under consideration that only the surface Pd(0) atoms of the nanoclusters are catalytically active, calculate the turnover number, TON, where TON = moles H2 consumed / moles of catalytically active Pd(0) and the turnover frequency, TOF, where TOF = moles H2 consumed / moles of catalytically active Pd(0)/ time (min) of the consumption. Density of cyclohexene, ρ = 0.81 g cm-3. 3 The polymer-protected Pd(0)n nanocluster catalyst are also used for the catalytic hydrogenation of hex-1-ene by H2. The experiment was performed under the conditions cited above, exept that the solvent was chloroform. It was found that Pd(0)n nanoclusters are an efficient catalyst for the hydrogenation of hex-1-ene.
CH3(CH2)4CH3 hex-1-ene hexane 1 The H-NMR spectra of the hex-1-ene and the reaction mixture after 30 min of reaction and after the removal of the catalyst are shown in the Fig. 5.
Pd(0)n nanocluster catalysts
6 5 4
0.0 5 4 3 2 1
(a) (b) 1 Figure 5. 300 MHz H-NMR spectra of (a) hex-1-ene and (b) the solution of the reaction after 30 min of reaction and after the elimination of the catalyst and the solvent. The relative integrals of the 1H-NMR spectra are given in the table below: δ / ppm relative integral Hex-1-ene 0.88-0.96 3 (Fig. 5a) 1.15-1.32 4 1.99-2.08 2 4.85-4.98 2 5.65-5.79 1 Solution of the reaction 0,88-0.96 9 (Fig. 5b) 1.12-1.37 12 1.99-2.08 2 4.85-4.98 2 5.65-5.79 1 Calculate the % conversion of hex-1-ene to hexane after 30 min. Problem 20: Drug kinetics The absorption of a drug by an organism often follows simple kinetics though the mechanism is very complicated. Consider a drug delivered orally in a common capsule. Let [A]s be its concentration in the stomach and assume that the rate of its introduction into the blood stream is first order with respect to this [A]s. Also assume that the rate by which the drug is metabolized or removed from the blood stream is proportional to its concentration in the blood, [A]b. Give the plot of [A]s as a function 12
of time and write the equation representing d[A]b/dt. After one hour 75% of [A]s have been removed from the stomach. What percentage of the initial [A]s will remain in the stomach two hours after taking the drug? Problem 21: Br2 + CH4 reaction mechanism The reaction of bromine with methane is represented by the following chemical equation: Br2 + CH4 → CH3Br + HBr The proposed mechanism for this reaction is as follows: k1 (1) initiation Br2 + M ??→ 2Br + M k2 (2) propagation Br + CH4 ??→ CH3 + HBr k3 (3) propagation Br2 + CH3 ??→ CH3Br + Br k4 (4) propagation HBr + CH3 ??→ CH4 + Br k5 (5) termination ??→ Br2 2Br + M M stands for some molecular species. k3 and k4 are of the same order of magnitude. 1. In the proposed mechanism of this reaction some very unstable species are involved, such as the radicals CH3 and Br. These very active species react as soon as they are formed, so their concentrations are very small compared to the other species. Shortly after the beginning of the reaction their concentrations remain approximately d[CH 3 ] d[Br ] = 0 and = 0 . This is called the “steady state” condition or constant, so: dt dt approximation for the CH3 and Br radicals. Find the expression for the rate of formation of CH3Br as a function of the concentration of the stable species that are involved in the reaction and the reaction rate constants, k1, k2, k3, k4, and k5. 2. The rate law you found may be simplified when we consider the reaction progress. The three expressions below refer to the form of the rate law at the start, the steady state condition of the CH3 and Br radicals and near the end of the reaction: 1/ 2 /2 [CH 4 ] k1 1 k 2 [Br2 ] v = 1/ 2 ? (I) k 4 [HBr ] k5 +1 k 3 [Br2 ] v' = v '' =
/2 k1 1/ 2 1 k2 ? [Br2 ] [CH 4 ] 1/ 2 k5 3/ 2
/2 k1 [CH 4 ] 1 k 2 k 3 [Br2 ] ? (III) 1/ 2 [HBr ] k5 k4 Enter a numeral (I, II, III) next to each stage of the reaction to indicate which expression corresponds to which stage. Start of the reaction
Steady state condition Near the end of the reaction 3. State the assumptions you need to make at each stage in order to simplify the rate law.