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Lesson 1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Fundamentals of Chemistry
Lesson 1
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Fundamentals of Chemistry
Prof. Roberto PaolesseDept. of Chemical Science and TechnologiesUniversity of Rome "Tor Vergata"Via della Ricerca Scientifica00133 Rome, Italy
Phone: 39.06.72594752Fax: 39.06.72594328e-mail: [email protected]
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Dr. Larisa LvovaDept. of Chemical Science and TechnologiesUniversity of Rome "Tor Vergata"Via della Ricerca Scientifica00133 Rome, Italy
Phone: 39.06.72594732Fax: 39.06.72594328e-mail: [email protected]
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ENGINEERING SCIENCES 1° Semestre (21/09/2020 - 16/01/2021)
1° ANNO
LUN MAR MER GIO VEN8.30-9.159.30-10.15 FCH FCH MAI FCH10.30-11.15 FCH FCH MAI FCH11.30-12.15 MAI MAI EE12.30-13.15 MAI MAI EE
14.00-14.45 EE ENGL MAI15.00-15.45 EE ENGL MAI16.00-16.4517.00-17.45
MAI Math. Anal. I Aula B3 ENGL English Aula B3FCH Fund. Chem. Aula B3EE Eng. Econom. Aula B3
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Exams:
Winter session17/01/2021 - 28/02/2021
Summer session06/2021 - 07/2021
Fall session08/2021 - 09/2021
Two exams each session
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The exam will consist in a written text and in a oral test
You can access to the oral test if you pass the written examination (grade 18)
The written test consists in five problems to be solved
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If you want to participate to the written test, you shouldreserve at the website: delphi.uniroma2.it
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You can find all the informations related to the course (syllabus, lecture slides, exercises, communications, etc. etc.)
in the website: didattica.uniroma2.it
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Textbook:
Raymond Chang:
General Chemistry: the essential concepts
7th edition ; ISBN 0071313680
McGraw-Hill Ed.
or
Digital book (Smartbook)
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The Study of ChemistryMacroscopic Microscopic
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The scientific method is a systematic approach to research
A hypothesis is a tentative explanation for a set of observations
tested modified
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A theory is a unifying principle that explains a body of facts and/or those laws that are based on them.
A law is a concise statement of a relationship between phenomena that is always the same under the same conditions.
Atomic Theory
Force = mass x acceleration
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Matter is anything that occupies space and has mass.
A substance is a form of matter that has a definite composition and distinct properties.
Chemistry is the study of matter and thechanges it undergoes
liquid nitrogen gold ingots silicon crystals
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A mixture is a combination of two or more substances in which the substances retain their distinct identities.
1. Homogenous mixture – composition of the mixture is the same throughout.
2. Heterogeneous mixture – composition is not uniform throughout.
soft drink, milk, solder
cement, iron filings in sand
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Physical means can be used to separate a mixture into its pure components.
magnet
distillation
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An element is a substance that cannot be separated into simpler substances by chemicalmeans.
• 117 elements have been identified
• 82 elements occur naturally on Earthgold, aluminum, lead, oxygen, carbon, sulfur
• 35 elements have been created by scientiststechnetium, americium, seaborgium
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A compound is a substance composed of atoms of two or more elements chemically united in fixed proportions.
Compounds can only be separated into their pure components (elements) by chemicalmeans.
lithium fluoride quartz dry ice – carbon dioxide
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Classifications of Matter
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Dalton’s Atomic Theory (1808)1. Elements are composed of extremely small particles
called atoms.
2. All atoms of a given element are identical, having the same size, mass and chemical properties. The atoms of
one element are different from the atoms of all other elements.
3. Compounds are composed of atoms of more than one element. In any compound, the ratio of the numbers of atoms of any two of the elements present is either an
integer or a simple fraction.
4. A chemical reaction involves only the separation, combination, or rearrangement of atoms; it does not
result in their creation or destruction.
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8 X2Y16 X 8 Y+
Law of Conservation of Mass
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J.J. Thomson, measured mass/charge of e-
(1906 Nobel Prize in Physics)
Cathode Ray Tube
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Cathode Ray Tube
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e- charge = -1.60 x 10-19 C
Thomson’s charge/mass of e- = -1.76 x 108 C/g
e- mass = 9.10 x 10-28 g
Measured mass of e-
(1923 Nobel Prize in Physics)
Millikan’s Experiment
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(uranium compound)
Types of Radioactivity
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Thomson’s Model
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1. atoms positive charge is concentrated in the nucleus2. proton (p) has opposite (+) charge of electron (-)3. mass of p is 1840 x mass of e- (1.67 x 10-24 g)
a particle velocity ~ 1.4 x 107 m/s(~5% speed of light)
(1908 Nobel Prize in Chemistry)
Rutherford’s Experiment
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atomic radius ~ 100 pm = 1 x 10-10 m
nuclear radius ~ 5 x 10-3 pm = 5 x 10-15 m
Rutherford’s Model of the Atom
“If the atom is the Houston Astrodome, then the nucleus is a marble on the 50-
yard line.”
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Chadwick’s Experiment (1932)(1935 Noble Prize in Physics)
H atoms - 1 p; He atoms - 2 p
mass He/mass H should = 2
measured mass He/mass H = 4
a + 9Be 1n + 12C + energy
neutron (n) is neutral (charge = 0)
n mass ~ p mass = 1.67 x 10-24 g
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mass p ≈ mass n ≈ 1840 x mass e-
A problem ….
According to the law of physics, a charged particle in motion should emit energyand have a spiral trajectory and finally collapse on the atomic nucleus ….
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Properties of Waves
Wavelength (l) is the distance between identical points on successive waves.
Amplitude is the vertical distance from the midline of a wave to the peak or trough.
Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s).
The speed (u) of the wave = l x n
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Maxwell (1873), proposed that visible light consists of electromagnetic waves.
Electromagnetic radiation is the emission
and transmission of energy in the form of
electromagnetic waves.
Speed of light (c) in vacuum = 3.00 x 108 m/s
All electromagnetic radiationl x n = c
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Mystery #1, “Heated Solids Problem”Solved by Planck in 1900
Energy (light) is emitted or absorbed in discrete units
(quantum).
E = h x nPlanck’s constant (h)
h = 6.63 x 10-34 J•s
When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths.
Radiant energy emitted by an object at a certain temperature depends on its wavelength.
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Light has both:1. wave nature
2. particle nature
hn = KE + W
Mystery #2, “Photoelectric Effect”Solved by Einstein in 1905
Photon is a “particle” of light
KE = hn - W
hn
KE e-
where W is the work function anddepends how strongly electrons
are held in the metal
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Line Emission Spectrum of Hydrogen Atoms
Il modello di Bohr
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1. e- can only have specific (quantized) energy
values
2. light is emitted as e-moves from one energy level to a lower energy
level
Bohr’s Model of the Atom (1913)
En = -RH( )1n2
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
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E = hn
E = hn
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Ephoton = DE = Ef - Ei
Ef = -RH ( )1n2f
Ei = -RH ( )1n2i
i fDE = RH( )1
n21n2
nf = 1
ni = 2
nf = 1
ni = 3
nf = 2
ni = 3
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De Broglie (1924) reasoned that e- is both particle and
wave.
Why is e- energy quantized?
u = velocity of e-
m = mass of e-
2pr = nl l = hmu
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Schrodinger Wave EquationIn 1926 Schrodinger wrote an equation that
described both the particle and wave nature of the e-
Wave function (y) describes:
1. energy of e- with a given y
2. probability of finding e- in a volume of space
Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its
solution for multi-electron systems.
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Schrodinger Wave Equationy is a function of four numbers called
quantum numbers (n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
n=1 n=2 n=3
distance of e- from the nucleus
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Where 90% of thee- density is foundfor the 1s orbital
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quantum numbers: (n, l, ml, ms)
angular momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … n-1
n = 1, l = 0n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
Shape of the “volume” of space that the e- occupies
l = 0 s orbitall = 1 p orbitall = 2 d orbitall = 3 f orbital
Schrodinger Wave Equation
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l = 0 (s orbitals)
l = 1 (p orbitals)
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l = 2 (d orbitals)
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quantum numbers: (n, l, ml, ms)
magnetic quantum number ml
for a given value of lml = -l, …., 0, …. +l
orientation of the orbital in space
if l = 1 (p orbital), ml = -1, 0, or 1if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
Schrodinger Wave Equation
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ml = -1, 0, or 1 3 orientations is space
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ml = -2, -1, 0, 1, or 2 5 orientations is space
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(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
Schrodinger Wave Equation
ms = -½ms = +½
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Existence (and energy) of electron in atom is described by its unique wave function y.
Pauli exclusion principle - no two electrons in an atomcan have the same four quantum numbers.
Schrodinger Wave Equationquantum numbers: (n, l, ml, ms)
Each seat is uniquely identified (E, R12, S8)Each seat can hold only one individual at a
time
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Schrodinger Wave Equationquantum numbers: (n, l, ml, ms)
Shell – electrons with the same value of n
Subshell – electrons with the same values of n and l
Orbital – electrons with the same values of n, l, and ml
How many electrons can an orbital hold?
If n, l, and ml are fixed, then ms = ½ or - ½
y = (n, l, ml, ½) or y = (n, l, ml, -½)An orbital can hold 2 electrons
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How many 2p orbitals are there in an atom?
2p
n=2
l = 1
If l = 1, then ml = -1, 0, or +1
3 orbitals
How many electrons can be placed in the 3d subshell?
3d
n=3
l = 2
If l = 2, then ml = -2, -1, 0, +1, or +2
5 orbitals which can hold a total of 10 e-
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Energy of orbitals in a single electron atom
Energy only depends on principal quantum number n
En = -RH( )1n2
n=1
n=2
n=3
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Energy of orbitals in a multi-electron atom
Energy depends on n and l
n=1 l = 0
n=2 l = 0 n=2 l = 1
n=3 l = 0n=3 l = 1
n=3 l = 2
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“Fill up” electrons in lowest energy orbitals (Aufbau principle)
H 1 electron
H 1s1He 2 electrons
He 1s2
Li 3 electronsLi 1s22s1Be 4 electronsBe 1s22s2B 5 electronsB 1s22s22p1
? ?
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The most stable arrangement of electrons in subshells is the one with the greatest number of
parallel spins (Hund’s rule).
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Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
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Electron configuration is how the electrons are distributed among the various atomic orbitals in an
atom.
1s1principal quantum
number nangular momentumquantum number l
number of electronsin the orbital or subshell
Orbital diagram
H1s1
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What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons
Abbreviated as [Ne]3s2 [Ne] 1s22s22p6
What are the possible quantum numbers for the last (outermost) electron in Cl?
Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electronsLast electron added to 3p orbital
n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½
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Outermost subshell being filled with electrons
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Paramagneticunpaired electrons
2p
Diamagneticall electrons paired
2p
