Chapter 6. Electronic Structure of Atoms.

6.1 The wave nature of Light.

Much of our present understanding of the electronic structure of atoms and molecules comes from light absorbed or emitted by substances. Light has a wave nature with wavelength and frequency

Frequency measures how many wavelengths pass a point per second:

1 s

4 x ÷ 1 s = 4 s-1 = 4 Hz

Light of different wavelengths and frequencies:

medium wavelength

short wavelengthhigh frequency

long wavelengthlow frequency

Relationship between speed (c), frequency (ν), and wavelength (λ) of light:

For light, we have:

c = νλ

where c is the speed of light (3 x 108 m/s), ν is the frequency, and λ is the wavelength. Frequency is expressed in cps (cycles per second), also called Hz (‘Herz’). 1 Hz = 1 s-1 or 1/s.

What is the wavelength, in m, of radiowaves transmitted by the local radio station WHQR at 91.3 MHz? (speed of light = 3 x 105 km/s)

91.3 MHz = 91.3 MHz x 106 Hz = 91.3 x 106 Hz 1 MHz

c = νλ, so λ = c/ν

λ = 3 x 105 km x 1000 m x ___1___ s s km 91.3 x 106 1

= 3.29 m

1 Hz =1/s

The electromagnetic spectrum:

Red Orange Yellow Green Blue Indigo Violet

wavelength ()

frequency ()

energy (E)

6.2. Quantized Energy and Photons:

Classical physics cannot explain:

1) black-body radiation

2) the emission of electrons from metal surfaces on which light shines (the photo-electric effect)

3) the emission of light from electronically excited gas-atoms.

Black-body radiation:

In physics, a black body is an object that absorbs all electromagnetic radiation that falls onto it. In the 19th century physicists could not explain why the frequency of light emitted by a black-body went up as the temperature went up. They expected that the frequency should remain the same and the intensity should just increase. We can see this in practice, because objects become red-hot, and then blue and finally white hot.

Iron heated in a furnace formaking wrought iron.

Blackbody radiation at different temperatures:

Planck’s Quantum theory:Planck solved the black-body problem by proposing that light was emitted in packets that he called quanta, and that the energy of a quantum of light was related to its frequency (v) by:

E = hν

where h is Planck’s constant, which has a value of 6.626 x 10-34 J-s. (Dimensional analysis: J-s x s-1 = J). Max Planck

What is the energy of a photon of electromagnetic radiation that has a frequency of 400 kHz?

hE

kHzJs 4001063.6 34

kHz

Hz1000Hz

s 1

= 2.65 x 10-28 J

Energy of a photon: Calculate energy of light of one photon with a wavelength of 589 nm.

We have c = νλ, so ν = c/λ

v = 3.00 x 108 m/s

589 x 10-9 m

= 5.09 x 1014 s-1.

E = 6.626 x 10-34 J-s x 5.09 x 1014 s-1

= 3.37 x 10-19 J

The photoelectric effect:

metalsurface

evacuatedchamber

currentindicator

positiveterminal

powersource

+ -

electromagnetic radiation

e

If energy of radiationis high enough, anelectron is excited tojump to the positiveterminal

+

The Photoelectric effect and Photons.

In 1905 Albert Einstein used Planck’s quantum theory to explain the photoelectric effect. For each metal there is a minimum frequency of light below which no electrons are emitted. He assumed light came in packets called quanta, and that these need a minimum energy E before they could dislodge an electron.

Albert Einstein

6.3. Line Spectra and the Bohr Model:

When a gas is placed in very low concentration in a vacuum, and excited, light at only a few specific wavelengths is produced. This is called a spectrum. Why this would be could not be explained.

line spectrum of the H-atom in the visible region, obtained from moist H2 gas in a tube with an AC current, due to H-atoms formed bybreak-up of H2 molecules.

400 700 nm

Fraunhofer (1815) lines in the sun:

The sun appears to produce a continuous spectrum, but this is only because it has so many different nuclei being excited. Careful examination shows that it is a superposition of many different spectra. Even a ‘white’ fluorescent lamp has spectral lines:

Emission-line spectrum from a standard fluorescent lightfixture. This shows the sharp, bright emission lines ofMercury plus broad bands from the organic phosphors coating the inside of the tube.

The Rhydberg Equation:

Sodium vapor, or H atoms, give characteristic lines. It had been discovered by Rydberg that the lines in the H-atom spectrum could be predicted by an empirical equation:

1/λ = (RH) 1 - 1 n1

2 n22

Where λ is the wavelength of the line in the H-atom spectrum, RH is the Rydberg constant (1.096776 x 107 m-1), and n1 and n2 are positive integers with n2 > n1.It took nearly thirty years to explain this simpleequation.

Example:Calculate the wavelength in nm of the band in the spectrum of

atomic hydrogen that occurs with n1 = 2 and n2 = 3.

1/λ = (RH) 1 - 1

n12 n2

2

= 1.096776 x 107 m-1 x (1/(22) – 1/(32))= 1.096776 x 107 m-1 x (1/4 – 1/9)

λ = 1/(1.096776 x 107 m-1 x (0.13889))= 6.56 x 10-7 m = 6.56 x 10-7 m x 1 nm

10-9 m= 656 nm

410 434 486 656 nm

The Bohr Model of the Atom.

According to classical physics, an electron moving in a circular orbit as in the Rutherford model should continuously radiate away energy and spiral into the nucleus. Bohr got around this problem with three postulates:1. Only orbits with definite radii are permitted2. An electron in a specific orbit has an allowed energy3. Energy is emitted or absorbed only when it changes from one state to another.

(1885-1962)

Niels Bohr

6.3 Line Spectra and the Bohr Model.

Recall that there are three important properties that classical physics cannot explain:

1) Black-body radiation. Why does the wavelength of light emitted shorten as the temperature of the light emitter goes up? This is explained by Planck’s relationship (before this time the relationship between the energy of light and its frequency was not understood). E = hν

2) The photoelectric effect. Why does each metal have a minimum frequency of light required to excite electrons from its surface? Light of lower frequency, no matter how intense, excites no electrons. This is answered by Einstein using Planck’s equation. Each photon must have a minimum energy to excite an electron.

3) The existence of line spectra. Why do atoms in e.g. Na vapor or H-atoms, emit light as line spectra? Why do the lines in the H-atom spectrum fit the Rhydberg equation?

1/λ = (RH) 1 - 1

n12 n2

2

Bohr calculated the value of RH in the Rydberg equation from the physics of motion and interacting charges, and found that he could reproduce it exactly.

The energies of transitions in the H-atom are given by:

ΔE = hν = hc/λ = (-2.18 x 10-18J)(1/nf2 – 1/ni

2)

The ni is the principal quantum number of the initial state, and nf of the final state. Thus transitions can occur when nf is greater than ni, i.e. energy is absorbed, or where nf is less than ni, when light is emitted.

Niels Bohr was the first to offer an explanation for line spectra:

electron orbits

Bohr’s Model of the Hydrogen Atom

n = 1n = 2n = 3n = 4n = 5n = 6

nucleus

In the Bohrmodel of theH-atom, theorbits closestto the nucleus(n = 1) are oflowest energy.As the energyincreases,n increases.The line spectracorrespond totransitions between theseorbits.

Energy levels in the Bohr model of the H-atom

Bohr’s Model of the Hydrogen Atom

n = 6n = 5n = 4

n = 3

n = 2

n = 1

Energy

Ground State

nucleus

eabsorption of a photon

hv

absorption ofa photon aselectron isexcited tohigher energy(n = 3) state

n = 6n = 5n = 4

n = 3

n = 2

n = 1

Energy

Ground State

nucleus

e “excited state”

Bohr’s Model of the Hydrogen Atom

n = 6n = 5n = 4

n = 3

n = 2

n = 1

Energy

Ground State

nucleus

e ground state

Bohr’s Model of the Hydrogen Atom

hv

emission ofa photon aselectronreturns toground state

Transitions corresponding tothe Balmer series

n=6 → n=2

n=5 → n=2

n=4 → n=2

n=3 → n=2

Figure 6.13.

Problem: using Fig. 6.13, predict which of the following transitions produces the longest wavelength. n = 2 to n = 1, n = 3 to n = 2, or n = 4 to n = 3? Ans: n = 4 to n = 3.

Figure 6.13.

Problem: using Fig. 6.13, predict which of the following transitions produces the longest wavelength. n = 2 to n = 1, n = 3 to n = 2, or n = 4 to n = 3? Ans: n = 4 to n = 3.

The energies of individual levels in the hydrogen atom

Note: One can calculate the energy of any level in the H-atom from the Bohr model. For a level with quantum number n, the energy is given by:

En = - RH x (1/n2)

Limitations of the Bohr model.

The Bohr model only works for atoms/ions with a single electron (e.g. H or He+). It cannot account for the more numerous lines in multi-electron atoms such as Na or He. It does, however, make two important postulates:

1) electrons exist only in discrete energy levels, and

2) energy is involved in moving an electron from one level to another. Energy is absorbed as photons to excite the electron from one level to a higher energy level, and is emitted as photons in dropping to a lower energy level.

6.4. The wave behavior of matter.

Louis de Broglie proposed that matter also had a wave property. For any particle

λ = h/mv (I have colored ‘v’ (velocity) red to distinguish it from ‘v’ (frequency)

(E = hv, E = mv2, so hv = mv2, λ = v/v, so λ = hv/mv2)

where m is its mass, and v is its velocity. The quantity mv is its momentum. The electron thus is not only a particle, but also a wave. This was confirmed in that electrons can be diffracted by crystals. Since electrons are waves, it is inappropriate to treat them only as particles, as in the Bohr model of the atom.

Problem:What is the wavelength of an electron moving at a speed of 5.97 x 106 m/s? (h = 6.63 x 10-34 J-s and 1 J = 1 Kg-m2/s2, mass of electron = 9.11 x 10-28 g)

λ = h/mvmass of electron in kg = 9.11 x 10-28 g x 1 kg

1000 g = 9.11 x 10-31 kg

λ = 6.63 x 10-34 J-s______ x 1 kg-m2/s2

9.11 x 10-31 kg x 5.97 x 106 m/s 1 J

= 1.22 x 10-10 m or 0.122 nmNote: for an ordinary object like a golf ball m is so large that λ is too shortto measure.

The uncertainty principle.

If the electron is a wave, we cannot state its position with any accuracy. Werner Heisenberg postulated the uncertainty principle, which states that we cannot precisely know the exact momentum of an electron and also its location in space.This means that we cannot use the Bohr model of the atom which has the electron as a particle circling the nucleus. We must instead consider it as a wave.

Werner Heisenberg (1901-1976)

6.5. Quantum mechanics and Atomic Orbitals.

In 1926 Erwin Schrodinger proposed the wave equation which led to quantum mechanics. All we can predict is a probability density (Ψ2) (psi-squared) for the electron.

Erwin Schrödinger (1887-1961)

Orbitals and Quantum numbers:The solution to the Schrodinger wave-

equation leads to a set of wavefunctions that yields 4 types of quantum numbers instead of the single quantum number yielded by the Bohr model.

These are:

1) The principal quantum number, n, which has values of 1,2,3,… This corresponds to the quantum number n in the Bohr model of the atom.

Quantum numbers (contd.)

2) The Azimuthal quantum number l, which has values of 0 to (n-1) for each value of n. The different values of l correspond to orbital types as follows:

l = 0 1 2 3letter used = s p d f

3) The magnetic quantum number ml, can have values of –l through 0 to +l for each value of l.

Value of l possible values of ml

0 01 -1,0,+12 -2,-1,0,+1,+23 -3,-2,-1,0,+1,+2,+3

4) The spin quantum number (ms). This can have values of +½ or –½. This means that for each value of ml there are two values of ms. It is this that leads to the occupation of each orbital by two electrons of opposite spin, i.e. with ms = +½ or –½.

These quantum numbers lead to the shells (different values of n) and subshells (different values of l) that lead to our modern understanding of chemistry. The number of orbitals in each sub-shell (1 for s, 3 for p, 5 for d, and 7 for f sub-shells) is determined by ml, and ms determines that only two electrons of opposite spin can occupy each orbital.

shells subshells orbitals

n l orbital ml

_________________________________________________= 0,1,..(n-1) = -l, -(l-1),..0,..(l-1), l

1 0 1s 02 0 2s 0

1 2p -1, 0, +13 0 3s 0

1 3p -1, 0, +12 3d -2,-1, 0, +1, +2

4 0 4s 01 4p -1, 0, +12 4d -2, -1, 0, +1, +23 4f -3, -2, -1, 0, +1, +2,

+3__________________________________________________

Schrödinger’s model:

z

x

y

s-orbital p-orbital(1 of 3)

d-orbital(1 of 5)

f-orbital(1 of 7)

Representation of orbitals:

1s, 2s, 3s, 4s,....

2pz 2py 2px

y

z z z

z

x

x

x x

y

yy

s-orbitals

p-orbitals

x

z z

xy

y

dxz dyz dxy

d(x2-y2) dz2

x

y z

d-orbitals

6.7 Many-electron atoms.

Electron spin and the Pauli exclusion principle.

The Pauli exclusion principle states that no two electrons in an atom can have all four quantum numbers, n, l, ml, and ms, the same. The result of this is that each orbital can hold a maximum of two electrons, with ms = +½ or -½.

For the H-atom, all the sub-shells within the same shell have the same energy. This is why the Bohr model of the atom works for the H-atom. It generates only a single quantum number, n, which predicts only the energy of shells, but not the existence of sub-shells. On the other hand, for a many-electron atom, the sub-shells within a single shell do not have the same energy. This is the cause of the failure of the Bohr model when applied to a multi-electron atom.

Failure of the Bohr model for many-electron atoms.

The H-atom compared to many-electron atoms:

ener

gy

1s

2s 2p

3s 3p 3d

n = 1

n = 3

n = 2

ener

gy

1s

2s

3s

n = 1

n = 3

n = 2

3d

3p

2p

4p

4s

H-atom. All subshells within Many-electron atom. Subshellssame shell have the same energy. within same shell have different energies.

n = 4n = 44s 4p 4d 4f

Section. 6.8 Electron Configurations.

The way in which the electrons are distributed among the various orbitals of the atom is know as the electron configuration. The orbitals are filled in order of increasing energy, two electrons of opposite spin per orbital. This is known as an orbital diagram:

Li

1s 2s

Electron configurations (contd.)

Each orbital is represented by a box. A half-arrow pointing up represents an electron with a positive spin (ms = +½), and arrow pointing down represents

an arrow with a negative spin (ms = -½ ). This is often

referred to as ‘spin-up’ and ‘spin-down’.

Electrons having opposite spins are said to be spin-paired when they are in the same orbital.

1s 2s

arrows pointingin opposite directions =opposite spins

box representsone orbital

Examples of electron configurations:

Electron configurations can be represented by little boxes (one per orbital) with arrows to indicate electron population of the orbitals. Alternatively, these can be written out with superscripts to show number of electrons in each orbital:

Oxygen atom: 1s22s22p4

Phosphorus: 1s22s22p63s23p3

Sodium: 1s22s22p63s1

indicate principal quantum number

indicate no of electrons in sub-shell

Comparison of different representations of electron configurations:

Oxygen atom: 1s22s22p4

Phosphorus: 1s22s22p63s23p3

Sodium: 1s22s22p63s1

1s 2s 2p

1s 2s 2p 3s 3p

1s 2s 2p 3s

Hund’s rule.Orbitals that have the same energy are known as degenerate. Hund’s rule states that for degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.

Thus with phosphorus the three electrons in the 3p sub-shell spread out to occupy it evenly, or the four electrons in the 2p shell of oxygen do the same:

phosphorus:

oxygen: Electrons in highestoccupied sub-shellspread out to occupyit as evenly as possible

3p

2p

Condensed Electron Configurations.

In order to avoid writing out all the electrons in an atom, the electronic configuration is abbreviated by writing only the electrons in the outermost occupied shell, the valence shell. This is called a condensed electron configuration. e.g.

Na: [Ne]3s1

Li: [He]2s1

P: [Ne]3s23p3

Transition metals:

The row ended by Ar marks the beginning of the 4th row (n = 4). In this row we encounter for the first time transition elements, where the 3d orbital is being filled up. Note that the 4s orbital is lower in energy than the 3d orbitals, so that the 3d orbital does not start to fill right away. The first two electrons, for K and Ca, go into this orbital, so we have:

K [Ar]4s1

Ca [Ar]4s2

Number of electrons in the 3d orbitals of transition metal ions:

After the 4s orbital has been filled, we start to fill the 3d orbital in accordance with Hund’s rule, since the 3d sub-shell is the highest energy subshell.

e.g.

Mn: [Ar]4s23d5

Zn: [Ar]4s23d10

Ni: [Ar]4s23d8

Cu: [Ar]4s23d9

Lanthanides and Actinides.

Here we are filling up f-orbitals, although the energies of the 4f and 5d orbitals are very close to each other, and the electrons can also occupy the 5d orbitals:

La: [Xe]6s25d1

Lu: [Xe]6s25f14

6.9. Electron configurations and the periodic table.

Each group of elements in the periodic table has a characteristic electronic configuration. e.g. halogens have the characteristic configuration ns2np5 (n = 2 = F, n = 3 = Cl, n = 4 = Br, n = 5 = I) for the highest energy occupied shell. The valence electrons determine the chemical properties of the elements, and occupy the highest energy shell.

Problem: Write the electronic configuration of Bi

In solving problems of this type use the periodic table as your guide, The orbitals fillup following exactly along the rows in the periodic table with increasing Z:

1s22s22p63s23p64s23d104p65s24d105p66s24f14

5d106p3

1s2s 2p

3s 3p

4s 3d 4p

5s 4d 5p6s 5d 6p

Lanthanides fit in here

Bid-orbitals

1s