1
Lecture 4 & 5 Hydrogen Atom, Atomic Structure, Periodic
Table and Bonds
Schrödinger Equation for the Hydrogen Atom
The hydrogen atom potential is given by: V(r)= -e2/r
Constraining potential leads to energy quantization
2
Applying Sch. eq≅ to find:
 What is the probability that the electron is found at a distance r
from the proton?
 What are the allowed energy levels
The Wave function is a function of time:
Ψ(r, t) = ψ(r)𝑤 (𝑡)
But if we are interested in how a hydrogen atom behaves
on the average |𝑤(𝑡)|2 = 1
 We need only to solve time independent sch. Eq.
−ℎ2
( 2𝑚 ∇2 + 𝑉) 𝜓 = 𝐸𝜓
−𝑒 2
Coulomb Potential
𝑉 = 4𝜋𝜖
0𝑟
ℎ2
2𝑚
𝑒2
2
∇ 𝜓 + (4𝜋𝐸 𝑟 + 𝐸) 𝜓 = 0
1
0
The term r does not lead to analytical solution
 The radial dependence of the potential suggests that we
should transform from Cartesian coordinates to spherical
coordinates:
(x, y, z)⟶ (𝑟, θ, φ)
3



0£r£¥
p
-p
£q £
2
2
0 £ f £ 2p Rotation ×in × xy
Rewriting sch eq ≅ in terms of spherical coordinates,
(𝑟, 𝜃, 𝜑) it gives 3 diff. eq≅ one in r, and the other two in
𝜃, 𝜑
 The solution is a product of the three solutions:
Ψ(𝑟, 𝜃, 𝜑 ) = 𝑅 (𝑟)𝜃(𝜃)𝜑(𝜑)
 Let’s look to spherically symmetrical case, when 𝜓 does
not depend on 𝜃, 𝜑
 So, we are interested in the radical part only:
4
2
∇ 𝜓=
𝛿𝜓
=
𝛿𝑥
𝛿2 𝜓
𝛿𝑥 2
𝛿2 𝜓
𝛿𝑦 2
𝛿2 𝜓
+ 𝛿𝑦 2 + 𝛿𝑧 2
∙
𝛿𝑟 𝛿𝑥
𝛿 𝛿𝜓 𝛿𝑟
= 𝛿𝑥 ( 𝛿𝑟 𝛿𝑥)
=

∇ 𝜓=
𝛿𝑧 2
𝛿2 𝜓
𝛿𝑥 2
𝛿𝜓 𝛿𝑟
2
𝛿2 𝑟
𝛿2 𝜓
𝛿2 𝜓
𝛿𝑟 2
𝛿2 𝜓
𝛿𝑧 2
𝛿𝑟 2
= 
𝛿𝑟 2
𝛿𝑟 2
[(𝛿𝑥) + (𝛿𝑦) + (𝛿𝑧) ] +
𝛿𝜓 𝛿 2 𝑟
𝛿𝑟
𝛿2 𝑟
(𝛿𝑥 2 + 𝛿𝑦 2 +
)
𝛿2 𝑟
𝛿𝑟
We need to find 𝛿𝑥 and 𝛿𝑥 2 and the other derivatives
Straight forward from
𝑟 = √𝑥 2 + 𝑦 2 + 𝑧 2
𝛿2 𝑟
𝛿𝑟
By sub. 𝛿𝑥,… and
2
∵∇ 𝜓=
𝛿2 𝜓
𝛿𝑟 2
𝛿𝑥 2
in the previous eq.
2 𝛿𝜓
+ 𝑟 𝛿𝑟
⟶ For spherically symmetry
Sch. Eq.
ℎ2
2𝑚
𝛿2 𝜓
2 𝛿𝜓
𝑒2
( 𝛿𝑟 2 + 𝑟 𝛿𝑟 ) + (𝐸 + 4𝜋𝜖 𝑟) 𝜓 = 0
0
One solution for this diff eq. ≅ could be 𝜓 = 𝑒 −𝐶0 𝑟
𝐶0 is a constant
5
By differentiating and sub. into equation 1. We can determine
the constant 𝐶0 and E.
me4
The energy E = − 8ϵ
E=
0
2 h2
9.1x1031 (1.6x10−19 )4
− 8(8.85x10−12 )2 (6.63x1034 )2
= −2.178x10−18 J
1𝑒V = 1.6𝑥10−19 𝐽
𝐸 = −13.6𝑒𝑉(It agrees with the lowest energy level of
Hydrogen known from spectrum
 Let’s go back to the other question: What is the
probability that the electron can be found between ri r +
dr

r 2 |ψ|2 = r 2 e−2C0 r
since ψ = e−C0 r
6
Corresponding to n= 1 First Bohr orbit
More solutions for spherically symmetrical case:
ψn = e−C0 r Ln (r)
Ln : polynomial
The allowed energies:
1
𝐸𝑛 = −13.6 𝑛2
𝑛 = 1,2,3 …
n=1 (ground state)
n>1 excited states (higher energy, more close to zero)
Wave functions for spherically symmetric solutions:
-The probability
distribution for n>1 have maximum further from the origin.
Why?
7
The general solutions (no spherical symmetry)
𝜓𝑛,𝑙,𝑚 (𝑟, 𝜃, 𝜑)
The dependence m 3 spatial dimensions lead to 3 quantum
numbers n, l, m.
n: How far the orbital is? (r)
l: How fast the orbit’s angular momentum
m: The angle of the orbit in the space
 They are required to specify the wave function
N=1, 2, 3…
L=0, 1, 2…(n-1)
MI = 0, ±1, ±2 … ± 𝑙
(2𝑙 + 1)
𝐿
𝐿
𝐿
𝐿
=0⟶𝑠
=1⟶𝑝
=2⟶𝑑
=3⟶𝑓
(The usual notation)
Sch. Equation for other elements and approximate models
Helium atom: 2 protons, 2 neutrons and 2 electrons
 Disregard neutrons
 2 protons, 2 electrons
 assume the 2 protons are together at the origin
 For the 2 electrons :∇2 𝜓 ⟶ ∇1 2 𝜓 + ∇2 2 𝜓
P.E. for electron 1:
P.E. for electron 2:
−2𝑒 2
4𝜋𝐸0 𝑟1
−2𝑒 2
4𝜋𝐸0 𝑟2
8
P.E. of the two electrons:
Sch. Eq.
𝑒2
4𝜋𝐸0 𝑟12
r12 : distance between electron 1 and 2
−ℎ2
1
−2𝑒 2 2𝑒 2 𝑒 2
2
2
(∇ 𝜓 + ∇2 𝜓) +
(
−
+ ) 𝜓 = 𝐸𝜓
2𝑚 1
4𝜋𝐸0
𝑟1
𝑟2
𝑟12
 No analytical solution!
For Sn Z=50
The diff eq≅ have 150 independent variables and 1,275
terms for P.E.
Approximate Solutions
Assume there are Z protons in the nucleus and Z electrons
around the nucleus. The electrons are unaware of each other.
The solution for each electron is the same as for Hydrogen
atom:____ in H- atom solution, we replace
𝑒 → 𝑍𝑒 2
𝑒 2 → 𝑍 2𝑒 4
The allowed energies are:
𝐸𝑛 =
−13.6𝑍 2
𝑛2
(Z, atomic number)
 We can explain the chemical properties of the elements
using this model and we can build up the periodic table
from sch. Solution of Hydrogen atoms (with the 3
quantum numbers)
9
Spin quantum number and Pauli exclusion principle
 Electron spin: electrons demonstrates inherent angular
momentum, it is defined by quantum numbers (𝑚𝑠 )
Pauli Exclusion Principle
 To understand the atomic structure and build the
−13.6𝑍 2
periodic table let’s use 𝐸𝑛 =
and the 4 quantum
𝑛2
numbers (n, l, 𝑚𝑙 , 𝑚𝑠 ) and the Pauli Principle.
* The lowest energy orbital is 1s
10
1
* Energy increase as 𝑛2
*Orbitals of the same n, but different l are considered to be of
equal energy, until electron interactions play a role
-Build the periodic table:
n=1, l=0, 𝑚𝑙 = 0, 𝑠 = ±1
One or two electrons may occupy the lowest energy level, we
get hydrogen & He.
Note: for He (Helium) shell is closed
→ This determines why He is chemically inert
 We go regularly until Potassium then, what?
 K(Z=19)
 Because of electron interactions, the energy increases
with increasing values of l.
 For K-element, the energy level of 3d (n=3, l=2) is higher
than 4s (n=4, l=0)
11
→ 4s shell is filled, then 3d
K 4s1
(Notes what happens here)
Ca 4s2
Sc 4s23d1
V 4s23d2
Cr 4s13d5
(Note what happens here)
Mn 4s23d5
Fe 4s23d6
Co 4s23d7
Cu 4s13d10
Zn 4s23d10
(Note what happens here)
Notes that the 4s is the outer shell
Rare earth metals: look to their electron
configurations and see why it’s hard to distinguish
between them chemically.
12
Bonds:
 What happened when two atoms (2 Hydrogen or 2 Cl)
come close to each other?
→They form bond to reduce the energy of the system
Chemical Bonds (General)
1. 𝐸 → 0
𝑟→∞
2. 𝑟 → 0
𝐸→∞
3. Opium Separation: 𝑟0 attractive force= repulsive force
13
Types of Bonding:
 Ionic
 Covalent
 Metallic
 Van der Waals
1. Ionic Bonds: ex: NaCl
Involves the transfer of electrons
Na→ loses electron and becomes Na+
Cl→ takes the electron and becomes Cl−
- They bond with attraction (covalent)
Na+ (positive→cations)
Cl− (negative→ anions)
(Group I & II): electro positive elements → lose electrons
14
NaCl Bulk Modulus
Madelung Energy
Q is the charge positive or negative
Electrostatic ±𝑞 2 /𝑟
The interaction energy 𝑈𝑖𝑗
𝑈𝑖 = ∑ 𝑈𝑖𝑗
𝑗
𝑟𝑖𝑗
𝑞2
𝑈𝑖𝑗 = 𝜆 exp (− ) ±
𝜌
𝑟𝑖𝑗
𝑅
𝜆 exp (− 𝜌) − 𝑞 2 /𝑟𝑖𝑗 Nearest neighbors
1 𝑞2
±𝑝
𝑖𝑗
𝑅
otherwise
𝑈𝑡𝑜𝑡 = 𝑁𝑈𝑖 =
−𝑅
𝑁(𝑧𝜆𝑒 𝜌
z is the nearest-neighbor contribution
𝛼𝑞 2
−
𝑅
𝜆 and 𝜌 the constants can be determined from gas phase
data.
15
𝜌 is the measure of the replusive interaction r= 𝜌 where it is
e-1 of the value of r=0.
𝛼= ∑
𝑗
±
≡ 𝑀𝑎𝑑𝑒𝑙𝑢𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑝𝑖𝑗
Evaluation of the Madelung Constant:
𝛼
±
= ∑ ≡ 𝑀𝑎𝑑𝑒𝑙𝑢𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑅
𝑟𝑗
𝑗
𝑟 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑗 𝑡ℎ 𝑜𝑛 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝑖𝑜𝑛 𝑖𝑠 𝑜𝑓 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑖𝑚𝑝𝑜𝑟𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑖𝑜𝑛𝑖𝑐 𝑠𝑜𝑙𝑖𝑑𝑠
𝛼
𝑅
1
= 2 [𝑅 −
1
1
1
+ 3𝑅 − 4𝑅 ∙∙∙∙∙∙∙]
2𝑅
or
𝛼 = 2 [1 −
1
1
1
+ 3 − 4 ∙∙∙∙∙∙∙]
2
The factor of 2 occurs because there are two ions left and
right at equal distance rj
𝑙𝑛(1 + 𝑥 ) = 𝑥 −
𝑥2
2
+
𝑥3
3
−
𝑥4
4
one-dimensional chain is 𝛼 = 2 ln 2
16
Three dimensions series is more difficult.
𝑒2
_ _ ..1 __ • ''"IT l' . '
Electrostatic energy = -M (4𝜋𝜖 𝑎)
0
lattice constant a= 2.8 Å
E=8.94
Solutions:
U(pot) = N (−
Pij =
𝑟𝑖𝑗
𝑟
𝑒2
𝑟
±1
𝛼
1
Σ 𝑃 + 𝑟𝑛 Σ 𝑃 )
𝑖𝑗
𝑖𝑗𝑛
r = 2.10
U(r) = N (−
𝑒2
𝑟
𝐶
α + 𝑟𝑚)
𝜕𝑢
𝑚𝐶
𝑚−1
= 0 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑟0
=
ð𝑟
𝛼𝑒 2
𝛼𝑒 2 𝑟0𝑚−1
𝐶=
𝑚
substituting
m ranges 6-9
𝑢0𝑡ℎ
= 𝑢(𝑟0 ) = −
𝛼𝑒 2 𝑚−1
𝑟0
(
𝑚
)
17
First Sum ( Na+ and Cl-)
By definition M = 1.748
Second Sum
1
𝛼Σ 𝑃 = C
𝑖𝑗𝑛
𝛿𝑢
P = -𝛿𝑣
V = 2Nr3
δV = 6Nr2δr
1
𝛿
1
P = -6𝑁𝑟 2 𝛿𝑟 𝑁(−𝑀𝑒 2 𝑟 +
𝑀𝑒 2
1
= -6𝑁𝑟 2 (
𝑟
𝐶
)
𝑟𝑛
𝑛𝐶
- 𝑟 𝑛+1 )
P = 0 = same as 1 atm
1
𝑀𝑒 2
𝑟0
𝑟02
2(0) = (
𝑛𝐶
1
- 𝑟 𝑛+1 ) 𝑟_2
Compressibility
B=-
𝑉𝜕𝑃
𝜕𝑉
1
=
2Nr3
𝛿
𝜕
1
𝑀𝑒 2
(
6𝑁𝑟 2 𝜕𝑟 6𝑟 2
𝑀𝑒 2
 = 18 𝑟 𝛿𝑟 (
1
𝑟4
𝑛𝐶
– 𝑟 𝑛+3 )
𝑟
𝑛𝐶
- 𝑟 𝑛+1 )
18
1
Using two Eg.** = 18 (
𝐶
1
𝑛(𝑛+3)
𝑟0𝑛+3
−
𝑀𝑒 2 4
𝑟04
)
2
=
𝑀𝑒
𝑛+3
𝑟
𝑛𝑟 4
0
B=
0
𝑀𝑒 2
(
18
𝑛(𝑛+3)
𝑛
𝑒 2𝐴
Φ coul. = 4𝜋Є
0 𝑟0
1
1
1
= 18 𝑀𝑒 2 𝑟 3 (𝑛 − 1)
− 4) 𝑟 4
0
0
= 8.53 𝑒𝑉
18𝑟03 𝐵
n = 1 = Φ coul
n=
1+18 (2.82𝑥10−8 𝑐𝑚)3 (2.4𝑥1011 𝑑𝑦𝑛/𝑐𝑚2 )
8.53𝑥10−12
𝑛𝑐
Φ coul = 𝑟 𝑛 → 𝐶 =
𝑀𝑒 2
𝑟2
0
𝑛𝐶
𝛼𝑒 2
0
𝑟04
= 𝑟𝑛 =
= 8.1
Φ coul 𝑟0𝑛
𝑛
C = 1.05eV (44.36Å)8.1
= 4660eV∙A8.1
1
1
1
Σ n = Є + 12 ( 4.05 ) + ( 4.05 ) = 6.81
eij
2
3
𝐶
α = 6.81 = 684eVÅ8.1
−𝛼𝑒 2
ucoul = 𝑟
α=M
𝑛𝐶
𝑟0𝑛−1 = 𝑒 2 𝛼
19
C=
𝑢0𝑡ℎ
−𝛼𝑒 2 𝑟0𝑚−1
𝑛
= 𝑢(𝑟0 ) =
−𝛼𝑒 2 (𝑛−1)
𝑛𝑟0
n =8.1
18𝐵𝑟03
n = 1 + 1𝑢
𝑐𝑜𝑢𝑙(𝑟0 )
20
E(n) = energy atom
Total E:
Na3 E(r0)
Na =
−𝛼𝑒 2 7.1
𝑟0
(8.0)
atoms/cm3
The cohesive energy: 8.13-1.53= 6.60eV with respect Na+,Cl-
21
:
The energy required to overcome bonding at T=0K
Electro negative elements→ accept electrons
Compressed (T)
22
Na a3 E(r0 – ∆r) = Total energy will increase
Expanding in a Taylors Series:
𝑑2 𝐸
E (r0) + ½ ( 𝑑𝑟 2 ) ∆𝑟 2 + ⋯
½ Na
a3(
𝑑2 𝐸
)
𝑑𝑟 2 𝑟=𝑟0
(∆𝑟)2
Eq. 5.2
23
𝑎
( ) ∆𝑟 → 𝑒𝑎𝑐ℎ 𝑓𝑎𝑐𝑒 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑒𝑑
2𝑟0
T is the stress increasing from 0 to T
6x(
1 𝑇 𝑎2
2
𝑎∆𝑟
Eq. 5.3
) ( 2𝑟 )
0
equality Eq. 5.3 and 5.2
Bulk Elastic Modulus
∆𝑎3
T = c 𝑎3 ≅
1
𝐶 3∆𝑎
𝑎
=𝑐
3∆𝑟
𝑟0
𝜕2𝐸
C = 9𝑟 ( 𝜕𝑟 2 )
0
𝑟=𝑟0
Hooke’s Law
Solutions (Approximate)
Z protons in the nucleus
Z electrons are the nucleus
Free and no interaction
24
𝑒2
𝑟
→
𝑍𝑒 2
𝑟
ucoul →
𝑒2
→𝑟 !
(𝑍 + 𝑒)(𝑍 − 𝑒)
4𝜋Є0 𝑟
=
nearly ideal ionic bonds are alkali halides
Metals:
(**) carriers of the negative charge are highly mobile
**electrons work as the glue that holds the lattice
together.
Strong cohesive energy ductile and malleable
Typical bonding per atom
Dominant Bonding:
Covalent diamond ,
Metallic
Ionic
Van der Waals:
Cohesive Energy (eV) C.E.
S ≈ 4.8 eV/atom
Na, Fe, La
≈ 1-9 eV/atom
NaCl, MgO ≈ 5 (eV) per ion pair
He, Ar, Cl2 ≈ .02 - .03
25
Several aspects of the van der Waals interaction and the resulting
bonding should be emphasized:
1. The interaction represented by U(r) is attractive and leads to the
formation of a potential well when the repulsive Coulomb interactions
between the atoms are also included.
2. The van der Waals interaction is quite weak and is also very short
range, due to the 1/r6 dependence of U(r).
3. When an empirical repulsive potential of the form +B/r12 is added to
the attractive van der Waals term, the result is known as the Lennardlones potential,
Here  and  are parameters characterizing the strength and
range of the interaction, respectively. This potential is often
used in calculations of the cohesive energies of inert-gas
crystals. ----
2. Covalent Bonds
Involves sharing of electrons, atoms like to have filled
shells
C 2s22p2
26
Si 3s23p2
Ge 4s24p2
These elements are tetravalent instead of divalent? Why?
 Look to diamond structure picture: each atom is
surrounded by 4 neighbors.
 Carbon in the form of diamond is: insulator-it has strong
covalent bond.
Si & Ge: The bonds are weaker, they are semiconductors
 Carbon in the diamond form is the hardest material (high
resistant to deformation)


 Diamond has a higher thermal conductivity than metals at room
temperatures!!!! Why?
 Metallic Bonds:
In metals, the electrons are highly mobile. Sea of electrons acts as
a glue that holds the positively charged metal ions from flying
apart.
 Delocalized electrons.
27
Electron charge density of valence band densities between the
pair of atoms
3. Van der Waals bonds. (graphite)
28
 Instantaneous asymmetric charge distribution in the atoms and
molecules→these distributions cause temporary dipole and the
atoms or molecules one attracted by electrostatic forces.
Derivative
A
𝐵
E(r) = n − 𝑚
r
𝑟
𝑚
𝐴=
𝑛
𝑚 𝐵
𝐵
𝐸𝑐 =
−
𝑛 𝑟0 𝑚 𝑟0 𝑚
𝐵 𝑚
𝐸𝑐 = 𝑚 ( − 1)
𝑟0
𝑛
𝑚<𝑛
Dipole-dipole bonds
29
 When the bond is not exactly ionic or covalent, shared
electrons one close to one atom (more electronegative) →bond is polar and the molecule has a dipole
moment.

Those molecular bonds due to higher electron density on
one side of the nucleus
30
Lennord-Jones potentials which describes interaction of two inert
Gas atoms such as Ne,Ar, Kr and Xe.
Ne Ar Kr Xe
R/
Derivative
A
𝐵
E(r) = n − 𝑚
r
𝑟
𝑚
𝐴=
𝑛
𝑚 𝐵
𝐵
𝐸𝑐 =
−
𝑛 𝑟0 𝑚 𝑟0 𝑚
𝐵 𝑚
𝐸𝑐 = 𝑚 ( − 1)
𝑟0
𝑛
𝑚<𝑛
Message:
Nearly ideal ionic bonds are alkali halides.
Metals:
** Carriers of the negative charge are highly malleable
** Electrons work as thee glue that holds the lattice
together. Strong cohesive energy ductile and malleable.
Dominant Bonding:
 Covalent: diamond, Si, Ge ≈ 4.8 eV/atom
31
 Metallic: Na, Fe, La ≈ 1-9eV/atom
 Ionic: NaCl, MgO≈ 5(eV) per ion pair
 Van der Waals: He, Ar, Cl2 ≈.02-.03