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CHEM 1500, Chapt er 7 1 The Quantum Mechanical Atom (pages 250-297)

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CHEM 1500, Chapter 7

1

The Quantum Mechanical Atom(pages 250-297)

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CHEM 1500, Chapter 7

Electromagnetic Radiation

Picture of em radiation wave, Kotz p 297

2

•All forms of radiation can be described in terms of wave-like electric and 

 magnetic fields. All radiation, such as visible light, X-rays, microwaves, and

cell phone signals are collectively known as electromagnetic radiation (EM).•All share common properties.

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CHEM 1500, Chapter 7

The Wave Nature of Light

An EM wave is often depicted as a sine wave that has an:

Amplitude

- height of the wave- related to the intensity or brightness of radiation

 

3

requency  ν    nu- number of cycles per second= s-1 = Hertz (Hz)

Wavelength (λ ) lambda- distance wave travels in 1 cycle

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CHEM 1500, Chapter 7

The Wave Nature of Light cont…

Speed of wave (v or c)

- for light, velocity is denoted c = 3.00 x 108

m/s- and c = λ ν

4

(Figure 7.2, page 252)

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CHEM 1500, Chapter 7

The Electromagnetic Spectrum

EM waves are characterized by frequency or wavelength.

(Figure 7.3, page 255)

Our eyes are only able tosense a very narrow bandof wavelengths fromabout 400 to 700 nm.

5

This band is called thevisible spectrum.

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CHEM 1500, Chapter 7

Practice Exercises (more on page 254)

An FM station broadcasts classical music at 103.4 MHz (103.4 x 106

Hz). Find the wavelength (in m, nm and Å) of these waves.(1 Å = 10-10 m)

c = λ ν3.00 x 108 m/s = λ  (103.4 x 106 s-1)

λ  = 2.90 mλ in nm = 2.90 m x 1 nm = 2.90 x 109 nm

6

 

10-9 mλ in Å = 2.90 m x 1 Å = 2.90 x 1010 Å

10-10 m

The organism that causes tuberculosis can be destroyed by irradiationwith UV light with a wavelength of 254 nm. What is the frequency?

c = λ ν

3.00 x 108 m/s = (254 x 10-9 m)  ν

 ν = 1.18 x 1015 s-1 (Hz)

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CHEM 1500, Chapter 7

Waves versus Particles

View in the late 19th century…

•Matter: Particles with location and momentum can be accurately

described.•Electromagnetic Radiation: No mass, exact position in space not fixed.

Considered distinct and unrelated: Classical view oint .

7

 

Quantum Theory unites the wave/particleproperties of EM radiation.

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CHEM 1500, Chapter 7

Quantized Energy of Light

Max Planck (German physicist) proposed that the energy of EM waves isquantized rather than continuous.

Thus when matter is heated it emits one or more quantum of energy.

quantum = a discrete bundle or packet of energy

8

Planck proposed that a quantum of energy (photon) can be described bythe following equation:

E = h ν

h = Planck’s constant = 6.626 x 10-34 J sec

So…what is the energy of light emitted by a red stop light (λ = 652 nm)?

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CHEM 1500, Chapter 7

Atomic Line Spectra

Atomic line spectra provide evidence that electrons in atoms havequantized energies.

Light bulbs and stars produce radiation containing many differentwavelengths. When the light is passed through a prism, it is separatedinto its component wavelengths.

 

9

This rainbow of colors is called a continuous spectrum.(Figure 7.3, page 255)

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CHEM 1500, Chapter 7

Atoms can absorb energy (when heated) and then emit energy as

electromagnetic radiation.

The emitted radiation can also be examined with a prism. Thepattern of emitted radiation is called the line spectrum of the atom.

(Figure 7.7, page 260)

10

Na atoms emityellow light

Hatoms emitpink light

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CHEM 1500, Chapter 7

• The line spectrum of hydrogen is very simple, only 4 lines in the visible

portion (other lines are in the UV and IR regions).

Rydberg (1888) :

came up with an equation, using 2 integer values,that predicted these wavelengths

what is the significance of this?

11

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CHEM 1500, Chapter 7

Energy of Electrons in Atoms is Quantized

Because atoms emit light of only certain specific frequencies, it must be truethat only certain energy changes are able to take place within the atoms.

How is it that atoms of a given element always undergo exactly the samespecific energy changes?

We sa that the electron is restricted to certain ener levels, and that the

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energy of the electron is quantized.

(Figure 7.8, page 261)

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CHEM 1500, Chapter 7

Niels Bohr accounted for these line spectra by assuming that the changesin energy of the H atom reflect energies of the H’s electron.

Bohr based his model on three postulates:1. Only orbits of certain radii are permitted.

2. An electron in a permitted orbit has a specific energy and is in an‘allowed’ energy state.3. Energy is emitted or absorbed by the electron only as the electron

chan es from one allowed ener to another.

The Bohr Model

13

 

Bohr was able to calculate the energies corresponding to each allowedorbit for the electron using the formula:

E = (-2.18 x 10-18 J)(1/n2)

where: h = Planck’s constantc = speed of lightn = energy level or quantum number (integer 1 to ∞∞∞∞)

n is also known as the principal quantum number.

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CHEM 1500, Chapter 7

Lowest energy level (n) = 1 = ground state

First excited state (n) = 2

n = 2 _______________ -0.545 x 10-18 J _________e-_________

light absorbedlight emitted

excited state

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n = 1 __________e-___ -2.18 x 10-18 J ____________________ ground state

Using Bohr’s equation for the Hydrogen atom:

∆E = Ephoton = -2.18 x 10-18 J 12 _ 12

nfinal2 ninitial

2

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CHEM 1500, Chapter 7

Energy Levels of the Hydrogen Atom

(Figure 7.9, page 263)

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Transitions:

n→1 = ultraviolet

n→2 = visiblen→3 = infrared

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CHEM 1500, Chapter 7

Problems:

1. Calculate the energy of the photon absorbed or emitted in thetransition n = 4 to n = 2 for a hydrogen atom. To what region of the electromagnetic spectrum does this photon belong? Was

the photon absorbed or emitted?

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CHEM 1500, Chapter 7

Problems cont…

2. Calculate the energy of the photon absorbed or emitted in thetransition n = 1 to n = 3 for a hydrogen atom. To what region of the electromagnetic spectrum does this photon belong? Was

the photon absorbed or emitted?

17

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CHEM 1500, Chapter 7

Limitations of the Bohr Model

1. Bohr’s model only worked for the H atom and other one-electronsystems. Could not explain the spectra of other multi-electron atoms.

2. Describes the electron as a small particle circling about the nucleus.De Broglie modified the model by considering the energy levels asstanding waves.

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3. The fixed orbit of each energy level is also incorrect.The electron in a given energy level isnot always at the same distance from

the nucleus.

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CHEM 1500, Chapter 7

Electrons: Properties of Particles and Waves

In 1905 Albert Einstein published his special theory of relativity, in whichhe related energy and mass in his, now famous, equation:

E = mc2

This shows that light, an EM wave, also has the properties typical of particles. This is known as the wave-particle duality.

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If light can be thought of as moving particles, can moving particles of matter such as electrons be thought of as waves? De Broglie suggested thatthe λ of matter is given by:

 λ = h mυ υυ υ 

Where υ = velocity of particle

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CHEM 1500, Chapter 7

Diffraction: Wave-Like Property

The reinforcement ( constructive interference) and cancellation ( destructiveinterference) of wave intensities is a phenomenon called diffraction.

Basis for ElectronMicroscope (like

(Figure 7.11, page 264)

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constructivedestructive

 microscope butwith much shorterwavelengths) !!!

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CHEM 1500, Chapter 7

What are Electrons (according to de Broglie)?

Electrons can be thought of as standing waves (vibrating string on a guitar)of only allowed wavelengths or energies .

(Figure 7.15, page 267)

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CHEM 1500, Chapter 7

Quantum Mechanics and Atomic Orbitals

A better theory of atomic structure :- the electron is viewed as a wave of certain allowed energy- assumes not possible to specify exact position of electron, but

one can calculate the probability of the electron being in someregion around the nucleus.

 

22

n   n

Schrödinger Equation

Wave functions (ψ) = the solutions to the wave equation

The electrons form a standing wave around the nucleus within an orbital.

- I think of it as an electron cloud 

Only certain wavelengths (solutions to the wave equation) are permitted.

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CHEM 1500, Chapter 7

Each orbital is associated with a complete set of quantum numbers.

These are analogous to defining a point on a 3D graph (x, y and z axes)

Quantum Numbers: n, llll, mllll

n (the principal quantum number) = 1, 2, 3 ..... ∞∞∞∞- related to the size of the orbital- related to the energy level of the orbital- all orbitals with the same n are in the same shell

23

as n ↑ , energy ↑ , size ↑

llll (the azimuthal quantum number) = 0, 1, 2 .... (n-1)- also know as angular momentum quantum #

- related to the shape of the orbital- divides the shells into subshells

when n = 1 llll is 0n = 2 llll can be 0 or 1

n = 3 llll can be 0, 1, or 2

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CHEM 1500, Chapter 7

llll orbital type

0 s

1 p

2 d

3 f 

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mllll ( the magnetic quantum number) = 0, ± 1, ± 2, ± 3…. ± l

- related to the orientation of the orbital

In Summary:

when n = 1 llll is 0 mllll

is 0n = 2 llll can be 0 or 1 m

llllis 0 or -1, 0, 1

n = 3 llll can be 0, 1, or 2 mllll

is 0 or -1, 0, 1 or -2,-1,0,1,2

s p d

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CHEM 1500, Chapter 7

(Table 7.1, page 271)

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(Figure 7.18, page 271)

Approximate energy leveldiagram for atoms withmore than 1 electron

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CHEM 1500, Chapter 7

Problem:Give all possible ml values for orbitals that have each of the following:a) l = 3

b) n = 2

c) n = 6, l = 1

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Problem:Are the following quantum number combinations allowed? If not, showtwo ways to correct them:

a) n = 1; l = 0; ml

= 0

b) n = 2; l = 2; ml

= +1

c) n = 7; l = 1; ml

= +2

d) n = 3; l = 1; ml

= -2

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CHEM 1500, Chapter 7

Problem:Which of the following is not an allowed set of quantum numbersand why?

(a) n = 4;l

= 5; ml = -5 (b) n = 4;l

= 2; ml = -3

= = = = = = -

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Problem:How many orbitals can have the following designation: 4p

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CHEM 1500, Chapter 7

Electron Spin and the Pauli Exclusion Principal

• Beam of H atoms get split into two when they go through aninhomogenous magnetic field

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• electron spin is responsible: when an electrically charged object spins,it creates a magnetic moment.

 It behaves like a magnet.

(Figure 7.19, page 272)

C C

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CHEM 1500, Chapter 7

Electron Spin is Quantized

ms (the spin magnetic quantum number): two possible values- indicates the direction an electron spins- either +½ or ↑  or -½ or ↓

• To fully define where an electron in an atom resides, it is required tohave a set of the four quantum numbers (n, l, m

l, and ms).

• Furthermore each electron in the atom must have a uni ue set of these

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four quantum numbers.Pauli Exclusion Principle:

no two electrons in the same atom can have the same four

quantum numbers; each electron has unique set of fourquantum numbers

Pauli Exclusion Principle:no two electrons in the same atom can have the same four

quantum numbers; each electron has unique set of fourquantum numbers

• This means that each orbital can have a maximum of 2 electrons

per orbital (one with ms = + ½ , the other -½).

CHEM 1500 Ch t 7

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CHEM 1500, Chapter 7

Electrons in Orbitals

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CHEM 1500 Chapter 7

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CHEM 1500, Chapter 7

Rules for Assigning Electrons:

• electrons are fed in to the lowest energy available orbitals ( Aufbau principle) for the ground state

1s first, then 2s, then 2p….

• each orbital can have at most two electrons ( Pauli exclusion principle)For example, consider Lithium:

32

• Electrons having opposite spins are said to be paired when they are in

the same orbital. An unpaired electron does not have a partner.• A substance is diamagnetic if all its electrons are paired (eg. He).These substances are not attracted to a magnet.

• A substance is paramagnetic if it has one or more unpaired electrons(eg. Li). These substances are attracted to a magnet.

CHEM 1500 Chapter 7

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CHEM 1500, Chapter 7

 Hund's Rule: when degenerate orbitals are available, the electron

configuration of lowest energy has the maximum number of unpairedelectrons with the same spin.

Violate

Hund’s Rule

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• Hund’s rule is based in part on the fact that electrons repel one another.By occupying different orbitals, the electrons remain as far apart as possible

to minimize electron-electron repulsions.

CHEM 1500, Chapter 7

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, p

Electron Configurations and the Periodic Table

(Figure 7.20, page 275)

We can use the structure of the periodic table to predict the fillingorder of the orbitals

34

(Figure 7.21, page 276)

Each row ( period ) represents anenergy level

Each region of the chart represents

a different type of sublevel

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CHEM 1500, Chapter 7

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Chemical Properties Depend on Valence Shell Configuration

• Following Argon is Potassium in the periodic table.• In all it’s chemical properties, Potassium is clearly a member of the

alkali metal group and thus it’s outermost (valence) electron is in as orbital.

• The electron has not gone into a 3d orbital as expected , but hasoccupied the 4s

36

Li [He]2 s1 or 1s22 s1

Na [Ne]3 s1 or 1s22s22 p63 s1

K [Ar]4 s1 or 1s22s22 p63s23 p64 s1

Rb [Kr]5 s1 or 1s22s22 p63s23 p64s23d 104 p65 s1

Cs [Xe]6 s1 or 1s22s22 p63s23 p64s23d 104 p65s24d 105 p66 s1

CHEM 1500, Chapter 7

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Transition and Rare Earth Metal Configurations Unexpected

Chromium and Copper expected to be:

Cr [Ar]4 s23 d 4Cu [Ar]4 s23 d 9

 

37

 

Cr [Ar]4 s13 d 5

Cu [Ar]4 s13 d 10

Similar things happen with silver and gold:

Ag [Kr]5 s14 d 10

Au [Xe]4 f 145 d 106 s1

CHEM 1500, Chapter 7

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Problems:

1) Write the full ground-state electron configuration for each and decidewhether it is diamagnetic or paramagnetic :

a) Br

38

b) Mg

c) Se

CHEM 1500, Chapter 7

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Problems:

2) Write the condensed ground-state electron configuration for each:

a) Ga

b) Zn

c) Sc

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*d) Cr

*e) Cu

* - These are exceptions. The 3d and 4s orbitals are very close inenergy and it is more stable to half-fill (as in Cr) or completely-fill 

(as in Cu), the degenerate set of d orbitals.

CHEM 1500, Chapter 7

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For a Summary of Valence Electron Configurations –

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CHEM 1500, Chapter 7

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Heisenberg said you could only predict the possibility of finding theelectron somewhere.

A 3-D graph of the space within which there is a certain probability of 

finding the electron is described by the square of the wavefunction, ψn2,and is called the probability electron density.

Ψn = solution to the wave equation

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- describes the motion of electronsin time and space

Ψ2n = orbital

- probability of finding an electron

in a particular region of space

(Figure 7.22)

CHEM 1500, Chapter 7

Representations and Shapes of Atomic Orbitals

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Representations and Shapes of Atomic Orbitals

An orbital shape is a picture of the highest probable electron locationaround the nucleus; represented by ψ2n.

The s Orbitals

1s, 2s, and 3s orbitals are spherical l = 0; ml = 0

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(Figure 7.23, page 282)

 Node = zero electron density

CHEM 1500, Chapter 7

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The p Orbitals (l = 1, ml = -1, 0, 1)

• Nodal plane results in a two-lobe shape; three p orbitals symmetricalong x, y and z axes; highly directional.

(Figure 7.24 & 7.25 & 7.26, page 283)

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• Electron density is concentrated in two regions on either side of the

nucleus. It is dumbbell shaped.

 Nodal Plane = surface with e- density of zero

CHEM 1500, Chapter 7

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The d Orbitals (l = 2; ml = -2, -1, 0, 1, 2)

• Two nodal planes result in a four-lobe shape. Shaped like a cloverleaf .(Figure 7.26, page 283)

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The f Orbitals (l = 3; ml = -3, -2, -1, 0, 1, 2, 3)

CHEM 1500, Chapter 7

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Atomic Properties Correlate with an Atoms Electron

ConfigurationThere are many chemical and physical properties that vary in a more or lesssystematic way according to an element’s position in the periodic table.

Mendeleev’s Periodic Law and Table (1869)

• Atomic numbers were not yet known, but atomic weight

46

generally increases with atomic number (Z).

• Now the periodic table has elements arranged by atomic

 number.

• Atomic numbers ( number of protons and electrons) arosefrom direct dependence between element’s nuclear chargeand its position in the periodic table (Moseley, 1913).

CHEM 1500, Chapter 7

Effective Nuclear Charge (Z )

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Effective Nuclear Charge (Zeff )

Coulomb’s Law: The strength of the interaction between two electrical chargesdepends on the signs and magnitudes of the charges and on thedistance between them.

Therefore the force of attraction increases as the nuclear charge increases

and decreases as the electron moves further from the nucleus.Shielding from the inner core electrons reduces the actual nuclear charge Zon the valence electrons to an effective nuclear charge.

47

Effective nuclear charge, Zeff : The positive nuclear charge actually experiencedby an electron

Zeff = Z – S

where: Z = Atomic number, nuclear charge, number of protonsS = Shielding constant (there are rules for its calculation)Usually close to the number of core electrons in an atom

Inner electrons shield outer electrons more effectively than doelectrons in the same sublevel (orbitals).

CHEM 1500, Chapter 7

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Example:

Consider: Li[He] 2s1

(1s2 2s1)

48

• There is some possibility of finding the 2s electron close to the nucleus,and therefore the 2s electron will experience a slightly greater attraction

than our simple model suggested.

(Figure 7.28, page 284)

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CHEM 1500, Chapter 7

Periodic Trends in Atomic Radii

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The atomic radius can be determined by measuring the bond length in a diatomicmolecule and dividing it in two (bonding atomic radius).

For example: Cl   Cl bond length is 199 pm, so the atomic radius assigned to

Cl is 99 pm.

 

Periodic Trends in Atomic Radii

53

 

atomic radii of the constituent atoms.

In a period:

In a group: Atomic radius increases down a column.This results from the increase in the principal quantum number, nof the valence electron (the number of electronic shells increases)

Atomic radius decreases from left to right in a row.This results from the increase in Zeff , which draws the valence

electrons closer to the nucleus.

CHEM 1500, Chapter 7

Trends in Atomic Radii

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Trends in Atomic Radii

(Figure 7.29, page 286)

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The atomic radius increases as you go down a column and as you move from

right to left in a period.

CHEM 1500, Chapter 7

Periodic Trends in Atomic Radii

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Periodic Trends in Atomic Radii

Exercise:

Using only a periodic table, rank the elements in the following sets in order of increasing size:

Same period: think Zeff

Same group: think n

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(a) Ga, K, Ca

(b) Cl, Br, Se

(c) I, Xe, Ba

Same period; order: K, Ca, GaZeff increases, atomic radius decreases Ga < Ca < K

Same group: Cl, BrSame period: Se, Br Cl < Br < Se

Same period: I, Xe Xe < I < Ba

CHEM 1500, Chapter 7

Periodic Trends in Ionic Radii

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Periodic Trends in Ionic Radii

• The radii of ions are based on the distances between ions in ionic compounds.

• The size of ions (like the size of atoms) depends on the:

1) nuclear charge, Z2) number of electrons

3) orbitals in which the valence electrons are found

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• Cations are smaller than their parent atoms.

• Anions are larger than their parent atoms.

• For ions carrying the same charge, size increases as we go down a group

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CHEM 1500, Chapter 7

• An isoelectronic series is a group of ions all containing the same number of 

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g p g f

electrons. (O2-, F-, Na+, Mg+2)

Because the number of electrons remain constant, the radius of the iondecreases with increasing effective nuclear charge.

e.g., Rank the Na+, Mg2+, F- ions in order of increasing size:

Z 12(Mg) 11(Na) 9(F)Ionic Radius (pm) 65 95 136

 

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Exercise:Write the condensed electronic configurations for the following ions:Al3+ (Z=13), O2- (Z=8), and Cl- (Z=17).

Which are isoelectronic? Which is the smallest ion?Al3+: [Ne]

O2-: [Ne]

Cl-: [Ar]

Isoelectronic

g < a < -

smallest ion →

CHEM 1500, Chapter 7

Ionization Energy

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gy

Ionization energy (IE): The minimum energy required to remove an electronfrom the ground state of an isolated gaseous atom orion.

Key factor in an element’s reactivity:

First ionization energy: Energy required to remove the least attractedelectron from a neutral atom in the as hase.

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Second ionization energy: Energy required to remove the second leastattracted electron, and so forth.

The greater the IE the harder it is to remove the electron.

A (g) →A+ (g) + e-

A+ (g) →A2+ (g) + e-

IE1 > 0 (always)

IE2 > IE1 (always)

CHEM 1500, Chapter 7Variations in Successive Ionization Energies

IE < IE < IE

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IE1 < IE2 < IE3

This trend is because with each successive removal, an electron is being pulledaway from an increasingly more positive ion.(Table 7.2, page 287) (Figure 7.23, page 289)

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Notice the sharp increase in ionization energy that occurs when an inner-shell(noble-gas core) electron is removed.

This supports the fact that outer electrons are used in bonding and reactions, whileinner electrons are too tightly bound to the nucleus to be lost or even shared.

Increases are not smooth:

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CHEM 1500, Chapter 7

Periodic Trends in First Ionization Energies

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In a period: IE generallyincreases with increasing Zeff 

The alkali metals show thelowest ionization energy andthe noble gases the highest.

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The representative elements show a larger range of values for IE1 than do thetransition-metal elements.

In general: Smaller atoms have higher IEs

As the atomic radius decreases(since Zeff increases), it isharder to remove the leastattracted electron in the atom.

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CHEM 1500, Chapter 7

Exercise

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Using only a periodic table, rank the elements in the following sets inorder of increasing IE1:

(a) Rb, K, Ca

(b) S, F, Cl

 

Rb < K < Ca

S < Cl < F

 

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, ,

Atomic size αααα 1/Zeff IE1 αααα Zeff 

 

CHEM 1500, Chapter 7

Electron Configurations of Ions

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• Ions are formed from atoms by gain or loss of electrons they want to achieve anoble gas electron configuration.

• Main group ions achieve ns 2

 np6 

 outer electron configurations.

 Electrons are removed rom the orbitals havin the

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largest principal quantum number (n) first.

Li: [He] 2s1 becomes Li+: [He]F: [He] 2s2 2p5 becomes F-: [He] 2s2 2p6 = [Ne]

When electrons are added to atoms to form an anion,

 they are added to the empty or partially filled orbital 

 having the lowest n.

CHEM 1500, Chapter 7

Ion Electron Configurations

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1) Cations (+) of main group elements have lost electrons in the reverse orderto that in which they were added under the Aufbau principle.

Consider sodium:

Na 1s2 2s2 2p6 3s1

Na+ 1s2 2s2 2p6

closed shell, noble gas e- configuration [Ne]

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2) Anions (-) have gained electrons until an ns2np6 outer electron configurationis obtained.

 noble gas

Consider sulfur:

S [Ne] 3s2 3p4

 add 2 electrons

S2- [Ne] 3s2 3p6

 closed shell, looks like [Ar] 

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CHEM 1500, Chapter 7

Electron Affinities

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• For most atoms, energy is released when the first electron is added (negative∆E)

A (g) + e- →A- (g) EA1 (can be + or -)

Electron affinity (EA): Energy associated with the process of adding anelectron to a gaseous atom.

First electron affinity: Energy change accompanying the addition of the firstelectron to a gaseous atom.

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because the e- is attracted to the atom’s nuclear charge.

Cl (g) + e– → Cl- (g) EA1 = ∆E = –349 kJ/mol

• Electron affinity can either be exothermic (as the above example) or endothermic:

Ar (g) + e– →Ar- (g) EA1, ∆E > 0

A- (g) + e- →A 2- (g) EA2

(always +)

•  EA2 is always positive because energy must be absorbed to overcome therepulsion between the two negative charges.

CHEM 1500, Chapter 7

Periodic Trends in Electron Affinity Energies

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• The greater the attraction between a given atom and the added e-,the more negative its EA.• Trends in EA are not as evident as they were in IE

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(Table 7.3, page 291)

EAs show a general, albeit, minor decrease when we go down a group.Consider the halogens: For F, the added electron goes into a 2p orbital,

for Cl a 3p orbital, for Br a 4p orbital.The average distance between the added e- and the nucleus increases.

The e- - nucleus attractions and e- / e- repulsion decrease.

CHEM 1500, Chapter 7

Summary

R ti t l (G 6A & 7A)

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Reactive non-metals (Groups 6A & 7A):

Reactive metals:

High IE →→→→ Lose e- with difficultyHigh (-) EA→→→→Attract e- strongly

⇒⇒⇒⇒ Form anions in ionic compounds

-

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Noble gases:

 Low (-) EA→→→→Attract e- weakly

⇒⇒⇒⇒ Form cations in ionic compounds

High IE & (+) EADo not tend to lose or gain e-

⇒⇒⇒⇒ Not very reactive

Trends not as regular as for IE

CHEM 1500, Chapter 7

Review Questions

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Page 293 – 297

7.5 - 7.14, 7.15, 7.16, 7.18, 7.21, 7.23, 7.27, 7.29, 7.30, 7.31, 7.33, 7.34, 7.35,

7.38 - 7.51, 7.53, 7.54, 7.55, 7.58 - 7.77, 7.80 – 7.83, 7.85, 7.87,

7.90, 7.91, 7.92, 7.93, 7.95, 7.97, 7.98, 7.99, 7.102, 7.104 – 7.108,

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7.110, 7.112, 7.114, 7.116, 7.118, 7.120, 7.122, 7.124, 7.126, 7.128,7.130, 7.132, 7.134, 7.136, 7.137, 7.139, 7.146, 7.147, 7.150, 7.151