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CHAPTER 8 Atomic Electron Configurations and Chemical Periodicity Road Map - Lesson 22 Recap of Ch 7 problems Area of Focus: Chapter 8 The Exclusion Principle – introduction of the 4th q.n. Electron Spin Subshell filling order Atomic electron configuration Kull Spring07 Lesson 22 Ch 7/8 1 Equations speed of light = wavelength x frequency c = λ X = 3.00 x 108 m/s E = nh = nh(c/) n= positive integer Planck’s constant(h) = 6.626 x 10–34 J s Eatom = Eemitted (or absorbed) radiation = nh Rydberg equation =R n2 > n1 R = 1.096776 x 107 m-1 ΔE = Efinal – Einitial = –2.18 x 10–18 J Ephoton = Estate A – Estate B = hν Kull Spring07 Lesson 22 Ch 7/8 2 Quantum mechanics Quantum number n– shell/level ℓ - subshell/ sublevel mℓ - orbital ms - spin Values Total number 1,2,3, …∞ 0, 1,… n-1 - ℓ, 0, + ℓ n=# subshells n-1 +½, n2 = # orbitals in a shell 2 ℓ +1 (orbitals in a subshell) -½ 2 possible Kull Spring07 Lesson 22 Ch 7/8 3 Practice Problem 22-1 According to the quantum mechanical treatment of the hydrogen atom, which set of quantum numbers is not allowed? A) n = 3, ℓ = 2, mℓ = 0 B) n = 3, ℓ = 0, mℓ = 0 C) n = 3, ℓ = 1, mℓ = 1 D) n = 3, ℓ = 1, mℓ = -1 E) n = 3, ℓ = 1, mℓ = 2 Kull Spring07 Lesson 22 Ch 7/8 4 Practice Problem 22-1 Answer According to the quantum mechanical treatment of the hydrogen atom, which set of quantum numbers is not allowed? A) n = 3, ℓ = 2, mℓ = 0 B) n = 3, ℓ = 0, mℓ = 0 C) n = 3, ℓ = 1, mℓ = 1 D) n = 3, ℓ = 1, mℓ = -1 E) n = 3, ℓ = 1, mℓ = 2 Kull Spring07 Lesson 22 Ch 7/8 5 Practice Problem 22-2 For an excited hydrogen atom with the quantum number n equal to 9, which of the following statements is true? A) The energy of the atom is less than the energy for the state in which n is equal to 8. B) If ℓ = 0, there are nine possible values for the magnetic quantum number mℓ. C) The smallest value of the magnetic quantum number mℓ is -9. D) The electron must be in one of the p orbitals. E) The angular momentum quantum number ℓ can have any of the values 0, 1, 2, 3, 4, 5, 6, 7, 8. Kull Spring07 Lesson 22 Ch 7/8 6 Practice Problem 22-2 Answer For an excited hydrogen atom with the quantum number n equal to 9, which of the following statements is true? A) The energy of the atom is less than the energy for the state in which n is equal to 8. B) If ℓ = 0, there are nine possible values for the magnetic quantum number mℓ. C) The smallest value of the magnetic quantum number mℓ is -9. D) The electron must be in one of the p orbitals. E) The angular momentum quantum number ℓ can have any of the values 0, 1, 2, 3, 4, 5, 6, 7, 8. Kull Spring07 Lesson 22 Ch 7/8 7 Practice Problem 22-3 According to the quantum-mechanical model, how many orbitals in a given atom have n = 3? A) 4 B) 7 C) 9 D) 10 E) 18 Kull Spring07 Lesson 22 Ch 7/8 8 Practice Problem 22-3 Answer According to the quantum-mechanical model, how many orbitals in a given atom have n = 3? A) 4 B) 7 C) 9 D) 10 E) 18 Kull Spring07 Lesson 22 Ch 7/8 9 Chapter 8 The Exclusion 4th q.n. Principle – introduction of the Hund’s rule Electron Spin Subshell filling order Atomic electron configuration Kull Spring07 Lesson 22 Ch 7/8 10 Chapter 8 Pauli Exclusion Principle - No two electrons in the same atom can have the same set of 4 quantum numbers. each electron has a unique address. HUND’S RULE - When placing electrons in a set of orbitals having the same energy, we place them singly as long as possible. aufbau principle is the building up process that results in ground-state electron configurations Kull Spring07 Lesson 22 Ch 7/8 11 Electron Configurations and the Periodic Table Active Figure 8.7 Kull Spring07 Lesson 22 Ch 7/8 12 Electron Filling Order Figure 8.5 Kull Spring07 Lesson 22 Ch 7/8 13 Quantum mechanics Quantum number n– shell/level ℓ - subshell/ sublevel mℓ - orbital ms - spin Values Total number 1,2,3, …∞ 0, 1,… n-1 - ℓ, 0, + ℓ n=# subshells n-1 +½, n2 = # orbitals in a shell 2 ℓ +1 (orbitals in a subshell) -½ 2 possible Kull Spring07 Lesson 22 Ch 7/8 14 Writing Atomic Electron Configurations Two ways of writing configs. 1) the spdf notation. 2) the orbital box notation. spdf notation for H, atomic number = 1 1 1s value of n no. of electrons value of l ORBITAL BOX NOTATION for He, atomic number = 2 Arrows 2 depict electron spin 1s 1s One electron has n = 1, l = 0, ml = 0, ms = + 1/2 Other electron has n = 1, l = 0, ml = 0, ms = - 1/2 Kull Spring07 Lesson 22 Ch 7/8 15 Phosphorus Yellow P Group 5A Atomic number = 15 1s2 2s2 2p6 3s2 3p3 [Ne] 3s2 3p3 All Group 5A elements have [core ] ns2 np3 configurations where n is the period number. Red P 3p 3s 2p 2s 1s Kull Spring07 Lesson 22 Ch 7/8 16 Kull Spring07 Lesson 22 Ch 7/8 17 PERIODIC TRENDS Kull Spring07 Lesson 22 Ch 7/8 18 General Periodic Trends Atomic and ionic size # electrons # shells – Larger orbitals, -electrons held less tightly Ionization energy Electron affinity energy involved when an atom gains an electron to form an anion. Effective nuclear charge When higher: Electrons held Kull Spring07 Lesson 22 Ch 7/8 more tightly 19 Trends in Atomic Size Radius (pm) 250 K 1st transition series 3rd period 200 Na 2nd period Li 150 Kr 100 Ar Ne 50 He 0 0 5 10 15 20 25 30 35 40 Atomic Number Kull Spring07 Lesson 22 Ch 7/8 20 Effective Nuclear Charge, Z* Z* is the nuclear charge experienced by the outermost electrons. Z* increases across a period owing to incomplete shielding by inner electrons. The 2s electron PENETRATES the region occupied by the 1s electron. 2s electron experiences a higher positive charge than expected. Estimate Z* by --> [ Z - (# inner electrons) ] Charge felt by 2s e- in Li Z* = 3 - 2 = 1 Be Z* = 4 - 2 = 2 B Z* = 5 - 2 = 3 and so on! Kull Spring07 Lesson 22 Ch 7/8 21 Ion Configurations To form cations, always remove electrons of highest n value first! P [Ne] 3s2 3p3 - 3e- ---> P3+ [Ne] 3s2 3p0 3p 3p 3s 3s 2p 2p 2s 2s 1s 1s Kull Spring07 Lesson 22 Ch 7/8 22 Ion Sizes + Li,152 pm 3e and 3p Li + , 78 pm 2e and 3 p Forming a cation. CATIONS are SMALLER than the atoms from which they come. The electron/proton attraction has gone UP and so size DECREASES. Kull Spring07 Lesson 22 Ch 7/8 23 Ion Sizes F, 71 pm 9e and 9p F- , 133 pm 10 e and 9 p Forming an anion. ANIONS are LARGER than the atoms from which they come. The electron/proton attraction has gone DOWN and so size INCREASES. Trends in ion sizes are the same as atom sizes. Kull Spring07 Lesson 22 Ch 7/8 24 Trends in Ion Sizes Active Figure 8.15 Kull Spring07 Lesson 22 Ch 7/8 25 Ionization Energy IE = energy required to remove an electron from an atom in the gas phase. Mg (g) + 738 kJ ---> Mg+ (g) + e- Mg+ (g) + 1451 kJ ---> Mg2+ (g) + e- Mg+ has 12 protons and only 11 electrons. Therefore, IE for Mg+ > Mg Kull Spring07 Lesson 22 Ch 7/8 26