CHAPTER 8 Atomic Electron Configurations and Chemical Periodicity

11Kull Spring07 Lesson 23 Ch 8

CHAPTER 8Atomic Electron Configurations

and Chemical Periodicity

OutlineOutline--Collect homeworkCollect homework--ReviewReview--Trends Trends -Ions-Ions

ReviewReview Spdf notationSpdf notation Orbital box notationOrbital box notation

Name Symbol Permitted Values Property

principal n positive integers(1,2,3,…) orbital energy (size)

angular momentum

l integers from 0 to n-1 orbital shape (The l values 0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.)

magnetic mlintegers from -l to 0 to +l orbital orientation

spin ms+1/2 or -1/2 direction of e- spin

Characteristics of Many-Electron Atoms: The Electron-Spin

Quantum Number

Consider this set of quantum numbers: n = 3, ℓ = 2, mℓ = -1, ms = +½ The maximum number of electrons in an atom which can share the above set of quantum numbers isA) 1 B) 14C) 3 D) 10E) none of the above

Practice Problem 23-1

Consider this set of quantum numbers: n = 3, ℓ = 2, mℓ = -1, ms = +½ The maximum number of electrons in an atom which can share the above set of quantum numbers isA) 1 B) 14C) 3 D) 10E) none of the above

Practice Problem 23-1 Answer

An atom in its ground state contains 30 electrons. How many of these are in sublevels with ℓ = 2?A) 2B) 4C) 6D) 8E) 10

What are the possible values for theangular momentum quantum number

ℓ ?A) integers from -ℓ to 0 to +ℓB) 1, 2, 3, etc.C) 2, 4, 6, etc.D) +½ , -½E) integers from 0 to n - 1

What are the possible values for the angular momentum quantum number (ℓ)?A) integers from -ℓ to 0 to +ℓB) 1, 2, 3, etc.C) 2, 4, 6, etc.D) +½ , -½E) integers from 0 to n - 1

The electron configuration of the outermost electrons of atoms of the halogen group is:A) ns2np7

B) ns1

C) ns2np5

D) ns2np6(n-1)d7

E) ns2np6

The electron configuration of the outermost electrons of atoms of the halogen group is:A) ns2np7

B) ns1

C) ns2np5

D) ns2np6(n-1)d7

E) ns2np6

The electronic configuration of the element whose atomic number is 26 is:A) 1s2 2s2 2p6 3s2 3p6 4s0 3d8

B) 1s2 2s2 2p6 3s2 3p6 3d6 4s2

C) 1s2 2s2 2p6 3s2 3p6 4s2 3d6

D) 1s2 2s2 2p6 3s2 3p6 4s2 3d4 4p2

E) none of the above

The electronic configuration of the element whose atomic number is 26 is:A) 1s2 2s2 2p6 3s2 3p6 4s0 3d8

B) 1s2 2s2 2p6 3s2 3p6 3d6 4s2

C) 1s2 2s2 2p6 3s2 3p6 4s2 3d6

D) 1s2 2s2 2p6 3s2 3p6 4s2 3d4 4p2

E) none of the above

The set of quantum numbers that correctly describes an electron in a 3p orbital isA) n = 3; ℓ = 0; mℓ = 0; ms = 0

B) n = 3; ℓ = 2; mℓ = -2, -1, 0, 1, or 2; ms = +½ or -½C) n = 3; ℓ = 1; mℓ = -1, 0, or 1; ms = +½ or -½

D) n = 4; ℓ = 0; mℓ = -1 ,0, or 1; ms = +½ or -½E) none of the above

The set of quantum numbers that correctly describes an electron in a 3p orbital isA) n = 3; ℓ = 0; mℓ = 0; ms = 0

B) n = 3; ℓ = 2; mℓ = -2, -1, 0, 1, or 2; ms = +½ or -½C) n = 3; ℓ = 1; mℓ = -1, 0, or 1; ms = +½ or -½D) n = 4; ℓ = 0; mℓ = -1 ,0, or 1; ms = +½ or -½E) none of the above

An atom in its ground state contains 18 electrons. How many of these are in orbitals with mℓ = 0?

A) 2B) 4C) 6D) 8E) 10

Practice Problem 23- 7 7

An atom in its ground state contains 18 electrons. How many of these are in orbitals with mℓ = 0?

A) 2B) 4C) 6D) 8E) 10

The configuration for the six outer electrons in ground state oxygen atoms isA) 2s3 2p3

B) 2p6

C) 2s2 2p-12 2p0

D) 2s2 2p-12 2p0

1 2p11

E) 2s4 2p2-1

The configuration for the six outer electrons in ground state oxygen atoms isA) 2s3 2p3

B) 2p6

C) 2s2 2p-12 2p0

D) 2s2 2p-12 2p0

1 2p11

E) 2s4 2p2-1

Which of the following is the electron configuration for chromium, element 24?

A) 1s2 2s2 2p6 3s2 3p6 4s2

B) 1s2 2s2 2p6 3s2 3p6 4s2 3d4

C) 1s2 2s2 2p6 3s2 3p6 3d6

D) 1s2 2s2 2p6 3s2 3p6 4s1 3d5

E) 1s2 2s2 2p6 3s2 3p6 4s2 3d1 3d1

Which of the following is the electron configuration for chromium, element 24?

A) 1s2 2s2 2p6 3s2 3p6 4s2

B) 1s2 2s2 2p6 3s2 3p6 4s2 3d4

C) 1s2 2s2 2p6 3s2 3p6 3d6

D) 1s2 2s2 2p6 3s2 3p6 4s1 3d5

E) 1s2 2s2 2p6 3s2 3p6 4s2 3d1 3d1

PERIODIC PERIODIC TRENDSTRENDS

Trends in Some Key Periodic Atomic Properties: Trends in

Electron AffinityAtomic and ionic sizeAtomic and ionic size# electrons# electrons# shells – # shells – Larger orbitals,-electrons held less tightly

Electron affinity: Electron affinity: energy energy involved when an atom gains involved when an atom gains an electron to form an anion.an electron to form an anion.

Effective nuclear chargeEffective nuclear chargeWhen higher: Electrons held more tightly

Atomic radii of the main-group and transition

elements.

Atomic Size

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 (Zeffective)

The Effect of Electron Repulsions (Shielding)

Characteristics of Many-Electron Atoms: Electrostatic Effects and the Splitting of

Energy Levels

Effective Nuclear Charge, Z*Effective Nuclear Charge, Z*Effective Nuclear Charge, Z*Effective Nuclear Charge, Z* Z* is the nuclear charge experienced by the outermost Z* is the nuclear charge experienced by the outermost

electrons. electrons. Z* increases across a period owing to incomplete shielding by Z* increases across a period owing to incomplete shielding by

inner electrons.inner electrons. The 2s electron PENETRATES the region occupied by The 2s electron PENETRATES the region occupied by

the 1s electron. the 1s electron. 2s electron experiences a higher positive charge than 2s electron experiences a higher positive charge than

expected. expected. Estimate Z* by --> [ Z - (# inner electrons) ]Estimate Z* by --> [ Z - (# inner electrons) ] Charge felt by 2s e- in Li Charge felt by 2s e- in Li Z* = 3 - 2 = 1 Z* = 3 - 2 = 1 Be Be Z* = 4 - 2 = 2Z* = 4 - 2 = 2 B B Z* = 5 - 2 = 3Z* = 5 - 2 = 3 and so on!and so on!

Ionization EnergyIonization EnergyIonization EnergyIonization Energy

IE = energy required to remove an IE = energy required to remove an electron from an atom in the gas phase.electron from an atom in the gas phase.

Mg (g) + 738 kJ ---> MgMg (g) + 738 kJ ---> Mg++ (g) + e- (g) + e-

Mg+ (g) + 1451 kJ ---> Mg2+ (g) + e-Mg+ (g) + 1451 kJ ---> Mg2+ (g) + e-

Mg+ has 12 protons and only 11 electrons. Therefore, IE for Mg+ has 12 protons and only 11 electrons. Therefore, IE for Mg+ > MgMg+ > Mg

Ionization Energy

Ion SizesIon SizesIon SizesIon Sizes

CATIONS are SMALLER than the atoms CATIONS are SMALLER than the atoms from which they come.from which they come.

The electron/proton attraction has gone The electron/proton attraction has gone UP and so size DECREASESUP and so size DECREASES..

Li,152 pm3e and 3p

Li+, 78 pm2e and 3 p

+Forming Forming a cation.a cation.Forming Forming a cation.a cation.

Ion SizesIon SizesIon SizesIon Sizes

ANIONS are LARGER than the atoms ANIONS are LARGER than the atoms from which they come.from which they come.

The electron/proton attraction has gone The electron/proton attraction has gone DOWN and so size INCREASES.DOWN and so size INCREASES.

Trends in ion sizes are the same as Trends in ion sizes are the same as atom sizes. atom sizes.

Forming Forming an anion.an anion.Forming Forming an anion.an anion.F, 71 pm

9e and 9pF-, 133 pm10 e and 9 p

Ion ConfigurationsIon ConfigurationsIon ConfigurationsIon Configurations

To form cations, always remove electrons of highest n value first!

P [Ne] 3sP [Ne] 3s22 3p 3p33 - 3e- ---> P - 3e- ---> P3+3+ [Ne] 3s [Ne] 3s22 3p3p00

Trends in Ion SizesTrends in Ion SizesTrends in Ion SizesTrends in Ion Sizes

Active Figure 8.15Active Figure 8.15

Which of the following has the largest radius?

Which of the following has the largest radius?A) FB) NC) CD) OE) Ne

Which of the following elements has the largest ionization energy?

Which of the following elements has the largest ionization energy?A) NaB) NeC) FD) KE) Rb

Which of the following has the greatest electron affinity (most negative value)?

In the periodic table, the most nonmetallic elements will be foundA) 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)

Which ion has the smallest radius?

A) Li+

B) Na+

D) Be2+

E) Mg2+

Which ion has the smallest radius?

A) Li+

B) Na+

D) Be2+

E) Mg2+

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.

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 Answer

Which of the following elements has the greatest ionization energy?

Practice Problem 23.16

Which of the following elements has the greatest ionization energy?

Which of the following elements has the greatest electron affinity?A) RbB) CaC) LiD) NaE) I

Which of the following ions and atoms has the largest radius?A) MgB) NaC) Na+

D) Mg2+

Which of the following ions and atoms has the largest radius?

C) Na+

D) Mg2+

Which of the following elements has the greatest metallic character?A) MgB) AlC) CaD) BaE) Cs

EquationsEquations speed of light = wavelength x frequencywavelength x frequency

c = λ X = 3.00 x 108 m/s E = nh = nh(c/= nh(c/) ) n= positive integerPlanck’s constant(h) = 6.626 x 10–34 J s Eatom = Eemitted (or absorbed) radiation = nh

Rydberg equationRydberg equation = R = R nn22 > n > n11

R = 1.096776 x 10R = 1.096776 x 1077 m m-1-1 ΔE = EΔE = Efinal final – E– Einitial initial = –2.18 x 10–18 J= –2.18 x 10–18 J Ephoton = Estate A – Estate B = hν

Quantum mechanicsQuantum mechanics

Quantum Quantum numbernumber

ValuesValues Total Total numbernumber

n – n – shell/levelshell/level

1,2,3, 1,2,3, … … ∞∞

n = # n = # subshellssubshells

nn2 2 = # orbitals = # orbitals in a shellin a shell

ℓ - subshell/ - subshell/ sublevelsublevel

0, 1,… 0, 1,… n-1n-1

n-1n-1

mℓ - orbital - - ℓ, 0, + ℓ 2 2 ℓ +1 (orbitals in a subshell)

mmss - spin - spin ++½, -½½, -½ 2 possible2 possible

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) 2 B) 4

C) 5D) 6 E) 7

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) 2 B) 4

C) 5D) 6 E) 7

Of the following four electron configurations, which two represent elements that would have similar chemical properties? 1. 1s22s22p5 2. 1s22s22p6

3. [Ar]4s23d104p4 4. [Ar]4s23d104p5

A) 1 and 3 B) 1 and 4 C) 2 and 3

D) 1 and 2 E) 2 and 4

Of the following four electron configurations, which two represent elements that would have similar chemical properties? 1. 1s22s22p5 2. 1s22s22p6

3. [Ar]4s23d104p4 4. [Ar]4s23d104p5

A) 1 and 3 B) 1 and 4 C) 2 and 3

D) 1 and 2 E) 2 and 4

The electron configuration for Fe2+ is 1s2 2s2 2p6 3s2 3p6 3d6. Therefore Fe2+ 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.

The electron configuration for Fe2+ is

1s2 2s2 2p6 3s2 3p6 3d6.

Therefore Fe2+ 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.

CHAPTER 8 Atomic Electron Configurations and Chemical Periodicity

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