Exam 2 Solutions Chem 6, 9 Section, Spring 2002

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1 1. Dartmouth s FM radio station, WDCR, broadcasts by emitting from its antenna photons of frequency 99.3 MHz ( Hz). (a) What is the energy of a single WDCR photon? The photon energy is simply Planck s constant times the photon frequency: E phot = hν = ( J) ( Hz) = J (b) Suppose one of these photons strikes a hydrogen atom in a particular quantum state with a principal quantum number n. What is the minimum value for n that would allow this photon to ionize the atom? The energy of the photon must equal or exceed the energy difference between the energy of the state with the n value we seek and the energy of the ionized atom (which is at our chosen zero of energy, or, equivalently, at the state with n = ): Ephot = J = En = En = J J or n = n J = 5756 (c) You now know n, but not l or m for this atom. Calculate the maximum average radius this atom could have, and in doing so, tell me what l and m values lead to this maximum value. (If you didn t get a value for n in part (b), assume n = 100, which isn t the answer to part (b).) If n is this huge number, then there are many, many possible l and m values for atoms with the same energy. But if we look at the expression for the average radius, we can see that as l increases from its minimum value of 0, the average radius decreases, which tells us that l = 0 has the largest radius, and if l = 0, then m = 0 as well: r nl = a 0 2 3n2 l(l +1) = a 0 2 3(5756)2 0(0 +1) = m 3(5756) 2 = m 2 This is one HUGE atom, with a 2.63 mm radius! 2. (a) Given the H atom orbital picture below, specify those quantum numbers that are determined by the picture, and give possible values for the quantum numbers not specifically determined. First, note that three views are needed to see the entire symmetry of the wavefunction. That tells us that it is NOT cylindrically symmetric about the z axis, meaning m 0. Next, note that there are three planar nodes: the xy plane, the xz plane, and the yz plane. This tells us that l = 3 (it is an f orbital). There are no radial nodes, so n = l + 1 = 4. This is one of the seven 4f orbitals. Finally, m can be anything allowed for l = 3 except zero: n = 4 l = 3 m = +3, +2, +1, 1, 2, 3 (b) One of the H atom s orbital radial probability distribution functions is

2 rpd r 2 2 r a 0 2 e r/a 0 For this orbital, n = 2. This is an s orbital, and thus there are only radial nodes. From the mathematical form of the rpd, we see that this function is zero at r = 0 (which doesn t count as a node), at r = (which also doesn t count), and when the factor in parentheses, (2 r/a 0 ), equals zero. There is only one value of r that meets this criterion, r = 2a 0, and thus there is only one radial node. Now we know n = 2, and this is the rpd for the 2s orbital. (c) The ratio of the wavelength of the n = 2 to n = 1 emission photon in H to the wavelength of the n = 4 to n = 2 emission photon in the one-electron ion He + is 1. In general, the wavelength λ of a one-electron atom emission is given by hc λ = ( J) Z2 n 2 Z2 f n 2 i If we write this equation once for the H atom case and once for the He + case, take the ratio of these two equations, we find that the ratio is 1: λ(h) λ(he) = (a) If I assign m s values to all the electrons in the ground state of the N atom, I find that 2 of them have one value while 5 of them have the other value. N has 7 electrons: 2 are paired in the 1s orbital, 2 are paired in the 2s orbital, and the other three are unpaired in separate 2p orbitals. = 1 5p. (b) The electrons lost by Sn when it forms the Sn 2+ cation come from the subshell(s) Sn has a Kr core of electrons plus the 5s 2 4d 10 5p 2 valence configuration. On ionization, the two 5p electrons are lost. (c) A Rb atom is in an excited state with the electron configuration [Kr] 9d 1. The effective nuclear charge seen by the 9d electron is closest to (circle one) 3e 2e 1e 0 1e 2e 3e The 9d orbital holds the electron far from the inner core of electrons. These core electrons shield all but one proton very well, making the effective nuclear charge close to the charge on a proton, which is just e, the elementary charge. (Electrons have the charge e.)

3 (d) The first four ionization energies of a particular element are IE 1 = J, IE 2 = J, IE 3 = J, and IE 4 = J. The element is F Ar Al C Na Li Ca The ionization energies are, in turn, one small amount, two larger but close to the same, and one last one quite large. This trend reveals the electron configuration shell sequence: the first electron to go is the Al 3p electron, then come two 3s electrons, and finally a 2p electron, requiring quite a large amount of energy due to the change in principal quantum number from 3 to 2. None of the other choices has the same shell sequence through the first four ionization steps. (e) Two elements in the range between Mg and Ar form stable anions that are spherical. These elements are: (circle them) Mg Al Si P S Cl Ar Si is isoelectronic to P and thus has a half-filled p subshell outside a closed core of electrons. Cl is isoelectronic to Ar with closed subshells. (In addition, both Al 2 and S 2 would be spherical, and if you circled these, you were not penalized as long as you also circled Si and Cl.) (f) The Cr 4+ ion in the compound CrO 2 is exploited in magnetic audio and video recording tape. The electron configuration of Cr 4+ is [Ar] 3d 2. The question here is simple: which four electrons does Cr with the [Ar] 4s 1 3d 5 configuration lose? There are two obvious possibilities for Cr 4+ that make sense following Hund s rules: [Ar] 4s 2 and [Ar] 3d 2. In the first, however, the two electrons would be spin-paired and the ion would be diamagnetic, a lousy choice for a component of a magnetic tape! In the other, however, the two electrons are not spin paired, and the ion is paramagnetic. (g) The smallest of the following atoms or ions is (circle one) Sc 3+ Ti K + S 2 Fe 2+ Zn Ar Highly positively charged ions that have lost all their valence electrons will be smaller than isoelectronic neutrals or isoelectronic positive ions with smaller nuclear charge. 4. The Li atom in its ground state can absorb many different photons, producing many different excited states. This question is concerned with the photon of longest wavelength that Li can absorb. This photon has an energy of J.

4 (a) What is the electron configuration of the Li atom produced when ground state Li absorbs this photon? Just as in the Na atom you studied in lab and we discussed in class, the longest wavelength excitation (smallest energy excitation) promotes the s valence electron to the p orbital of the same principal quantum number. For Li, this means the 1s 2 2s 1 ground state becomes the 1s 2 2p 1 excited configuration. (b) The first ionization energy of Li is J. Draw an energy level diagram that locates the Li ground state, the excited state, and the state corresponding to Li + + e below. Label each of your energy levels. The ionization limit is at the zero of energy, and the ground state is J lower (i.e., at J). The excited state is above the ground state by an amount equal to the photon energy, placing it at J J = J. The diagram below summarizes these energies. 0 Li + + e 2 Energy/10 19 J Excited State Ground State (c) Calculate the effective nuclear charge felt by the valence electron in the excited state. We note that the excited state valence electron has n = 2, and we equate the energy of this state, J, to the expression for the energy in terms of the effective nuclear charge: E n,l = J = ( J) Z 2 eff, n, l (4) ( Z eff, n, l = 19 J) = J 5. (a) Which of the following factors appears in the complete wavefunction for an H atom in the 2p z state? Circle the correct choice and explain your answer. cos θ sin θ sin φ sin θ sin φ sin θ

5 The 2p z wavefunction has a nodal plane (i.e., is zero) everywhere in the xy plane. This is the plane for which the spherical polar angle θ = 90 for any value of the spherical polar angle φ. When θ = 90, cos θ = 0. (b) A particle in a box of length l is in a state in which the probability of finding the particle in the region 0 x L/4 is equal to the probability of finding the particle in the region L/2 x 3L/4. The quantum number for the particle could be (circle all that are possible): First, n = 0 is not an allowed quantum number value for the particle in a box system. Next, we look for a wavefunction with the correct symmetry to satisfy the specified criterion. This symmetry shows up for any state with an even quantum number. (c) An electron is in a one-dimensional box of length 10.0 Å in the quantum state n = 2. What is the electron s de Broglie wavelength? Embarrassingly easy way to find the answer: for n = 2, the wavefunction is a single, full wave spanning the full length of the box. In other words, the wavelength of the wavefunction equals the box length. But the wavelength of the wavefunction is the de Broglie wavelength! So λ = 10.0 Å! Equally good, straightforward way to find the answer: equate the energy of the state to the expression for kinetic energy in terms of momentum, express the momentum in terms of the de Broglie wavelength, and solve for the de Broglie wavelength: E = h2 n 2 8m e L 2 = p2 = h/λ 2 so λ = 2m e 2m 2L e n = L = 10.0 Å