Chapter 7 Problems: 16, 17, 19 – 23, 26, 27, 30, 31 , 34, 38 – 41, 45, 49, 53, 60, 61, 65, 67, 75, 79, 80, 83, 87, 90, 91, 94, 95, 97, 101, 111, 113, 115 – 117, 121, 122, 125a

Chapter 7 – Atomic Structure and Periodicity

4 primary characteristics of waves:

1) Wavelength ( λ) – the distance between successive peaks in a wave.

2) Frequency ( υ) – The number of waves passing a point per unit of time.1 Hertz (Hz) = 1 wave per second.

3) Amplitude – height of wave.

4) Speed – sound travels around 1000 feet/ second light travels 186,000 miles/ second(3 x 108 m/sec)

Light – Electromagnetic Radiation – Energy that trave ls through space in the form of waves, and having fluctuating electrical and mag netic fields.

High energy Low Energy

All forms of light travel the same speed. What is the relationship between wavelength and frequency?

C = λ υ where c = speed of light (meters/second)

(λ) = wavelength (meters)

(υ) = frequency ( 1 )second

C = λ υ where c = speed of light (meters/second)

(λ) = wavelength (meters)

(υ) = frequency ( 1 )second

Problem: Calculate the wavelength of Alice, 105.9 MHz radio waves.

Problem: Calculate the wavelength of KOA, 850 KHz radio waves, and compare to Alice.

Max Planck (1900) – found that heated objects didn’t emit energy of any quantity, but rather, gave off whole number amounts of energy.

Planck coined the term quantum – the smallest, whole number amount of light energy that can be emitted or absorbed.

Examples of a quantum of:

pop in a pop machine?

Tires at the tire store?

Babies at the hospital?

Gasoline at the gas pump?

Light?

The energy of multiple waves of light can be calcul ated:

E = n h υ where E = energy (Joules)n = number of waves (whole number)h = Plancks constant = 6.63 x 10 -34 J .secυ = frequency (sec -1)

And the energy of a single wave:

E = h υ

Light is quantized – it comes in whole number multip les of the basic unit.

Problem: Determine the energy of a photon (1 wave) of x-ray (5.00 x 10 -2 nm) light. Compare to a photon of Infrared radiation (5 .00 x 104 nm)

Problem: H 2O(l) + light � H2(g) + ½ O2(g) ∆H = 285.8 kJ

Determine the minimum wavelength of light to photo- dissociate a molecule of water. Assume that 1 photon will dissociate 1 m olecule of water, if it has enough energy.

The Photoelectric Effect: (Einstein won a Nobel Pr ize in 1921 for this discovery)

Determined that light acts like a particle.

When light of short enough wavelength is shone in a lkali metals, electrons are ejected from the metal surface.

The number of electrons ejected are proportional to the light intensity, or volume of light.

The energy of the electrons ejected is proportional to the wavelength of light, where the shorter the wavelength, the higher the energy of the ejected electrons.

Einstein reasoned that light was acting like a part icle, which he called a photon. If the photon had sufficient energy in a collision wit h an electron, the photon would knock the electron out of its orbital.

Wave/Particle Duality of Light

Einstein suggested particle-like properties of light could explain the photoelectric effect.

But diffraction patterns suggest photons are wave-like.

Einstein also determined that mass and energy were related, through his famous equation, E = mc 2 where E = energy (Joules)

m = mass (kg)c = speed of light (m/sec)

This re-arranges to m = Ec2

And E = hcλ

We can calculate the apparent mass of a photon of l ight..

m = h/λc

The dual nature of light – light has both wave and pa rticle like properties.

As a wave:

As a stream of particles:

Louis DeBroglie (1923) reasoned that if light could behave like a particle, then electrons, which until now were considered to be a particle, could have wavelike properties.

By re-arranging Einstein and Plancks Equations:

λ = hmc where c is the speed of light for a photon, or th e velocity of an object.

This equation is used to calculate the wavelength o f any moving particle!

Problem: Compare the wavelength of:

a) electron (9.11 x 10 -31 kg) traveling 1.0 x 10 7 m/sec.

b) ball (.10 kg) traveling 35 m/sec.

Note the relationship:

The smaller the object, the larger the wavelength

The larger the object, the smaller the wavelength.

Energy is a form of matter – large objects have un-m easurable wavelengths, and display predominantly particle lik e behavior.

Small objects, like photons of light, exhibit predo minantly wavelike behavior, and barely perceptible particle like beha vior.

Medium size objects (electrons) display both wave a nd particle like behavior.

Very Cool Stuff!

Continuous spectrum – all colors of light are present

Bright Line spectrum – only certain colors of light are present

Absorption Spectrum – opposite of a bright line spectrum

These 2 types of spectra are used to identify one element or compound from another.

These bands of light are caused when electrons absorb external energy and jump to a higher energy level. This is unstable, so they fall back to thei r lower energy level, emitting a photon of light.

Neils Bohr noticed that Hydrogen gas emitted only 4 bands of light in its bright line spectra, and th ese lines were always absent in the absorption spectra.

Since the type of light emitted didn’t shift, the energy levels that the electrons exist in must be quantized…the electrons only exist in certain energy levels, and can only be excited to other given energy levels.

Decided that the Hydrogen atom looked like this:

Ground State – the lowest energy level an electron c an exist in.

Excited State – an electron that has absorbed extern al energy, and exists in a higher energy state.

To calculate the energy of an electron in Hydrogen at different energy levels:

En = -RH ( 1 ) where R H = Rydberg Constant n2 = 2.178 x 10-18 J

Problem: Calculate the energy of an electron in th e third energy level.

Calculate the energy of an electron in the second energy level.

Calculate the energy released when an electron fall s from the third to second energy level. Determine t he color of light.

An easier formula for the previous problem: (this solves energy per photon!)

Ephoton = -RH ( 1 - 1 )n f

2 n i2

Problem: How much energy is required to remove an electron in the first energy level?

Unfortunately, the only element that Bohr’s model s uccessfully predicts is Hydrogen.

Erwin Schrodinger (mid 1920’s) determined formulas t o describe electrons as waves rather than particles.

Schrodingers formulas calculated regions in space ar ound a nucleus where an electron is likely to be found, and diagramed as an electron density map.

An orbital is constructed by enclosing the region where the electron is found 90% of the time. This is an s orbital.

3 orientations of p orbitals

5 orientations of d orbitals

7 orientations of f orbitals

It is very difficult to determine the “path” of an e lectron. The smaller an object, the more influenced the particle is by the device you use to measure it.

This is called The Uncertainty Principle, developed by Werner Heisenberg

“It is impossible to know, simultaneously, the position and momentum of a particle with certainty.”

A set of 4 quantum numbers is used to describe each electron.

1) Principal Quantum number: n = 1,2,3,..describes the energy level, or the shell of an electron, and the size of the orbital.

2) Angular Momentum Quantum Number: l = 0 … (n – 1) (whole numbers)Describes the type of the sublevel.

l = 0 s sublevell = 1 p sublevell = 2 d sublevell = 3 f sublevel

3) Magnetic Quantum Number: m l = -l to +l (whole numbers)Relates the position of the orbital to an x, y, z a xis, and determines the # of orbitals in a sublevel.

where l = 0 m l = 0 indicates 1 orientationl = 1 m l = -1, 0, 1 indicates 3 orientations.

4) Spin Quantum Number: m s = +1/2 or -1/2

Describes the spin of an electron. Scientists noticed that an electron has a magnetic moment with two possible orientations when place in a magnetic field. Since magnetic moments are caused by spinning objects, there are 2 spin states.

More detailed look at orbitals:

How many orientations are available for a 1s sublevel? 2s? 2p?

Radial Probability Density Graph: plots radius (x) vs. rpd (y)

Radial probability vs. radius for a 2p sublevel.

What would a 3p orbital look like?

How many orientations are available for a 2p sublev el? 3p?

p orbital

l = 1

d orbital l = 2 How many orientations?

f orbital l = 3 How many orientations?

Problem: How many sublevels can the 5 th energy level have?

Problem: What is the total number of orbitals in th e third energy level?

For H, the energy of an electron is determined by n

1s < 2s = 2p < 3s = 3p = 3d .. (called d egenerate orbitals – same energy orbitals)

For all other atoms, the energy of an electron is d etermined by n + l

1s < 2s < 2p < 3s < 3p < 3d

Pauli Exclusion Principle – No 2 electrons in an atom can have identical quantum numbers.

Hunds Rule – The most stable arrangement of electrons in a subshell is with the greatest number of parallel spins.

Problem: Write the 4 quantum numbers for all elect rons in an atom of calcium. Recall, electrons will fall to the lowest available energy level possible.

Aufbau Principle – as protons are added to an atom, s o are electrons, 1 for 1

Why is 2s preferable (lower in energy) than 2p in a multiple electron atom?

Shielding: inner electrons shield outer electrons from protons, 1 for 1outer electrons shield outer electrons from protons ineffectively.

While the 2p has its maximum probability closer to the nucleus, the small hump of the 2s orbital makes this 2s orbital more penetrating toward the nucleus than the 2p orbital. As distance increases between 2 charged particles, attraction dramatically decreases.

Penetration Effect

Restated: The more effectively an orbital allows i ts electrons to penetrate the shielding electrons, the lower the energy of that e lectron orbital.

Link to laser

The Periodic Table

Dmitri Mendeleev (1834 – 1907) * Youngest of 17 children* Cut hair once per year* Founder of the modern periodic

table

* Organized elements according to their chemical and physical properties.

* Left spaces for elements that were unknown at the time, and accurately predicted properties of those soon to be discovered elements.

Henry Mosley (1913) – arranged atoms by atomic number. (the ability to determine the number of protons was now available) Tweaked Mendeleev’s table to what we have now.

Problem: Write electron configurations for: F As

Problem: Write electron configurations for ions of : N Ca

Problem: Write a noble gas shorthand configuration for: Mo

Problem: Write an orbital diagram for S (hint..w rite the electron configuration first.)

Some irregularities:

Sc: [Ar]4s 2 3d1

Ti: [Ar]4s 2 3d2

V: [Ar]4s 2 3d3

Cr: [Ar]4s 1 3d5

.

.Cu: [Ar]4s 1 3d10

½ and completely filled d orbitals are stable

Which other elements in the periodic table will be similar to Cr and Cu?

4s is filled before 3d because 4s is more penetrati ng , and therefore, less shielded.

After La (Lanthanum) [Xe]6s 25d1 , the lanthanides are filled with 4f electrons. The energy of 5d and 4f are very simila r (5 + 2 = 7, 4 + 3 = 7). The same is true for Ac.

Trends in the Periodic Table

1) Ionization energy – energy required to remove an el ectron from a gaseous atom or ion.

X(g) + energy � X+(g) + e-

Al (g) � Al+(g) + e- I.E.1 = 580 kJ/mole

Al+(g) � Al+2

(g) + e- I.E.2 = 1815 kJ/mole

Al+2(g) � Al+3

(g) + e- I.E.3 = 2740 kJ/mole

Al+3(g) � Al+4

(g) + e- I.E.4 = 11,600 kJ/mole

Ionization energy trends in the Periodic Table:

1) Going Left to Right, first ionization energy gene rally increases.

a) inner electrons completely shield outer electro ns from the nucleusb) outer electrons incompletely shield other outer electrons from the nucleus.

Exceptions: Be (1s 22s2 )� B (1s 22s22p1) This electron is added to a well shielded 2p orbital. Note that this trend occurs 3 periods down group 13.Note that N (1s 22s22p3) � O (1s22s22p4) This electron is being added to a half filled 2p orbital, and must pair up with an electro n already in the orbital. The repulsion causes the electron to be more easily rem oved. Occurs 3 periods down the group.

Ionization energy decreases going down a group. As the distance between 2 oppositely charged particles increases, the force of attraction decreases.

Different view of the same thing.

2. Electron Affinity – The energy change associated w ith adding an electron to an atom.

X(g) + e- � X-(g)

Generally,

EA increases going left to right through a period.

EA decreases going down a group.

Electron Affinity exceptions:

Period 2: Be 1s 2 2s2 – added electron goes into 2p, and completely shield ed, therefore, has no attraction to the nucleus, and th e electron will not stay.

Period 15: N 1s 2 2s2 2p3 added electron goes into a half filled orbital, and the repulsion encountered causes excess shielding.

Period 18: Ne 1s 2 2s2 2p6 added electron goes into 3s, where the electron is completely shielded, and will not stick.

These trends continue down each group to some extent.

3) Atomic Radius (size of an atom) - defined by t he size of the electron cloud.

Trend: Size increases going down a group, because a new shell of electrons is added each time.

Size decreases going Left to Right across a period, for the samereason that ionization energy increases…less shield ing.

Also note that as electrons are added, size increases. As electrons are removed, size decreases.

Remove electrons

Add electrons

Problem: Arrange from largest to smallest:

1) Ca Mg Sr

2) Li O C

3) Cl- Ar K +

4) Fe Fe+2 Fe+3

5) S Se Cl

Inner Transition Elements

Group 1, 2, 13-18 – Representative Elements

Each family has elements in it with similar chemical properties.

Alkali Metal Family:

All have ns 1 electron configuration, and therefore, have low fir st ionization, high second ionization energy.

All are very reactive, and never found pure in natu re.

React with O 2

Li + O2 � Li 2O lithium oxide (all alkali metals form o xides)

Na + O2 � Na2O2 sodium peroxide (sodium and below form peroxides)

K + O2 � KO2 potassium superoxide (potassium and below form superoxides.

All alkali metals react with water:

Na(s) + H2O(l) � NaOH(aq) + H2(g)

Problem: Write an equation of Potassium + water