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Honors Chemistry. Chapter 7: Quantum Mechanics. 7.1 Wave Properties. 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

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Honors Chemistry

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## Honors Chemistry

Chapter 7: Quantum Mechanics

### 7.1 Wave Properties

• 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

### 7.1 The Wave Equation

• 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!)

### 7.1 Electromagnetic Spectrum

• Radio, micro, IR, ROYGBIV, UV, X, g

• long l ------------------------------- short l

• low n -------------------------------- high n

• Gamma ray end is high energy radiation

• Wave theory fails to account for this!

### 7.1 Quantum Theory

• 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!

### 7.1 Quantum Theory

• 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!

### 7.2 The Photoelectric Effect

• 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

• Bright light ejects more electrons

• Quantum theory explains results

• Light is made of quanta called photons

• ### 7.3 Spectroscopy

• 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

### 7.3 Bohr’s Model

• 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, ….

### 7.3 Bohr’s Model

• DE = Ef – E0

• -RH -RHDE = ----- - ----- nf2 n02

• Factor out RH

• 1 1 DE = RH (----- - ----- ) n02 nf2

### 7.3 Bohr’s Model

• 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.

### 7.4 Duality

• 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

### 7.4 Duality

• 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)

### 7.5 Uncertainty Principle

• 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

### 7.5 Uncertainty Principle

• 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

### 7.5 Uncertainty Principle

• Try this…

• Find the uncertainty in position of a 20.0 mg fly whose position is known to within ±0.5 mm.

• Uncertainty limits are not significant for macroscopic objects, but they are significant to subatomic particles

• Cannot know the position and momentum of an electron at the same time!

• Concept of orbits will not work

• ### 7.5 Quantum Mechanics

• 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

### 7.5 Copenhagen Interpretation

• 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

### 7.6 Quantum Numbers

• 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

• Describes energy level of the shell

• Defines the size of the electron cloud

• ### 7.6 Quantum Numbers

• Angular Momentum Quantum Number (l)

• l = 0, 1, 2, … , n – 1

• There are a total of n values

• Sublevels of the energy level

• Angular distribution of electron cloud

• For hydrogen, sublevels are degenerate

• Correspond to fine structure spectral lines

• l = 0 is s orbital l = 2 is d orbital

• l = 1 is p orbital l = 3 is f orbital

• ### 7.6 Quantum Numbers

• Magnetic Quantum Number (ml)

• ml = -l, … , 0, …, +l

• There are a total of 2l + 1 values

• Number of degenerate orbitals in sublevel

• Spatial orientation of the orbital

• Zeeman Effect

• Electron Spin Quantum Number (ms)

• ms = +½, -½

• Two possible electron spin states

• Spin up, spin down

• ### 7.7 Atomic Orbitals

• s, p, d, f orbitals

• distance from nucleus of high e- probability

### 7.7 Atomic Orbitals

• Angular probability distributions

• Show regions of high e- probability

• Cool 3d pictures

### 7.7 Orbital Energies

• n determines energy

• For H, all subshells are degenerate

• Multielectron atomseach subshell liesat a different energy

• Shielding effect

• Fill lowest energyorbitals first

### 7.7 The Diagonal Rule

• Rule of thumb

• Shows the orderin which orbitalsare filled

• Paramagnetic

• Unpaired e-

• Attracted to mag

• Diamagnetic

• Paired e-

• Not attracted

• ### 7.8 Electron Configurations

• H has 1 electron

• Put it in 1s

• Write it 1s1

• You got it…1s2

• Keep filling 1s until it is full

• But when is it full?

### 7.8 Pauli Exclusion Principle

• 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!

### 7.8 Electron Configurations

• What is the electron configuration of Li?

• 1s2 2s1

• 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

### 7.9 Aufbau Principle

• 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

### 7.9 Exceptions

• Particularly stable configurations

• Full sublevel

• Half-full sublevel

• Some transition metals rearrange

• Cr = [Ar] 4s2 3d4

• Just missed the stable half-full 3d5

• Kick one e- up to d to get [Ar] 4s1 3d5

• What other family would do this?

• Cu = [Ar] 4s2 3d9 Ar 4s1 3d10