unit 3: atomic structure
DESCRIPTION
Outline. Chemistry. Unit 3: Atomic Structure. PowerPoint Presentation by Mr. John Bergmann. Basics of the Atom. Location in the Atom. Charge. Particle. Mass. proton. 1+. in nucleus. ~1 a.m.u. neutron. 0. in nucleus. ~1 a.m.u. electron. 1–. orbiting nucleus. ~0 a.m.u. - PowerPoint PPT PresentationTRANSCRIPT
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Unit 3: Atomic Structure
Chemistry
Outline
PowerPoint Presentation by Mr. John Bergmann
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Basics of the Atom
Particle Charge Location inthe Atom Mass
a.m.u.: unit used to measure mass of atoms
proton
neutron
electron
1+
0
1–
in nucleus
in nucleus
orbiting nucleus
~1 a.m.u.
~1 a.m.u.
~0 a.m.u.
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atomic number:
--
--
To find net charge on an atom, consider____ and ____.
mass number:
# of p+
the whole numberon Periodic Table determines identity of atom
(# of p+) + (# of n0)
10
Ne20.1797
p+ e–
(It is NOT on “the Table.”)
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“When I see a cation, I see a positive ion;
ion:
anion: a (–) ion cation: a (+) ion
-- --
a charged atom
more e– than p+
formed whenatoms gain e–
-- -- more p+ than e–
formed whenatoms lose e–
anions negative ions.areI think that
that is, I… C ion.”A +
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19
Description NetCharge
AtomicNumber
MassNumber
IonSymbol
15 p+
16 n0
18 e– 38 p+
50 n0
36 e–
18 e– 1+ 39
Te2– 128
K1+
Sr2+
P3–
88
31153–
38
522–
2+
76 n0
52 p+
54 e–
20 n0
19 p+
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protium
Isotopes: different varieties of an element’s atoms
-- -- have diff. #’s of n0; thus, diff. masses
some are radioactive; others aren’t
IsotopeH–1
Mass p+ n0 Common Name
H–2H–3 tritium
deuterium123
1 01 11 2
C–12 atoms C–14 atoms
All atoms of an element react the same, chemically.
6 p+ 6 n0 stable
6 p+ 8 n0 radioactive
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a- or b-particles, g rays
Radioactive Isotopes: Nucleus attempts to attain a lower energy state by releasing extra energy as __________.
e.g.,
half-life: the time needed for ½ of a radioactive sample to decay into stable matter
have too many or too few n0
radiation
e.g., C–14: -- half-life is 5,730 years-- decays into stable N–14
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Years from now
0
g of N–14present
5,73011,460
1206030157.5
0
Say that a 120 g sample of C-14 is
found today.
g of C–14present
17,19022,920
6090
105112.5
= C–14= N–14
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Complete Atomic Designation
I53
125 1–
…gives precise info about an atomic particle
mass #
atomic #
charge (if any)elementsymbol
Goiter due to lack of iodine
iodine is now added to salt
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ProtonsComplete
Atomic Designation
92
Neutrons Electrons
34
11
146
45
12
92
36
10
5927
3+Co
3717
1–Cl
55 7+Mn25
23892 U
2311
1+Na
7934
2–Se
20 1817
30 1825
32 2427
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Historical Development of the Atomic Model
Greek modelof atom
Greeks (~400 B.C.E.)
Matter is discontinuous (i.e., “grainy”).
Democritus & Leucippus
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Hints at the Scientific Atom
** Antoine Lavoisier: law of conservation of mass
** Joseph Proust (1799): law of definite proportions: every compound has a fixed proportion
e.g., water……………………..
chromium (II) oxide…….
8 g O : 1 g H
13 g Cr : 4 g O
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Hints at the Scientific Atom (cont.)** John Dalton (1803):
8 g O e.g., water……………………..
chromium (II) oxide…….
hydrogen peroxide..…….
chromium (VI) oxide……
1 g H : 16 g O 1 g H :
13 g Cr 4 g O : 13 g Cr 12 g O :
23
law of multiple proportions: When two different compounds have same two elements, equal mass of one element results in integer multiple of mass of other.
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1. Elements are made of indivisible particles called atoms.
2. Atoms of the same element are exactly alike; in particular, they have the same mass.
Isotopes!
John Dalton’s Atomic Theory (1808)
3. Compounds are formed by the joining of atoms of two or more elements in fixed, whole number ratios.
e.g., 1:1, 2:1, 1:3, 2:3, 1:2:1
Dalton’s model
of atom
Atoms are not
indivisible.
NaCl, H2O, NH3, Fe2O3, C6H12O6
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** William Crookes (1870s): Rays causing shadow were emitted from cathode.
Maltese cross CRT
radar screen television computer monitor
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The Thomsons (~1900)J.J. Thomson discoveredthat “cathode rays” are……deflected by electric and magnetic fields
J.J. Thomson
… (–) particles electrons
+ + + + + +
– – – – – –
electric field lines“cathode rays”
phosphorescentscreen
Crooke’s tube
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William Thomson (a.k.a., Lord Kelvin):Since atom was known to beelectrically neutral, he proposedthe plum pudding model.
Lord Kelvin
Thomson’s plum pudding model
-- Equal quantities of (+) and (–) charge distributed uniformly in atom.
-- (+) is ~2000X more massive than (–)
(plumpudding)
+–
+
++
+
+
+ ++
++
–
–
–
– –
–
–– –
–
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And now we know of many other subatomic particles:
Chadwick
** James Chadwick discovered neutrons in 1932.
--
-- n0 have no chargeand are hard to detect purpose of n0 = stability of nucleus
quarks,muons,positrons,neutrinos, pions, etc.
photo from liquid H2 bubble chamber
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Ernest Rutherford (1909) Gold Leaf Experiment
Beam of a-particles (+) directed at gold leaf surrounded by phosphorescent (ZnS) screen.
a-source
lead block
ZnS screen
goldleaf
particle beam
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Most a-particles passed through, some angled slightly, and a tiny fraction bounced back.
Conclusions: 1. Atom is mostly empty space.
(+) particles are concentrated at center. nucleus = “little nut”
(–) particles orbit nucleus.
2.
3.
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–
–
–
–
–
–
N
Rutherford’s ModelThomson’s Plum Pudding ModelDalton’s (also the Greek) Model
–+
+
+
+
+
+
+ ++
++–
–
–– –
–
–– –
–
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Recent Atomic Models
Max Planck (1900): Proposed thatamounts of energy are quantized
only certain values are allowed
Niels Bohr (1913): e– can possess only certain amounts of energy, andcan therefore be only certain distances from nucleus.
e– foundhere
e– neverfound here
planetary(Bohr) model
N
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Biology ExperimentTo conduct a biology experiment, you need 100 mL of cola per trial, and you plan to conduct 500 trials.
If 1 can contains 355 mL of cola, and there are 24 cans in a case, and each case sells for $4.89, and there is 7.75% sales tax…
A. How many cases must you buy?
B. How much will the cola cost?
cases 5.86 cans 24
case 1 mL 355
can 1 mL 50,000 cases X
$31.61385 1.0775 case 1
$4.89 cases 6 $ X
6 cases
$31.61
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quantum mechanical modelelectron cloud modelcharge cloud model
Schroedinger, Pauli, Heisenberg, Dirac (up to 1940):According to the QMM, we never know for certain where the e– are in an atom, but the equations of the QMM tell us the probability that we will find an electron at a certain distance from the nucleus.
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Average Atomic Mass (Atomic Mass, AAM)This is the weighted average mass of all atoms of an element, measured in a.m.u.
For an element withisotopes A, B, etc.:
AAM = Mass A (% A) + Mass B (% B) + …
(use the decimal form of the %;
e.g., use 0.253 for 25.3%)
% abundance
Ti has five naturally-occurring isotopes
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Lithium has two isotopes.Li-6 atoms have mass 6.015 amu;Li-7 atoms have mass 7.016 amu.Li-6 makes up 7.5% of all Li atoms.Find AAM of Li.
AAM = Mass A (% A) + Mass B (% B)
AAM = 6.015 amu (0.075) + 7.016 amu
AAM = 6.94 amu
(0.925)
AAM = 0.451 amu + 6.490 amu
Li batteries
** Decimal number on Table refers to…molar mass (in g) OR AAM (in amu).
6.02 x 1023 atoms 1 “average” atom
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Isotope
Si-28
% abundance
27.98 amu 92.23%
Mass
Si-29Si-30
28.98 amu 4.67%
AAM = MA (% A) + MB (% B) + MC (% C)
3.10%?
= 27.98 (0.9223) + 28.98 (0.0467) + X (0.031)28.086
28.086 = 25.806 + 1.353 + 0.031X
28.086 = 27.159 + 0.031X0.927 = 0.031X
0.0310.031X = MSi-30 = 29.90 amu
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Electron Configurations “e– Jogging” Rules1. Max. of two e– per jogging track (i.e., orbital).
2. Easier orbitals fill up first. s orbital(level)
p orbital(rolling hills)
d orbital(steep hills)
3. e– must go 100X around. 4. All orbitals of equal difficulty must have one
e– before any doubling up.5. e– on same orbital must go opposite ways.
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1s orbital(1 of these, 2 e–)
2s orbital(1 of these, 2 e–)
3s orbital(1 of these, 2 e–)
4s orbital(1 of these, 2 e–)
2p orbitals(3 of these, 6 e–)
3p orbitals(3 of these, 6 e–)
4p orbitals(3 of these, 6 e–)
3d orbitals(5 of these, 10 e–)
1s2 2s2 3p62p6 4s2 3d103s2 4p6…1,2 3,4 5-10 11,12 13-18 19,20 21-30 31-36
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Writing Electron Configurations:
Where are the e–? (probably)
As
HHe
N
Al
Li
Ti
Xe
1s2 2s2 3p62p6 4s2 3d103s2 4p6…
1s1
1s2
1s2 2s1
1s2 2s2 2p3
1s2 2s2 3p12p6 3s2
1s2 2s2 3p62p6 4s2 3d23s2
1s2 2s2 3p62p6 4s2 3d103s2 4p3
1s2 2s2 3p62p6 4s2 3d103s2 4p6 5s2 4d10 5p6
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Sections of Periodic Table to Know
f-block
s-block
d-block
p-block
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Three Principles about Electrons
Aufbau Principle:
for equal-energy orbitals,each must have one e– before
any take a second
Pauli Exclusion Principle:
e– will take lowest-energyorbital available
Hund’s Rule:
two e– in same orbitalhave different spins
1s22s2
3p6
2p6
4s23d10…
3s2
Friedrich Hund
Wolfgang Pauli
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O
Orbital Diagrams
…show spins of e– and which orbital each is in
1s 2s 2p 3s 3p
1s 2s 2p 3s 3pP
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S
Shorthand Electron Configuration (S.E.C.)
To write S.E.C. for an element:
1. Put symbol of noble gas that precedeselement in brackets.
2. Continue writing e– config. from that point.
Co
In
Cl
Rb
[ Ne ] 3s2 3p4
[ Ar ] 4s2 3d7
[ Kr ] 5s2 4d10 5p3
[ Ne ] 3s2 3p5
[ Kr ] 5s1
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HAVE MOREENERGY
ARE FARTHERFROM NUCLEUS
AND
The Importance of Electrons
In “jogging tracks” analogy,the tracks representorbitals:
In a generic e– config (e.g., 1s2 2s2 2p6 3s2 3p6…):
regions of space where an e– may be found
# of energy level
# of e– in those orbitals
coefficient
superscript
In general, as energy level # increases, e–…
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INVOLVED INCHEMICALBONDING
kernel electrons: valence electrons: in inner energy level(s);
close to nucleusin outer energy level
He: Ne:
Ar: Kr:
1s2 [ He ] 2s2 2p6
[ Ne ] 3s2 3p6
[ Ar ] 4s2 3d10 4p6
Noble gas atoms have FULL valence shells. They are stable, low-energy, and unreactive.
(2 v.e–) (8 v.e–)
(8 v.e–) (8 v.e–)
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octet rule: the tendency for atoms to “want” 8 e– inthe valence shell
Other atoms “want” to be like noble gas atoms.
-- doesn’t apply to He, Li, Be, B (which want 2) or to H (which wants either 0 or 2)
fluorine atom, F 9 p+, 9 e–
** They give away or acquire e–.
steal 1 e–
9 p+, 10 e–
F atom would ratherbe F1– ion.
F1–
chlorine atom, Cl 17 p+, 17 e– steal 1 e–
17 p+, 18 e–
Cl atom would ratherbe Cl1– ion.
Cl1–
How to be likea noble gas…?
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lithium atom, Li 3 p+, 3 e– lose 1 e–
3 p+, 2 e– Li atom would rather
be Li1+ ion.
Li1+
sodium atom, Na 11 p+, 11 e– lose 1 e–
11 p+, 10 e– Na atom would rather
be Na1+ ion.
Na1+
How to be likea noble gas…?
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1+
Know charges on these columns of Table:
Group 1:Group 2:Group 13:Group 15:Group 16:Group 17:Group 18:
2+3+3–2–1–0
1+2+ 3+ 3– 2– 1–
0
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Naming Ions
e.g., Ca2+ Cs1+
Al3+
Cations use element name and then say “ion”
e.g., S2–
P3–
N3–
O2–
Cl1–
Anions change ending of element name to “ide”and then say “ion”
calcium ion cesium ionaluminum ion
sulfide ionphosphide ionnitride ionoxide ionchloride ion
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ENERGY(HEAT, LIGHT,ELEC., ETC.)
LightWhen all e– are in lowest possible energy state, an atom is in the ____________.ground state
e.g., He: 1s2
If “right” amount of energy is absorbed by an e–, it can“jump” to a higher energy level. This is an unstable,momentary condition called the ____________. excited state
e.g., He: 1s1 2s1
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When e– falls back to a lower-energy, more stableorbital (it might be the orbital it started out in, but itmight not), atom releases the “right” amount ofenergy as light.
EMITTED LIGHTAny-old-value of energy to beabsorbed or released isNOT OK. This explainsthe lines of color in anemission spectrum.
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Emission Spectrum for a Hydrogen Atom
Lyman series:
Balmer series:
Paschen series:
e– falls to 1st energy level
e– falls to 2nd energy level
e– falls to 3rd energy level
typical emission spectrum
H discharge tube, with power
supply and spectroscope
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1ST E.L.
2ND E.L.
3RD E.L.
4TH E.L.5TH E.L.6TH E.L.
Lyman(UV)
Paschen(IR)
Balmer(visible)
~ ~ ~~ ~ ~
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Resources - Atomic Structure
Objectives
Worksheet - vocabularyWorksheet - development of atomic theoryWorksheet - atomic number and mass numberWorksheet - ions and subatomic particlesLab - isotopesWorksheet - orbital diagramsWorksheet - electron configuration
Outline (general)
Worksheet - light problemsWorksheet - half-lifeTextbook - questions
Episode 6 - Atom