4.1 Are there patterns in the molar volumes of the elements? This will be discussed in Week 9 lecture.

4.2 How do you interpret patterns in the boiling point data?

(a) Metals generally have higher boiling points than non-metals, with Group I metals (Li, Na, Kä) having the lowest boiling points. Group 8 elements (noble gases) and many of the transition metals are missing from the graph.

(b) Adding in the noble gas data will add another point at each major ådipπ in the curve. This would make the periodic patterns even more obvious.

4.3 How do you predict the properties of an undiscovered element? Weπll see some of this in Week 9 lecture. From Figure 4.2, we see that the boiling points in Group 4 elements are: 3200/?/2900K for Si/Ge/Sn. Assuming that Ge follows the periodic trend, it should have a boiling point between 3200 and 2900K, or about 3050K. (The CRC Handbook says that the boiling point is actually 3093 K.) From Figure 4.1, the molar volumes of the Group 4 elements are: ~12/?/17 cc/mol for Si/Ge/Sn. We expect Geπs molar volume to fall someplace between 12 and 17 cc/mole, and the CRC handbook says itπs ~13.57 cc/mol. A similar line of reasoning will help us predict that Ge has a boiling point of ~2600K and a molar volume of ~12cc/mol.

4.4 How do you interpret patterns in the ionization energy data?

(a) Ionization energy increases going from group 1 (alkali metals) to group 8 (noble gases) and then drops again at the next group 1 element. There are small deviations (zig-zags) in the pattern. We also see that ionization energy seems to decrease for a particular element in a group as the atomic number increases (so, the ionization energies of the noble gases decrease from He to Ne to Ar).

(b) A shell is populated starting with the first element in the row (group 1) until it is filled with electrons. This happens with the last element in the row (group 8). It appears that it is åharderπ to remove an e- from the shell if it is the last electron used to fill the shell (e.g., it is hard to remove an e- from a noble gas). It is easier to remove an e- from a group 1 element (e.g., Na has a much lower ionization energy than Ne).

4.5, 4.6a Is the visible spectrum affected by substances in the light beam? This is part of Week 9 lecture.

4.6b A solution of NiCl2 will åsubtractπ the violet portion of the spectrum. Remember, the color absorbed is complementary to the color transmitted (seen).

4.7 How do you interpret the results shown in Figure 4.6? This is similar to a topic in the Week 9 lecture. Figure (a) shows an emission experiment (Na atoms emit certain lines of light). Figure (b) shows an absorption experiment (Na atoms absorb certain lines of light from a white light source). The lines absorbed or emitted are at the same wavelength ≠ that is, they have the same energy. The energy change for emission is equal to the energy change for absorption.

4.8 Does light absorption explain the results in Investigate This 4.5?

(a) Some substances appear colored since they absorb some wavelengths of visible light (which is a mixture of all visible wavelengths). Light that passes through a colored solution and makes it to our eyes is unabsorbed or transmitted light. Light that doesnπt make it through the solution has been absorbed by the solution.

(b) The observations in 4.5 (absorption of certain wavelengths of light by the KMnO4 solution) are an example of an absorption spectrum. An absorption spectrum is a plot of absorption as a function of wavelength. Since wavelength determines color, the absorption spectrum could be plotted instead as absorption as a function of color.

4.9 Elemental identification by absorption spectroscopy. The major dark lines (the absorption lines) occur at 435 and 546 nm. In Figure 3.5, the only element with lines close to these wavelengths is mercury. However, the lines in mercury are at ~440 and 551 nm. There seems to be a 5 nm calibration error somewhere!

4.10, 4.11 How is light affected by passing through narrow openings?

(a) KDL cannot get this to work, but you should see a diffraction pattern when you look at the gap between your fingers.

(b) This will be demonstrated in Week 9 lecture. The single beam from the laser pointer is passed through the diffraction grating, and this creates an pattern of light spots around the central light spot. The light spots and dark regions are due to reinforcement and cancellation, respectively, of light waves (see p. 224).

4.12, 4.13 What are the characteristics of waves? This will be discussed in Week 9 lecture. Click on http://www.falstad.com/ripple/ to open a wave simulation applet. Set to Setup: Single Source; 1 Src, 1 Freq; Color Scheme 3. The waves before the opening (slit) are not shown, but you can envision that they would look like parallel lines. The wave coming out of the slit shows the familiar concentric rings of dark and light rings. All this is also shown in the figure at the bottom of p. 223.

4.15 Wave properties.

(a) Velocity = distance/time = (8.7 mm/cycle)*(1.87 cycles/s) = 16.mm/s

4.16, 4.17 How do water waves interact? This will be discussed in Week 9 lecture. Click on http://www.falstad.com/ripple/ to open a wave simulation applet. Set to Setup: Double Slit, Source: 1 Plane Src, 1 Freq. You see that the waves starting from each of the two slits look like the ones from the single slit in 4.12. These overlap to make a complex pattern of light and dark areas (reinforced) and gray areas (cancelled waves).

4.19 How are Investigate This 4.10 and 4.16 related?

(a) The light spots observed in 4.10 are due to waves reinforcing (adding together). The dark regions are caused by light waves canceling each other (subtraction of waves).

4.24, 4.25, 4.26 Are the temperature and color of a hot filament related?

The filament glows faintly orange-red (emits only limited wavelengths) at low voltage, and the lamp is fairly cool. At higher voltages, the filament glows bright white (emits all wavelengths) and the lamp is very hot. The emission intensity (amount of light produced) apparently is related to temperature. The smaller curve in Figure 4.14 corresponds to the faintly glowing filament since it has the lower temperature and since it shows emission occurring over only part of the visible spectrum (closer to the red end of the spectrum). The larger curve shows emission over the entire visible region (which is the case for white light).

4.27 Are emission and absorption of E by oscillators related?

(a) Since DE = Dn(hn) and Dn = n_final ≠ n_initial = 2-4 = -2, DE must be negative. The process is exothermic and E is emitted.

(b) Now, Dn = 4-2 = +2, DE must be positive. It is equal in magnitude but opposite in sign to the DE in (a).

(c) In (a), the process is emission as in the top part of Figure 4.6. In (b), the process is absorption. The same amount of energy is involved, but in (a) it is coming out of the system and in (b) it is going into the system.

4.29, 4.30 What is the effect of light on silver chloride?

Hereπs a photo of the results. Spots of white AgCl were put onto filter paper with various filters covering them; then we exposed to white light. We used red, green, blue, and colorless filters as indicated in the text instructions, but we also used a yellow filter. You can see that the white AgCl has sometimes been converted to a gray product (solid Ag metal). The colorless filter has no effect vs. no filter at all. The lightest spots are from the red and green filters. Conclusion: Not all colors of light reaching the silver chloride have the same effect. These results are not consistent with wave theory which says that any light (frequency) should cause the reaction (AgCl conversion to Ag) so long as there is enough of that light (it is intense enough). In this experiment, all the AgCl spots got the same intensity of light, but the amount of product differed depending on the color of the light getting through the filter to the AgCl.

4.31 How does photoelectric current depend on light intensity?

(a) Intensity just means the number of photons. In Figure 4.17, no current occurs until the light reaches a minimum frequency (minimum energy). The number of photons (intensity) apparently has no effect. It is the energy of the photon that matters.

(b) Since increasing the frequency means the energy of the photon is increasing, there must be excess energy associated with the photons above the minimum frequency. So, some of the photon energy is used to åejectπ the e-, and the excess energy is just transferred to the e-. The energy of the ejected e- increases as the frequency of the photon increases (this is shown in 4.18).

4.32 How is the darkening of AgCl related to photons?

(a) The different colors of light passing through the filters have different energies or frequencies. In our experiment red and green light were much less effective in causing the reaction than blue or yellow light. Photons of blue or yellow light apparently can be absorbed by the AgCl (are the årightπ energy), causing the conversion to silver metal.

(b) The wavelength of the light has to be at most

E=hc/wavelength

4 ¥ 10¹⁹ J = (6.6256 ¥ 10^-34 J^.s)(3 x 10⁸m/s)/wavelength

wavelength = ~500 nm (green light)

An energy of about 4 ¥ 10¹⁹ J corresponds to a wavelength of 500 nm. This wavelength of light is in the green region of the visible spectrum. So, we would expect that the green filter would show a very dark Ag spot, but that isnπt what was observed on the filter paper!

4.34 Wave and photon phenomena

This question requires that you identify the underlying physics of the phenomena (e.g., what causes a rainbow). Donπt worry if you couldnπt predict these.

(a) Light is diffracted by the grooves on the CD surface. Diffraction is a wave phenomenon, so wave theory explains the color.

(b) An e- in Ne is excited (e.g., by high voltage). When the e- loses energy, photons of light with a specific energy are released. So, photon theory.

(c) Photons of UV light are absorbed by molecules in the skin, causing the sunburn. So, photon theory.

(d) A rainbow is caused by diffraction of light by small water droplets. Diffraction is a wave phenomenon, so wave theory explains the color.

4.35 What property of atoms causes line emission (and absorption)?

This is part of Week 9 lecture (Plankπs basic quantum theory)

(a) if an e- can have any energy, then it could undergo an infinite number of energy changes. That is, it could emit or absorb light of any energy/wavelength. Its spectrum would be a continuous spectrum.

(b) If the e- were limited to only certain energies, then it could absorb/emit only certain energies of light as it moved between those energy levels. Its spectrum would be a line spectrum.

4.43 Probability in three dimension

A sphere would be a good 3D figure to represent the probability distribution in Figure 4.25(b). The figure itself is just a cross section of a spherical shape. You have to keep in mind that the sphere would not fully represent the distribution. That is, the probability of finding an e- around the nucleus is higher near the nucleus than at the åedgeπ of the sphere.

4.46 What is the effect of nuclear charge on the potential energy?

This will be discussed in Week 10 lecture. Coulombπs Law (see p.88) says that the attraction of charged particles is directly proportional to their individual charges. (Attraction also depends on how far apart the charges are, but that distance is fixed here.) The attraction would be larger for the 2+ nucleus and e-. He+ should be smaller than H. Both have one e-, but He+ has a +2 nuclear charge while H has only +1. The larger nuclear charge åpullsπ the e- toward it (has greater attraction for the e- than the +1 H nucleus has for its e-).

4.47 How can you interpret Figure 4.27?

(a) Potential energy is proportional to -1/R where R=wave radius (see p. 248). So, it is always negative and it drops as the e- gets closer to the nucleus. If this were the only energy in the atom system, the e- would be pulled into the nucleus (atom would åcollapseπ).

(b) The kinetic energy is proportional to +1/R², so as the e- moves closer to the nucleus, the kinetic energy increases. If this were the only energy in the atom system, the atom energy would become so large (unstable) that the atom would not form at all.

(c) The size of the e- wave is determined by the balance between potential energy of attraction of e- for the nucleus (negative) and the kinetic energy of the e- wave (always positive). The system is most stable (when total energy, E, is at a minimum) when the radius of the wave is 2R.

4.48 Ionization energy and nuclear charge

This will be discussed in Week 10 lecture. H and He⁺ each have one e-, but He⁺ has a greater nuclear charge (+2 compared to +1 on H). So, the e- in He⁺ feels a greater attraction toward the He⁺ nucleus. The energy required to overcome this attraction and remove the e- will be greater than for removing the e- from H.

4.49 What patterns do you find in atomic ionization energies?

This will be discussed in Week 10 lecture.

4.50 How do you interpret patterns in these ionization energy data?

This will be discussed in Week 10 lecture. It helps to remember that the plot if Figure 4.30 is upside down.

(a) Group 1 elements (alkali metals) are the most stable with respect to loss of an e-. Group 8 (noble gases) are the least stable. These groups are at opposite ends of the periodic table.

(b) Generally, the higher the nuclear charge the greater the magnitude of the ionization energy, as expected. There are some little deviations, however.

(c) This is essentially the same question as in (b). The attraction for the last e- (which is proportional to ionization energy) increases going across each section of the plot (from Group 1 to Group 8).

4.52 What are the patterns in atomic size?

This will be discussed in Week 10 lecture.

(a) Across a period, radii decrease, though radius increases slightly at the noble gases. Nuclear charge is increasing, and this means each e- feels a greater attraction to the nucleus, decreasing the radius (Coulombπs Law).

(b) Atomic radius increases going down a family. There are more åshellsπ of electrons, making the atom bigger.

(c) The effects seem to be at odds, so you have to consider both Coulombic attraction and number of e- shells.

(d) Figure 4.1 is a plot of molar volume as a function of atomic number. Atomic volume (proportional to molar volume) is a function of atomic radius. Elements with high molar volume (alkali metals) have large radii.

4.53 What are the patterns in electronegativity?

This will be discussed in Week 10 lecture.

(a) Electronegativity increases going across the table and decreases going down a group.

(b) Electronegativity is a measure of the attraction that an atom has for its bonding electrons. The bonding electrons are in the outermost åshellπ (inner shell electrons are the core electrons). If electrons are arranged in shells of increasing energy, we would expect the attraction of the atom for its bonding electrons should decrease as the shells are further and further away from the nucleus.

(c) These graphs all show relationships that depend on the distance an e- is from the nucleus. Ionization energy and electronegativity increase as the e- is closer to the nucleus while atom size decreases as the e- is closer to the nucleus.

4.54 What are some further properties of electron shells in atoms?

(a) Six shells (corresponding to six rows of elements) are shown in these figures. These contain 2, 8, 8, 18, 18, 32 e-, respectively.

(b) The shells begin in the center and each new one begins with the light blue wedge containing the symbols of the Group 1 elements. So, each shell is a different concentric ring of the spiral. The transition metal and lanthanide/actinide subshells (d and f subshells, respectively) are represented as åspokesπ on the spiral. The s and p subshells are not distinguished at all in this kind of table.

4.55 Number of electron subshells in each shell In period 2 (Li to Ne), the 8 e- subshells (s and p) build the elements. In period 3, the same is true. In period 4 (K to Kr), the 8 e- subshells and the 10 e- subshell (d subshell) build the elements. Period 5 is the same as period 4. In period 6, the 8 e- subshells, the 10 e- subshell, and the 14 e- subshell (f subshell) build the elements. The spiral periodic table shows the åinvolvmentπ of the 10 e- and 14 e- subshells as separate spokes on the spiral.

4.56 How can successive ionization energies be interpreted? This will be discussed in Week 11 lecture.

(a) Table 4.2 shows that successive ionization energies always increase. Li is the first element with both valence (1) and core e- (2). You can see that the first ionization energy (for the loss of the valence e-) is relative small compared to the second ionization energy (for the loss of the first core e-) ≠ 520 vs. 7300 kJ/mol.

(b) The ionization energies for all e- in 8 subshells are shown for the group 8 elements (Ne and Ar). Group 8 elements are those that have filled 8e- shells.

(c) We will look at this plot in Week 11 lecutre.

4.57 Electron spin pairing in second period elements This will be discussed in Week 11 lecture.

4.58 How are orbital nodes described?

(a) The boundary surfaces for both orbitals are spherical in shape. The 2s orbital is larger than the 1s.

(b) Nodes are the åplacesπ in a wave where wave amplitude is zero, so in the sketches of Figure 4.36, the nodes are where the dot density is zero. A 2s orbital has a spherical nodal surface. A 2p orbital has a planar nodal surface centered between the lobes of the orbital.

4.59 How are principal quantum numbers, nodes, and orbitals related?

(a) The number of nodes is always 1 less than the principal quantum number (e.g., when n=2, there is 1 node in the orbital as shown in Figure 4.36). The 3s orbital will be similar to the 2s orbital shown in 4.36 except that it will have an added concentric ring (so, the orbital is larger and has 2 nodes).

(b) There is lots of evidence for d orbitals in the various periodic tables. Perhaps the most important is that there are 10 elements in the transition metal block (headed by Sc through Zn). There must be 5 equivalent orbitals that are being filled with e-, and these are the d orbitals.

(c) When n =1, there is 1 type of orbital (s orbital). When n=2, there are 2 types of orbitals (s and p) and a total of four orbitals (one s and three p). When n=3, there are 3 types of orbitals (s, p, d) and a total of nine orbitals (one s, three p, and five d). The relationship is number of orbitals = n². So, for n=4, the number of orbitals should be n⁴ = 16. These would be one s, three p, five d, and seven f orbitals.

4.61 The e- configuration for B

There are 5 e- in boron, so the configuration is 1s²2s²2p¹.

4.62 How are e- configurations related to ionization energies and e- spin?

(a) This will be discussed in Week 11 lecture.

		Ionization E (kJ/mol)
Na	[Ne]^103s^1	500
Mg	[Ne]^103s^2	738
Al	[Ne]^103s^23p^1	577
Si	[Ne]^103s^23p^13p^1	787
P	[Ne]^103s^23p^13p^13p^1	1012
S	[Ne]^103s^23p^23p^13p^1	1000
Cl	[Ne]^103s^23p^23p^23p^1	1251
Ar	[Ne]^103s^23p^23p^23p^2	1520

Nuclear charge increases going from Na to Ar, so the attraction of the nucleus for its outermost e- should increase. This explains the general increase in IE from Na to Ar.

The energy levels from Figure 4.37 increase in the order 3s<3p<4s. Removal of the first e- from the 3s orbital occurs at Na and removal of the first 3p e- occurs at Al. The IE_Na < IE_Al, as expected.

Half-filled subshells are particularly stable because of spin pairing energy (see p 267) which explains why IE_P > IE_S.

Filled subshells and shells are also stable which explains the relatively high IE for both Ar and Mg.

(b) This will be discussed in Week 11 lecture

		IE	Ionization Energy (kJ/mol)
Ne	[He]^22s^22p^22p^22p^2	1	2080
Ne^+	[He]^22s^22p^22p^22p^1	2	3950
Ne^2+	[He]^22s^22p^22p^12p^1	3	6120
Ne^3+	[He]^22s^22p^12p^12p^1	4	9370
Ne^4+	[He]^22s^22p^12p^1	5	12200
Ne^5+	[He]^22s^22p^1	6	15000
Ne^6+	[He]^22s^2	7	20000
Ne^7+