Video AP 2.1 Quantum Mechanical Model Schrodinger, de Brogli, and Heisenberg solved mathematical equations to describe the behavior of e - in the H atom.

ATOMICTHEORY AND STRUCTURE

QMMVideo AP 2.1

Quantum Mechanical ModelSchrodinger, de Brogli, and Heisenberg solved mathematical equations to describe the behavior of e- in the H atom as being both particle and wavelike, which lead to the quantum mechanical model. The QMM specifies that each e- has a specific energy, however, they do not follow a specific path. Instead, there are areas of probable e- location, which are called orbitals.

Quantum Mechanical Model

The scientists chose to study the hydrogen e- with the lowest energy (ground state), which they labeled 1s. They found that the e- is moving but not necessarily in circles.

Heisenberg Uncertainty Principle

It’s impossible to know both the location of an e- and its velocity (speed) at the same time. It is more probable to find an e- near the nucleus. The size of the 1s orbital is described as the radius of a sphere that encloses 90% of the total e- probability.

Quantum Mechanical Model These calculations continued until they could

describe any e- from any element. The first number for the electron represents the row that the element can be found in. This is the energy level.

The letter represents the sublevel the electron is in, based on area of the periodic table. It will tell you the shape of the orbital. (s is spherical, p is lobes and d has 2 lobes).

The superscript represents the number of electrons in that sublevel.

S p d orbitals

P orbitals

D orbitals

Check Your Understanding…Give the row number and sublevel of each of the following elements:

a. Fluorine

b. Carbon

c. Manganese

d. Sodium

e. Phosphorous3s

2p

3d

2p

3p

Aufbau Principle

As protons are added one by one to the nucleus to build up the elements, so are e-. A new and more specific e- configuration can be written using all the first three quantum letters. Here is the order to fill the orbitals:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6

6s2 4f14 5d10 6p6 7s2 5f14 6d10 7p6

Examples

The configuration for Be ends at 2s and the second element in. S it ends at 2s2. Write everything before 2s2:Be is 1s2 2s2

The configuration for Sulfur ends at 3p4.S is 1s2 2s2 2p6 3s2 3p4

Problems?

You will notice that it seems to skip around a lot and this is because this is the order of the periodic table. This shows that 4s is in fact a lower energy level than 3d and 4f is higher energy than many other sublevels in energy level five.

Orbital Energies

Using a diagram like the one to your left, it is easy to show the way e- fill the orbitals.

Hund’s Rule

Notice that Carbon’s 6th arrow is in the second p orbital. Hund’s Rule states the lowest energy configuration is one having the maximum number of unpaired electrons allowed by Pauli Principle in a particular set of degenerate orbitals. They should have parallel spins.

In english? Put one up arrow in each box before any get two.

Examples

How many unpaired electrons does nitrogen have?

How many unpaired electrons does nitrogen have?

Valence electrons

Valence e- are e- in the outermost principal energy level. The rule still holds: the elements in the same group have the same number of valence e-. Therefore, Nitrogen with a configuration of 1s22s22p3

and Phosphorous with a configuration of 1s22s22p63s23p3 both have 5 valence e- because they are both in group 15.

Electron Configurations Part 2

Video AP 2.2

Ions

Remember, when ions from electrons are added and subtracted to the valence shell!Fluorine is 1s2 2s2 2p5

The F- is 1s2 2s2 2p6

Manganese is 1s2 2s2 2p6 3s2 3p6 4s2 3d5

The Mn+2 ion is 1s2 2s2 2p6 3s2 3p6 3d5

Noble Gas Short Cut

Larger elements will have extremely long configurations. A shortcut is to use noble gas configurations. So Sodium has 11e- and a configuration of 1s22s22p63s2 or [Ne]3s1.

Exceptions

*Half filled sublevels are not as stabled as filled sublevels, but they are more stable than other configurations.

Ex. Cr looks like it should be1s22s22p63s23p64s23d4

But it is 1s22s22p63s23p64s13d5

Ex. Cu looks like it should be 1s22s22p63s23p64s23d9

But it is 1s22s22p63s23p64s13d10

This rule is for all transition metals in groups 6 and 11.

Examples

Give the electron configurations using the noble gas short cut for gold. [Xe] 6s2 5d9

[Xe] 6s1 5d10

Isoelectronic

When two ions or atoms have the same number of electrons. Example: Ar and K+1.Argon: 1s2 2s2 2p6 3s2 3p6

Potassium: 1s2 2s2 2p6 3s2 3p6 4s1

Potassium Ion: 1s2 2s2 2p6 3s2 3p6

Magnetism

Paramagnetism: A type of induced magnetism associated with unpaired electrons that cause a substance to be attracted to the inducing electric field.

Diamagnetism: a type of magnetic field associated with paired electrons that cause a substance to be repelled from the inducing electric plate.

MagnetIsm

Nodes The diagram shows 1s, 2s, and 3s orbitals. The

colored areas are areas of high probability of finding an e-. The areas that are white are areas of zero probability of finding e-, which are called nodes.

The number of nodes and the size of the orbital increase as the principal energy level increases. For s orbitals the number of nodes equals n-1 where n is the principle energy level.

P, d and f orbitals have a more complicated probability distributions but it is important to remember that it is more probable to find an e- near the nucleus.

Electromagnetic Radiation

Video AP 2.3

ER ER is energy that

exhibits wave like behavior and travels through space at the speed of light

(c = 3x108m/s) Wavelength(λ):distance

between 2 peaks. Frequency(v): waves per

second

Which wave is more frequent? Which has a longer wavelength?

ER examplesc=λv

1. If the wavelength is 650nm, calculate the frequency of light with units.

2. If the frequency is 200.s-1, calculate the wavelength.

(3.0x108m/s) = (650x10-9m)(v) v = 4.6x1014s-1

(3.0x108m/s) = (λ)(200./s) λ = 1.5x106m

Planck

Matter can transform into energy but only in packets of energy called quantum. Planck found that these energy packets are in multiples of hv.

∆E=hv h=Planck’s constant = 6.63x10-34Js

PlanckExample 3: CuCl in fireworks give off blue light with a

wavelength of 450nm. What is the amount of energy emitted?

c = λv (3.0x108 m/s) = (450x10-9m)(v) v = 6.7x1014/s

E = hvE = (6.626x10-34J/s)(6.7x1014/s)E = 4.4x10-19J

Einstein

Einstein stated that ER or light is a stream of particles called photons.

Between Plank and Einstein, light is both wave-like and particle-like.

E= hv = hc/λBecause v=c/λ

deBroglie

If light is both wave and particle-like, can matter be both as well?

deBroglie found electrons (which are particles) are so small that they have a measurable wavelength! So, YES!

λ=h/mυ υ= velocity

Remember Light Spectra and Bohr? Energy is released in quanta (packets) to produce

light. When light is passed through a prism, colors may

be seen at various wavelengths. Bohr measured the energy emitted to create his

quantum model of the atom.

Light Spectra and Bohr

En= -2.178x10-18J

n2

n= energy level

Classroom NotesAtomics

1s 2s 2p 2p 2p 3s 3p 3p 3p 4s 3d 3d 3d 3d 3d

O __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

O2- __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Na __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Na+ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Cl __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Cl- __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

S __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

S6+ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

K __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

K+ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Mn __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Mn2+ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Mn4+ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Mn7+ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Cu __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Cu+ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Cu2+ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Zn __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

Fe __ __ __ __ __ __ __ __ __ __ __ __ __ __ __