The color of the light emitted depends upon the  E as the electron(s) move from higher to lower energy levels. He Ne

The shorter the wavelength ( ), the higher the frequency ( ). Energy of the wave increases as frequency increases.

Electromagnetic spectrum Converting from wavelength (  to frequency (   Ex: What is the frequency of a 400 nm EM wave? = c/ = (3.00 x 10 8 m/s) / (4.00 x m) = 7.50 x Hz c =

Ground state: electron is at its lowest possible energy level + + n = 1 n = 2 n = 3 n = 4 n = 5 Excited state: electron is above the lowest possible energy level H 1 1 Photon absorbed Photon emitted

A few definitions: Quantum: Photon: Monochromatic radiation: Continuous spectrum: Line spectrum: The smallest quantity of energy that can be absorbed or emitted as electromagnetic radiation (1) A whole number multiple of a quantum; (2) a ‘particle’ of light (from Einstein’s photoelectric effect) Electromagnetic radiation that consists of a single wavelength Spectrum:EM radiation that has been split into its separate wavelengths A spectrum that does not contain all possible wavelengths of EM radiation A spectrum containing all possible wavelengths of EM radiation

Hydrogen Atomic emission (bright line) spectrum of hydrogen

Bohr atomic orbit Quantum number Orbit radius (nm) Atomic energy level Relative Energy Firstn = E1E1 Secondn = E 2 = 4E 1 Thirdn = E 3 = 9E 1 Fourthn = E 4 = 16E 1 Fifthn = E 5 = 25E 1 Sixthn = E 6 = 36E 1 Seventhn = E 7 = 49E 1 For emission:  E = E higher – E lower = E photon = h = frequency (Hz)h = Planck’s constant (6.626 x J·s) = wavelength (m) c = speed of light (3.00 x 10 8 m/s) E photon = hc/

The Rydberg equation: used to calculate the wavelength of a photon emitted or absorbed by a hydrogen atom for specific values of n. R H = x 10 7 m -1 n 2 > n 1 and both are positive Bohr modified Rydberg’s equation to calculate the energy levels available to an electron around a hydrogen atom: The more negative the energy value, the lower the energy of the electron in the orbit.

The  E for an electronic transition can be found using: Notice that  E is negative when the electron goes from a higher to a lower energy level. Ex: What is the energy of the photon associated with an n = 4 to n = 2 electronic transition? = x J Q: Is energy absorbed or emitted? emitted Q: What is the wavelength of this photon? 4.86 x m = 486 nm = 4860 Å

If light can have characteristics of BOTH particles and waves… …can particles of matter also behave like waves? De Broglie’s equation h = x J·s m = mass in kg v = velocity in m/s Ex: What is the wavelength of an electron traveling at 5.97 x 10 6 m/s? (The mass of an electron is x kg) = 1.22 Å

Heisenberg’s Uncertainty Principle: it is impossible to know precisely both the velocity (or energy) and the electron’s position at the same time. Given: m e = x kg v e = 5 x 10 6  1% m/s Q: What is the uncertainty in the location of the electron?  v= 1%(5 x 10 6 m/s) = 5 x 10 4 m/s = 1 x m That’s an order of magnitude bigger than a hydrogen atom!

The Four Quantum Numbers In the quantum mechanical model of the atom, each electron is described by four quantum numbers, and no two electrons in an atom can have the same numbers. The quantum numbers are: n: the principle quantum number. n = 1, 2, 3, 4… l: the angular quantum number. It describes the ‘shape’ of the orbital…the electronic distribution about the nucleus. l = n-1, n-2,… to 0 If l = 0, it’s an s orbital If l = 1, it’s a p orbital If l = 2, it’s a d orbital If l = 3, it’s an f orbital (and so on) m l : the magnetic quantum number. It describes the orientation of the orbital on an xyz-coordinate axis. m l = -l…0…+l in integral values. m s : the spin quantum number. It equals + ½ or - ½. Conventionally, the + ½ is always given first.

Within an energy level (n = 1, 2, 3, 4…), there exists n types of orbitals and n 2 sublevels. nOrbital types One s-orbital Three p-orbitals One s-orbital Three p-orbitals Five d-orbitals One s-orbital Seven f-orbitals Three p-orbitals Five d-orbitals

A few examples: 1 nlmlml 00 (all ms’s are ± ½, so won’t be shown here) Notes The 1s orbital , 0, 1The three 2p orbitals. 2 The 2s orbital 32-2, -1, 0, +1, +2The five 3d orbitals. 31-1, 0, +1The three 3p orbitals. 3 00The 3s orbital Notice that n = # of subshells present and that n 2 = total number of orbitals for a particular energy level.

x y z x y z x y z pxpx pzpz pypy The three p-orbitals are oriented along the x, y and z axes and can hold a maximum of 2 electrons each. It’s been suggested that the p-orbitals look like peanuts if that helps you to remember their shape.

The d-orbitals d xy d xz d yz Even though the d-orbitals look like two p-orbitals, it is important to remember that each orbital can hold a maximum of 2 electrons regardless of how many lobes it has. d x 2 -y 2 dzdz 2

The freaky (or flowery) f-orbitals

1. (a) What is the designation for the subshell with n = 5 and l = 1? (b) How many orbitals are in this subshell? (c) Indicate the values of m l for each of these orbitals. Examples (a) l = 1 is a p-orbital, so it is a 5p subshell (b) There are three p-orbitals(c) m l = -1, 0, 1 3. (6.54) Which of the following are permissible sets of quantum numbers for an electron in a hydrogen atom: (a) n = 2, l = 1, m l = 1; (b) n = 1, l = 0, m l = −1; (c) n = 4, l = 2, m l = −2; (d) n = 3, l = 3, m l = 0? For those combinations that are permissible, write the appropriate designation for the subshell to which the orbital belongs (that is, 1s, and so on). (a) 2p x (c) 4d x2-y2

And Electron Configurations Orbital Diagrams

H: 1 electron, 1s 1 Electron configuration Electrons behave as though they are spinning on their axis. A half-arrow is used to indicate if the direction of the spin is up (, m s =+½) or down (, m s = - ½ ). m s = spin quantum number

He: 2 electrons, 1s 2 Pauli Exclusion Principle: No two electrons in the same orbital can have identical quantum numbers. Hence, each orbital can contain 2 electrons iff the electron spins are in opposite directions.

Li: 3 electrons, 1s 2 2s 1 Auf bau principle: the electrons fill the orbitals starting with the lowest energy level and working their way up the energy ‘ladder’

B: 5 electrons, 1s 2 2s 2 2p 1

C: 6 electrons, 1s 2 2s 2 2p 2 Hund’s Rule: must fill energetically degenerate orbitals so that maximum multiplicity results  “Fill ‘em up with singles before you start pairing ‘em up.”

n n -1 n -2 d-block metals f-block metals s-block metals p-block elements

Hydrogen Helium

LithiumBoronBeryllium

LithiumBeryllium Boron

Carbon Nitrogen Oxygen

Fluorine Neon

Chlorine (Cl) Nickel (Ni) Rubidium (Rb)

Chemical reactivity depends upon the number of electrons in the highest energy level (the largest n). Electron dot structures show only the outermost electrons so that reactivity can be easily predicted. The outermost electrons are also called the valence electrons. Electron Dot Structures Rules to Drawing Electron Dot Structures 1.Electrons 1-4 are added singly. 2.Exception: If there only two electrons, they’re shown as a pair. 3.Electrons 5-8 are paired with the single electrons that are already there.

Xe Sr N Examples: 1. What is the electron configuration of strontium (Sr)? 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 3p 6 5s 2 The highest energy level is n=5 and there are 2 electrons in it. Therefore, Sr has 2 valence electrons. 2. What is the electron configuration of nitrogen (N)? 1s 2 2s 2 2p 3 The highest energy level is n=2 and there are 5 electrons in it. Therefore, N has 5 valence electrons. 3. What is the electron configuration of xenon (Xe)? 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 3p 6 5s 2 4d 10 5p 6 The highest energy level is n=5 and there are 8 electrons in it. Therefore, Xe has 8 valence electrons.

# Valence electrons

Excluding a few exceptions, how many valence electrons would all of the d-block and f-block elements have?

Electron Configuration Exceptions Hund’s Rule (part 2): For n ≥ 4, the energy of the electrons decreases (and stability increases) when all of the boxes for a sublevel are either half-filled OR completely filled. 4s 3d All of the orbitals are half-filled Cr 4s 3d Cu [Ar] All of the 3d orbitals are filled and the 4s orbital is half-filled

Where would the exceptions be in the 5 th through 7 th periods?