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Chemistry 1000 (Fall 2011) Problem Set #2: Orbitals and Electrons Solutions Answers to Questions in Olmsted (only those w/out answers at the back of the book) 93. In the real periodic table, screening effects only compete with the effect of increasing quantum number when n > 2 (i.e. with d and f orbitals). As such, the first two rows of the periodic table would not change if the effect of increasing n was always more significant than the effect of screening. We would, however, see the 3d subshell filled before the 4s subshell (and the 4d before 5s, 5d before 6s, etc.) we would see the 4f subshell filled before the 5s (and the 5f before 6s). In effect, the order in which electrons would fill the subshells would be: 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, 5s, 5p, 5d, 5f, 5g(!), 6s, 6p, etc. (a) The periodic table would therefore look something like: H Li Na K Rb He Be Mg Ca Sr (b) (c) (d) 97. (a) (b) (c) B Al Ga In C Si Ge Sn N P As Sb O S Se Te F Cl Br I Ne Ar Kr Xe Sc Y La Ti Zr Hf V Nb Ta Cr Mo W Mn Tc Re Fe Ru Os Co Rh Ir Ni Pd Pt Cu Ag Au Zn Cd Hg lanthanides actinides g-block!!!! This is, of course, assuming that the name of each element is the same based on the identity and number of electrons in the last subshell. In the crazy world described by this question, who knows if that would be the case!!! Note that it is due to the competition between the effects of increasing quantum number and screening that we have the periodic table we do (without a g-block!) If you used atomic numbers for your periodic table (or our current elements in order of atomic number), that’s fine too. It’s the overall shape change that’s important. Given the table above, the elements we’d expect to behave as “halogens” (adding an electron easily) would be those one electron away from a complete valence shell: H, F, Cu, the second last lanthanide and the second last element in the period 5 g-block. Given the table above, the elements we’d expect to behave as “noble gases” (chemically unreactive) would be those with a complete valence shell: He, Ne, Zn, the last lanthanide and the last element in the period 5 g-block. Any opinion with a reasoned rationale is acceptable. Morspin element 18 would be expected to have a larger first ionization energy than Morspin element 30. They are both in the same Group and therefore have analogous valence electron configurations (3s3 for Morspin element 18 and 4s3 for Morspin element 30 ); however, the valence electrons in Morspin element 18 have a lower n quantum number which means that they are closer to the nucleus and therefore more strongly attracted to it. Thus, a removal of a valence electron from Morspin element 18 will require more energy. Morspin element 15 would be expected to have a larger radius than Morspin cation 172+. They both have the same electron configuration (since they both have 15 electrons); however, Morspin cation 172+ has two more protons. As such, the electrons in Morspin cation 172+ are more strongly attracted to the nucleus and its radius is therefore smaller. Morspin element 47 would be expected to have a more negative enthalpy of electronic attraction (i.e. “larger electron affinity”) than Morspin element 48. The valence electron configuration for Morspin element 47 is 4s3 4p2 while the valence electron configuration for Morspin element 48 is 4s3 4p3. Since the quantum number laws *other* than that for electron spin appear to be the same on Morspin, it is reasonable to conclude that there are still three p orbitals in each subshell (as evidenced by there being nine Groups in the Morspin p-block). Adding one electron to the third (empty) 4p orbital in Morspin element 47 should therefore release more heat than adding a second electron into one of the three (partially filled) 4p orbitals in Morspin element 48. This is because there will be enough electrostatic repulsion between the two electrons in the same p-orbital to counteract the fact that Morspin element 48 has one more proton than Morspin element 47 (and therefore a greater nuclear charge). This effect can be observed on the real periodic table (Earth version). Nitrogen (analogous to Morspin element 48) is considered to have no electron affinity (enthalpy of electronic attraction = 0) whereas carbon does release heat when adding an electron to make C-. 99. (a) (b) The first three noble gases on Morspin would be elements 3, 15 and 27 (each having the same number of electrons as the atomic number). Morspin element 4: valence electron in 2s therefore: 2s n = 2, l = 0, ml = 0, ms = 0 ms = +½ or -½ would also be acceptable instead of ms = 0 Morspin element 7: valence electrons in 2s (3 e-) and 2p (1 e-) therefore: 2s n = 2, l = 0, ml = 0, ms = 0 n = 2, l = 0, ml = 0, ms = +½ n = 2, l = 0, ml = 0, ms = -½ 2p n = 2, l = 1, ml = 0, ms = 0 ml = +1 or -1 would also be acceptable instead of ml = 0 ms = +½ or -½ would also be acceptable instead of ms = 0 Morspin element 32: valence electrons in 4s (3 e-) and 3d (2 e-) therefore: 4s n = 4, l = 0, ml = 0, ms = 0 n = 4, l = 0, ml = 0, ms = +½ n = 4, l = 0, ml = 0, ms = -½ 3d n = 3, l = 2, ml = 0, ms = 0 n = 3, l = 2, ml = +1, ms = 0 values for ml may be +1, 0 or -1; however, the two *must* be different! values for ms may be +½, 0 or -½; however, the two *must* be the same! Additional Practice Problems 1. Give the orbital label (1s, 2p, 6f, etc.) for one orbital that corresponds to each set of quantum numbers. Also, sketch the orbital using Cartesian axes to show its orientation. Note that you can rotate the axes if it makes your picture clearer/easier to draw. See your notes/text for pictures of the different types of orbitals. (a) n = 4, l = 2, ml = -2 4dxy, 4dxz, 4dyz, 4dx2-y2, or 4dz2 (c) n = 3, l = 1, ml = +1 3px, 3py, or 3pz (b) n = 6, l = 0, ml = 0 6s (d) n = 2, l = 1, ml = 0 2px, 2py, or 2pz 2. Sketch the full set of 3d orbitals and label them with their Cartesian labels. The orbitals are 3dxy, 3dxz, 3dyz, 3dx2-y2, and 3dz2. See your notes/text for pictures. 3. What is the maximum number of electrons (in a single atom) that can be associated with each of the following combinations of quantum numbers? (a) n=2 8 (b) n = 3, l = 2 10 2 (c) n = 2, l = 0, ml = 0 (d) n = 6, l = 3, ml = -3, m s = -½ 1 4. Does each of the following set of quantum numbers describe a possible atomic orbital? If so, give the label for this orbital. If not, explain why an electron with that set of quantum numbers isn’t possible. (a) n = 4, l = 3, ml = -3, m s = ½ YES 4f (b) n = 1, l = 0, ml = 0, m s = -1 NO m s must be either +½ or -½ (c) n = 0, l = 0, ml = 0, m s = -½ NO n cannot be zero (d) n = 2, l = 3, ml = 0, m s = ½ NO l cannot be greater than n (no 2f orbital exists) (e) n = 3, l = -2, ml = 3, m s = -½ NO l cannot be negative and, since ml cannot be greater than l and l cannot be greater than or equal to n, ml cannot be equal to n 5. (a) Explain why each of the following statements is true. Hydrogen has a more negative enthalpy of electronic attraction than helium. H-1 = 1s2 He = 1s2 He-1 = 1s22s1 Electron configurations: H = 1s1 Enthalpy of electronic attraction is the enthalpy change when a neutral gaseous atom gains an electron. When gaining an electron, hydrogen goes from having an incomplete valence shell (1s1) to a full valence shell (1s2). Thus, H-1 is relatively stable and can form. If helium were to gain an electron, it would go from having a full valence shell (1s2) to having an incomplete valence shell (1s22s1). Thus, He-1 does not form because it is much less stable than He. Since H-1 can form and He-1 cannot, hydrogen has a more negative enthalpy of electronic attraction than helium (which is one of the few elements to have a positive enthalpy of electronic attraction). (b) (c) (d) 6. (a) (b) (c) 7. (a) Potassium has a greater second ionization energy than calcium. Second ionization energy is the energy required to remove an electron from X+1 where X is the element of interest. For calcium, this is the energy required to form Ca2+ from Ca+. Ca2+ has a complete valence shell whereas Ca+ does not. As such, Ca2+ is relatively stable and forms readily from Ca+. For potassium, this is the energy required to form K2+ from K+. K+ has a complete valence shell whereas K2+ does not. As such, K+ is more stable and very difficult to convert to K2+. The atomic radius of Br -1 is larger than the atomic radius of Br. Br and Br -1 have the same number of protons, but Br -1 has one more electron. This extra electron partially shields the other electrons in the valence shell from the positive charge of the nucleus. As such, the effective nuclear charge felt by the valence electrons of Br -1 is smaller than that felt by the valence electrons of Br, and they are not pulled as strongly toward the nucleus (increasing the atomic radius). The atomic radius of nitrogen is smaller than the atomic radius of boron. Nitrogen has two more protons in its nucleus than boron does. It also has two more electrons, but these do not completely shield the valence electrons from the increased positive charge of the nucleus. As such, the effective nuclear charge felt by the valence electrons of nitrogen is higher than that felt by the valence electrons of boron. The valence electrons of nitrogen are therefore pulled more strongly toward the nucleus (decreasing the atomic radius). Rank the following sets of atoms/ions according to the number of valence electrons (from least to most). Sn2+ (2) Sb (5) Se (6) S2- (8) (i) S2-, Sb, Se, Sn2+ (ii) F-, Fe2+, Fe3+, Fr Fr (1) Fe3+ (5) Fe2+ (6) F- (8) (iii) C, Ca, Cl, Co Ca (2) C (4) Cl (7) Co (9) List all of the atoms/ions from part (a) that are paramagnetic. Se [Ar]4s23d104p4 Sb [Kr]5s24d105p3 Fr [Rn]7s1 Fe3+ [Ar]3d5 Fe2+ [Ar]3d6 C [He]2s22p2 Cl [Ne]3s23p5 Co [Ar]4s23d7 List all of the atoms/ions from part (a) that are diamagnetic. Sn2+ [Kr]5s24d10 S2[Ne]3s23p6 F[He]2s22p6 Ca [Ar]4s2 Consider H and He+ in the ground state. Which would you expect to have a higher ionization energy? He2+ (Both are one-electron atoms/ions, but He2+ has two protons in the nucleus so the electron should be more strongly attracted to the nucleus therefore more difficult to remove.) (b) Calculate the ionization energy for H. First ionization energy is the energy required to remove an electron from an atom (far enough away that it no longer feels any attraction to the nucleus. Mathematically, we can approximate this as the energy required to excite an electron from n = 1 to n = ∞ (where ∞ stands for infinity). Recall that 1/0 = ∞ therefore 1/∞ = 0. For a hydrogen atom, Z = 1 (1 proton in the nucleus), n1 = 1 and n2 = ∞ I1 ( H ) = En =∞ − En =1 ⎛ ⎛ 1 Z2 ⎞ ⎛ Z2 ⎞ 1 ⎞ 1⎞ ⎛ 1 I1 ( H ) = ⎜⎜ − RH 2 ⎟⎟ − ⎜⎜ − RH 2 ⎟⎟ = − RH ⋅ Z 2 ⎜⎜ 2 − 2 ⎟⎟ = ( −2.179 × 10 −18 J )(1) 2 ⎜ 2 − 2 ⎟ = 2.179 × 10 −18 J n2 ⎠ ⎝ n1 ⎠ n1 ⎠ ⎝∞ 1 ⎠ ⎝ ⎝ n2 (c) Calculate the ionization energy for He+. For a helium cation, Z = 2 (2 protons in the nucleus), n1 = 1 and n2 = ∞ I 1 ( He) = E n =∞ − E n =1 ⎛ ⎛ 1 1 ⎞ 1⎞ Z2 ⎞ Z2 ⎞ ⎛ ⎛ 1 I 1 ( He) = ⎜⎜ − RH 2 ⎟⎟ − ⎜⎜ − RH 2 ⎟⎟ = − RH ⋅ Z 2 ⎜⎜ 2 − 2 ⎟⎟ = ( −2.179 × 10 −18 J )(2) 2 ⎜ 2 − 2 ⎟ = 8.716 × 10 −18 J n1 ⎠ n1 ⎠ n2 ⎠ ⎝ ⎝∞ 1 ⎠ ⎝ ⎝ n2 (d) Did your calculations in parts (b) and (c) support your answer to part (a)? Yes. The ionization energy for He+ was four times higher than that for H. 8. Give the name and symbol for the neutral element matching each of the following descriptions. “I have 8 neutrons and 6 valence electrons.” oxygen (O) 6 valence electrons = group 6 or 16; the only element in either group likely to have only 8 neutrons (i.e. mass of about 2×8 = 16 g/mol) is oxygen “I have a full octet and belong to the same period as nickel.” krypton (Kr) full octet = noble gas (group 18); nickel is in period 4 “Half of my electrons are valence electrons.” beryllium (Be) If half the electrons are valence, then half are core. There are only a few possible values for the number of core electrons. Any element in period 1 has 0 core electrons; any element in period 2 has 2 core electrons (1s2); any element in period 3 has 10 core electrons (1s22s22p6); any element in period 4 has 18 core electrons (1s22s22p63s23p6); etc. Beryllium has 2 core electrons plus 2 valence electrons. Note that there is no other element that meets these criteria (calcium has 20 electrons, but 18 are core – not 10). “I have twice as many valence electrons as core electrons.” see logic for 6(c) carbon (C) 2 core electrons and 4 valence electrons. (a) (b) (c) (d) (e) “I have twice as many core electrons as valence electrons.” (bonus: name all three elements fitting this description) see logic for 6(c) lithium (Li) 2 core electrons + 1 valence electron phosphorus (P) 10 core electrons + 5 valence electrons cobalt (Co) 18 core electrons + 9 valence electrons ***note that xenon (Xe) has 8 valence electrons – not 18!*** 9. (a) For a neutral ground state nickel atom, Write the complete electron configuration. 1s2 2s2 2p6 3s2 3p6 4s2 3d8 Write the electron configuration using the noble gas abbreviation. [Ar] 4s2 3d8 Draw an orbital occupancy diagram showing the valence electrons. Label each subshell. (b) (c) 3d 4s (d) Use the table below to list a set of quantum numbers describing the valence electrons. Use as many rows as necessary; the correct answer may include one or more empty rows. Electron ms n l ml 1 (4s) 4 0 0 +½ 2 (4s) 4 0 0 –½ 3 (3d) 3 2 +2 +½ 4 (3d) 3 2 +1 +½ 5 (3d) 3 2 0 +½ 6 (3d) 3 2 -1 +½ 7 (3d) 3 2 -2 +½ 8 (3d) 3 2 +2 –½ 9 (3d) 3 2 +1 –½ 10 (3d) 3 2 0 –½ 11 12