1

ELECTRON STRUCTURE

I. Experimental Approach A. Periodic Law.

1. The physical and chemical properties of elements are periodic functions of their atomic numbers. Since the atomic numbers are equal to the number of electrons in a neutral atom, the properties are functions of the number of electrons. a. Group 17 (Halogens): 1) Reactive nonmetals. 2) Form -1 ions. 3) Exist as diatomic molecules in their elemental forms (F2, Cl2, etc.) b. Group 18 (Inert Gases): 1) Unreactive monatomic gases. 2) No compounds are known for the lighter elements (He and Ne). 3) Heavier members, such as Xe, form compounds only with strong nonmetals such as O and F. c. Group 1 (Alkali Metals): 1) Soft, reactive metals. 2) Form +1 ions.

2. Chemical and physical properties (except for mass) are due to electron structure. Those elements with similar properties, that is, that are in the same Group in the Periodic Table, must have similar electron structures. B. Emission Spectra of elements. 1. Line vs. continuous spectra. a. Spectrum 1) When electromagnetic radiation (light) passes through a prism or impinges on a grating, the light is separated into its different wavelengths (colors), 2) The spectrum is the display of the radiation after it has been separated into its component wave lengths. (Plural = spectra) b White light gives a continuous spectrum. That is, one that contains all wave lengths (colors) merging into one another. c. A line spectrum contains only several discrete wavelengths. 2. Light obtained from energized atoms exhibits a line spectrum. Lines are independent of the particular isotope of the element. Therefore, line spectra are functions of the electron structures of atoms.

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C. Consideration of Light. Transmission of energy through space. Two ways that energy can be transferred, through wave disturbance that is propagated through space or by a particle that moves through space. 1. Light as a Wave Phenomenon. a. Radiant energy (light) - an oscillating electric and magnetic field (force) that is propagated through free space at a speed of 2.9979x108 m/s. b. Some wave properties. At some instant in time, the wave resembles a series of crests and valleys that are evenly spaced.

Distance

+A

–A

0

λ λ λ λ λ

c. Some parameters that describe a wave.

1) Wavelength (λ) - distance between repeating units of the wave. This can be expressed in any convenient unit of length. The SI unit is the meter. 2) Amplitude (A) - value of the wave disturbance. Light is an oscillating electric and magnetic field, therefore, the amplitude is the field strength. Since the field is a vector quantity, it can be positive or negative. The intensity of the radiation is directly proportional to A2. 3) Frequency (ν) - Number of cycles per second. The wave packet moves through space at its speed of propagation (C). Therefore, if one were to monitor the field at one point over a period of time, it would appear as if the field was going through a series of complete cycles as the wave passed the point. The number of complete cycles per second that one would observe is the frequency. The dimension of frequency is s-1 or Hertz (Hz).

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4) The wavelength, frequency and speed of propagation are related by the equation: λν = C ( if C is in m/s, λ must be in meters)

For light, (λ in m)(ν) = 2.9979x108 m/s d. Electromagnetic Spectrum Mode of Type of Radiation Wavelength Range Frequency Range (Hz) Generation AM radio waves greater than 1m less that 3x108 Electrical oscillations Microwaves 1m to 1mm 3x108 to 3x1011 Radar, FM radios Infrared (IR) 1mm to 750 nm 3x1011 to 4x1014 Heat Visible light 750 nm to 400 nm 4x1014 to 7.5x1014 Electronic red (610 - 750 nm) green (500 - 570 nm) orange (590 - 610 nm) blue (450 - 500 nm) yellow (570 - 590 nm) violet (400 - 450 nm) Ultraviolet (UV) 400 nm to 10 nm 7.5x1014 to 1x1016 Electronic X-rays 50 nm to 0.01 nm 6x1015 to 3x1019 Electronic γ - rays less than 0.05 nm greater than 6x1018 Nuclear 2. Light as a corpuscular phenomenon. a. Max Planck-black body radiation. 1) Black body = perfect absorber and emitter of radiation. 2) To account for the energy vs. wavelength distribution emitted from a black body, Planck had to assume that radiant energy was not emitted in a continuous fashion but in "bundles" of energy, called quanta. (Singular is quantum from the Latin for "amount") 3) The energy, E, of a quantum of radiant energy is proportional to the frequency of the radiation.

E = hν h = Planck's constant = 6.62618x10-34 Js (or J/Hz) b. Albert Einstein-photoelectric effect. 1) Photoelectric effect-when light of a particular wavelength shines on certain metals, electrons, called photoelectrons, are emitted. 2) Measurements showed that the average kinetic (KE) of the photoelectrons was directly

4

proportional to the frequency of the light and the number of photoelectrons depended on the intensity of the radiation.

KE ∝ (ν - νo) where νo = the minimum frequency for electrons to appear. 3) Einstein used Planck’s quantum theory to explain. - an electron in the metal absorbs a quantum of energy = hν. -some of the energy, equivalent to hνo, goes into bringing the electron to the surface of the metal. This energy is called the work function of the metal. - the rest of the energy, hν - hνo, goes into the kinetic energy of the photoelectron. - ∴ KE = hν - hνo and the slope of a plot of KE vs. ν should equal h. c. Radiant energy is emitted and absorbed by matter in term of quanta. 1) Photon = a quantum of electromagnetic radiation. 2) The energy of the photon, E = hν = hC

!

D. Line Spectrum of Hydrogen. 1. The frequencies of the photons emitted form energized hydrogen atoms can be grouped into families, called series. A simple relationship was found to exist for the photons in the series. 2. The frequencies of the major lines in the spectrum of hydrogen could be generated from the equation

! = 3.29x1015

n1

2

1 -

n2

2

1

This equation is called the Rydberg equation. It is an empirical equation that was found to

reproduce the correct frequencies.

Series n1 n2 Spectral Region Lyman 1 2,3,4,5... Ultraviolet Balmer 2 3,4,5,6... Visible Paschen 3 4,5,6,7... Infrared Brackett 4 5,6,7,8... Infrared 3. Examples a. First three lines in the Lyman series n1 n2 ν,s–1 λ, nm E, J 1 2 2.47x1015 121.6 1.63x10–18 1 3 2.92x1015 102.6 1.94x10–18 1 4 3.08x1015 97.3 2.04x10–18 b. First three lines in the Balmer series. n1 n2 ν,s–1 λ, nm E, J 2 3 4.58x1014 656.5 3.03x10–19 2 4 6.17x1014 486.3 4.09x10–19

2 5 6.91x1014 434.2 4.58x10–19

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c. First three lines in the Paschen series. n1 n2 ν,s–1 λ, nm E, J 3 4 1.60x1014 1875.8 1.06x10–19 3 5 2.34x1014 1282.3 1.55x10–19 3 6 2.74x1014 1094.2 1.82x10–19 Any theory of the electron structure of hydrogen must be able to explain this equation.

II. Bohr Theory - old quantum theory A. Bohr theory of the electron structure of hydrogen. 1. When the hydrogen atom is energized, the electron absorbs energy. The electron will later lose that extra energy by emitting photons of light. a. Since only certain specific photons are emitted, the electron can have only certain, restricted energies in the atom. b. The electron can be thought of as existing in certain energy states or energy levels. It changes its energy by going from one energy state to another. The only way for an electron to change its energy is for it to go from one energy state to another. The transition must be complete and instantaneous. c. The energies of the photons emitted in the line spectrum of hydrogen are the same as the differences in energies between the different electron energy states of the

hydrogen atom. We can use the Rydberg equation to calculate the frequencies of these photons.

2. Bohr assumed that the electron in the hydrogen atom moves about the nucleus in circular orbits. The orbits are the energy states of the electron.

a. In a stable orbit the electron moves at a constant speed and has a constant energy. The centrifugal force of the electron is just balanced by the force of attraction between the

electron's charge and the nuclear charge so that the radius of the orbit, r, does not change. b. Only certain orbits are stable, that is, exist. Bohr had to assume that in a stable orbit, the angular momentum of the electron,

msr = nh

2π

h = Planck’s constant and n is an integer number, called a quantum number. n = 1, 2, 3, 4, .... any positive integer.

1st Bohr orbit n = 1 msr = h

2π

2nd Bohr orbit n = 2 msr = 2 h

2π

3rd Bohr orbit n = 3 msr = 3 h

2π

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c. The total energy, E, of the electron is equal to the sum of its kinetic energy, KE, and its potential energy, PE.

One can combine the equations relating m, s, r and E to give the results:. r = n2 ao

ao is the radius of the first Bohr orbit = 5.29x10-11m or 52.9 pm

E = -2.18x10

-18

n2

J

Note that the only way the electron can change its energy is to go from its orbit to one with a different n.

B. The Bohr model and the line spectrum of hydrogen. 1. Shown below are sketches of the first three Bohr orbits and the general relationship between n, the radius, and the energy.

3. Suppose an electron absorbs energy and goes to a high energy orbital whose quantum

+

n=1

r1=a0

E1=-2.18x10-18

n=2

r2=4a0

E2=-2.18x10-18

4

n=3

r3=9a0

E3=-2.18x10-18

9

n=n

rn=n2a0

En= -2.18x10-18

n2

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number is n2. When the electron then falls back to a lower energy state whose quantum number is n1, the change in energy, ΔE, is

ΔE = E n -

2 E

n

1 =

-2.18x10-18

n22

- -2.18x10-18

n21

ΔE = 2.18x10-18 ( 1n

21 -

1n

22 )

This energy will be lost by emitting a photon whose energy is ΔE = hν ∴ ν = ΔEh

ν = 2.18x10-18

h ( 1n

21 -

1n

22 ) =

2.18x10-18

6.63x10-34 ( 1n

21 -

1n

22 )

ν = 3.29x1015 ( 1n

21 - 1

n22 )

This is the Rydberg equation for hydrogen. (n1 is the quantum number of the orbital where the electron ende up, n2 is the quantum number of the orbit the electron leaves)

3. Origin of the line spectrum of hydrogen.

EN

ER

GG

Y

n = 1

n = 2

n = 3

n = 4

n = 5

Lyman Series

Balmer Series

Paschen Series

Rydberg Series for hydrogen

n = 6

121.6

nm

102.6

nm

97.3

nm

95.0

nm

93.8

nm

656.3

nm

486.1

nm

434.1

nm

410.1

nm

1875.8

nm

1282.3

nm

1094.2

nm

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C. Limitations of the Bohr theory. 1. Cannot account for the line spectra of atoms with more than one electron. Spectra are too complex to be explained on the basis of electrons existing in circular orbits. a. Many attempts were made to modify the theory to take into account the extra lines. For example, the orbits were assumed to be elliptical instead of circular and a second quantum number, l , called the azimuthal quantum number, was introduced to specify the eccentricity of the ellipse. b. Despite these modifications, a satisfactory theory based on electrons in orbits could not be sustained. 2. Could not be extended to give the correct periodicity of the elements. a. Assume that the inert gases have completely filled orbits. He, the first inert gas, has two electrons. Therefore, the first orbit is large enough to accommodate two electrons. b. The circumference of an orbit = 2πr. Since the Bohr orbit radii increase as n2 (r = n2ao), the circumference and the maximum population of the orbits should increase as n2 increases. Therefore, the maximum populations of the orbits should be equal to 2n2. n shell 2n2

1 K 2 2 L 8 3 M 18 4 N 32 c. On this basis, the following elements should be inert gases: 2He, 10Ne, 28Ni, 60Nd ... Only the first two are inert gases. 3. The theory is not in accord with the Heisenberg Uncertainty Principle. a. Heisenberg Uncertainty Principle = it is impossible to measure simultaneously and with complete accuracy both the position (q) and momentum (ms) of a particle in motion. Measurement of position requires that the particle be observed, that is, interacted with a photon. Since the photon has energy, the act of observation effects the momentum of the particle. To determine the position more precisely, photon of shorter wavelength must be used. This effects the momentum more since the photon now has a higher energy. b. Thus, in an experiment, the more precisely we know the position, the less precisely we

know the momentum. This is can be expressed mathematically as (Δq)(Δms) > h

Where Δq is the uncertainty in the position and Δms is the uncertainty in the momentum. c. The Bohr theory attempts to describe the electron as moving in precisely known orbits

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and having a definite momentum This is not consistent with the Uncertainty Principle. d. At best, by making a series of single observations, one can determine the energy of the electron and the probability, or chance, of finding the electron at different points in space. The details of motion of the electron cannot be determined. Therefore, a theory need only give information about energy and probability. IV. Wave Mechanics. Our present theory of electron structure. A. Basis of the new theory. 1. Compton - scattering of photons by electron. a. Compton Effect - When a beam of photons encounters a beam of electrons, the photons are scattered. The scattered photons have different wavelengths from the original photons. The wavelength shift of the scattered photons depends on the angle of scattering. b. To account for the wavelength shift-scattering angle dependence Compton treated the problem as one would treat the problem of two beams of particles colliding, that is, he used the Law of Conservation of Momentum. To do so he had to ascribe a momentum to the photons. c. Momentum of a photon. 1) When a photon of energy E is formed, an amount of mass, m, is lost in accordance with Einstein's special theory of relativity, E = mC2 , where C is the speed of light. 2) The energy of the photon, according to Planck's equation is

E = hν = hCλ

3) Equating these two expressions for the photon energy, one obtains hCλ = mC2 or

mC = hλ = momentum of the photon. Using

hλ as the momentum of the

photon and the Law of Conservation of Momentum, Compton was able to account for the details of the photon scattering experiments. 4) m = mass equivalent of the photon = mass lost when the photon was created. 2. De Broglie - wave character of particles. a. Nature is symmetric. If photons, which have wave properties, also have particle-like properties, then particles should exhibit wave-like properties. This idea was proposed by de Broglie in 1925. He proposed that one can ascribe a wavelength, called the de Broglie wavelength to any particle in motion such that

de Broglie wavelength = Λ = h

ms .

Note that this is just a generalization of the Compton expression for the momentum of a

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photon. b Examples: 1) Calculate the de Broglie wavelength of a 1 mg gnat flying at a speed of 1 m/s. m = 1.0x10-6kg v = 1.0 m/s h = 6.63x10-34 Js = 6.63x10-34 kg m2/s

Λ = h

mv = 6.63x10-34 Js

(1.0x10-6 kg)(1 m/s) = 6.6x10-28 m.

This length of is too small ever to be detected.(The radius of a nucleus is 10-15m) 2) Calculate the wavelength of an electron ( mass = 9.1x10-31 kg) moving at a speed of 2.2x106 m/s (speed of the electron in the first Bohr orbit).

Λ = 6.63x10-34kg m2/s

(9.1x10-31 kg)(2.2x106 m/s) = 3.3x10-10 m = 0.33 nm

This is the same wavelength as light in the X-ray region and should be detectable. 3. Davisson and Germer measured the wavelength of the electron by studying its diffraction from a Nickel crystal, acting as a grating. The wavelength was that predicted by the de Broglie relation. 4. The de Broglie wavelength and the Bohr model of the hydrogen atom. a. Bohr made the arbitrary assumption that in a stable orbit the angular momentum,

msr = n h

2π

b. This is a logical requirement if we think of the electron as existing around the nucleus as a standing wave. The circumference of the orbit, 2πr, must be an integer number of wavelengths.

∴ 2πr = nΛ = n h

ms or msr = n h

2π . This was Bohr's assumption.

5. In Wave Mechanics we treat electrons in atoms and molecules as standing matter waves. B. Schrödinger Equation.

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1. All waves (sound, ocean, light, etc.) obey an equation called the general wave equation, which gives the amplitude of the wave as a function of position and time. Erwin Schrödinger was able to deduce the form of the wave equation for matter waves. This equation is

- h2

8π2m [ ∂2Ψ∂2x +

∂2Ψ∂2y +

∂2Ψ∂2z ] + V(x, y, z) Ψ = EΨ

This is a second order differential that can, in theory, be solved. In this equation a. x, y, and z are the spatial coordinates. b. V(x, y, z) is the potential energy of the electron. This is just the electrostatic potential energy of the electron due to the nucleus, and, for many electron atoms, the other electrons. c. Ψ is the amplitude function of the matter wave. It is a function of the coordinates x, y, and z. Ψ is called the wave function. It is associated (or describes) the energy state of the electron.. 2. The solution of the equation yields a series of Ψ, E pairs, one for each energy state of the electron. a. Each Ψ corresponds to a different energy level of the electron and E is the total energy of the electron in that energy state. b. The intensity of a wave is proportion to the square of the amplitude. Therefore, Ψ2 evaluated at any point in space = the probability per unit volume (the probability density) of finding the electron at that point. c. From wave mechanics one can obtain the energies of the electron and, through Ψ2, the probability of finding the electron. This is all a theory need give.

3. In wave mechanics, the basic energy state of the electron is characterized by a particular Ψ. Thus,

they replace the old Bohr orbits. Therefore, the energy state described by Ψ is called an orbital. If

the orbital is known the energy and probability profiles of the electron is known. Note that in wave

mechanics we must abandon any attempt to describe the details of motion of the electron.

V. Orbitals and Quantum Numbers A. The energy state of an electron is designated by four quantum numbers, n, l, ml, ms. Of

these the first three (n, l, ml) specify the orbital. The exact form of depends on the values

of n, l, ml, ms, an orbital = Ψ nl ml

1. Principal Quantum Number, n. a. Similar to the Bohr quantum number. n = 1, 2, 3,… b. n is important in specifying the energy of the orbital and also the size of the orbital. c. All orbitals having the same value of n are said to be in the same main shell. We use a letter designation for the main shells.

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n Shell max. population 1 K 2 [ = (2)12] 2 L 8 [ = (2)22] 3 M 18 [ = (2)32] 4 N 32 [ = (2)42] Note that the maximum main shell population = 2n2. 2. Azimuthal (or subshell) quantum number, l. a. Values of l depend on the value of n. l = 0, 1, 2, 3, … n-1 b. All orbitals with the same values of n and l are said to be in the same subshell. Think of each main shell being subdivided into subshells. A letter designation is used to specify l. l subshell max. population 0 s 2 ( = 2x1) 1 p 6 ( = 2x3) 2 d 10 ( = 2x5) 3 f 14 ( = 2x7) 4 g 18 ( = 2x9) … up by letters 2x(odd number sequence) of the alphabet c. In the absence of strong magnetic fields, the energy of an electron in a many electron atom will depend on both n and l, that is, all electrons in the same subshell have the same energy. We specify the subshell by giving n and l, using letter designations. Examples 1s(n = 1, l = 0), 3p(n = 3, l = 1), 5f(n = 5, l = 3). d. Summary of main shells and subshells. Main Sub- Subshell Main shell n shell l shell population population 1 K 0 1s 2 2 _________________________________________________________________________ 2 L 0 2s 2 1 2p 6 8 _________________________________________________________________________ 3 M 0 3s 2 1 3p 6 18 2 3d 10 _________________________________________________________________________ 4 N 0 4s 2 1 4p 6 2 4d 10 32 3 4f 14 3. Magnetic Quantum Number, ml .

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a. Values depend on l. ml = l, l-1, l-2, l-3 … 0 … -l (or ml = 0, ±1, ±2, ±3, … ±l) There are 2l+1 value of ml. b. In the presence of strong magnetic fields the energy of an orbital depends on n, l, and ml, that is, each orbital will have a different energy. c. No more than two electrons can be in the same orbital (occupy the same orbital). The spatial dependence of Ψ depends only on n, l, ml. Therefore, the two electrons in the same orbital have identical probability profiles, that is, they occupy the same regions in space d. A summary of the orbitals in the first three main shells is given in the table on the

following page.

4. Spin Quantum Number, ms.

a. Ψ (n l ml) , is the solution to the Schrödinger equation and is the fundamental energy

state of the electron. The two electrons in the same orbital have identical energies and occupy the same regions in space. Do they have any distinguishing properties ? Stated in another way, is there any property of substances that cannot be explained on the basis of our model of nuclear structure and electrons existing in orbitals? There is an unexplained property and that is the phenomenon of magnetism. b. In order to explain magnetic behavior one has to assume that associated with each electron there is an intrinsic magnetic moment (and angular momentum) that is independent of its orbital motion. Stated another way, each electron acts as a tiny bar

magnetic with a north and a south pole. The two electrons in the same orbitals have their moments (poles) pointing in opposite directions so that they cancel out one another.

c. A fourth quantum number, called the spin quantum number, ms, is introduced to specify the relative orientation of the intrinsic moment of the electron. Since there are only two relative orientation of the magnetic moments, parallel and antiparallel, there are only

two values of ms, +12 and -

12 .

Arrows (or half arrows) are many times used to show the relative orientations of the

moments, that is, ↑ for ms = 12 and ↓ for ms = -

12 .

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TABLE OF QUANTUM NUMBERS (n, l, and ml) AND ORBITALS

n Shell l Subshell Max pop ml Orbital (ψn l ml)

1 K 0 1s 2 0 ψ 100 = ψ

1s

--------------------------------------------------------------------------------------------------------------------------------------

2 L 0 2s 2 0 ψ 200 = ψ

2s

0 ψ 210 ψ

2p z

1 2p 6 1 ψ 211 ψ

2p x

-1 ψ 211 ψ

2p y

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

3 M 0 3s 2 0 ψ 300 = ψ

3s

0 ψ 310 ψ

3p z

1 3p 6 1 ψ 311 ψ

3p x

-1 ψ 311 ψ

3p y

.

0 ψ 320 ψ

3d z2

1 ψ 321 ψ

3d x2

- y2

2 3d 10 -1 ψ 321 ψ

3d xy

2 ψ 322 ψ

3d xz

-2 ψ 322 ψ

3d yz

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VI. Probabilities - Shapes of orbitals. A. General.

1. Ψ2(n l ml ) = the probability density (probability per unit volume) of finding the electron

in space, a. We are interested in seeing how the probability of finding the electron changes as one moves away from the nucleus. The are two questions to be answered. How does the probability change with distance from the nucleus ? Does it make any difference what direction you go as you move away ? b. The wave function is made up of two parts, a radial part that only depends on the distance from the nucleus, r, and an angular part that only depends on direction. c. We will consider the radial part and the angular parts separately. 2. It turns out that the radial part of the wave function depends only on the quantum numbers n and l, while the angular part depends only on l and ml. B. Radial Part - Probability and distance.

1. R depends on n and l 2. As n increases the best chance of finding the electron occurs at greater and greater

distances from the nucleus. 3. Size of the orbital increases as n increases. His was also seen in the Bohr theory. 4. Some plots of the radial parts of the wavefunction are shown below.

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C. Angular Part - Directional preference. 1. l = 0, ml= 0 s orbitals. a. All s orbitals are spherically symmetric. This means that the probability is the same irrespective of what direction you move away from the nucleus. b. Using the x, y, and z Cartesian axes to specify direction, the angular part of an s orbital wave function would plot as a sphere centered at the nucleus.

++

++

Z

X

Y

2. Plots of the angular parts of the p and d orbitals are shown on the following pages. The best physical analogue to use in describing these probability plots is a cloud. Just as the probability may be large in some regions in space and small in others, a cloud can be dense in some regions and sparse in others. We will refer to the "shape of the electron cloud" or "the extension of the electron cloud in space".

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3. l = 1, ml = 0, ±1 p orbitals.

pz px py

ZZ

Z

X

YX

Y

X

Y

ZZ Z

XXXY Y Y

! ! !

18

4. l = 2, ml = 0, ±1, ±2 d orbitals.

dxzdxy dyz

dz2 dx2- y2

ZZ Z

ZZ

X

Y

XY

X

Y

XY

X

Y

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Electron Structure Problems

1. Fill in the missing information about the following photons. Energy (J) λ (nm) ν (Hz) Momentum (kg m/s) Mass Equivalent (kg) _______ 50.3 ________ ______________ _______________ _______ _______ 5.87x1016 ______________ _______________ 5.69x10-18 _______ _________ ______________ _______________ __________ ________ _________ 7.90x10-28 _______________ __________ ________ _________ ______________ 5.01x10-36

2. The Pfund series in the line spectrum of hydrogen lies in the far infrared region of the spectrum and is due to transitions of the electron to the fifth main energy state of hydrogen. a) Calculate the three longest wavelengths of the photons in this series. b) Calculate the frequency of the highest energy photon in this series.

3. Use the Bohr to equation to calculate the ionization energy of hydrogen in kJ/mol. (The ionization energy is the energy required to remove an electron from it lowest energy state to the state where n=∞)

4. Calculate the de Broglie wavelength of a neutron moving at a speed of 3.8x102 m/s. 5. The average speed of H2 molecules at room temperature is about 1.9x103 m/s. Calculate the de Broglie wavelength of H2. 6. How fast must an electron be moving so that it has numerically the same wavelength as a photon whose frequency is 3.0x1019 ? 7. Show the filling of the electrons in the 3d subshell of Fe. Use arrows to indicate the direction of the magnetic moments of the electrons. What type of magnetism will Fe exhibit ?

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8. Shown below are sketches of the shapes of some orbitals in the main shell where n=4. Identify each orbital.

Z

X

Y

Z

Y

Z

X

Y

Z

X

Y

X

9. Consider the main shell where n=5. a. List the values of l for the subshells in this main shell and the letter designation and maximum populationof each subshell. b. For the subshell with the highest l list the different of ml.

21

Electron Structure

Answers to Problems 1. Fill in the missing information about the following photons. Energy (J) λ (nm) ν (s-1) Momentum (kg m/s) Mass Equivalent (kg) 3.95x10-18 50.3 5.96x1015 1.32x10-26 4.39x10-35 3.89x10-17 5.11 5.87x1016 1.30x10-25 4.32x10-34 5.69x10-18 35.0 8.58x1015 1.90x10-26 6.32x10-35 2.37x10-19 839 3.57x1014 7.90x10-28 2.63x10-36 4.51x10-19 441 6.80x1014 1.50x10-27 5.01x10-36

2. The Pfund series in the line spectrum of hydrogen lies in the far infrared region of the spectrum and is due to transitions of the electron to the fifth main energy state of hydrogen. a) Calculate the three longest wavelengths of the the photons in this series. b) Calculate the frequency of the highest energy photon in this series. a) 7.46x10-6m, 4.65x10-6m, 3.74x10-6m b) 1.32x1014 s-1 3. Use the Bohr to equation to calculate the ionization energy of hydrogen in kJ/mol. (The ionization energy is the energy required to remove an electron from it lowest energy state to the state where n=∞ (6.63x10-34)(3.29x1015)[ 1/12 - 1/∞] = 2.18x10-18 J 4. Calculate the de Broglie wavelength of a neutron moving at a speed of 3.8x102 m/s. 1.0x10-9 m or 1.0 nm

5. The average speed of H2 molecules at room temperature is about 1.9x103 m/s. Calculate the de Broglie wavelength of H2. 1.0x10-10 m or 0.10 nm

6. How fast must an electron be moving so that it has numerically the same wavelength

as a photon whose frequency is 3.0x1019 ? wavelength = 1.0x10-11m ; speed = 7.3x107 m/s 7. Show the filling of the electrons in the 3d subshell of Fe. Use arrows to indicate the direction of the magnetic moments of the electrons. What type of magnetism will Fe exhibit ? 3d6 ↓↑ ↑ ↑ ↑ ↑ paramagnetic with 4 unpaired electrons

22

8. Shown below are sketches of the shapes of some orbitals in the main shell where n=4. Identify each orbital. 4dz2 4s

Z

X

Y

Z

Y

Z

X

Y

Z

X

Y

X

4dyz 4pY 9. Consider the main shell where n=5. a. List the values of l for the subshells in this main shell and the letter designation and maximum populationof each subshell. l = 0 1 2 3 4 5s 5p 5d 5f 5g max. population 2 6 10 14 18 b. For the subshell with the highest l list the different of ml. 4,3,2,1,0,-1,-2,-3,-4