Kull Spring07 Lesson 23 Ch 8 1 CHAPTER 8 Atomic Electron Configurations and Chemical Periodicity Outline -Collect homework -Review -Trends -Ions

Kull Spring07 Lesson 23 Ch 8 2 Review  Spdf notation  Orbital box notation

Kull Spring07 Lesson 23 Ch 8 3 NameSymbolPermitted ValuesProperty principalnpositive integers(1,2,3,…)orbital energy (size) angular momentum l integers from 0 to n-1orbital shape (The l values 0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.) magnetic mlml integers from - l to 0 to + l orbital orientation spin msms +1/2 or -1/2direction of e - spin Characteristics of Many-Electron Atoms: The Electron-Spin Quantum Number

Kull Spring07 Lesson 23 Ch 8 4

5   Consider this set of quantum numbers: n = 3, ℓ = 2, m ℓ = -1, m s = +½ The maximum number of electrons in an atom which can share the above set of quantum numbers is A) 1 B) 14 C) 3D) 10 E) none of the above Practice Problem 23-1

Kull Spring07 Lesson 23 Ch 8 6   Consider this set of quantum numbers: n = 3, ℓ = 2, m ℓ = -1, m s = +½ The maximum number of electrons in an atom which can share the above set of quantum numbers is A) 1 B) 14 C) 3D) 10 E) none of the above Practice Problem 23-1 Answer

Kull Spring07 Lesson 23 Ch 8 7   An atom in its ground state contains 30 electrons. How many of these are in sublevels with ℓ = 2? A) 2 B) 4 C) 6 D) 8 E) 10 Practice Problem 23-2

Kull Spring07 Lesson 23 Ch 8 8   An atom in its ground state contains 30 electrons. How many of these are in sublevels with ℓ = 2? A) 2 B) 4 C) 6 D) 8 E) 10 Practice Problem 23-2 Answer

Kull Spring07 Lesson 23 Ch 8 9 What are the possible values for the angular momentum quantum number ℓ ? A) integers from -ℓ to 0 to +ℓ B) 1, 2, 3, etc. C) 2, 4, 6, etc. D) +½, -½ E) integers from 0 to n - 1 Practice Problem 23-3

Kull Spring07 Lesson 23 Ch 8 10   What are the possible values for the angular momentum quantum number (ℓ)? A) integers from -ℓ to 0 to +ℓ B) 1, 2, 3, etc. C) 2, 4, 6, etc. D) +½, -½ E) integers from 0 to n - 1 Practice Problem 23-3 Answer

Kull Spring07 Lesson 23 Ch 8 11   The electron configuration of the outermost electrons of atoms of the halogen group is: A) ns 2 np 7 B) ns 1 C) ns 2 np 5 D) ns 2 np 6 (n-1)d 7 E) ns 2 np 6 Practice Problem 23-4

Kull Spring07 Lesson 23 Ch 8 12   The electron configuration of the outermost electrons of atoms of the halogen group is: A) ns 2 np 7 B) ns 1 C) ns 2 np 5 D) ns 2 np 6 (n-1)d 7 E) ns 2 np 6 Practice Problem 23-4 Answer

Kull Spring07 Lesson 23 Ch 8 13   The electronic configuration of the element whose atomic number is 26 is: A) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 8 B) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 C) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 D) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 4p 2 E) none of the above Practice Problem 23-5

Kull Spring07 Lesson 23 Ch 8 14   The electronic configuration of the element whose atomic number is 26 is: A) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 8 B) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 C) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 D) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 4p 2 E) none of the above Practice Problem 23-5 Answer

Kull Spring07 Lesson 23 Ch 8 15   The set of quantum numbers that correctly describes an electron in a 3p orbital is A) n = 3; ℓ = 0; m ℓ = 0; m s = 0 B) n = 3; ℓ = 2; m ℓ = -2, -1, 0, 1, or 2; m s = +½ or -½ C) n = 3; ℓ = 1; m ℓ = -1, 0, or 1; m s = +½ or -½ D) n = 4; ℓ = 0; m ℓ = -1,0, or 1; m s = +½ or -½ E) none of the above Practice Problem 23-6

Kull Spring07 Lesson 23 Ch 8 16   The set of quantum numbers that correctly describes an electron in a 3p orbital is A) n = 3; ℓ = 0; m ℓ = 0; m s = 0 B) n = 3; ℓ = 2; m ℓ = -2, -1, 0, 1, or 2; m s = +½ or -½ C) n = 3; ℓ = 1; m ℓ = -1, 0, or 1; m s = +½ or -½ D) n = 4; ℓ = 0; m ℓ = -1,0, or 1; m s = +½ or -½ E) none of the above Practice Problem 23-6 Answer

Kull Spring07 Lesson 23 Ch 8 17   An atom in its ground state contains 18 electrons. How many of these are in orbitals with m ℓ = 0? A) 2 B) 4 C) 6 D) 8 E) 10 7 Practice Problem 23- 7

Kull Spring07 Lesson 23 Ch 8 18   An atom in its ground state contains 18 electrons. How many of these are in orbitals with m ℓ = 0? A) 2 B) 4 C) 6 D) 8 E) 10 Practice Problem 23-7 Answer

Kull Spring07 Lesson 23 Ch 8 19   The configuration for the six outer electrons in ground state oxygen atoms is A) 2s 3 2p 3 -1 B) 2p 6 C) 2s 2 2p p 0 2 D) 2s 2 2p p 0 1 2p 1 1 E) 2s 4 2p 2 -1 Practice Problem 23-8

Kull Spring07 Lesson 23 Ch 8 20 Practice Problem 23-8 Answer  The configuration for the six outer electrons in ground state oxygen atoms is A) 2s 3 2p 3 -1 B) 2p 6 C) 2s 2 2p p 0 2 D) 2s 2 2p p 0 1 2p 1 1 E) 2s 4 2p 2 -1

Kull Spring07 Lesson 23 Ch 8 21   Which of the following is the electron configuration for chromium, element 24? A) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 B) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 C) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 D) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 E) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 1 3d 1 Practice Problem 23-9

Kull Spring07 Lesson 23 Ch 8 22   Which of the following is the electron configuration for chromium, element 24? A) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 B) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 C) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 D) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 E) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 1 3d 1 Practice Problem 23-9 Answer

Kull Spring07 Lesson 23 Ch 8 23

Kull Spring07 Lesson 23 Ch 8 24

Kull Spring07 Lesson 23 Ch 8 25 PERIODIC TRENDS

Kull Spring07 Lesson 23 Ch 8 26 Trends in Some Key Periodic Atomic Properties: Trends in Electron Affinity Atomic and ionic size # electrons # shells – Larger orbitals, -electrons held less tightly Electron affinity: energy involved when an atom gains an electron to form an anion. Effective nuclear charge When higher: Electrons held more tightly

Kull Spring07 Lesson 23 Ch 8 27 Atomic radii of the main- group and transition elements. Trends in Some Key Periodic Atomic Properties: Trends in Atomic Size

Kull Spring07 Lesson 23 Ch 8 28 Factors Affecting Atomic Orbital Energies Additional electron in the same orbital An additional electron raises the orbital energy through electron-electron repulsions. Additional electrons in inner orbitals Inner electrons shield outer electrons more effectively than do electrons in the same sublevel. Higher nuclear charge lowers orbital energy (stabilizes the system) by increasing nucleus-electron attractions. The Effect of Nuclear Charge (Z effective ) The Effect of Electron Repulsions (Shielding) Characteristics of Many-Electron Atoms: Electrostatic Effects and the Splitting of Energy Levels

Kull Spring07 Lesson 23 Ch 8 29 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* = = 1  Be Z* = = 2  B Z* = = 3and so on!

Kull Spring07 Lesson 23 Ch 8 30 Ionization Energy IE = energy required to remove an electron from an atom in the gas phase. Mg (g) kJ ---> Mg + (g) + e- Mg+ (g) kJ ---> Mg2+ (g) + e- Mg+ has 12 protons and only 11 electrons. Therefore, IE for Mg+ > Mg

Kull Spring07 Lesson 23 Ch 8 31 Trends in Some Key Periodic Atomic Properties: Trends in Ionization Energy

Kull Spring07 Lesson 23 Ch 8 32 Ion Sizes  CATIONS are SMALLER than the atoms from which they come.  The electron/proton attraction has gone UP and so size DECREASES. Li,152 pm 3e and 3p Li +, 78 pm 2e and 3 p + Forming a cation.

Kull Spring07 Lesson 23 Ch 8 33 Ion Sizes  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. Forming an anion. F, 71 pm 9e and 9p F -, 133 pm 10 e and 9 p -

Kull Spring07 Lesson 23 Ch 8 34 Ion Configurations To form cations, always remove electrons of highest n value first! P [Ne] 3s 2 3p 3 - 3e- ---> P 3+ [Ne] 3s 2 3p 0

Kull Spring07 Lesson 23 Ch 8 35 Trends in Ion Sizes Active Figure 8.15

Kull Spring07 Lesson 23 Ch 8 36   Which of the following has the largest radius? A) F B) N C) C D) O E) Ne Practice Problem 23-10

Kull Spring07 Lesson 23 Ch 8 37   Which of the following has the largest radius? A) F B) N C) C D) O E) Ne Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 38   Which of the following elements has the largest ionization energy? A) Na B) Ne C) F D) K E) Rb Practice Problem 23-11

Kull Spring07 Lesson 23 Ch 8 39   Which of the following elements has the largest ionization energy? A) Na B) Ne C) F D) K E) Rb Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 40   Which of the following has the greatest electron affinity (most negative value)? A) Cl B) K C) He D) Na E) Rb Practice Problem 23-12

Kull Spring07 Lesson 23 Ch 8 41   Which of the following has the greatest electron affinity (most negative value)? A) Cl B) K C) He D) Na E) Rb Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 42   In the periodic table, the most nonmetallic elements will be found A) at the top of Group 3A (13) B) at the top of Group 1A (1) C) at the top of Group 7A (17) D) at the bottom of Group 1A (1) E) at the bottom of Group 7A (17) Practice Problem 23-13

Kull Spring07 Lesson 23 Ch 8 43   In the periodic table, the most nonmetallic elements will be found A) at the top of Group 3A (13) B) at the top of Group 1A (1) C) at the top of Group 7A (17) D) at the bottom of Group 1A (1) E) at the bottom of Group 7A (17) Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 44   Which ion has the smallest radius? A) Li + B) Na + C) K + D) Be 2+ E) Mg 2+ Practice Problem 23-14

Kull Spring07 Lesson 23 Ch 8 45   Which ion has the smallest radius? A) Li + B) Na + C) K + D) Be 2+ E) Mg 2+ Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 46   Which of the following statements about periodic properties is incorrect? A) Both electron affinity and ionization energy decrease down a group. B) Atomic size increases to the right across a period. C) Ionization energy increases to the right across a period. D) Atomic size increases down a group. E) Electron affinity increases to the right across a period. Practice Problem 23-15

Kull Spring07 Lesson 23 Ch 8 47   Which of the following statements about periodic properties is incorrect? A) Both electron affinity and ionization energy decrease down a group. B) Atomic size increases to the right across a period. C) Ionization energy increases to the right across a period. D) Atomic size increases down a group. E) Electron affinity increases to the right across a period. Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 48   Which of the following elements has the greatest ionization energy? A) Ga B) As C) K D) Pb E) Na Practice Problem 23.16

Kull Spring07 Lesson 23 Ch 8 49   Which of the following elements has the greatest ionization energy? A) Ga B) As C) K D) Pb E) Na Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 50   Which of the following elements has the greatest electron affinity? A) Rb B) Ca C) Li D) Na E) I Practice Problem 23.17

Kull Spring07 Lesson 23 Ch 8 51   Which of the following elements has the greatest electron affinity? A) Rb B) Ca C) Li D) Na E) I Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 52   Which of the following ions and atoms has the largest radius? A) Mg B) Na C) Na + D) Mg 2+ E) Al Practice Problem 23.18

Kull Spring07 Lesson 23 Ch 8 53   Which of the following ions and atoms has the largest radius? A) Mg B) Na C) Na + D) Mg 2+ E) Al Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 54   Which of the following elements has the greatest metallic character? A) Mg B) Al C) Ca D) Ba E) Cs Practice Problem 23.19

Kull Spring07 Lesson 23 Ch 8 55   Which of the following elements has the greatest metallic character? A) Mg B) Al C) Ca D) Ba E) Cs Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 56Equations  wavelength x frequency  speed of light = wavelength x frequency c = λ X = 3.00 x 10 8 m/s  = nh(c/ )  E = nh = nh(c/ ) n= positive integer Planck’s constant(h) = x 10 –34 J s    E atom = E emitted (or absorbed) radiation =  nh  Rydberg equation = R  n 2 > n 1  R = x 10 7 m -1  ΔE = E final – E initial = –2.18 x 10–18 J   E photon = E state A – E state B = hν

Kull Spring07 Lesson 23 Ch 8 57 Quantum mechanics Quantum number Values Total number n – shell/level 1,2,3, … ∞ n = # subshells n 2 = # orbitals in a shell - subshell/ sublevel ℓ - subshell/ sublevel 0, 1,… n-1 n-1 m ℓ - orbital - - ℓ, 0, + ℓ 2 2 ℓ +1 (orbitals in a subshell) m s - spin + ½, -½ 2 possible

Kull Spring07 Lesson 23 Ch 8 58   When electrons in helium atoms fall from the 3s orbital down to the 1s orbital, how many different energies could be released if all possible pathways of decreasing energy are considered? A) 2B) 4C) 5 D) 6E) 7 Practice Problem 23-20

Kull Spring07 Lesson 23 Ch 8 59   When electrons in helium atoms fall from the 3s orbital down to the 1s orbital, how many different energies could be released if all possible pathways of decreasing energy are considered? A) 2B) 4C) 5 D) 6E) 7 Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 60   Of the following four electron configurations, which two represent elements that would have similar chemical properties? 1. 1s 2 2s 2 2p s 2 2s 2 2p 6 3. [Ar]4s 2 3d 10 4p 4 4. [Ar]4s 2 3d 10 4p 5 A) 1 and 3 B) 1 and 4 C) 2 and 3 D) 1 and 2 E) 2 and 4 Practice Problem 23-21

Kull Spring07 Lesson 23 Ch 8 61   Of the following four electron configurations, which two represent elements that would have similar chemical properties? 1. 1s 2 2s 2 2p s 2 2s 2 2p 6 3. [Ar]4s 2 3d 10 4p 4 4. [Ar]4s 2 3d 10 4p 5 A) 1 and 3 B) 1 and 4 C) 2 and 3 D) 1 and 2 E) 2 and 4 Practice Problem Answer

Kull Spring07 Lesson 23 Ch 8 62   The electron configuration for Fe 2+ is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6. Therefore Fe 2+ is A) paramagnetic with two unpaired electrons. B) paramagnetic with one unpaired electron. C) paramagnetic with three unpaired electrons. D) paramagnetic with four unpaired electrons. E) diamagnetic. Practice Problem 23-22

Kull Spring07 Lesson 23 Ch 8 63   The electron configuration for Fe 2+ is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6. Therefore Fe 2+ is A) paramagnetic with two unpaired electrons. B) paramagnetic with one unpaired electron. C) paramagnetic with three unpaired electrons. D) paramagnetic with four unpaired electrons. E) diamagnetic. Practice Problem Answer