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Honors Chemistry

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Honors Chemistry

Chapter 7: Quantum Mechanics

- Wavelength (l) = distance between two in-phase points
- Measured in meters
- Frequency (n) = number of waves per second
- Measured in Hertz (Hz)
- Amplitude (y) = distance of maximum displacement from rest position
- Amplitude corresponds to wave energy

- v = ln
- Find the wavelength of a 256 HZ (middle C) sound wave traveling at 343 m/s.
- v = ln
- 343 m/s = l (256 Hz)
- l = 1.34 m
- Try this….
- Find the frequency of a 25.0 cm wave traveling at 0.75 m/s.

- James Clerk Maxwell (1873)
- Mathematical description of light waves
- Light is an electromagnetic wave
- Speed of light (c) is constant
- c = 2.99792458 x 108 m/s
- To 3 sig dig, 3.00 x 108 m/s is fine
- Try this….
- Find the frequency of a 250 nm light wave. (Don’t forget about the “nano” prefix!)

- Radio, micro, IR, ROYGBIV, UV, X, g
- long l ------------------------------- short l
- low n -------------------------------- high n
- Radio wave end of the spectrum is low energy radiation
- Gamma ray end is high energy radiation
- Black body radiation
- Wave theory fails to account for this!

- Max Planck (1900)
- Energy is emitted and absorbed only in small, discrete packets called quanta
- Energy of a quantum of energy given byE = hn
- h = 6.626 x 10-34 Js (Planck’s constant)
- Correctly accounts for blackbody curves
- Planck has no idea why it works!

- Find the energy of a 2.50 x 1014 Hz light wave.
- E = hn
- E = (6.626 x 10-34 Js)(2.50 x 1014 Hz)
- E = 1.66 x 10-19 J
- A quantum holds a tiny amount of energy!
- Try this….
- Find the energy of a 475 nm light wave.
- Hint: Use the wave equation first!

- Albert Einstein (1905)
- Electrons ejected from surface of metal exposed to light
- Depends on frequency of light
- Electrons ejected at a certain cutoff frequency
- Above cutoff n, electrons leave with more energy

- Emission spectra – light given off by glowing objects
- Can be continuous or discontinuous
- Line spectra – series of bright lines emitted by gas phase atoms
- Pattern of bright lines is characteristic of the element that is glowing
- Absorption spectra – dark lines in spectrum as light passes through a gas

- Niels Bohr (1913)
- Electron energies are quantized
- Only certain orbits are allowed
- - RHEn = ------ n2
- RH = 2.18 x 10-18 J (Rydberg constant)
- n = 1, 2, 3, 4, ….

- DE = Ef – E0
- -RH -RHDE = ----- - ----- nf2 n02
- Factor out RH
- 1 1 DE = RH (----- - ----- ) n02 nf2

Link to Hydrogen energy states

- Find the energy of a photon of light emitted by an electron jumping from level 5 down to level 2.
- DE = RH (1/n52 – 1/n22)
- DE = (2.18 x 10-18 J)(1/25 – 1/4)
- DE = -4.58 x 10-19 J
- Try this….
- Find the energy of the jump from level 1 to level 4.
- Find the frequency of the light produced.

- Louis de Broglie (1924)
- Electrons can be treated as waves
- Each orbit must contain a whole number of waves…explains orbit quantization!
- hl = ---- mv
- mv is momentum (p), so we can write l = h/p
- Verified by Davisson, Germer, and Thomson

Link to quantum atom model

- Find the wavelength of a 3.00 kg duck flying at 5.00 m/s.
- l = h/mv
- l = (6.626 x 10-34 Js) / (3.00 kg)(5.00 m/s)
- l = 4.42 x 10-35 m
- Try this….
- Find the wavelength of an electron traveling at 500,000 m/s. (me = 9.11 x 10-31 kg)

- Werner Heisenberg (1926)
- Complementary variables cannot be known to arbitrary precision
- dp dq ≥ħ/2
- Minimum limits to uncertainties in values are inversely proportional
- Position and momentum are an important complementary pair
- dx dpx≥ħ/2

- Find the uncertainty in velocity of an electron confined to a hydrogen atom (dx = 0.037 nm).
- dx dpx≥ħ/2
- (3.7 x 10-11 m) dp ≥ 5.27 x 10-35 Js
- dp ≥ 1.4 x 10-24 kg m/s
- p = mv
- 1.4 x 10-24 kg m/s = (9.11 x 10-31 kg) dv
- dv = 1.5 x 106 m/s

- Try this…
- Find the uncertainty in position of a 20.0 mg fly whose position is known to within ±0.5 mm.

- Erwin Schrödinger (1926)
- Schrödinger equation – treat electron as a standing wave surrounding the nucleus
- Schrödinger equation is ugly!
- Solve for amplitude function (y)
- Remember – amplitude is energy
- Produces an energy diagram like Bohr’s, but this one actually works
- Wave function has no physical meaning

- Max Born (1926)
- y2 denotes probability
- Electron is delocalized
- Wave function collapses on observation
- electron density refers to magnitude of the probability wave for the electron
- Orbital = spatial probability distribution
- Electron clouds
- Objections: Schrödinger’s Cat
- Other interpretations

- Set of numbers that describe the distribution of electrons in the atom
- Principal Quantum Number (n)
- n = 1, 2, 3, 4, …
- Corresponds to the n value used by Bohr

- Angular Momentum Quantum Number (l)
- l = 0, 1, 2, … , n – 1
- There are a total of n values

- l = 0 is s orbital l = 2 is d orbital
- l = 1 is p orbital l = 3 is f orbital

- Magnetic Quantum Number (ml)
- ml = -l, … , 0, …, +l
- There are a total of 2l + 1 values

- ms = +½, -½
- Two possible electron spin states
- Spin up, spin down

- s, p, d, f orbitals
- Radial probability distributions
- distance from nucleus of high e- probability

- Angular probability distributions
- Show regions of high e- probability
- Cool 3d pictures

- n determines energy
- For H, all subshells are degenerate
- Multielectron atomseach subshell liesat a different energy
- Shielding effect
- Fill lowest energyorbitals first

- Rule of thumb
- Shows the orderin which orbitalsare filled
- Paramagnetic
- Unpaired e-
- Attracted to mag

- Paired e-
- Not attracted

- H has 1 electron
- Put it in 1s
- Write it 1s1
- Read “one-s-one”
- What about He?
- You got it…1s2
- Keep filling 1s until it is full
- But when is it full?

- No two electrons may share the exact set of quantum numbers
- Consider Helium’s 1s2 configuration
- First electron: n = 1, l = 0, ml = 0, ms = +½
- Second electron: same n, l, ml
- ms must be -½
- No room for more electrons in 1s orbital
- Each orbital can hold only two electrons!

- What is the electron configuration of Li?
- 1s2 2s1
- What about N?
- 1s2 2s2 2p3
- But how are p electrons organized?
- Hund’s Rule – arrange electrons in such a way as to maximize total spin state
- Put e-’s in separate orbitals, same spin

- Build up on previous e- configurations
- Each atom adds one more e-
- Express configuration with noble gas core
- Al = 1s2 2s2 2p6 3s2 3p1
- First 3 terms are the same as Ne config.
- Write it as [Ne] 3s2 3p1

- Particularly stable configurations
- Full sublevel
- Half-full sublevel

- Just missed the stable half-full 3d5
- Kick one e- up to d to get [Ar] 4s1 3d5