Quantum Theory and the Electronic Structure of Atoms Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Schrodinger Wave Equation Y = fn(n, l, ml, ms) Existence (and energy) of electron in atom is described by its unique wave function Y. Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time 7.6

For a certain value of l , there are (2l +1) integral values of ml And For a certain value of n, there are n2 orbital and 2n2 electron. 7.6

Schrodinger Wave Equation Y = fn(n, l, ml, ms) Shell – electrons with the same value of n (2s and 2p) Subshell – electrons with the same values of n and l (2px , 2py and 2pz ) Orbital – electrons with the same values of n, l, and ml (2px) How many electrons can an orbital hold? If n, l, and ml are fixed, then ms = ½ or - ½ Y = (n, l, ml, ½) or Y = (n, l, ml, -½) An orbital can hold 2 electrons 7.6

How many 2p orbitals are there in an atom? If l = 1, then ml = -1, 0, or +1 2p 3 orbitals l = 1 How many electrons can be placed in the 3d subshell? n=3 If l = 2, then ml = -2, -1, 0, +1, or +2 3d 5 orbitals which can hold a total of 10 e- l = 2 7.6

Worked Example 7.6

Worked Example 7.7

( ) Energy of orbitals in a single electron atom 1 En = -RH n2 Energy only depends on principal quantum number n n=3 n=2 En = -RH ( ) 1 n2 n=1 7.7

Energy of orbitals in a multi-electron atom Energy depends on n and l n=3 l = 2 n=3 l = 1 n=3 l = 0 n=2 l = 1 n=2 l = 0 n=1 l = 0 7.7

“Fill up” electrons in lowest energy orbitals (Aufbau principle) ? ? Li 3 electrons Be 4 electrons C 6 electrons B 5 electrons B 1s22s22p1 Be 1s22s2 Li 1s22s1 H 1 electron He 2 electrons He 1s2 H 1s1 7.9

The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). Ne 10 electrons Ne 1s22s22p6 F 9 electrons O 8 electrons N 7 electrons C 6 electrons N 1s22s22p3 C 1s22s22p2 O 1s22s22p4 F 1s22s22p5 7.7

Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7

in the orbital or subshell Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. number of electrons in the orbital or subshell 1s1 principal quantum number n angular momentum quantum number l Orbital diagram 1s1 H 7.8

What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½ 7.8

Outermost subshell being filled with electrons 7.8

7.8

Paramagnetic Diamagnetic unpaired electrons all electrons paired 2p 2p 7.8

Worked Example 7.10

Worked Example 7.11

Chemistry Mystery: Discovery of Helium In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines Mystery element was named Helium In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay).

1. What is the wavelength (in nanometers) of light having a frequency of 8.6 x 1013 Hz?   A) 3.5 nm B) 3.5 x 103 nm C) 3.5 x 106 nm D) 2.9 x 105 nm 2. What is the frequency of light having a wavelength of 456 nm?   A) 1.37 x 102 Hz B) 6.58 x 105 Hz C) 6.58 x 1014 Hz D) 1.37 x 1014 Hz 3. A photon has a wavelength of 624 nm. Calculate the energy of the photon in joules.   A) 3.19 x 10–16 J B) 3.19 x 10–19 J C) 1.24 x 10–22 J D) 3.19 x 10–28 J

4. The blue color of the sky results from the scattering of sunlight by air molecules. The blue light has a frequency of about 7.5 x 1014 Hz. Calculate the energy, in joules, of a single photon associated with this frequency.   A) 2.6 x 10–31 J B) 2.6 x 10–22 J C) 5.0 x 10–19 J D) 5.0 x 10–16 J 5. A photon has a frequency of 6.0 x104 Hz. Calculate the energy (in joules) of 1 mole of photons all with this frequency.   A) 2.4 x 10–5 J/mol B) 4.0 x 10–10 J/mol C) 6.6 x 10–15 J/mol D) 4.0 x 10–20 J/mol 6. What is the wavelength, in nm, of radiation that has an energy content of 1.0 x 103 kJ/mol? In which region of the electromagnetic spectrum is this radiation found?   A) 1.2 x 10–1 nm, X-ray B) 2.0 x 101 nm, ultraviolet C) 1.2 x 102 nm, ultraviolet D) 2.0 x 103 nm, infrared

7. Some copper compounds emit green light when they are heated in a flame. How would you determine whether the light is of one wavelength or a mixture of two or more wavelengths?   A) Observe the emitted light with green tinted glasses. B) Pass the emitted light through a beaker of water. C) Pass the emitted light through a prism. D) Pass the emitted light through green tinted glasses. 8. Calculate the wavelength (in nanometers) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state.   A) 1.28 x 10–6 nm B) 1.46 x 10–6 nm C) 1.46 x 103 nm D) 1.28 x 103 nm 9. The first line of the Balmer series occurs at a wavelength of 656.3 nm. What is the energy difference between the two energy levels involved in the emission that results in this spectral line?   A) 3.367 x 10–36 J B) 3.027 x 10–28 J C) 1.299 x 10–22 J D) 3.027 x 10–19 J

10. Calculate the frequency (Hz) of the emitted photon when an electron drops from the n = 4 to the n = 2 level in a hydrogen atom.   A) 2.74 x 1014 Hz B) 6.17 x 1014 Hz C) 1.62 x 1015 Hz D) 3.65 x 1015 Hz 11. An electron in the hydrogen atom makes a transition from an energy state of principal quantum numbers ni to the n = 2 state. If the photon emitted has a wavelength of 434 nm, what is the value of ni?   A) 3 B) 4 C) 5 D) 6 12. An electron in a certain atom is in the n = 2 quantum level. List the possible values of l, and ml, that it can have.   A) l = 0, ml = 0; l = 1, ml = –1, 0, 1; l = 2; ml = –2, –1, 0, 1, 2 B) l = 0, ml = 0; l = 1, ml = –1, 0, 1 C) l = 0, ml = –1, 0, 1 D) l = 1, ml = –1, 0, 1

13. Give the values of the quantum numbers associated with the 2p subshell.   A) n = 2, l = 2, ml = –2, –1, 0, 1, 2 B) n = 2, l = 1, ml = 0 C) n = 2, l = 1, ml = 1 D) n = 2, l = 1, ml = –1, 0, 1 14. Calculate the total number of electrons that can occupy: (A) one s orbital, (B) three p orbitals, (C) five d orbitals, (D) seven f orbitals.   A) (A)2; (B)9; (C)10, (D)14 B) (A)2; (B)6; (C)8, (D)14 C) (A)2; (B)6; (C)10, (D)14 D) (A)2; (B)6; (C)10, (D)16 15. What is the total number of electrons that can be held in all orbitals having the same principal quantum number n?   A) 4n2 B) 2n2 C) 2n D) 2

16. Determine the maximum number of electrons that can be found in each of the following subshells: 3s, 3d, 4p, 4f, 5f.   A) 3s(2); 3d(8); 4p(6); 4f(14); 5f(14) B) 3s(2); 3d(10); 4p(6); 4f(14); 5f(16) C) D) 3s(2); 3d(10); 4p(6); 4f(14); 5f(14) 17. State the total number of: p electrons in N (Z = 7); s electrons in Si (Z = 14); and 3d electrons in S (Z = 16).   A) N, 3p electrons; Si, 6s electrons; S, 5d electrons B) N, 2p electrons; Si, 6s electrons; S, 5d electrons C) N, 3p electrons; Si, 6s electrons; S, 0d electrons D) N, 6p electrons; Si, 6s electrons; S, 0d electrons 18. Why do the 3s, 3p, and 3d orbitals have the same energy in a hydrogen atom but different energies in a many-electron atom?   A) Many-electron atoms have shielding from the lower orbitals. B) The many electrons excite each other to higher energies. C) The orbitals are shaped differently with many-electron atoms. 19. For each of the following pairs of hydrogen orbitals, indicate which is higher in energy: (A) 1s, 2s; (B) 2p, 3p; (C) 3dxy, 3dyz; (D) 3s, 3d; (E) 4f, 5s.   A) (A) 2s; (B) 3p; (C) equal; (D) 3d; (E) 5s B) (A) 2s; (B) 3p; (C) equal; (D) equal; (E) 5s C) (A) 2s; (B) 2p; (C) equal; (D) equal; (E) 5s D) (A) 2s; (B) 3p; (C) equal; (D) equal; (E) 4f

20. Indicate which of the following sets of quantum numbers in an atom are unacceptable: (A) (1, 0, ½, ½); (B) (3, 0, 0, +½); (C) (2, 2, 1, +½); (D) (4, 3, –2, +½); (E) (3, 2, 1, 1).   A) (A) and (E) are unacceptable. B) (B), (C) and (E) are unacceptable. C) (A), (B), (C) and (E) are unacceptable. D) (A), (C) and (E) are unacceptable. 21. Indicate the number of unpaired electrons present in each of the following atoms: B, Ne, P, Sc, Mn, Se.   A) B(1); Ne(0); P(3); Sc(1); Mn(5); Se(2) B) B(0); Ne(0); P(3); Sc(1); Mn(5); Se(2) C) B(1); Ne(0); P(2); Sc(2); Mn(5); Se(2) D) B(1); Ne(0); P(3); Sc(2); Mn(4); Se(2) 22. Use the Aufbau principle to obtain the ground-state electron configuration of selenium.   A) Se: [Ar]4s23d104p3 B) Se: [Ar]4s23d104p4 C) Se: [Ar]4s23d104p5 D) Se: [Ar]4s23d104p6

23. What is the maximum number of electrons in an atom that can have the following quantum numbers: (1) n = 2, ms = +½ ; (2) n = 4, ml = +1; (3) n = 3, l = 2; (4) n = 2, l = 0, ms = -½ ; (5) n = 4, l = 3, ml = -2.   A) (1)4; (2)5; (3)8; (4)2; (5)2 B) (1)4; (2)6; (3)8; (4)1; (5)2 C) (1)4; (2)6; (3)10; (4)1; (5)2 D) (1)4; (2)6; (3)10; (4)2; (5)2 24. The He+ ion contains only one electron and is therefore a hydrogen-like ion. Calculate the wavelengths of the first four transitions in the Balmer series of the He+ ion. (The Rydberg constant for He+ is 8.72 ´ 10-18 J.) Which of the following is not one of these transitions?   A) n = 3 to 2; l = 164 nm; UV B) n = 4 to 2; l = 121 nm; UV C) n = 5 to 2; l = 107 nm; UV D) n = 6 to 2; l = 103 nm; UV 25. The electron configurations described in this chapter all refer to gaseous atoms in their ground states. An atom may absorb a quantum of energy and promote one of its electrons to a higher-energy orbital. When this happens, we say that the atom is in an excited state. The electron configurations of some excited atoms are given. Identify the species. (A) 1s12s1 ; (B) 1s22s22p23d1 ; (C) 1s22s22p64s1   A) (A) He; (B) C; (C) Ne B) (A) He+; (B) N+; (C) Na+ C) (A) He; (B) N; (C) Na D) (A) He; (B) O; (C) Na

26. Scientists have found interstellar hydrogen atoms with quantum number n in the hundreds. Calculate the wavelength of light emitted when a hydrogen atom undergoes a transition from n = 236 to n = 235. In what region of the electromagnetic spectrum does this wavelength fall?   A) 0.596 nm, Xray B) 9.12 ´ 101 nm, Ultraviolet C) 5.96 ´ 108 nm, Microwave D) 9.12 ´ 1012 nm, Radiowave Answer Key 1- B 2- C 3- B 4- C 5- A 6- C 7- C 8- D 9-D 10- B 11- C 12- B 13- D 14- C 15- B 16- D 17- C 18- A 19- B 20- D 21- A 22- B 23- C 24- C 25- C 26- C

Problems 7.3 – 7.8 – 7.16 – 7.18 7.32 – 7.34 – 7.120 7.56 – 7.58 – 7.62 – 7.66 – 7.70 7.76 – 7.78 – 7.79 – 7.84 – 7.88 – 7.90 – 7.124