Download 1 Dalton`s Atomic Theory THE NATURE OF ATOMS (p.44)

Dalton’s Atomic Theory
-each element has characteristic
atoms, all of which are identical.
Atoms of different elements differ
in some fundamental way.
Compounds are formed when
atoms of different elements
combine and each compound has
the same relative number of atoms
of each element. Reactions are
reorganisations of atoms and the
way they are bound together that do
not alter the atoms themselves
John Dalton
19th Century – atomic masses determined accurately for known
elements
THE NATURE OF ATOMS (p.44)
-ca. 1900 J.J. Thompson created ‘cathode rays’ – a beam of negatively
charged particles streaming from negative electrode. All metals seemed to
contain these ‘electrons’
-proposed negative electrons were
dispersed amongst cloud of postitive
charge – plum pudding model of the
atom:
“a positive pudding with negative plums”
Thompson determined charge-to-mass
ratio of an electron as
-1.76 x 108 C/g
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THE NATURE OF ATOMS (p.44)
-ca. 1908-1917 R.A. Millikan measured charge on a single electron by
suspending charged oil droplets in an electric field
-(voltage adjusted to balance droplet’s upwards electrostatic attraction
with its own weight)
determined electron charge
= -1.6 x 10-19 C
Using Thompson’s chargeto-mass ratio:
⇒ gives electron mass as:
9.1 x 10-28 g (very small!)
Rutherford’s Nuclear Atom (p.47)
-1906-09 Rutherford set up beam of alpha particles
at thin gold foil.
-Expected most particles to pass through foil with
only slight deflections due to the ‘negative plums’
-Big surprise: 1 in 20,000 alpha particles were strongly deflected by > 90º or
even ‘reflected’ :
⇒ atom’s positive charge was very small and concentrated
nucleus 100,000 times smaller than atom! (atom is mostly empty space)
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The Spectrum of Atomic Hydrogen (p.274)
-model (theory) of structure of atom based on negative
electrons orbiting a compact, positive nucleus – electrostatic
attraction
-spectrum of hydrogen: excited hydrogen gas becomes
atomised (H-H bonds break). Energy then lost by emission of
light
-problem: spectrum of atomic hydrogen shows emissions at
certain wavelengths only – not continuous range of
wavelengths like a rainbow
Wavelength, Energy and Frequency Reminder
long wavelength (λ)
low energy (E)
low frequency (f)
short wavelength (λ)
high energy (E)
high frequency (f)
3
4
-several series of converging lines e.g. Balmer series lies in visible
region:
Lines follow pattern according to:
ν = R[(1/a2 ) – (1/b2 )]
where a & b are whole numbers. a is characteristic of the series
e.g.a = 2 for Balmer series
b identifies individual line
R = Rydberg constant 3.3 x 1015 Hz
-photon of light emitted as atom loses energy (when electron falls from
higher energy state to lower)
E = hc = hν
λ
i.e. larger the energy drop, the higher the energy of the emitted photon
energy (the shorter its wavelength)
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The Bohr Atom (p. 274)
Bohr: hydrogen’s electron only adopts certain orbits of particular energy (and
distance from nucleus) – electron’s energy is quantised
-energy of electron can be regarded as zero when proton and electron widely
separated
E = -hR/n2
n = whole number (quantisation)
And E = lower (more negative) for small n, electrons more tightly held
-deduced that electron’s orbit radius r = kn2
higher energy orbits ⇒ bigger radius
r = 52.9 x 10-12 m (52 pm) for ground state hydrogen
– the Bohr radius
Quantisation – only particular
discrete energy levels are allowed.
Like climbing a ladder – you can’t
stand between rungs!
high
energy
Bohr calculated energy gaps
between his allowed levels - they
corresponded exactly to emission
lines of excited atomic hydrogen
low
energy
(Lyman series)
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Lyman series (UV)
electrons returning to ground state
Balmer series (visible)
electrons returning to n = 2 state
Paschen series (IR)
electrons returning to n = 3 state
∆E = hc/λ
= hR[(1/nL2 ) – (1/nU2 )]
We can now justifiably describe
quantised energy levels as
lower nL and upper nU
∆E = hc/λ = hR[(1/nL2 ) – (1/nU2 )]
Q: What wavelength is the line emitted as
an electron drops down from the third
quantum level to the second (i.e. for the
transition n = 3 → n = 2 ?)
A: hc/λ = hR[(1/nL2 ) – (1/nU2 )]
656 nm
rearrange: c/λ = R[(1/nL2 ) – (1/nU2 )]
= R[(1/22 ) – (1/32 )]
X UV Vis
IR Rad
= 0.139R
400
800 nm
rearrange: λ = c/0.139R
= 3 x 108/(0.139 x 3.29 x 1015) = 6.56 x 10-7 m
(656 nm - in the visible range of electromagnetic spectrum )
Bohr’s model only works for one-electron atoms: H, He+, Li2+
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Q5: Using the Rydberg equation:
∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )]
for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz,
h = 6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023).
a) Calculate the photon energy and wavelength corresponding to i) the
three lowest energy lines of the Lyman series and ii) the convergence
limit of the Lyman series.
Lowest energy line: nL = 1, nU = 2 so [(1/nL2 ) – (1/nU2 )] = 0.75
∆E = 1.636 x 10-18 J
λ = 121.6 nm
second lowest energy line: nL = 1, nU = 3 so [(1/nL2 ) – (1/nU2 )] = 0.8889
∆E = 1.939 x 10-18 J
λ = 102.6 nm
third lowest energy line: nL = 1, nU = 4 so [(1/nL2 ) – (1/nU2 )] = 0.9375
∆E = 2.045 x 10-18 J
λ = 97.3 nm
the convergence limit: nL = 1, nU = ∞ so [(1/nL2 ) – (1/nU2 )] = 1.0000
∆E = 2.181 x 10-18 J
λ = 91.2 nm
Q5: Using the Rydberg equation:
∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )]
for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h =
6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023).
b) sketch the appearance of the Lyman series of lines in relation to the
visible region of the electromagnetic spectrum
UV
UV
100
200
visible
400
IR
800
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Q5: Using the Rydberg equation:
∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )]
for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h =
6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023).
c) By calculating the highest energy line of the Balmer series,
determine if the two series of lines overlap
Highest energy Balmer line: nL = 2, nU = ∞ so [(1/nL2 ) – (1/nU2 )] = 0.25
∆E = 5.453 x 10-19 J
λ = 364.7 nm
(no overlap with Lyman series)
Q5: Using the Rydberg equation:
∆E/h = c/λ = ν = R[(1/nL2 ) – (1/nU2 )]
for the atomic spectrum of hydrogen (where R = 3.29 x 1015 Hz, h =
6.63 x 10-34 Js, c = 3 x 108 m/s, NA = 6.022 x 1023).
d) By considering the convergence limit of the Lyman series, calculate
the ionisation energy (in kJ/mol) of the hydrogen atom. Why is the
energy required to remove an electron from He much greater (2370
kJ/mol)?
the Lyman convergence limit:
∆E = 2.181 x 10-18 J/photon
∆E = 2.181 x 10-18 x 6.022 x 1023 = 1.31 x 106 J/mol
= 1310 kJ/mol
1st I.E. of He is higher due to greater nuclear charge and smaller
atomic radius
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pra
ctise
Q
7.30 s
,7
& 7 .32
.34
Particles as Waves: Orbitals (p.279)
-1920s ‘particle-wave duality’ theory ⇒ waves behave as
particles and particles behave as waves
e.g. light behaves as photons (packets of energy)
beams of electrons can be diffracted through crystals (wave
behaviour)
-electrons as waves inside atoms?
-Heisenberg Uncertainty Principle (p.283) – the position and
speed of a moving particle cannot be simultaneously defined
Bohr model (based on classical physics) therefore not valid
– cannot describe precise orbits ⇒ quantum mechanics needed
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-describe electron in quantum mechanics ⇒
can talk only about the probability of finding
the particle in a particular place (p. 283)
-Schrodinger Equation (1926, p. 285)
solutions are wavefunctions (ψ) called atomic
orbitals – these are the quantised energy levels
of a one-electron atom.
ψ
in ground state (n = 1)
wavefunction describing
the single electron of H
-orbitals are region where
there’s a high chance of
finding the electron. The
probability of finding it at
any given point is given
by ψ2
The Hydrogen 1s orbital (p. 288)
- cross-section of ψ2 (probability
distribution) 3-D plot shows highest
electron density close to nucleus:
-small chance of finding electron far
away (0 chance at infinity)
to draw ‘shape’ of orbital we must draw
contour e.g. can draw region within
which there is 90% chance of finding electron
– region is spherical so hydrogen’s occupied
orbital is spherical:
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orbitals are defined by three quantum numbers:
n (shell number)
l (determines orbital shape)
ml (determines orbital orientation)
Naming atomic orbitals (p.279)
- there are n2 orbitals per shell i.e.
one orbital for n = 1
four orbitals for n = 2
nine orbitals for n = 3
n
l
ml
n
l
ml
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2s orbital – higher in energy than 1s – positive and negative
regions of wavefunction
1s
2s
For 2s orbital there is a nodal surface where ψ = 0 and changes
from +ve to -ve.
n.b. probability (ψ2) always positive.
2p orbitals (p. 289)
for the other three orbitals in n = 2 shell l = 1.
these feature a nodal plane (ψ changes sign at the
nucleus) – there are two lobes - these are called
p orbitals
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p orbitals are drawn like:
or
-the three p orbitals within shells n = 2, n = 3, n = 4 etc…
are at right angles to each other e.g. 2px, 2py, 2pz
but within each shell all are same energy (degenerate) in one electron atoms.
name indicates the shell, shape of orbital and the orientation of its lobes….
e.g.
2py
indicates radius of
orbital. Determined by
principal quantum
number n
indicates shape (symmetry) of orbital.
Determined by quantum number l
indicates orientation of
orbital. Determined by
quantum number ml
- other orbitals occupied by
electron in excited states
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3d orbitals (p. 289)
in n = 3 shell there are
nine orbitals): one s, three
p, and five d orbitals
in this and higher energy
shells l can equal 2 giving
rise to d orbitals which
have four lobes:
Occupancy of d orbitals is
important in explaining the
colourful compounds and
complexes formed by transition
metals e.g. Fe and Cu
Electron Spin and ms (p.292)
Each orbital can hold two electrons. These must be of opposite spin
(Pauli Exclusion Principle).
Fourth quantum number –
electron spin ms = +½ or -½
Pauli Exclusion Principle – each electron in any atom has a unique set
of four quantum numbers
-an electron’s set of quantum numbers is like a unique address within the
atom - tells us shell no., type of orbital, orientation of orbital, electron
spin
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Shielding (p.295)
Schrödinger equation very powerful but can’t be solved for manyelectron atoms due to repulsions between electrons
-In many-electron atoms, each electron doesn’t feel full nuclear
charge – they are shielded by other electrons
-e.g. in He, Z = +2, but electrons repel each other so:
‘effective nuclear charge’ Zeff = +1.7 (hard to measure)
-electrons in same orbital exert a small shielding effect on each
other
Shielding (p.295)
radial distribution function
= 4πr2ψ2
-electrons in inner orbitals shield outer
electrons very effectively - (largely
cancelling out nuclear charge)
e.g. in carbon (2p2), the 2px electron is
shielded slightly by 2py electron but
much more by 1s electrons which lie
closer to nucleus (see right)
Zeff = +1.3 for Li 2s electron despite
nuclear charge of +3
-We can still apply Schrödinger equation if we consider each electron in
turn moving in a field resulting from a nuclear charge of +Zeff rather than
+Z
⇒ Can then describe many-electron atoms using hydrogen-like orbitals
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Building Up Principle
(p.298)
-to explain electron
arrangement in all elements:
-add successive electrons to
hydrogen to form heavier
elements
-orbitals fill from lowest
energies to highest – the
Aufbau principle
-in many-electron atoms,
orbitals within each shell no
longer degenerate e.g. 2p
orbitals higher in energy than
2s orbitals
- for element of atomic no. Z, we add Z electrons to fill its orbitals
- Hydrogen has a half-full 1s orbital - we write its electron configuration
as 1s1
He has a full 1s orbital – 1s2
Li is 1s2 2s1 or because of its He core we can write [He] 2s1
Similarly, Be is [He] 2s2
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-Only outermost shell contains valence electrons – those available to
take part in reactions
-At boron (Z = 5) one 2p orbital is half occupied
-For C there is a choice – sixth electron can either pair up with fifth or
else go into another p orbital
Hund’s Rule – the lowest energy configuration for an atom is to
have the maximum no. of unpaired electrons all adopting parallel spin
reason for maximising electron spin:
mutual repulsion of electrons is minimised
-Only outermost shell contains valence electrons – those available to
take part in reactions
-At boron (Z = 5) one 2p orbital is half occupied
-For C there is a choice – sixth electron can either pair up with fifth or
else go into another p orbital
Hund’s Rule – the lowest energy configuration for an atom is to
have the maximum no. of unpaired electrons all adopting parallel spin
⇒ carbon’s two 2p electrons have parallel spins.
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-similarly N has three unpaired electrons
at oxygen electrons begin to pair up so O has only two unpaired electrons:
At Ne, n = 2 shell is full - atom has eight valence electrons - a stable octet
⇒ Ne forms no compounds
After Ne third period elements (n = 3) starts to fill same as 2nd period
- 3s orbital starts to fill first at sodium (Na):
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The Building-up Principle has a surprise for us after Argon (Z = 40)
4s orbital is lower in energy than 3d so fills first. Therefore K has
config [Ar] 4s1 rather than [Ar] 3d1.
The Building-up Principle has a surprise for us after Argon (Z = 40)
4s orbital is lower in energy than 3d so fills first. Therefore K has
config [Ar] 4s1 rather than [Ar] 3d1.
Radial distribution function of
4s and 3d orbitals explains why:
most probable distance lies
further from nucleus for 4s
BUT
4s electron density is greater
than 3d close to nucleus
(4s ‘penetrates’ nucleus more
than 3d)
penetration by 4s lowers its energy
20
filling of d orbitals (10 e) corresponds to transition metals (d-block):
(p.285)
Periodicity (p.50)
Groups (vertical columns) and periods (horizontal rows) of periodic table
are named after orbitals that are partially filled:
e.g. S is in group VI (six valence electrons) [Ne] 3s2 3p4
and period 3 (valence electrons are in n = 3 shell)
e.g. Ca is in group II (two valence electrons) [Ar] 4s2
and period 4 (valence electrons are in n = 4 shell)
repetition of orbital filling s > p > d > s > p > d
leads to periodicity (repeating chemical properties)
of the elements
e.g. halogens all have s2p5 configuration
⇒ similar chemistry ⇒ all form HX gases, react
vigorously with metals to form MX salts
- Mendeleev in Siberia ca. 1870 ⇒ periodic table
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Atomic Number (Z) (p.49)
Moseley (1913) fired electron beam (cathode ray) at elements. Emitted Xray wavelength depended on atomic number.
Since no. of protons in nucleus of atom = no. of electrons in neutral atom
He found way to count electrons in an atom
☺
Neutrons and Isotopes (p.49)
Dalton’s original assumption that all atoms of the
same element have same mass – WRONG!
E.g. Sn atoms can differ in mass by over 10%!
Some H atoms weighed twice as much as others - Why?
Today we ‘weigh’ atoms and molecules directly using mass spectrometer
Mass Spectrometry (p.86)
-sample is vapourised and ionised (e.g. bombarded by electrons
to give negative charge)
-then accelerated towards electromagnet that deflects ions
amount of deflection gives mass-to-charge (m/z) ratio
e.g. most Ne atoms are about 20 times the mass of hydrogen atom
but a few are 21 or 22 times as heavy – why?
22
- three different masses of Ne seen…...we say there are three isotopes of neon
isotopes are atoms of the same element with differing mass
isotopes contain differing numbers of neutrons – neutral particles found in all
atomic nuclei except hydrogen-1 (1H)
A
ZX
20
10 Ne
21
10 Ne
22
10 Ne
atomic mass unit (a.m.u. or Dalton) – defined as one-twelfth the mass of
carbon-12 atom or 1.66 x 10-24 g
-the proton and neutron both ‘weigh’ about 1 a.m.u.
-differing numbers of neutrons change mass of atom but not the number
electrons required to stay neutral
-atoms of different isotopes have the identical chemistry – mass difference
affects physical properties only
e.g. 2H2O ice (solid D2O, density = 1.1 g/cm3) cubes sink in H2O water
(density = 1.0 g/cm3)
23
Most elements are occur as a mixture of isotopes
(fluorine is an exception):
mass number (A)– the total number of nucleons (protons + neutrons)
in nucleus of the atom of an isotope
e.g. an isotope of nitrogen (Z = 7) with eight neutrons has a mass number
of 15
an isotope of phosphorus (Z = 15) with 16 neutrons has a mass number
of 31
A
- mass number written above element symbol, atomic number below: Z
e.g. carbon-12 (six protons, six neutrons) ⇒ 126C
….natural isotopes of Ne (Z = 10) ⇒ 2210Ne, 2110Ne, 2010Ne
X
Q: The most abundant isotope of tin has a mass number of 120?
How many neutrons are in 120Sn nuclei?
A: First find atomic number of Sn from periodic table. No. of
protons = 50.
120 – 50 = 70 neutrons
There are more neutrons than protons – typical of heavy atoms.
24
relative atomic mass (atomic weight) of an element is a weighted
average of isotope masses of that element
e.g. chlorine consists of 75% chlorine-35 and 25% chlorine-37. Its
R.A.M. is 35.5
e.g. the Mr of hydrogen is 1.008
(more than 1.000 because of traces of the heavier isotopes:
deuterium 2H and tritium 3H)
Q: Fill in every blank box regarding the atoms or ions of the following
isotopes:
Species
239
No. of
neutrons
4+
1
Np
[Rn]5f
3
1
mercury-201
53
2+
Cr
[He]
1
2
1s
201
24
43
bismuth-211
6
34
[Kr]
83
answers at the end of this slide sequence - try without looking first!!
25
Ions (p.53)
-Atoms can donate or accept electrons to form atomic ions (ionic
bonding)…..
-or share electrons with other atoms (covalent bonding)
similarly, a molecule can also become a charged molecular ion if
it gains or loses electrons e.g. NH4+, NO2-,
Atoms become ions when they accept or donate electron(s)
Al
Na
F
O
+
+
Al3+
Na+
e2e-
+
+
3eeFO2-
Positive ions are called cations
Negative ions are called anions
Cation and Anion formation
Postive ions are called cations
Negative ions are called anions
26
Molecules (p.53)
-Atoms form bonds in which electrons are
shared between atoms > bonded groups form
molecules e.g. H2O, PCl3, N2O
the molecular formula of a substance tells us the exact number
of atoms in the smallest unit of a substance
e.g. H2O is the molecular formula of water
a triatomic molecule with two covalent O-H bonds.
-N2 is a diatomic molecule with a triple covalent bond between
the two atoms
N≡N
27
-most molecules are polyatomic e.g.
P4
estradiol
-molecules can become ions if a covalent bond is broken
unevenly leaving positive and negative fragments:
-e.g.
CH3COOH
H2SO4
SO42-
CH3COO- + H+
+
2H+
Structural Formulae (p.55)
the structural formula tells us which atoms are bonded
to which in a molecule
e.g. the structural formula of i) water is H–O–H
ii) hydrogen is H–H
iii) ammonia is H–N–H
H
Empirical Formulae (p. 55)
the empirical formula tells us which atoms are present in
the substance in their simplest whole-number ratios
e.g. hydrogen peroxide is H–O–O–H ⇒ empirical formula is HO
the empirical formula of i) ethanol (C2H5OH) is C2H6O
ii) glucose (C6H12O6) is CH2O
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The Mole (p.77)
-Atoms too small to count
-Atoms weigh different amounts so need way to
count atoms/molecules involved in reactions
-Chemists invented the mole – no. of particles in
a sample that translates its atomic/molecular
weight into grams
1 mole = 6.023 x 1023 particles
-6.02 x 1023 = Avogadro’s number (NA)
-defined as the number of atoms in 12.00 g of 12C.
-mass of a mole of an element (g) = R.A.M. of
element (Daltons)
-1 mole Ne = 20.18 g
-1 mole Cu = 63.6 g
-1 mole H2 = 2.016g
-1 mole O3 = 48.0g
-n.b. 1 mole carbon = 12.01 g not 12.00g due to 13C and 14C
-therefore 6.02 x 1023 a.m.u. = 1g
-Q: A silicon chip weighs 5.68 mg. How many atoms does it
contain?
-A: convert to moles:
5.68 mg = 5.68 x 10-3 g
R.A.M. Si = 28 g/mol
no. moles = 5.68 x 10-3/28 = 2.02 x 10-4 mol
no. atoms = 2.02 x 10-4 mol x 6.02 x 1023 = 1.22 x 1020 atoms
29
A:
Species
239
Np4+
6 +
Li
3
He+
mercury-201
53
Cr2+
77
Se2bismuth-211
No. of
neutrons
146
3
1
121
29
No. of
protons
93
3
2
80
24
No. of
nucleons
239
6
3
201
53
Electron
configuration
[Rn]5f1
[He]
1s1
[Xe]6s25d10
[Ar]3d4
43
128
34
83
77
211
[Kr]
[Xe]6s25d106p3
30