Chapter 6 Lecture- Electrons in Atoms

Waves

• To understand the electronic structure of atoms, one must understand the nature of electromagnetic radiation.

• The distance between corresponding points on adjacent waves is the wavelength ().

Waves• The number of waves

passing a given point per unit of time is the frequency ().

• For waves traveling at the same velocity, the longer the wavelength, the smaller the frequency.

Electromagnetic Radiation

• All electromagnetic radiation travels at the same velocity: the speed of light (c), 3.00 108 m/s.

• Therefore,

c =

Visible light is only one form of electromagnetic radiation; all visible light is electromagnetic radiation, but not all electromagnetic radiation is visible light.

Both visible light and X rays travel at the same speed, about 186,000 miles per second.

SAMPLE EXERCISE 6.1 Concepts of Wavelength and Frequency

Two electromagnetic waves are represented in the image below. (a) Which wave has the higher frequency? (b) If one wave represents visible light and the other represents infrared radiation, which wave is which?

(b) The electromagnetic spectrum (Figure 6.4) indicates that infrared radiation has a longer wavelength than visible light. Thus, the lower wave would be the infrared radiation.

Solution (a) The lower wave has a longer wavelength (greater distance between peaks). The longer the wavelength, the lower the frequency ( = c/ ). Thus, the lower wave has the lower frequency, and the upper one has the higher frequency.

Answer: The expanded visible-light portion of Figure 6.4 tells you that red light has a longer wavelength than blue light. The lower wave has the longer wavelength (lower frequency) and would be the red light.

PRACTICE EXERCISE

If one of the waves in the image above represents blue light and the other red light, which is which?

SAMPLE EXERCISE 6.2 Calculating Frequency from Wavelength

The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation?

Solve: Solving Equation 6.1 for frequency gives = c/ . When we insert the values for c and , we note that the units of length in these two quantities are different. We can convert the wavelength from nanometers to meters, so the units cancel:

Check: The high frequency is reasonable because of the short wavelength. The units are proper because frequency has units of “per second,” or s–1.

PRACTICE EXERCISE

(a) A laser used in eye surgery to fuse detached retinas produces radiation with a wavelength of 640.0 nm. Calculate the frequency of this radiation. (b) An FM radio station broadcasts electromagnetic radiation at a frequency of 103.4 MHz (megahertz; MHz = 106 s–1). Calculate the wavelength of this radiation.

Answers: (a) 4.688 1014 s–1, (b) 2.901 m

The Nature of Energy• The wave nature of light

does not explain how an object can glow when its temperature increases.

• Max Planck explained it by assuming that energy comes in packets called quanta.

As temperature increases, the average energy of the emitted radiation increases. Blue-white light is at the shorter wavelength (higher energy) end of the visible spectrum while red light is at the longer wavelength (lower energy) end of the visible spectrum.

The Nature of Energy

• Einstein used this assumption to explain the photoelectric effect.

• He concluded that energy is proportional to frequency:

E = hwhere h is Planck’s constant, 6.63 10−34 J-s.


• Therefore, if one knows the wavelength of light, one can calculate the energy in one photon, or packet, of that light:

c = E = h

SAMPLE EXERCISE 6.3 Energy of a Photon

Calculate the energy of one photon of yellow light whose wavelength is 589 nm.

This is the magnitude of enthalpies of reactions (Section 5.4), so radiation can break chemical bonds, producing what are called photochemical reactions.

Plan: We can use Equation 6.1 to convert the wavelength to frequency:

We can then use Equation 6.3 to calculate energy:

Solve: The frequency, , is calculated from the given wavelength.

The value of Planck’s constant, h, is given both in the text and in the table of physical constants on the inside front cover of the text, and so we can easily calculate E:

Comment: If one photon of radiant energy supplies 3.37 10–19J, then one mole of these photons will supply

SAMPLE EXERCISE 6.3 continued

Answers: (a) 3.11 10–19 J, (b) 0.16 J, (c) 4.2 1016 photons

PRACTICE EXERCISE(a) A laser emits light with a frequency of 4.69 1014 s–1. What is the energy of one photon of the radiation from this laser? (b) If the laser emits a pulse of energy containing 5.0 1017 photons of this radiation, what is the total energy of that pulse? (c) If the laser emits 1.3 10–2 J of energy during a pulse, how many photons are emitted during the pulse?

Electrons would be ejected, and they would have greater kinetic energy than those ejected by yellow light.


Another mystery involved the emission spectra observed from energy emitted by atoms and molecules.


• One does not observe a continuous spectrum, as one gets from a white light source.

• Only a line spectrum of discrete wavelengths is observed.

Photons of only certain allowed frequencies can be absorbed or emitted as the electron changes energy state.


• Niels Bohr adopted Planck’s assumption and explained these phenomena in this way:1. Electrons in an atom can only

occupy certain orbits (corresponding to certain energies).


• Niels Bohr adopted Planck’s assumption and explained these phenomena in this way:2. Electrons in permitted orbits

have specific, “allowed” energies; these energies will not be radiated from the atom.


• Niels Bohr adopted Planck’s assumption and explained these phenomena in this way:3. Energy is only absorbed or

emitted in such a way as to move an electron from one “allowed” energy state to another; the energy is defined by

E = h

The Nature of EnergyThe energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation:

E = −RH ( )1nf

2

1ni

2-

where RH is the Rydberg constant, 2.18 10−18 J, and ni and nf are the initial and final energy levels of the electron.

SAMPLE EXERCISE 6.4 Electronic Transitions in the Hydrogen Atom

Using Figure 6.13, predict which of the following electronic transitions produces the spectral line having the longest wavelength: n = 2 to n = 1, n = 3 to n = 2, or n = 4 to n = 3.

Solution The wavelength increases as frequency decreases ( = c/. Hence the longest wavelength will be associated with the lowest frequency. According to Planck’s equation, E = h, the lowest frequency is associated with the lowest energy. In Figure 6.13 the shortest vertical line represents the smallest energy change. Thus, the n = 4 to n = 3 transition produces the longest wavelength (lowest frequency) line.

Answers: (a) emits energy, (b) requires absorption of energy

PRACTICE EXERCISEIndicate whether each of the following electronic transitions emits energy or requires the absorption of energy: (a) n = 3 to n = 1; (b) n = 2 to n = 4 .

It absorbs energy.

Considering only the n = 1 to n = 5 states in the hydrogen atom, which transition will emit

the most energy?

• n = 5 to n = 4• n = 5 to n = 2• n = 3 to n = 1• n = 3 to n = 2• n = 4 to n = 2

• n = 5 to n = 4• n = 5 to n = 2• n = 3 to n = 1• n = 3 to n = 2• n = 4 to n = 2


the most energy?


the longest wavelength?

• n = 5 to n = 4

• n = 5 to n = 2

• n = 3 to n = 1

• n = 3 to n = 2

• n = 4 to n = 2

• n = 5 to n = 4

• n = 5 to n = 2

• n = 3 to n = 1

• n = 3 to n = 2

• n = 4 to n = 2


the longest wavelength?

What is the wavelength for radiation transmitted by WGR, 550 AM (550 kHz)?

1. 0.545 m2. 5.45 m3. 545 m4. 0.0018 m5. 1.8 m

Correct Answer:

1. 0.545 m2. 5.45 m3. 545 m4. 0.0018 m5. 1.8 m

)s10(5.5

)ms10(3.015

18

c

545=

Predict which of the following electronic transitions will produce the longest wavelengthspectral line.

1. n = 4 to n = 22. n = 5 to n = 23. n = 5 to n = 34. n = 6 to n = 4

Correct Answer:

1. n = 4 to n = 22. n = 5 to n = 23. n = 5 to n = 34. n = 6 to n = 4

The wavelength increases as frequency decreases. The lowest frequency (longest wavelength) is associated with the lowest energy, and the smallest energy difference here is between n = 6 and n = 4.

Observations that led to the development of quantum mechanics

1. Hot objects emit electromagnetic radiation. (blackbody radiation)

2. Metals eject electrons when struck by a minimum frequency/energy of electromagnetic radiation.(the photoelectric effect)

3. Excited atoms emit electromagnetic radiation of certain frequencies. (atomic emission spectra)

Bellwork- The Photoelectric Effect

A photon with a minimum energy of 4.41x10-19J will cause emission of an electron from sodium metal.

a)What frequency and wavelength of light does this correspond to?

b)What type of light is it?

c)Describe what happens if yellow light hits sodium metal.

d)Describe what happens when UV light irradiates Na(s)

E=hv v=6.66x1014s-1

c=λν λ=4.50x10-7m λ=450nm

Visible, turquoise!

Sodium electrons may absorb some photons/energy, but it is not enough to help the e- escape

Sodium electrons may absorb some photons, and then they will have more than enough energy to escape (n= ∞). The electron is FREE! Extra energy = KE.

The Wave Nature of Matter

• Louis de Broglie posited that if light can have material properties, matter should exhibit wave properties.

• He demonstrated that the relationship between mass and wavelength was

=h

mv

The Uncertainty Principle

• Heisenberg showed that the more precisely the momentum of a particle is known, the less precisely is its position known:

• In many cases, our uncertainty of the whereabouts of an electron is greater than the size of the atom itself!

(x) (mv) h4

Quantum Mechanics

• Erwin Schrödinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated.

• It is known as quantum mechanics.

Quantum Mechanics

• The wave equation is designated with a lower case Greek psi ().

• The square of the wave equation, 2, gives a probability density map of where an electron has a certain statistical likelihood of being at any given instant in time.

Quantum Numbers

• Solving the wave equation gives a set of wave functions, or orbitals, and their corresponding energies.

• Each orbital describes a spatial distribution of electron density.

• An orbital is described by a set of three quantum numbers.

Principal Quantum Number, n

• The principal quantum number, n, describes the energy level on which the orbital resides.

• The values of n are integers ≥ 0.

Azimuthal Quantum Number, l

• This quantum number defines the shape of the orbital.

• Allowed values of l are integers ranging from 0 to n − 1.

• We use letter designations to communicate the different values of l and, therefore, the shapes and types of orbitals.

Azimuthal Quantum Number, l

Value of l 0 1 2 3

Type of orbital s p d f

Magnetic Quantum Number, ml

• Describes the three-dimensional orientation of the orbital.

• Values are integers ranging from -l to l:

−l ≤ ml ≤ l.

• Therefore, on any given energy level, there can be up to 1 s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, etc.

Magnetic Quantum Number, ml

• Orbitals with the same value of n form a shell.• Different orbital types within a shell are

subshells.

s Orbitals

• Value of l = 0.• Spherical in shape.• Radius of sphere

increases with increasing value of n.

s Orbitals

Observing a graph of probabilities of finding an electron versus distance from the nucleus, we see that s orbitals possess n−1 nodes, or regions where there is 0 probability of finding an electron.

p Orbitals

• Value of l = 1.• Have two lobes with a node between them.

d Orbitals• Value of l is 2.• Four of the five

orbitals have 4 lobes; the other resembles a p orbital with a doughnut around the center.

What is the maximum number of orbitals described by the quantum numbers:

• 1• 3• 5• 8• 9

n = 3 l = 2

What is the maximum number of electrons described by the quantum number:

• 7• 14• 16• 32• 48

n = 4

How many nodal planes does a d orbital have?

• 0• 1• 2• 3• 4

Which of the following is not an allowed set of quantum numbers?

1. n = 3, l = 1, ml = 1

2. n = 3, l = 3, ml =23. n = 4, l = 0, ml = 0

4. n = 3, l = 1, ml =1

Correct Answer:

1. n = 3, l = 1, ml = 1

2. n = 3, l = 3, ml =23. n = 4, l = 0, ml = 0

4. n = 3, l = 1, ml =1

The value of l can be no larger than n 1.

How many electrons maximum can exist in the 4d orbital?

1. 2 2. 63. 104. 14

Correct Answer:

1. 2 2. 63. 104. 14

There are five 4d orbitals, each of which can contain up to 2 electrons each for a total of 10 maximum.

The electron subshell 3p represents the principal quantum number n = ___ and azimuthal quantum number l = ___.

1. 3; 2 2. 2; 33. 3; 14. 1; 3

Correct Answer:

1. 3; 2 2. 2; 33. 3; 14. 1; 3

The principal quantum number n is 3, and a p orbital indicates l = 1.

The maximum number of electrons in an atom that can exist in the 4f subshell and have ml = 1 is

1. 2 2. 73. 144. 32

Correct Answer:

1. 2 2. 73. 144. 32

For any value of ml, the maximum number of electrons is 2.

Energies of Orbitals

• For a one-electron hydrogen atom, orbitals on the same energy level have the same energy.

• That is, they are degenerate.

Energies of Orbitals

• As the number of electrons increases, though, so does the repulsion between them.

• Therefore, in many-electron atoms, orbitals on the same energy level are no longer degenerate.

Spin Quantum Number, ms

• In the 1920s, it was discovered that two electrons in the same orbital do not have exactly the same energy.

• The “spin” of an electron describes its magnetic field, which affects its energy.

Spin Quantum Number, ms

• This led to a fourth quantum number, the spin quantum number, ms.

• The spin quantum number has only 2 allowed values: +1/2 and −1/2.

Pauli Exclusion Principle

• No two electrons in the same atom can have exactly the same energy.

• For example, no two electrons in the same atom can have identical sets of quantum numbers.

Electron Configurations

• Distribution of all electrons in an atom.

• Consist of Number denoting the

energy level




energy levelLetter denoting the type

of orbital.




energy level.Letter denoting the type

of orbital.Superscript denoting the

number of electrons in those orbitals.

QuickTime™ and aSorenson Video decompressorare needed to see this picture.

Orbital Diagrams

• Each box represents one orbital.

• Half-arrows represent the electrons.

• The direction of the arrow represents the spin of the electron.

Hund’s Rule

“For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.”

Periodic Table

• We fill orbitals in increasing order of energy.

• Different blocks on the periodic table, then correspond to different types of orbitals.

What are the valence electrons of vanadium?

• 4s2

• 3d3

• 4s2 3d3

• 3d5

• 4d5

What are the valence electrons of gallium?

• 4s2

• 4p1

• 4s2 3d10 4p1

• 4s2 3d10

• 4s2 4p1

How many unpaired electrons does selenium have?

• 0• 2• 4• 6• 8

The following orbital diagram represents the electron configuration of which element?

1. Carbon2. Nitrogen3. Oxygen4. Neon

1s 2s 2p

Correct Answer:

1. Carbon2. Nitrogen3. Oxygen4. Neon

1s 2s 2p

What is the characteristic outer shell electron configuration of the noble gases?

1. s2p6

2. s2p5

3. s2p4

4. s2p3

Correct Answer:

1. s2p6

2. s2p5

3. s2p4

4. s2p3

Noble gases have completely filled s and p orbitals, hence their lack of reactivity.

Some Anomalies

Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row.

Some Anomalies

For instance, the electron configuration for copper is

[Ar] 4s1 3d5

rather than the expected

[Ar] 4s2 3d4.

Some Anomalies

• This occurs because the 4s and 3d orbitals are very close in energy.

• These anomalies occur in f-block atoms, as well.

How many unpaired electrons does chromium have?

• 0• 2• 4• 6• 8

Which of the following orbital diagrams obeys Hund’s rule for a ground-state atom?1.

2.

3.

3s 3d

3s 3d

3s 3d

Correct Answer:

1.

2.

3.

3s 3d

3s 3d

3s 3d

Chapter 6 Lecture- Electrons in Atoms

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Transcript of Chapter 6 Lecture- Electrons in Atoms