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 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. Phosphorous 3s 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: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 7s 2 5f 14 6d 10 7p 6

Examples  The configuration for Be ends at 2s and the second element in. S it ends at 2s 2. Write everything before 2s 2 : Be is 1s 2 2s 2  The configuration for Sulfur ends at 3p 4. S is 1s 2 2s 2 2p 6 3s 2 3p 4

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 6 th 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?

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 1s 2 2s 2 2p 3 and Phosphorous with a configuration of 1s 2 2s 2 2p 6 3s 2 3p 3 both have 5 valence e - because they are both in group 15.

Video AP 2.2

Ions  Remember, when ions from electrons are added and subtracted to the valence shell! Fluorine is 1s 2 2s 2 2p 5 The F - is 1s 2 2s 2 2p 6 Manganese is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 5 The Mn +2 ion is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5

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 1s 2 2s 2 2p 6 3s 2 or [Ne]3s 1.

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 be 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 But it is1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 Ex.Cu looks like it should be 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 9 But it is1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 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] 6s 2 5d 9  [Xe] 6s 1 5d 10

Isoelectronic  When two ions or atoms have the same number of electrons. Example: Ar and K +1. Argon: 1s 2 2s 2 2p 6 3s 2 3p 6 Potassium: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 Potassium Ion: 1s 2 2s 2 2p 6 3s 2 3p 6

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.

MagnetIsmMagnetIsm

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.

Video AP 2.3

ER  ER is energy that exhibits wave like behavior and travels through space at the speed of light (c = 3x10 8 m/s)  Wavelength( λ ):distance between 2 peaks.  Frequency(v): waves per second Which wave is more frequent? Which has a longer wavelength?

ER examples c=λ v 1. If the wavelength is 650nm, calculate the frequency of light with units. 2. If the frequency is 200.s -1, calculate the wavelength. (3.0x10 8 m/s) = (650x10 -9 m)(v) v = 4.6x10 14 s -1 (3.0x10 8 m/s) = ( λ )(200./s) λ = 1.5x10 6 m

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 = 6.63x Js

Planck Example 3: CuCl in fireworks give off blue light with a wavelength of 450nm. What is the amount of energy emitted? c = λ v (3.0x10 8 m/s) = (450x10 -9 m)(v) v = 6.7x10 14 /s E = hv E = (6.626x J/s)(6.7x10 14 /s) E = 4.4x J

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/ λ Because 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/m υυ= 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  E n = x J n 2 n= energy level

Atomics

1s2s2p 2p 2p 3s3p 3p 3p 4s3d 3d 3d 3d 3d O______ __ __ ____ __ ______ __ __ __ __ O 2- ______ __ __ ____ __ ______ __ __ __ __ Na______ __ __ ____ __ ______ __ __ __ __ Na + ______ __ __ ____ __ ______ __ __ __ __ Cl______ __ __ ____ __ ______ __ __ __ __ Cl - ______ __ __ ____ __ ______ __ __ __ __ S______ __ __ ____ __ ______ __ __ __ __ S 6+ ______ __ __ ____ __ ______ __ __ __ __ K______ __ __ ____ __ ______ __ __ __ __ K + ______ __ __ ____ __ ______ __ __ __ __ Mn______ __ __ ____ __ ______ __ __ __ __ Mn 2 +______ __ __ ____ __ ______ __ __ __ __ Mn 4+ ______ __ __ ____ __ ______ __ __ __ __ Mn 7+ ______ __ __ ____ __ ______ __ __ __ __ Cu______ __ __ ____ __ ______ __ __ __ __ Cu + ______ __ __ ____ __ ______ __ __ __ __ Cu 2+ ______ __ __ ____ __ ______ __ __ __ __ Zn______ __ __ ____ __ ______ __ __ __ __ Fe______ __ __ __ __ __ ______ __ __ __ __