chapter 4-dmc 101-arrangement of electron.pdf
TRANSCRIPT
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
1/33
CHAPTER 4: THE ARRANGEMENT OF
ELECTRONS IN ATOMS
Reaction happens WHEN electronin atoms
interact.
Electromagnetic Radiation
When electron interacts produces
electromagnetic radiation
Electromagnetic radiation carries energy Example:
visible light, x-rays, radio waves
Audi Majdan - DMC 101 - KLIUC
Chemical
Reaction
Interaction of
electrons in atoms=
1
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
2/33
Properties of electromagnetic radiation:
(a)
Move through vacuum at speed oflight, c = 3.00 x 108 m/s
(b) Wave-like properties
Properties Symbol Unit Pronunciation
frequency (Hz) nuwavelength (m) lambda
Audi Majdan - DMC 101 - KLIUC 2
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
3/33
Relationship:
= c
c (speed of light) = 3.00 x 108 m/s
Electron interaction depends on
arrangement of electron in atoms.
The arrangements of electronin atoms
provides information on:(a) Number of electron
(b) Position of electron
(c) Energypossess by atom
Audi Majdan - DMC 101 - KLIUC 3
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
4/33
Quantum and Photons
Electron orbiting the nucleus have it's
own energy Energy can be released orabsorbed by the
atoms during the interaction
Quantum = a unit of energy released or
absorbed by atoms
A photon = a quantum of energy
E = h or E = hc
E = Energy of a quantum
= frequencyh = Planck's constant
= 6.63 x 10-34 Joule seconds (Js)
Example:
Calculate the smallest amount of energy (one
quantum) that an object can absorb from yellow
light with a wavelength of 589 nm.
Audi Majdan - DMC 101 - KLIUC 4
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
5/33
Solution
1) Frequency:
= c
= c/
= (3.00 x 108 m/s)
(589 x 10-9 m)
= 5.09 x 1014 s-1
2) Energy:
E = h
E = (6.63 x 10-34 Js) x (5.09 x 1014 s-1)
E = 3.37 x 10-19 J
Exercise 1
A laser emits light with a frequency of 4.69 x
1014 s-1. What is the energy of one quantum ofthe laser?
Electrons Arrangement in Atom
Audi Majdan - DMC 101 - KLIUC 5
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
6/33
Several systems are created to know the
LOCATION ofELECTRONs in ATOM
Bohr's Model of Atom
Dalton's model of atomprovides the basisof atom structure
Bohr's model of atom is an advanced
model of atom structure
Characteristics of Bohr's Model:
(a) Electrons orbiting the nucleus(b) The electrons are restricted only to orbits
of certain radius and energies
Audi Majdan - DMC 101 - KLIUC
ELECTRONS
ARRANGEMENT
Single electron atom Multiple electrons atom
Bohrs Model Schrodingers Model
6
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
7/33
(c) An electron orbiting in an orbit has a
fixed energy state and will not release or
absorb energy
The energy possess by the electron:
RH = Rydberg constant
= 2.18 x 10-18 J (for calculation of E)
or
= 1.097 x 107 m-1 (for calculation of )
n = principle quantum number
= orbits for the electron
Energies always in negative value
n=1 an electron in the first allowed orbit
(closest to the nucleus)
n=2 an electron in the next allowed orbit
further from the nuclei n=3,4,5
Audi Majdan - DMC 101 - KLIUC 7
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
8/33
The energies of allowed orbits for the
electrons:
The lower the energy (more negative), the
more stable the atom
Audi Majdan - DMC 101 - KLIUC 8
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
9/33
Ground state = electron is in the lowest
energy state (n=1) the most stable state
Electrons can jump from one orbit to
another
Excited state = electron is in other state
than n=1 (n 2)
When n = infinity
Electron completely separated from the
atom
E = (-2.18 x 10-18 J) x (1/)
= 0
Reference or zero energy state = the state
in which the electron is separated from the
nucleus
Audi Majdan - DMC 101 - KLIUC 9
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
10/33
Electron moves to higher state (higher n)
= Energyabsorbed
Electron moves to lower state (lower n)
= Energyemitted
The difference in energy levels between the
final and initial orbits:
E = Ef - Ei
or
ni = initial orbit
nf = final orbit
E=+ve when energy absorbed:nf> ni = electron moves to higher n
Audi Majdan - DMC 101 - KLIUC 10
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
11/33
E=-vewhen energy emitted:ni > nf = electron falls to lower n
Example:
Calculate the energy of light that corresponds to
the transition of the electron from the n=4 to the
n=2 state of the hydrogen atom. Is the lightabsorbed or emitted by the atom?
Solution
ni = 4
nf = 2
RH = 2.18 x 10-18 J (for calculation of E)
E = (2.18 x 10-18 J) x ((1/16)-(1/4))= -4.09 x 10-19 J
(light is emitted)
Audi Majdan - DMC 101 - KLIUC 11
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
12/33
Exercise 2
Determine the energy of light that corresponds
to the transition of electron from n=3 to n=5
state for hydrogen atom. Identify whether the
energy is absorbed or released.
Bohr's Model for Hydrogen atom
Line Spectra
Line spectra = spectrum of coloured line
with different and separated wavelengths
when electron fall (move to lower orbit)
Example:
A rainbow represents the spectrum of
wavelengths of light contained in the light
emitted by the sun
Forhydrogen atom: 5 series ofline spectra
observed depending on the type of electrontransition
Audi Majdan - DMC 101 - KLIUC 12
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
13/33
Audi Majdan - DMC 101 - KLIUC 13
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
14/33
Series nf ni Spectrum
Region
Lymann 1 2,3,4.. Ultra violet
Balmer 2 3,4,5.. Visible light
Paschen 3 4,5,6.. Infrared
Brackett 4 5,6,7.. Infrared
Pfund 5 6,7,8.. Infrared
To determine wavelengths of the spectrum:
=
22
111
fi
H
nnR
Example:
Calculate the wavelength of energy released
during a transition from the n=2 to n=4.
=
22
111
fi
H
nnR
RH = 1.097 x 107 m-1 (for calculation of )
=
22
7
4
1
2
110097.1
1
( )1875.010097.11
7
=
Audi Majdan - DMC 101 - KLIUC 14
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
15/33
16
1006.21
= m
m710862.4 =
Exercise 3
Calculate the wavelength of the hydrogen
emission line that corresponds to the transition
of the electron from the n = 3 to the n =1 state.
Schrodinger's Model of Atom
Difference between Bohrs Model and
Schrodingers Model
Properties Bohr Schrodinger
Type of atom Single electron
atom
Multiple
electrons atom
Audi Majdan - DMC 101 - KLIUC 15
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
16/33
Electrons
Location
In a defined
orbit
Probability
electron
distribution
only
Notation
system
One quantum
number:
n
Four quantum
numbers:
n, l, ml, ms
Probability distribution of an electron in the
ground state in a hydrogen atom:
Audi Majdan - DMC 101 - KLIUC 16
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
17/33
Uncertainty Principle: it is impossible to
simultaneously know both the position
and momentum of an object as small asan electron
Schrodingers Notation System
The Bohrs model: one quantum number (n)to describe an orbit
The Schrdinger model: four quantum
numbers:
Audi Majdan - DMC 101 - KLIUC 17
More probability
to find an
electron
Less probability to
find an electron
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
18/33
(a) Principal quantum number (n),
(b) Orbital quantum number (l),
(c) Magnetic quantum number (ml)
(d) Spin quantum number (ms)
to describe an orbital.
(a) The principle quantum number (n)
n = shell (the energy level/orbits) in the
atom
n = 1, 2, 3, ...
Example: n = 1 (electron is at 1st shell)
As n increases, the electron density is
further away from the nucleus
As n increases, the electron has a higher
energy and is less tightly bound to the
nucleus
(b) The azimuthal quantum number (l)
Audi Majdan - DMC 101 - KLIUC 18
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
19/33
l= subshell in a shell
The value ofldepends on n
For every n, there are n value of l
l= (n-1)
lis referred to by a letter:
s = 0, p =1, d =2, f =3
l 0 1 2 3 4
notation s p d f g
Example:
n = 1 l= (n-1)
= 1-1= 0
(n = 1: one subshelllwith value of 0)
Notation: 1s
n = 2 l= 2-1
Audi Majdan - DMC 101 - KLIUC 19
n = 1 l= 0
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
20/33
= 1
(n = 2: two subshelllwith value of 0 and 1)
Notation: 2s (n = 2 and l= 0 )
Notation: 2p (n = 2 and l= 1)
n lSubshell
notation
1 0 1s2
0 2s
1 2p
3
0 3s
1 3p
2 3d
40 4s1 4p
2 4d
3 4f
(c) The magnetic quantum number (ml)
ml = orbitals in a subshell Orbital is the place where electrons located
ml = all values between 'l', 0, -'l'
Audi Majdan - DMC 101 - KLIUC 20
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
21/33
nValues
ofl
Subshell
Designat
ion
Values
ofml
Number of
Orbitals in
Subshell
Total
number
of orbitals
in shell
1 0 1s 0 1 1
2 0 2s 0 14
1 2p 1,0,-1 3
3 0 3s 0 1
91 3p -1,0,1 3
2 3d-2,-
1,0,1,25
4 0 4s 0 1
16
1 4p -1,0,1 3
2 4d-2,-
1,0,1,25
3 4f
-3,-2,-
1,0,1,2,
3
7
Audi Majdan - DMC 101 - KLIUC 21
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
22/33
(d) The electron spin quantum number (ms)
ms = direction of spin of the electron
ms = + or -
Explanation on The Quantum Number
Electron shell = consisting ofsubshell
Subshell = consisting oforbitals
Orbitals = consisting ofelectrons
Audi Majdan - DMC 101 - KLIUC 22
+
-
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
23/33
Example:
All orbitals that have principle quantum
number of 3 (n=3) is said to be in third shell.
The third electron shell ('n'=3) consists of the
3s, 3p and 3dsubshells
The 3s subshellcontains 1 orbital,
the 3p subshellcontains 3 orbitals
the 3d subshellcontains 5 orbitals
The third electron shellis comprised of
nine different orbitals, but all orbital with
the same energy (true for hydrogen only)
Each subshell is divided into orbitals:
Subshell Number of orbitals
s 1
p 3
d 5
f 7
Audi Majdan - DMC 101 - KLIUC 23
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
24/33
Energies of all hydrogen electron orbitals
until n=3:
At room temperatures all hydrogen atoms
are in their ground states
The electron may be promoted to an excitedstate by the absorption of a photon
Example:
a) Predict the number of subshells in the
fourth shell, n=4.
b) Give the label for each of the subshells
in (a)
c) How many orbitals are there in each of
the subshells in (c)?
Audi Majdan - DMC 101 - KLIUC 24
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
25/33
Answer:
a) n = 4 l= n-1
= 3
Therefore: 0, 1, 2, 3
4 subshells
b) s = 0, p = 1, d = 2, f = 3
Therefore: 4s, 4p, 4d, 4f
c) 4s l= 0, value forml ( from -l to l)
= 0 one 4s orbital
4p three 4p orbitals;
4d five 4d orbitals;
4f seven 4f orbitals
Exercise 4
a) What is the designation for the subshell
with n =5 and l=1?
b)How many orbitals are in this subshell?
c) Indicate the values ofml for each ofthese orbitals.
Orbitals in Multiple-Electron Atoms
Audi Majdan - DMC 101 - KLIUC 25
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
26/33
The hydrogen atom only 1 electron.
All subshells ofhydrogen atoms with the
same principle quantum number (n) have thesame energetic level
In multiple electrons atom: electron-
electron repulsion influences the energy
levels of the orbitals. Example:
Energy state 2s orbital < 2p orbital in a
many-electron atom
Electron spin and the Pauli Exclusion
Principle
Audi Majdan - DMC 101 - KLIUC 26
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
27/33
Pauli exclusion principle = No two
electrons in an atom can have the same
set of four quantum numbers (n, l, mland
ms)
For any given orbital has fixed values of:
n
l
ml
ms quantum number is use to differentiate
between 2 electron in an orbital (+ and
- )
Only two electrons at most can occupy the
same orbital, and they have opposite valuesfor magnetic spin
Example:
2 electron in one orbital can have quantum
number of
(1, 0, 0, +) & (1, 0, 0, -)
Electron Configurations
Audi Majdan - DMC 101 - KLIUC 27
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
28/33
Electron configuration = the way electrons
are distributed among the orbitals
Orbitals are occupied in order of
increasing energy, with no more than two
electrons per orbital
Example
Lithium
1. Lithium has 3 electrons.2. Start : Place two electrons in the 1s,
lowest energy orbital. These two electrons
have opposite ms (+ & -).
3. Place the third electron in the next
highest energy level orbital = the 2s orbital:
Audi Majdan - DMC 101 - KLIUC 28
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
29/33
The arrows indicate the value of the
magnetic spin (ms) quantum number
(up for + and down for - )
Diagram representation:
Audi Majdan - DMC 101 - KLIUC 29
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
30/33
1s2 2s1
Audi Majdan - DMC 101 - KLIUC 30
Li
1s 2s
Orbital
Diagram
Electronic
Configuration
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
31/33
Writing electronic configurations: Period 1, 2,
3
All 2p orbitals are equal energy any 2p
orbital can be filled
Carbon has 6 electrons. 3 possible
arrangements can be made in 2p orbital.
Which one is true?
Audi Majdan - DMC 101 - KLIUC 31
(a) (b) (c)
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
32/33
Hund's rule:
For degenerate orbitals, the lowest energy is
attained when the number of electrons with thesame spin is maximized
The second 2p electron in Carbon is placed
in another 2p orbital, but with the same
spinas the first 2p electron
Core electrons: the electrons in the stable
(Noble gas) configuration
Valence electrons: the electrons in the outer
shell (beyond the stable core)
Audi Majdan - DMC 101 - KLIUC 32
-
7/30/2019 Chapter 4-DMC 101-Arrangement of Electron.pdf
33/33
Exercise 5
1. Draw the orbital diagram
representation for the electron
configuration of oxygen, atomic
number 8.
2. Write the electron configuration of
phosphorus, element 15.