Solutions Manual For Inorganic Chemistry, 6th Edition (Alen Hadzovic, 2024) Chapters 1-27, Latest Guide A+.

Solutions Manual For Inorganic Chemistry, 6th Edition (Alen Hadzovic, 2024) Chapters 1-27, Latest Guide A+. TABLE OF CONTENTS Preface, v Acknowledgments, vii PART 1 Foundations Chapter 1 Atomic Structure Chapter 2 Molecular Structure and Bonding Chapter 3 The Structures of Simple Solids Chapter 4 Acids and Bases Chapter 5 Oxidation and Reduction Chapter 6 Molecular Symmetry Chapter 7 An Introduction to Coordination Compounds Chapter 8 Physical Techniques in Inorganic Chemistry PART 2 The Elements and Their Compounds Chapter 9 Periodic Trends Chapter 10 Hydrogen Chapter 11 The Group 1 Elements Chapter 12 The Group 2 Elements Chapter 13 The Group 13 Elements Chapter 14 The Group 14 Elements Chapter 15 The Group 15 Elements Chapter 16 The Group 16 Elements Chapter 17 The Group 17 Elements Chapter 18 The Group 18 Elements Chapter 19 The d-Block Elements Chapter 20 d-Metal Complexes: Electronic Structure and Properties Chapter 21 Coordination Chemistry: Reactions of Complexes Chapter 22 d-Metal Organometallic Chemistry Chapter 23 The f-Block Metals PART 3 Frontiers Chapter 24 Materials Chemistry and Nanomaterials Chapter 25 Catalysis Chapter 26 Biological Inorganic Chemistry Chapter 27 Inorganic Chemistry in Medicine 1 13 27 43 61 79 89 101 107 111 119 123 127 137 145 153 159 171 175 181 193 201 213 217 223 233 237 Solutions Manual For Inorganic Chemistry, 6th Edition (Alen Hadzovic, 2024) Chapters 1-27, Latest Guide A+. Self-Test Exercises S l .l For the Paschen series n = 3 and m = 4, 5, 6,... The second line in the Paschen series is observed when «2 = 5. Hence, starting from equation 1.1, we have = 1.097xl07m -'|-! -----! -| = l .097x 1 O’ m-i x 0.071 = 779967m-i, 32 5^ The wavelength is the reciprocal value of the above-calculated wavenumber: 1 779967m-1 = 1.28 X10-6 m or 1280 nm. 81.2 The third shell is given by « = 3, and the subshell for / = 2 consists of the d orbitals. Therefore, the quantum numbers « = 3, / = 2 define a 3d set of orbitals. For / = 2, m/ can have the following values: -2, -1, 0, 1, 2. Thus, there are five orbitals in the given set. Figure 1.15 shows the electron density maps for 3d orbitals. 51.3 The number of radial nodes is given by the expression: n-l-. For the 5s orbital, « = 5 and/ = 0. Therefore: 5-0-1 = 4. So there are four radial nodes in a 5s orbital. Remember, the first occurrence of a radial node for an s orbital is the 2s orbital, which has one radial node, the 3s has two, the 4s has three, and finally the 5s has four. If you forget the expression for determining radial nodes, just count by a unit of one from the first occurrence of a radial node for that particular “shape” of orbital. Figure 1.9 shows the radial wavefuntions of Is, 2s, and 3s hydrogenic orbitals. The radial nodes are located where the radial wavefunction has a value of zero (i.e., it intersects the xaxis). 81.4 There is no figure showing the radial distribution functions for 3p and 3d orbitals, so you must reason by analogy. In the example, you saw that an electron in a p orbital has a smaller probability of close approach to the nucleus than in an s orbital, because an electron in a p orbital has a greater angular momentum than in an s orbital. Visually, Figure 1.12 shows this. The area under the graph represents where the electron has the highest probability of being found. The origin of the graph is the nucleus, so one can see that the 2s orbital, on average, spends more time closer to the nucleus than a 2p orbital. Similarly, an electron in a d orbital has a greater angular momentum than in a p orbital. In other words, /(d) /(p) /(s). Therefore, an electron in a p orbital has a greater probability than in a d orbital of close approach to the nucleus. 81.5 The configuration of the valence electrons, called the valence configuration, is as follows for the four atoms in question: Li: 2s‘ B: 2s^2p' Be: 2s‘ C: 2s^2p^ When an electron is added to the 2s orbital on going from Li to Be, Zeff increases by 0.63, but when an electron is added to an empty p orbital on going from B to C, Zgfr increases by 0.72. The s electron already present in Li repels the incoming electron more strongly than the p electron already present in B repels the incoming p electron, because the incoming p electron goes into a new orbital. Therefore, increases by a smaller amount on going from Li to Be than from B to C. However, extreme caution must be exercised with arguments like this because the effects of electron-electron repulsions are very subtle. This is illustrated in period 3, where the effect is opposite to that just described for period 2. 81.6 Following the example, for an atom of Ni with Z= 28 the electron configuration is: Ni: ls-2s^2p^3s^3p^3d*4s“ or [Ar]3d*4s^ Once again, the 4s electrons are listed last because the energy of the 4s orbital is higher than the energy of the 3d orbitals. Despite this ordering of the individual 3d and 4s energy levels for elements past Ca (see Figure 1.19), interelectronic repulsions prevent the configuration of an Ni atom from being [Ar]3d‘®. For an Ni^^ ion, with two fewer electrons than an Ni atom but with the same Z as an Ni atom, interelectronic repulsions are less important. Because of the higher energy 4s electrons as well as smaller Zeff than the 3d electrons, the 4s electrons are removed from Ni to form Ni^^, and the electron configuration of the ion is: Ni: ls^2s^2p^3s^3p*’3d“ or [Ar]3d* and Ni^'": ls^2s"2p‘’3s-3p*’3d“ or [Ar]3d“ 51.7 The valence electrons are in the « = 4 shell. Therefore the element is in period 4 of the periodic table. It has two valence electrons that are in a 4s orbital, indicating that it is in Group 2. Therefore the element is calcium, Ca. 51.8 When considering questions like these, it is always best to begin by writing down the electron configurations of the atoms or ions in question. If you do this routinely, a confusing comparison may become more understandable. In this case the relevant configurations are: F: ls^2s^2p^ or [He]2s^2p^ Cl: ls^2s^2p^3s^3p^ or [Ne]3s^3p^ The electron removed during the ionization process is a 2p electron for F and a 3p electron for Cl. The principal quantum number, n, is lower for the electron removed from F (w == 2 for a 2p electron), so this electron is bound more strongly by the F nucleus than a 3p electron in Cl is bound by its nucleus. A general trend: within a group, the first ionization energy decreases down the group because in the same direction the atomic radii and principal quantum number n increase. There are only a few exceptions to this trend, and they are found in Groups 13 and 14. 51.9 When considering questions like these, look for the highest jump in energies. This occurs for the fifth ionization energy of this element: I4 = 6229 kJ m of', while I5 = 37838 kJ mof^ indicating breaking into a complete subshell after the removal of the fourth electron. Therefore the element is in the Group 14 (C, Si, Ge, etc.). 51.10 The electron configurations of these two atoms are: C: [He]2s^2p^ andN: [He]2s-2p^ An additional electron can be added to the empty 2p orbital of C, and this is a favourable process (Ze = 122 kJ/mol). However, all of the 2p orbitals of N are already half occupied, so an additional electron added to N would experience sufficiently strong interelectronic repulsions. Therefore, the electron-gain process for N is unfavourable (Ae -8 kJ/mol). This is despite the fact that the 2p Zefr for N is larger than the 2p Zeff for C (see Table 1.2). This tells you that attraction to the nucleus is not the only force that determines electron affinities (or, for that matter, ionization energies). Interelectronic repulsions are also important. 51.11 According to Fajan’s rules, small, highly charged cations have polarizing ability. Cs^ has a larger ionic radius than Na^. Both cations have the same charge, but because Na'^ is smaller than Cs^, Na^ is more polarizing. 2+ . End-of-Chapter Exercises ELI The energy of a hydrogenic ion, like He^ or Be^"^, is defined by equation 1.3: Z^Rhc E „ = - Both He" and Be^"^ have ground state electronic configuration Is'; thus for both, the principal quantum number n in the above equation equals 1. For the ratio £’(He'*')/£'(Be^‘^), after cancelling all constants, we obtain: E1.2 £(Hei/£:(Be^^ = ZHeZ^(Be^^) = 2^/4^ = 0.25 (a) The ground state of hydrogen atom is Is*. The wavefunction describing Is orbital in H atom is given with ^ . The values of this function for various values of r/ao are plotted on Figure 1.8. From the plot we can see that the most probable location of an electron in this orbital is at nucleus because that is where this function has its maximum. (b) The most probable distance from the nucleus for a Is electron in H atom can be determined from the radial distribution function, P(r), for Is orbital. The radial distribution function for Is orbital is given 1 4r^ by P(r) = 4;zr2/?(r)2 = x ------e-inan = ------g-iriM . Remember that the wavefunction for Is orbitals depends m l a only on radius, not on angles, thus R^{r) = Pis- To find the most probable distance we must find the value of r where the value for the radial distribution function is maximum. This can be done by finding the first derivative of P{r) by r, making this derivative equal to zero and solving the equation for r: P{r) = ± | dr at ao7 The exponential part of this derivative is never zero, so we can take the part in the brackets, make it equal to zero, and solve it for r. 2 r - 2r2 2r = ao y 2r2 ao = 0 r = ao Thus, the most probable distance from the nucleus is exactly at Bohr radius, ao. This value and the value determined in part (a) of this question are different because the radial distribution function gives the probability of finding an electron anywhere in a shell of thickness dr at radius r regardless of the direction, whereas the wavefianction simply describes the behaviour of electron. (c) Similar procedure is followed as in (b) but using the R(r) function for 2s electron. The final result is 3+^f5ao. E1.3 The expression for E given in Equations 1.3 and 1.4 (see above) can be used for a hydrogen atom as well as for hydrogenic ions. For the ratio £(11, n = 1)/E(H, n = 6), after canceling all constants, we obtain: E(H, n = 1)/£(H, « = 6) = {IV)l{ie^) = 36 The value for £'(H, n= ) has been given in the problem (13.6 eV). From this value and above ratio we can find E(H, « = 6) as: E(H, « = 6) = (E(H, n = l))/36 = 13.6eV/36 = 0.378 eV, and the difference is: P(H ,«= 1)-P (H ,« = 6)= 13.6 eV-0.378 eV= 13.2 eV. E1.4 Both rubidium and silver are in period 5; hence their valence electrons are in their respective 5s atomic orbitals. If we place hydrogen’s valence electron in 5s orbital, the ionization energy would be: 1 = -E,s = hcRZ^ _ 13.6eV «2 52 = 0.544eV This is significantly lower value than for either Rb or Ag. First the difference for H atom only: the energy of an electron in a hydrogenic atom is inversely proportional to the square of the principal quantum number n. Hence we can expect a sharp decrease in / with an increase in n (i.e. from 13.6eV for « = 1 to 0.544eV for n = 6). The difference between H(5s*) on one side and Rb and Ag on the other lies in Z (atomic number or nuclear charge; or better Zeff- effective nuclear charge); as we increase Z (or Zgff) the ionization energy is exponentially increasing (Z(Rb) = 37 and Z(Ag) = 47). And finally, the difference in ionization energies for Rb and Ag is less than expected on the basis of difference in their nuclear charges; we would expect significantly higher I value for Ag than is actually observed. The discrepancy is due to the shielding effect and Zefr; in comparison to Rb, silver has an additional ten 4d electrons placed between the nucleus and its 5s valence electron. These 4d electrons shield the 5s electron from the nucleus (although, being d electrons, not very efficiently). E1.5 When a photon emitted from the helium lamp collides with an electron in an atom, one part of its energy is used to ionize the atom while the rest is converted to the kinetic energy of the electron ejected in the ionization process. Thus, the total energy of a photon (^v) is equal to the sum of the first ionization energy (/i) and the kinetic energy of an electron hv = I + nieVi From here the ionization energy is: I = h v - rrievi Since both krypton and rubidium atoms are ionized with the same radiation, we can calculate the energy of one photon emitted from the helium discharge lamp: c 2 998x108ms-i hv = h x — = 6.626x10-34 Js x —^ ------------------- 3.40xl0-‘8J A 58.4x10-9 m Now we can calculate the first ionization energies using given velocities of respective electrons: = /z V - = 3.40 X10-18 y 9.109X 10-31 k g x (l.59x1 Q6ms-')^ ^2.25xl0-'8j LRb = h v - meV,e,Rb= 3.40x 10-18 J jg 9.109xl0-3'kgx(2.45xl06ms-')^ _ = 6.68 x l 0-18 J 2 2 Note that calculated energies are the first ionization energies per one atom because we used only one photon. To calculate the first ionization energies in eV as asked in the exercise, the above energies must be multiplied by the Avogadro’s constant (to obtain the energies in J m of') and then divided by the conversion factor 96485 J m of' e V ‘ to obtain the values in eV: TKr -.-1C 1A.OT 6.022x1023 mol-i = 2.25x10-18Jx - ■ — ^— ^ - = 14.0eV 96485 Jmol-i eV-i TRb /CO iA tsT 6.022x1023mol-1 = 6.68xlO-i8Jx _____— :— ___ = 4. i6 eV 96485 Jmol-i eV-' E l.6 To solve this exercise we are going to recall equation 1.1 and substitute the given values for ri and 2 ri. - = R A -------!-l = 1.097 X107 m-i f-!-----!- I = 1.097 x 10’ m-i x 0.899 = 9752330 m-i. 1,12 32 The corresponding wavelength is (9752330 m~') ' = 102.5 nm. The energy is: hv = hx — = 6.626x 10-34Jsx 2.998x 10®ms-* x9752330/w-i = 1.937x 10-'®J A E l.7 The visible region starts when wi = 2. The next transition is where «2 equals 3. This can be determined using the Rydberg equation (Equation 1.1). 1 f l - = R - A U 3"; = 1.524x10 ^nm”^ And from the above result, X = 656.3 nm.