Chapter 7 Electrons in Atoms and Periodic Properties

Chapter 7

Electrons in Atoms and

Periodic Properties

Light is electro/magnetic (EM) radiation

Terms

• Wavelength () is the distance from crest to crest or trough to trough on a wave.

• The frequency () of a wave expresses the number times a wave passes a given point in some unit of time.

• Amplitude of a wave is the height of the crest or depth of the trough with respect to the center line of the wave.

Visible light is a small portion of the entire spectrum of EM radiation

Increasing Energy and Frequency(decreasing wavelength)

EM Spectrum

EM Spectrum and Classification

Electromagnetic Radiation

The mathematical relationship between wavelength and frequency for EM radiation is:

c = ۰

c = 2.998E8 m/s (speed of light in a vacuum)

= wavelength (in meters) (Greek letter, lambda)

= frequency (in Hertz or s-1) (Greek letter, nu)

Electromagnetic Radiation

Using the mathematical relationship between wavelength and frequency:

c =

Calculate the wavelength associated with Montana Tech’s student radio station, KMSM-FM which broadcasts at a frequency of 107.1 MHz.

KMSM-FM broadcasts at a frequency of 107.1 MHz

• What is the frequency of light, in hertz, if it has a wavelength of 1.05 x 10-7 m and is traveling in vacuum? What portion of the electromagnetic spectrum does this “light” belong to?

• Electromagnetic radiation slows down as it travels through matter. What fraction and percentage of “c” is the velocity of 475 nm light, if its frequency is 6.00 x 1014 Hz?

Behavior of Waves

• Waves refract or bend when they pass from one medium to another with different densities.

• Diffraction is the bending of electromagnetic radiation as it passes around the edge of an object or through narrow openings.

• Interference is the interaction of waves that results in either reinforcing their amplitudes or canceling them out.

Refraction

Refraction – the change in direction of a beam of Electromagnetic Radiation (light) as it passes from one medium into another.

Diffraction and Interference

Laser light is monochromatic (one wavelength) and spatially and temporally coherent

Wavelength (nm)

He-Ne Laser Spectrum

632.8 nm

Missing lines of light (dark lines) are found in solar spectra. In distant stars these lines are shifted towards longer wavelengths (red shift). These red shifts are caused by the “Doppler effect”.

What other places in your regular life to do find examples of the Doppler effect?

Fraunhofer Lines (dark) in the Solar Spectrum

Shifts to lower energies (red-shift) of these lines suggested to Hubble et al. that more distance galaxies were moving away more rapidly. This would be the expected result assuming the universe began with a Big Bang

Redshift calculations

• Using wavelength = v/c

• Using frequency = v/c

’ represents the longer wavelength and ’ represents the lower frequency

'

'

'

Try this example problem:

• A spectral line for atomic hydrogen (H) is known to occur at 485 nm. Studying the stars in a distant galaxy, it is noted that the spectral line now appears at 558 nm ( a shift to longer wavelength). At what percentage of “c” is the galaxy moving away from earth? What is the velocity of the galaxy relative to earth?

Atomic Emission Spectra…the light from atoms

Types of Spectra

• Atomic emission spectra consist of bright lines on a dark background.

• Atomic absorption spectra consist of characteristic series of dark lines produced when free gaseous atoms are illuminated by external sources of radiation.

Emission versus Absorption Spectra

Absorption Spectra

How do the spectra change in going from H to Ne? Why?

Black Body Radiation and the end of classical physics…the UV catastrophe.

Quantum Theory• From work on BB radiation, Max Planck

proposed that light can have both wavelike and particle-like properties.

• A quantum is the smallest discrete quantity of a particular form of energy.

• Particles of radiant energy are known as Photon.

• Quantum theory is based on the idea that energy is absorbed and emitted in discrete quanta…at least in small (nm) spaces.

Quantum Theory

• Something that is quantized has values that are restricted to whole-number multiples of a specific base value.

• The energy of a quantum of radiation is: E = h where h is Planck’s constant h = 6.6260755 x 10-34 J•s Or E = hc/

Particle Nature of Light

• Each packet of electromagnetic radiation energy is called a quantum.

• Einstein called the packets photons. A mole of photons is called an Einstein.

• What is the wavelength of a photon, in vacuum, with an energy of

1.25 x 10-20 J?

What portion of the electromagnetic spectrum does this photon belong to?

E = h = hc/

= hc/E = (6.626E-34 J s)(2.998E8 m/s)/(1.25E-20 J)

= 1.59E-5 m

EM Spectrum

Planck and Einstein (1929) in happier days.

Plank was among the few that recognized the significance of Special Relativity. But did not accept that the “quantum” was a real phenomenon.

A new scientific truth does not triumph by convincing its opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it.Max Planck

Photoelectric Effect• The photoelectric effect is

the release of electrons from a metal as a result of electromagnetic radiation.

• The photoelectric effect can be explained if the electromagnetic radiation is treat as being composed of tiny particles (wave packets) called photons.

Only electrons with sufficient energy will displace electrons. This energy (or threshold frequency) is known as the Work Function ()

= h0 = Ebound electron

The work function depends on the type of metal. If the element surface is irradiated with light of frequency greater than the threshold, the excess energy appears in the kinetic energy of the electron.

KEelectron = h – h0 = hv -

Work Function for various elements (eV)

Problem

When light of frequency 1.30E15 s-1 shines on the surface of cesium metal, electrons are ejected with a maximum kinetic energy of

5.2E-19J.

a) Calculate the wavelength of this light.

b) Calculate the work function for cesium.

c) Calculate the longest wavelength of light that will displace electrons.

The Hydrogen Spectrum

• Johannes Rydberg revised Balmer’s equation to describe the complete hydrogen spectrum.

22

21

1-2-

n

1 -

n

1)(nm) 10 x (1.097 =

1

N1 is a whole number that remains fixed for a series of calculations in which n2 is also a whole number with values of n1+1, n1+2,… for successive line in the spectrum.

Problem What is the wavelength of the line in the visible

spectrum corresponding to n1 = 2 and n2 = 4?

1/ = 1.097E-2 nm-1(1/22 – 1/42)

= 1.097E-2 nm-1(3/16)

= 2.057E-3 nm-1

= 486 nm

The Bohr Model for Electrons in the Hydrogen Atom

• The electron in a hydrogen atom occupies a discrete energy level and may exist only in the available energy levels.

• The electron may move between energy levels by either absorbing or emitting specific amounts of energy.

• Each energy level is designated by a specific value for n, called the principal quantum number.

The Bohr model of the hydrogen atom places electrons in concentric orbits with certain “allowed” orbital energies for the electrons in the field of the nucleus (~Ze).

Z is the atomic number, and e is the fundamental charge (1.6E-19 C)

Balmer Series of The Hydrogen Emission Spectrum

Energy of Electronic Transitions

• Neil Bohr derived the following formula for the energy levels of hydrogen-like orbitals

En = - Z2e4m

80n2h2

Z is the atomic number, is the vacuum electric permittivity

m and e is the mass and charge of the electron.

Hydrogen Spectrum

• An energy level is an allowed state that an electron can occupy in an atom.

• Movements of electrons between energy levels are called electronic transitions.

Mathematically in the Bohr model, the energy of each orbital is:

En = (- 2.18E-18J) (1/n2)

Where n= 1, 2, 3,…∞

The constant in the equation equals: Rhc

Where R = 1.0974E7 m-1 (Rydberg constant)

Note that the orbital energies are mathematically all negative (< 0) in energy ( corresponding to bound electronic states)

The Bohr model of the hydrogen atom places electrons in concentric orbits with certain “allowed” orbital energies for the electrons in the field of the nucleus (~Ze).

Note pattern of orbital spacings…

Problem

What is the minimum wavelength of light that can ionize a hydrogen atom…in the gas phase? (Big Hint…assume nf = ∞).

Electronic States• The lowest energy level (n) available to

an electron in an atom is its ground state.

• An excited state of an electron in an atom (or molecule) is any energy state above the ground state.

Particle or Waves?

• If electromagnetic radiation behaves as a particle, de Broglie reasoned, why couldn’t a particle in motion, such as an electron, behave as a wave?

• de Broglie’s Equation = h/mu (m in kg and u in m/s)

De Broglie Wave Equation

De Broglie Wave Equation

Wave Equations

Wave equation for a standing wave:

L = n

The wave equation for electrons is called the Schrödinger Equation

Ĥ = E

Where (psi) is a wave function, and Ĥ is the Hamiltonian operator:

Ĥ = -iħ (t)

Wave equations describe a “quantized” electron (sec 3.5 & 3.6)

• Mathematical equations known as wave equations are use to describe probabilities of finding electrons around the nucleus.

• The wave equations for electrons yield three quantum numbers (n, l, ml) that define the energy, shape, and orientation for the electron orbitals.

• A fourth quantum number (ms) gives the quantized (relativistic) spin of an electron in an orbital.

Quantum Numbers (QN)

QN Restrictions Range

n (principal) positive integers 1, 2, …, l (ang. momentum) positive less than n 0, 1,..(n-1)

ml (magnetic) integers between –l and l

s (spin) half-integers –½, ½ ; –½, ½

Possible sets of quantum number for n = 1, 2, 3

As n increases the possible number of orbitals increases by n2. Only two electrons can occupy a single orbital, thus there are 2n2 electrons per n-shell

Quantum Numbersn l m l # orbitals

1 0 0 1

2 0 0 1

1 -1,0,+1 3

3 0 0 1

1 -1,0,+1 3

2 -2.-1,0,+1,+2 5

4 0 0 1

1 -1,0,+1 3

2 -2.-1,0,+1,+2 5

3 -3,-2.-1,0,+1,+2,+3 7

Value of l 0 1 2 3 4

Orbital Name(s) s p d f g

Shell and Orbital Names

Value of n 1 2 3 3 5

Shell Name* K L M N O

*Primarily used in X-ray spectroscopy

Practice

What orbitals and how many electrons are identified by the following combinations of quantum numbers?

a) n=3, l=2

b) n=3, l=0

c) n=4, l=3

d) n=5, l=2, ml =1

Problem What are the letter designations of all

the subshells in the n = 5 energy level or shell? What is total number of orbitals in the n = 5 shell?

Problem

For the following sets of quantum numbers, determine which describe actual orbits, and list why others are non-existent.

n l ml s(a) 5 3 -2 -1

(b) 5 3 -3 ½

(c) 3 3 -3 ½

(d) 3 0 0 -½

Shape and Sizes of Orbitals

• Psi squared, 2, defines the probability of an electron in some region of space around the nucleus .

• A radial distribution plot is a graphical representation of the probability of finding an electron in a thin spherical layer near the nucleus of an atom.

The shape of an atomic orbital is determine by “l” the angular momentum quantum number

l=0; s-orbital

l=1; p-orbital

l=2; d-orbital

node+

-

Probability Electron Density for 1s (l=0) Orbital

Electron density in the 1s orbital of the hydrogen atom

The probability of the electron density is a function of the square of the wave-function ()

rmp

r90

Probability Density of s-Orbitals

The 2p (l=1) Orbitals

The Five 3d (l=2) Orbitals

Assigning Quantum Number for Electrons

• Pauli’s exclusion principle - no two electrons in an atom may have the same set of four quantum numbers. An orbital can only hold two electrons and

they must have opposite spins.

Aufbau Principle

• As protons are added one by one to the nucleus to build up the elements, electrons are similarly added to these hydrogen-like orbitals.

Orbital Energy Levels for Hydrogen-Like Atoms

E

3s 3p 3d

2s 2p

1s

Many Electron Atoms The presence of more than one electron in

an atom effect the relative orbital energies which only depend on n in the one-electron orbital (hydrogen-like atom).

In the many electron atom, orbital energies depend on both n and l (n+l rule).

In general, for a given value of n, the lower the value of l, the lower in energy the orbital subshell (4s < 3d

The effective nuclear charge (Zeff) felt by the 2s electron in lithium is less than the effective nuclear charge in hydrogen or helium.

The presence of electron density in the 1s orbital “screens” the outer electron in the 2s orbital from the nuclear charge. This results a lowering in the outer shell. orbital energy

Orbital Energies in Multi-electron Atoms

Note the spacing of the orbitals and that the ordering of the energies depends on n + l.

The 2s orbital is lower in energy than the 2p orbital because there is some 2s density closer to the nucleus

Terminology• Orbitals that have the exact same energy

level are said to be degenerate (e.g 2px and 2py).

• Core electrons are those in the filled, inner shells in an atom and are not involved in chemical reactions.

• Valence electrons are those in the outermost shell of an atom and have the most influence on the atom’s chemical reactivity.

Electron Configuration1s 2s 2p1s 2s 2p

H: 1sH: 1s11

He: 1sHe: 1s22

Li: 1sLi: 1s222s2s11

Be: 1sBe: 1s222s2s22

B: 1sB: 1s222s2s222p2p11

Orbital diagrams and electron configurations describe how the electrons fill orbitals

Hund’s rules state that the lowest energy electron configuration will have a maximum in unpaired electrons. Note the filling of the 2p orbitals in C, O and N.


oror

C: 1sC: 1s222s2s222p2p22

C: 1sC: 1s222s2s222p2p22

Hund’s Rule tells us which configuration is correctHund’s Rule tells us which configuration is correct..

Hund’s Rule

• The lowest energy configuration for an atom is the one having the maximum number of unpaired electrons allowed by the Pauli principle in a particular set of degenerate orbitals.


C: 1sC: 1s222s2s222p2p22

N: 1sN: 1s222s2s222p2p33

O: 1sO: 1s222s2s222p2p44

F: 1sF: 1s222s2s222p2p55

Ne: 1sNe: 1s222s2s222p2p66

Electron Configurations of the Fourth Period

K K 1s1s222s2s222p2p663s3s223p3p664s4s1 1 or [Ar]4sor [Ar]4s11

Ca Ca 1s1s222s2s222p2p663s3s223p3p664s4s2 2 or [Ar]4sor [Ar]4s22

Sc Sc 1s1s222s2s222p2p663s3s223p3p664s4s223d3d1 1 or [Ar]4sor [Ar]4s223d3d11

Ti Ti 1s1s222s2s222p2p663s3s223p3p664s4s223d3d2 2 or [Ar]4sor [Ar]4s223d3d22

V V 1s1s222s2s222p2p663s3s223p3p664s4s223d3d3 3 or [Ar]4sor [Ar]4s223d3d33

Cr Cr 1s1s222s2s222p2p663s3s223p3p664s4s113d3d55 or [Ar]4sor [Ar]4s113d3d55

Mn Mn 1s1s222s2s222p2p663s3s223p3p664s4s223d3d55 or [Ar]4sor [Ar]4s223d3d55

Cu Cu 1s1s222s2s222p2p663s3s223p3p664s4s113d3d1010 or [Ar]4sor [Ar]4s113d3d1010

••••••

4s

3d

Anomalies in Configurations

• Chromium and Copper do not follow the pattern of the other elements. You should remember these two families,

because other elements in these families exhibit the same types of configurations

• You can use the Periodic Table to guide you in writing electron configurations.

The energies of orbitals in multi-electron atoms are different than for hydrogen due to electron-electron interactions (repulsion and exchange).

The organization of the periodic table is based on the chemical properties of the elements which is determined by their electron configurations.

n

Zeff

Mendeleev’s Periodic Table organized elements according to their chemical combining properties.

Electron Configurations of Ions

• Start with the configuration for the neutral atom, then add or remove electrons from the valence shells to make the desired ion.

• Atoms or ions that are isoelectronic with each other have identical numbers and configurations of electrons.

Write the electron configurations for the following:

a) C

b) S

d) Ti

e) Ti4+

How many unpaired electrons are in the following

Sc

Ag+

Cd2+

Sizes of Atoms and Ions

Orbital Penetration and Effective Nuclear Charge

• Orbital penetration occurs when an electron in an outer orbital has some probability of being close to the nucleus Penetration ability follows this order: s

> p > d > f.

• Effective nuclear charge (Zeff) is the attractive force toward the nucleus experienced by an electron in an atom.

Penetration Ability of s Orbitals

Radii of Atoms and Ions

Ionization Energy

The quantity of energy required to remove 1 mole of electrons from 1 mole of the gaseous atom or ion.

X(g) ---> X+(g) + e-(g)

Ionization Energy Trends

Ionization Energies

Periodic Trends

• First ionization energy:

increases from left to right across a period;

decreases going down a group.

Successive Ionization Energies (kJ/mol)

Elements IE1 IE2 IE 3 IE4 IE5 IE6

H 1312

He 2372 5249

Li 520 7296 12040

Be 897 1758 15050 21070

B 801 2426 3660 24682 32508

C 1087 2348 4617 6201 37926 46956

N 1402 2860 4581 7465 9391 52976

O 1314 3383 5298 7465 10956 13304

The Uncertainty Principle

• Quantum mechanics allows us to predict the probabilities of where we can find an electron.

• We cannot map out on the path an electron travels. The Heisenberg’s uncertainty principle

says that you cannot determine the position and momentum of an electron at the same time.

Heisenberg’s Uncertainty Principle (the quantum mechanical rules of the road)

Example…

The position and momentum of an electron in an atom can not be simultaneously determined.

x*p ≥ h/4

ChemTour: Electromagnetic Radiation

Click to launch animation

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This ChemTour explores the relationship of frequency, wavelength, and energy using animations, interactive graphs, and equations. The quantitative exercises include graph reading and calculations using Planck’s constant and the speed of light.

ChemTour: Light Diffraction


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This animation recreates Thomas Young’s double-slit experiment and demonstrates how constructive and destructive interference occur.

ChemTour: Doppler Effect


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A boat moving with or against the direction of wave movement demonstrates the motion-induced shifts in wavelengths and frequency that are examples of the Doppler effect.

ChemTour: Light Emission and Absorption


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This ChemTour examines the emission and absorption spectra for sodium and hydrogen and relates them to energy level transitions.

ChemTour: Bohr Model of the Atom


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This ChemTour explores the idea that energies of electrons surrounding atomic nuclei are quantized. In Practice Exercises, students learn to calculate the energies of specific states of hydrogen, and the energies involved in electronic transitions.

ChemTour: de Broglie Wavelength


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In this ChemTour, students learn to apply the de Broglie equation to calculate the wavelength of moving objects ranging from baseballs to electrons. Includes Practice Exercises.

ChemTour: Quantum Numbers


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In this ChemTour, students explore the rules for designating quantum numbers. Includes Practice Exercises.

ChemTour: Electron Configuration


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This ChemTour explains how electrons are distributed within atomic orbitals. Students learn how to determine an element’s electron configuration and learn how to complete an orbital box diagram. Includes practice exercises.

Suppose two photons combine in a crystal to form a single photon of green light or "green photon."

Which of the following could be the colors of the two combining photons?

A) Green & green B) Blue & yellow C) Infrared & infrared

Combining Two Photons

Combining Two Photons

Please consider the following arguments for each answer and vote again:

A. A green photon can only be produced by the combination of two other green photons of the same wavelength.

B. The color green is the result of combining the colors blue and yellow, just as a green photon will result from the combination of blue and yellow photons.

C. Only two infrared photons have the proper total energy needed to form a green photon.

An electron in the ground state absorbs a single photon of light and then relaxes back to the ground state by emitting an infrared photon (1200 nm) followed by an orange photon (600 nm).

What is the wavelength of the absorbed photon? A) 400 nm B) 600 nm C) 1800 nm

Absorption and Fluorescence of Light

Absorption and Fluorescence of Light


A. The wavelength is inversely proportional to the energy,

so for energy to be conserved the absorbed photon must have a wavelength of 400 nm.

B. The wavelength of the absorbed photon is the difference of the wavelength of the two emitted photons, which is 600 nm.

C. For the energy to be conserved, the sum of the wavelengths must be conserved. So the wavelength of the absorbed photon is 1800 nm.

The diagram to the left depicts the interference pattern that results from the constructive and

destructive interference of light waves that are diffracted as they pass through two slits. If the pattern is the result of green light passing through two slits, which of the following patterns would be the result of blue light passing through the same two slits?

A) B) C)

Two-Slit Diffraction and Interferometry

Two-Slit Diffraction and Interferometry

Please consider the following arguments for each

answer and vote again:

A. The wavelength of blue light is shorter than that of green light, so constructive and destructive interference occurs at smaller intervals.

B. The interference pattern is dependent only on the width of and distance between the two slits. Therefore, the interference pattern should not change.

C. Blue light is higher in energy than green light and therefore would be less affected by the two slits.

When a photon of red light hits metal X, an electron is ejected. Will an electron be ejected if a photon of yellow light hits metal X?

Photoelectric Effect: Red and Yellow Light

A) Yes B) No C) Can't tell

Photoelectric effect: Red and Yellow Light


A. Photons of yellow light possess more energy than photons of red light, so a yellow photon also must eject an electron.

B. Each metal has a specific wavelength of light that will cause electrons to be ejected. If red light has the correct wavelength, yellow cannot.

C. Whether a yellow photon will eject an electron from the metal will depend on how tightly the electron is bound to the metal.

When a photon of blue light hits metal X, an electron is ejected. Will an electron be ejected if a photon of green light hits metal X?

Photoelectric Effect: Blue and Green Light

A) Yes B) No C) Can't tell

Photoelectric effect: Blue and Green Light


A. So long as enough photons of light hit the metal, an electron will always be ejected, regardless of the wavelength of the light.

B. The energy of a blue photon is higher than the energy of a green photon so an electron removed with blue light will not be removed with green light.

C. Whether a green photon will eject an electron from the metal will depend on how tightly the electron is bound to the metal.

Photoelectric Effect: Kinetic Energies of Electron

A 300-nm photon can eject an electron from a metal surface with a certain kinetic energy. What photon wavelength would be required to eject an electron from the same metal surface with twice the kinetic energy?

A) 150 nm B) 200 nm C) 600 nm

Photoelectric Effect: Kinetic Energies of Electrons

Please consider the following arguments for each

answer and vote again:

A. To eject an electron with twice the kinetic energy, twice the energy must be provided by the photon, so the photon wavelength must be halved.

B. A photon with a wavelength of 200 nm will overcome the work function and provide twice the kinetic energy.

C. To double the kinetic energy of the ejected electron, the wavelength of the impacting photon also must be doubled.

De Broglie Wavelengths of H20 Molecules

Suppose a hydrogen molecule (1H2) is traveling at 800 m/s and a deuterium molecule (2H2) is traveling at 400 m/s. What can be said of the de Broglie wavelengths of the two molecules?

A) λH > lD B) λH < lD C) λH = lD

De Broglie Wavelengths of H2O Molecules


A. The kinetic energy of the deuterium molecule is twice that of the hydrogen molecule. Therefore, the deuterium molecule will have a shorter de Broglie wavelength.

B. Because the speed of the hydrogen molecule is greater than the speed of the deuterium molecule, the de Broglie wavelength of the hydrogen molecule will be shorter.

C. The hydrogen molecule and the deuterium molecule have the same momentum and therefore will have the same de Broglie wavelength.

Laser Cooling of Sodium Atoms

One method for decreasing the temperature of atoms, known as laser cooling, involves bombarding an atom with photons of light, decreasing its overall momentum and thus its kinetic energy (just like one could slow a fast-moving car by colliding it with another car).

A sodium atom at a temperature of 60 K has a de Broglie wavelength of 66 pm (6.6x10-11 m). Approximately how many photons of red light (at λ = 660 nm) would it take to stop a sodium atom at 60 K?

A) ~1 B) ~102 C) ~104

Laser Cooling of Sodium Atoms


A. A photon travels ~105 times faster than a sodium atom. Therefore, only one photon is required.

B. The kinetic energy of a sodium atom is ~100 times less than the kinetic energy of a red photon.

C. The de Broglie wavelength of a sodium atom at 60 K is ~104 times shorter than the wavelength of a red photon, so it will take 104 photons to stop a single sodium atom.

Transmission of Light through a Color Filter

What color will a yellow object appear when it is seen through a filter with the absorption spectrum shown to the left?

A) Yellow B) Blue C) Black

Transmission of Light through a Color Filter


A. The filter absorbs no yellow light, so the object will appear yellow.

B. Blue light is absorbed by the filter, so an object seen through the filter will appear blue.

C. No yellow light is absorbed by the filter, so the object will appear black.

Emission Spectra

Photon emission from a system possessing the energy level diagram to the left would produce which of the following spectra?

A) B) C)

Emission Spectra

Consider the following arguments for each answer and vote again:

A. The photon wavelength depends only on the energy of the lowest state, so only 1 wavelength is possible.

B. There are 2 possible transitions—one from each of the 2 upper levels. Thus, 2 wavelengths of light are emitted.

C. The 3 energy levels lead to 2 high-energy transitions and 1 low-energy transition. Therefore, 3 different photon wavelengths are possible.

Energy Levels

Emission from which of the following energy level diagrams would produce the spectrum shown to the left?

A) B) C)

Energy Levels


A. The arrangement of the energy levels reflects the arrangement of the lines in the emission spectrum.

B. This energy level diagram allows only 1 low-energy transition, consistent with the emission spectrum.

C. Only this energy level diagram allows 3 high-energy transitions and 1 low-energy transition.

Transition in H and He+

The diagram to the left shows the spacing of the first five energy levels for a hydrogen atom. Which of the following transitions in He+ has the same wavelength as the 4→2 transition in H?

A) 4→2 B) 8→4 C) 16→8

Transition in H and He+


A. He+ has the same electron configuration as H; therefore, the energy level diagram will be the same.

B. The atomic number of He+ is twice that of H. Therefore, to produce the same energy splitting, the energy levels must be twice that of H.

C. The energy of the electron is proportional to Z2, which is 4 for He+. Therefore, the two levels, 4 and 2, must be increased by a factor of 4 to 16 and 8, respectively.

Electron Configurations

Which atom or ion can have the electron configuration 1s22s22p1?

Periodic Table

A) Li B) Be- C) B+

Electron Configurations


A. The answer must be lithium because it is the first element in row 2 to possess only one unpaired electron.

B. Beryllium in its ground state has the electron configuration 1s22s2, so Be- in its ground state will have the configuration 1s22s22p1.

C. In its ground state, boron has the electron configuration 1s22s22p1, so B+ must also have this configuration.

Ionization Energies

Which of the following has the lowest ionization energy?

A) H(1s1) B) He(1s13p1) C) He+(4p1)

Ionization Energies


A. Hydrogen has a lower nuclear charge than helium, so it always has a lower ionization energy than any helium atom or ion.

B. He(1s13p1) has almost the same ionization energy as H(3p1), which has a lower ionization energy than either H(1s1) or He+(4p1).

C. Because the electron in He+(4p1) is in the fourth shell, the ionization energy of He+(4p1) is the lowest.

Ionization Energies of He(1s2)

How does the ionization energy of He(1s2) compare to the ionization energies of H(1s1) and He+(1s1)?

A) Higher B) Lower C) In-between

Ionization Energies of He(1s2)


A. It is harder to remove an electron from a doubly occupied orbital than from a singly occupied orbital.

B. Each electron offsets the charge of one of the protons, giving an effective nuclear charge of zero.

C. Each electron partially shields the other, leading to an effective nuclear charge that is between 1 and 2.

Atomic and Ionic Radii

Which of the following atoms (or ions) has the smallest radius?

A) K+ B) Ar C) Cl-

Atomic and Ionic Radii


A. K+ has the highest nuclear charge and so has the smallest atomic radius.

B. Because it is a noble gas, Ar has the smallest atomic radius.

C. Cl- has the nucleus with the lowest mass, so it has the smallest atomic radius.

Electron Affinity of Halogen Atoms

Suppose an electron is transferred from a potassium atom to an unknown halogen atom. For which of the following halogen atoms would this process require the least amount of energy?

A) Cl B) Br C) I

Electron Affinity of Halogen Atoms


A. Chlorine has the greatest affinity for electrons and so would release the most energy when an electron is added.

B. Electron donation is most favorable energetically when it occurs between atoms on the same row of the periodic table.

C. Because of its massive nuclear charge and large electron cloud, an iodine atom can most easily accept an additional electron.

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Chapter 7 Electrons in Atoms and Periodic Properties