Download Lecture 1 Atomic Structure

91536105
Fundamentals of Organic and Inorganic Chemistry
Tosapol Maluangnont, Ph.D.
Lecture 1 – November 05, 2013
Lecture 1 Atomic Structure
Required reading: Shriver & Atkins pp8-12, 14-20
In this Chapter, we will be covering :
• Hydrogenic atoms & Rydberg formula
• Basic principles of quantum mechanics
• Four quantum numbers and atomic orbitals
• Many-electron atoms: shielding and Zeff
• Pauli’s exclusion principle, Hund’s rule, and Aufbau principle
Page 1 of 33
What kind of information can we gather from these symbols?
19
F
9
• Fluorine
12
13
14
C
C
C
6
6
6
• Three isotopes of Carbon.
• Atomic number Z =
number of proton = 9
• Z=6
• The number of e- = 9
• The number of e- = 6.
• Atomic mass = 19. (This is
what you use in calculation
of the molar mass etc.)
• Atomic mass is 12, 13 and 14
respectively.
• The number of neutron =
19 – 9 = 10
• The number of neutron is 6, 7, and 8
respectively.
Page 2 of 33
What kind of information can we gather from these symbols?
14
C
6
14
vs
N
7
• These two elements have
the same atomic mass
(p+n) = 14.
1
2
H
H
1
1
proton deuterium
3
H
1
tritium
• Hydrogen is the only element
where each isotope has its own
symbol: H, D, and T.
• For carbon, Z = 6,
whereas for nitrogen, Z =
7.
• Different elements can
have the same number of
(p+n). Elements with
different atomic number
are different elements.
Page 3 of 33
Try this Fill in this Table.
Symbol
1
H
1
7
Li
3
No. of proton
4
No. of electron No. of neutron
5
Page 4 of 33
Hydrogen-like atom
• Hydrogen atom has one electron (1s1).
• The presence of only one atom makes it free from complicating
effects of electron-electron repulsions.
• Hydrogen-like atom is also called hydrogenic atoms.
• Examples are H, He+ and C5+ (why?)
H
He
C
Z=1
Z=2
Z=6
1s1
1s2
1s22s22p2
So, He+ 1s1
So, C5+ 1s1
Reminder :
Z is the atomic number, which is the number of
proton. The number of electron (in a neutral atom) is equal to that of
proton.
http://www.youtube.com/watch?v=OJzW2RoZq1Y
Page 5 of 33
Rydberg Formula
R = Rydberg constant
= 1.097×107 m-1
l = wavelength (m)
n1 = 1, 2, …
n2 = n1+1, n1+2, …
Series with n1 = 1  Lyman series (in UV)
Series with n1 = 2  Balmer series (in visible)
Series with n1 = 3  Paschen series (IR)
Series with n1 = 4  Brackett series (IR)
The energy is not
continuous, but
dependent on n
which is an integer.
Don’t forget E = hn = hc/ l
(h = Planck’s constant = 6.626×10-34 m2 kg s-1
n = frequency, c = speed of light = 2.997×108 m s-1, l = wavelength)
All info in this slide will be given in the exam.
Page 6 of 33
Example Calculate the wavelength of the first transition in the visible
region in the atomic spectrum of H.
จงคำนวณควำมยำวคลืน
่ ทีส
่ อดคลองกั
บกำรเปลีย
่ นระดับพลังงำนแรก
้
ในช่วง visible light ในสเปคตรัมของอะตอมไฮโดรเจน
In the visible region, this is the Balmer’s series with n1 = 2. The first
transition therefore will begin at n2 = 3. Substituting these values
together with R (= 1.097×107 m-1) gives
1/l = 1.524 x 10-3 nm-1
l = 656.3 nm (recall 1 nm = 10-9 m)
(Don’t just read this – make sure you actually get this number.)
Page 7 of 33
Try this Calculate the wavelength of the other transitions (in addition
to that at 656 nm) in the visible region in the atomic spectrum of H.
จงคำนวณควำมยำวคลืน
่ ทีส
่ อดคลองกั
บกำรเปลีย
่ นระดับพลังงำนอืน
่ ๆ
้
(นอกเหนือจำกที่ 656 nm) ในช่วง visible light ในสเปคตรัมของอะตอม
ไฮโดรเจน
Page 8 of 33
Basic principles of quantum mechanics
• Electrons can behave as particles or as waves “wave-particle duality”.
Here, we will concentrate on the fact that they are wave.
• A wave can be described by a mathematical equation (recall a sin-wave,
cos-wave, etc.).
• Wavefunction y (psi) – a function of the position x, y, z which describes
the behavior of an electron
• Schrödinger equation
where
me = the mass of an electron
V = the potential energy of the electron
E = the total energy of the electron
Solution can be
obtained only for
certain values of E.
That is, the energy is
quantized.
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• Like other waves, wavefunctions can interact.
• There are regions of positive and negative
amplitude (sign).
• Constructive interference - Enhanced
probability of finding the electrons in that
region.
• Destructive interference - Reduced
probability of finding the electrons in that
region.
• This will be used to explain chemical bonding
later.
Page 10 of 33
• The probability of finding an electron at a given
location is proportional to y2 at that point.
• There is a high probability of finding the electron
where y2 is large.
• The electron will not be found where y2
is zero
• y2 is called probability density.
Page 11 of 33
Quantum Number
n – principal quantum number
• This is the shell of electrons. n = 1 is called K-shell, n = 2 is called Lshell, and so on.
• The allowed energies are specified by n. For a hydrogen-like atom with
the atomic number Z,
with
n
R
hcR
=
=
=
Compare this n vs
the n in slide #6.
1, 2, 3, …
1.097×107 m-1
13.6 eV
• The zero of energy (E = 0) means that electron and nucleus is widely
separated.
• The negative energy means that, electron in a bound state is lower
than widely separated electron and nucleus.
Page 12 of 33
Example What is the ratio of the energy of a
ground-state H atom to that of a He+ ion?
จงคำนวณอัตรำส่ วนของพลังงำนที่สภำวะพืน้ ของ
อะตอมไฮโดรเจนเทียบกับอะตอมของ He+
Note that H atom and He+ ion are both
hydrogen-like atoms. Therefore, we can use the
equation En = -hcRZ2/n2.
For the ground state, n = 1. Therefore,
E(H)/E(He+) = (-hcRZH2)/ (-hcRZHe2)
= ZH2/ ZHe+2
= 12 / 22
= 0.25
Page 13 of 33
Example The ionization energy of H atom is 13.6 eV. What is the
difference in energy between n = 1 and n = 6 levels?
กำหนดให้ พลังงำนไอออไนเซชันของอะตอมไฮโดรเจนเท่ ำกับ 13.6 eV
จง
คำนวณควำมแตกต่ ำงของพลังงำนที่ระดับ n = 1 เทียบกับระดับที่ n = 6.
The question gives E(n=1) = -13.6 eV, and asks for what is
E(n=1) – E(n=6)?
Similar to the previous example,
E(n=1)/E(n=6) = (-hcRZ2/12)/(-hcRZ2/62)/ = (1/12)/(1/62)/ = 36
This gives
E(n=6) = E(n=1)/36 = -13.6 eV/ 36 = -0.378 eV
Therefore
E(n=1) – E(n=6) = (-13.6) – (-0.378) eV = -13.2 eV
Page 14 of 33
Atomic orbital – wavefunction of an electron in an atom. Electrons in an
atom are classified by a set of four quantum number:
n, l, ml, ms
We have seen n already when we talked about the electronic structure of
hydrogenic atom. But, what’s about the rest?
No electrons in the same atom will have all four of the quantum
numbers the same!
Page 15 of 33
Radial and angular part of wavefunctions
• Orbitals are easily described with spherical polar coordinates.
• The wavefunction of hydrogenic atom is the product of the part with
radius, and another part with the angle.
• R(r) of the radius depends on n and l.
• Y(q,F) of the angular depends on l and ml.
Page 16 of 33
n – principal quantum number: the shell
of electrons. n = 1 is called K-shell, n = 2 is
called L-shell, and so on
l=0 l=1
l=2 l=3
l – orbital angular momentum quantum
number: angular dependence of orbitals
 the subshell of electrons (s, p, d, etc.)
ml – magnetic quantum number:
orientation in space (component of l
around the axis)
ms – magnetic quantum number:
Electrons can be either
spin up ms = +1/2 or ↑ or a
spin down ms = -1/2 or↓ or b
(The fifth quantum number is the spin of an
electron s, and is always +1/2.)
n
l
0
1
2
3
Subshell
s
p
d
f
l = 0, ml = 0 (s)
l = 1, ml = -1, 0, +1 (there
are 3 p orbitals)
l = 2, ml = -2, -1, 0, +1, +2
(there are 5 d orbitals)
Page 17 of 33
Q: What are the orbitals having n = 4 and l = 1. How many orbitals are
there in this set?
ออร์ ปิตัลที่ n = 4 และ l = 1 คือออร์ ปิตัลใด และออร์ ปิตัลนีป้ ระกอบด้ วยกี่ออร์
ปิ ตัลย่ อย
A: n = 4 is the number of the shell. It can have l =
0, 1, 2 and 3.
l = 1 means that these are the p orbitals.
For l = 1, there are three values of ml (-1, 0, +1)
So, these orbitals are 4p orbitals (there are 3 of
them, namely 4px, 4py, and 4pz.)
ออรปิ์ ตล
ั ที่ n = 4 และ l = 1 คือ ออรปิ์ ตล
ั 4p ซึง่
ประกอบดวยสำมออร
ปิ์ ตล
ั คือ 4px, 4py และ 4pz
้
Page 18 of 33
n, l, ml, ms
Q: What is this orbital: 3, 2, 0, +1/2
A: n = 3 is the number of the shell. It can have l = 0, 1, and 2
l = 2 means that these are the d orbitals.
For l = 2, there are five values of ml (-2, -1, 0, +1, +2)
So, the all five orbitals below are the correct answer to this question. (In the
exam, giving just one answer is ok.)
3dxy, 3dxz, 3dyz, 3dz2, or 3dx2-y2
(Note that sometimes it is not defined with ml belongs to which orbitals.)
Page 19 of 33
• The quantum number l describes the angular variation of orbitals (= shape)
• The quantum number ml describes the orientation in space (direction).
• Boundary surface – region with high probability of finding an electron
• Spherically symmetrical
• Equal probability of finding electrons within
the sphere
• For an s-orbital, l = 0 and ml = 0.
s orbital (spherical)
I will expect that you can sketch the shape of s, p and d orbitals
with correct labeling of the axis.
Page 20 of 33
p orbitals (dumb-bell shape)
• px directs itself along the x-axis (and so on for py and pz).
• Nodal plane – the plane cutting through a nucleus, separating the
region of + and – sign of the wavefunction. (Note the bright and
dark lope which is opposite to each other.)
• No electrons at nodal plane
• There are three ml values (-1, 0, and +1) for p-orbitals, representing
px, py and pz.
I will expect that you can sketch the shape of s, p and d orbitals
with correct labeling of the axis.
Page 21 of 33
dzx
dxy
dyz
dx2-y2
dz 2
d orbitals
• dzx is on the zx plane (and so is dyzand dxy).
• dx2-y2 is along the x and y axis
• dz2 looks like a cone along the z-axis.
• There are two nodal planes for each d orbital.
• There are five values of ml for d orbitals: -2, -1, 0, +1, and +2.
I will expect that you can sketch the shape of s, p and d orbitals
with correct labeling of the axis.
Page 22 of 33
• The d orbitals (l = 2) can be found for elements of period
n > 2. The d orbitals mostly describe the chemistry of
transition metals.
Reminder: n = 1  l = 0 (s)
n = 2  l = 0, 1 (s and p)
n = 3  l = 0, 1 and 2 (s, p and d)
• The absence of d orbitals in the first member of group
15, 16, and 17, makes them different from the rest. This
is called anomalies. (Anomaly แปลวำ่ สิ่ งทีแ
่ ปลก
ประหลำด)
• Example: Coordination number (= the number of ions
(atoms) surround that ion (atom).)
neutral:
NCl3
vs PCl3, PCl5
ion:
NH4+ vs PCl6-
Page 23 of 33
Page 24 of 33
Many-electron atoms
Consider the ground state e- configuration of the following atoms
H: 1s1
He: 1s2
1s orbital of He is more
compact than that of H.
(There are 2 protons in
He’s nucleus which draw ecloser.)
Li: 1s22s1 (NOT 1s3!) “Pauli exclusion principle”
• no more than two
electrons may occupy a
single orbital.
• If they do, their spins
must be paired.
Page 25 of 33
• An electron at different shell feels reduced (and unequal) amount of charge
at the nucleus due to shielding by electrons closer to nucleus. (Think
about the pressure you experience in the classroom, which is probably a
function of where you sit!)
• The magnitude of shielding is depending on which electrons you are
looking at.
• Because of shielding, the “effective nuclear charge Zeff” is lower than the
true nuclear charge.
• Zeff (usually) increases across the period.
• Zeff = Z – s, where Z is the atomic number, and s is the shielding
constant
Page 26 of 33
I will expect that you can explain the trend in Zeff.
Page 27 of 33
• 1s e- of H atom: it is attracted by the proton at the
nucleus, and it does not interact with any other
electrons. In this case, Zeff = Z = 1.
• 1s e- of Li atom: Because of interaction between
2 e- in the 1s orbital (Li 1s2), one electron shields
another electron. Each electron does not fully
feel the presence of Z = 3. Instead, Zeff is 2.69.
H 1s2
Li 1s22s1
• 2s e- of Li atom: This is electron is further away
from the nucleus. Even more, it is shielded by 2ein the 1s orbital. So, instead of Z= 3, Zeff is only
1.28.
Page 28 of 33
Example Use data from Table 1.2 to calculate the shielding constants for
the outermost electron in the elements Li and Be.
จงใช้ ข้อมูลจำกตำรำง 1.2 คำนวณ shielding constant ของอิเลคตรอนวงนอก
สุดของอะตอมตัง้ แต่ Li และ Be
Li:
Be:
s = Z – Zeff
s = 3 - 1.28 = 1.72
s = 4 – 1.91 = 2.09
Try this Use data from Table 1.2 to calculate the shielding constants for
the outermost electron in the elements from Be to F.
Page 29 of 33
I will expect that
you know this in
the exam.
Aufbau principle
1. General order of occupation: 1s 2s 2p 3s 3p 4s 3d 4p
2. Hund’s rule – when more than one orbital has the same energy,
electrons occupy separate orbitals, with parallel spins
Z=1
Z=2
Z=3
Z=4
Z=5
Z=6
Z=7
Z=8
2p1
H
He
Li
Be
B
C
N
O
1s1
1s2
1s22s1, or [He]2s1
[He]2s2
[He]2s22p1
[He]2s22p2
[He]2s22p3
[He]2s22p4
2p2

2p3
closed shell
2p4
Page 30 of 33
Aufbau principle (continued)
3.1 For incompletely filled d subshell, e-configuration is mostly in the form
4s23dn
3.2 Spin correlation  a case where half-filled or filled shell is more
stable for 4s23dn.
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Z = 18 1s22s22p63s23p6
or [Ne]3s23p6  closed shell [Ar]
Z = 19 [Ar]4s1
Z = 20 [Ar]4s2
Z = 21 [Ar]3d14s2
Z = 22 [Ar]3d24s2
Z = 23 [Ar]3d34s2
Z = 24 [Ar]3d54s1 (not 3d44s2!)  spin correlation
Z = 25 [Ar]3d54s2 (not 3d64s1)  spin correlation
Z = 26 [Ar]3d64s2
Z = 27 [Ar]3d74s2
Z = 28 [Ar]3d84s2
Z = 29 [Ar]3d104s1 (not 3d94s2!)  spin correlation
Page 31 of 33
C: [He]2s22p2
N: [He]2s22p3
O: [He]2s22p4
Q: We previously mentioned that Zeff increases across the period. However,
going from C to N, the increase in Zeff is 0.69, while going from N to O, the
increase in Zeff is only 0.62. Why?
A: From C to N, electrons add into an empty p orbital. From N to O, the eadd into occupied p orbital. There is more repulsion in the latter case,
making the increase in Zeff less.
Page 32 of 33
Exercises
Do the exercises in Shriver & Atkins on p32.
1.8
1.11
1.12
1.13
1.14
1.15
Page 33 of 33