Fayetteville Technical Community College Electron Configuration Lab Report

Experiment 9 – Electron ConfigurationsIntroduction
The model of an atom has evolved since it was first conceived by John Dalton in 1805.
This
experiment takes a closer look at Neils Borh’s model of atom (1913) in terms of energy levels
and electron configurations.
He has shown from the hydrogen atom, electron orbiting about the
nucleus is stabilized at certain distance from its nucleus. This volume of space associate with a
specific energy according to the principle quantum number, n.
According to Borh, this energy
(in unit of J or Joule) can be calculated for the hydrogen atom,
1
En = −RH ( 2 )
n
Where RH = 2.18 × 10−18 J is the Rydberg constant and n = 1, 2, 3, …
In Borh’s atomic model, electron at its stable orbit or lowest energy level is called the ground
state. Electron can be excited by absorbing energy from an external source. When electron
acquired enough energy, it can move from the ground state to a higher energy state or excited
state, as long as the energy excited the electron equals to the energy difference between the
higher and lower states.
The energy difference can be calculated by
∆E = RH (
1
1
− 2)
2
ni nf
Where ni is the initial state of the electron and nf is the final state of the electron.
Electron in the excited state would release its extra energy by a process called relaxation. The
emitted energy is in the form of photon associated with a particular frequency of the
electromagnetic radiation according to Max Planck’s equation.
|ΔE| = h × ν
Where h = 6.63 × 10−34 J·s is the Planck’s constant and ν (the Greek letter nu) is the associated
frequency in unit of Hz (Hertz or /s).
Electromagnetic radiation is a form of light and travels at one speed in vacuum, c = 3.00 × 108
m/s. When light travels through the atmosphere on Earth, it would slow down a bit, but it is not
significant enough for us to consider here.
Light travels in the form of wave and its speed is
related to its wavelength (λ, Greek letter lambda) and frequency (ν, Greek letter nu),
c=λ×ν
The unit of wavelength can be in unit of m (meter) or nm (nanometer), 1 nm = 10−9 m.
Experiment 9 – Electron Configurations
In 1924, Louis de Broglie theorized electron in the stable orbital must be moving in wave like
fashion and its wavelength is associated with its mass (m, in unit of kg) and speed (u, in unit of
m/s) according to the equation:
λ=
h
mu
In 1926, Erwin Schrödinger formulated an equation that can calculate the energy of an electron
in a hydrogen atom base on the wave behavior of the electrons which he called “wave
functions”. These functions used a set of four quantum numbers to describe the probable
location and the nature of the electron in the hydrogen atom. The electronic wave functions can
also be used to approximate probable locations and nature of electrons in atoms other than
hydrogen.
The set of four quantum numbers are in the form of (n, l, ml, ms)
n is the principle quantum number, it described the main energy level and associated with
the distance from the nucleus, where n = 1, 2, 3, ….
It also referred as the shell.
l is the angular momentum quantum number, it associated with the subshell inside the
shell. Thus, the quantum number l depends on n, or l = 0, 1, 2, to n – 1. It also describe the
type and shape of orbitals in the shell.
It is traditionally to associate the quantum number l with
a letter, where l = 0, 1, 2, and 3 are s, p, d, and f, respectively.
There are n number of subshell l
in each shell n.
ml is the magnetic quantum number, it referred to the orbitals in the subshell. The
values of ml = −l, …, 0, …, +l. There are (2l + 1) orbitals in each subshell l. The magnetic
quantum number also associated with the orientation of the orbitals in the subshell.
For each
shell n, there are n2 number of orbitals in the shell.
ms is the spin quantum number. This number referred to the nature of electron spin in
the orbital. The values of ms can be ½ (up spin) or −½ (down spin).
The set of four quantum numbers is used as identification of an electron in an atom. Each
electron has its own unique set of four quantum numbers. This was proposed by Wolfgang
Pauli in 1925, also known as Pauli Exclusion Principle – “No two electrons in the same atom
can have the same set of four quantum numbers.”
Under this principle, each orbital can have a
maximum occupancy of two electrons of opposite spins.
(2)
Experiment 9 – Electron Configurations
Purpose
This experiment is aimed to learn and practice the following concepts:
1) The concept of energy levels of electronic orbitals
2) The relaxation of excited electrons from excited states to lower states
3) Concept of quantum numbers
4) The concept of placing electron according to the Aufbau Principle and electron configurations
5) Hund’s rule
6) Half-filled rule
Concept of the Experiment
The evident of electronic levels comes from the emission of light at different frequencies when
an atom is excited by burst of energy.
For example, when hydrogen atoms are excited by
radiation, colors of the emitted light can be separated by a spectrophotometer. These color lines
are referred as light spectrum. The observed color lines in the visible spectrum for hydrogen are
known as the Balmer series, where the excited electron at electronic levels greater or equal than
3 (ni ≥ 3) relaxed to electronic level of 2 (nf = 2).
They occur at wavelengths of 656.3 nm (red),
486.1 nm (aqua), 434.0 nm (violet), and 410.2 nm (violet).
determined by its relationship with the speed of light, c.
The frequency of these color can be
For example, the frequency of the
wavelength at 656.3 nm is
c
ν= =
λ
3.00 × 108 m/s
−9
10 m
(656.3 nm × 1 nm )
= 4.57 × 1014 /s (or Hz)
The energy associated with this frequency is
|∆E| = h × ν = (6.63 × 10−34 J∙s)(4.57 × 1014 /s) = 3.03 × 10−19 J
The excited electronic level can be determined by
∆E = RH (
1
1
− 2)
2
ni nf
Since light is emitted, ΔE = −3.03 × 10−21 J, and
1
∆E 1
−3.03 × 10−19 J 1
=
+ =
+ 2 = 0.111
RH n2f
n2i
2.18 × 10−18 J
2
n2i =
1
=9
0.111
(3)
and ni = √9 = 3
Experiment 9 – Electron Configurations
Therefore, the color occurs at 656.3 nm is when the electron at the excited electronic level (ni =
3) relaxed to lower electronic level (nf = 2).
The speed of the excited electron having wavelength of 656.3 nm can also be found by de
Broglie’s relation, given the mass of the electron is 9.109 × 10−31 kg.
u=
h
=

6.63 × 10−34 J∙s
−31
(9.109 × 10
10−9 m
kg) (656.3 nm × 1 nm )
= 1109 m/s
where the unit J = kg∙m2/s2.
Although the model of the Bohr atom describes the main electronic energy levels, Schrödinger’s
equation has extended the atomic model further using a set of four quantum numbers: n, l, ml,
and ms.
The concepts of shells (n), subshells (l), orbitals (ml), and electron spins (ms) have
shaped how the atomic model is used today in arranging electrons in an atom.
Electrons are
placed in an atom according to the Aufbau Principle, where the arrangement of electrons
starting from the lowest electronic energy level.
The Aufbau principle can be simplified by the
following diagram of the subshells.
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
7s
7p
8s
The subshell energy levels are increasing from 1s, 2s, 2p, 3s, 3p, 4s, 3d, … by following the
direction of the arrow.
The number of electrons can be in each subshell depends on the number
of orbitals (ml) in the subshell.
there are 3 different orbitals.
For example, for a p type subshell (l = 1, then ml = −1, 0, or +1),
To satisfy Pauli’s exclusion principle, each orbital can only
occupied by maximum of 2 electrons of opposite spins (ms = −½ and +½).
subshell can have a maximum of 6 electrons.
(4)
Hence, a p type
Experiment 9 – Electron Configurations
The expression of the electrons placed in an atom of an element or an ion is referred as electron
configuration. For example, the atomic number of aluminum is 13, its neutral atom would
have 13 electrons. The order of filling these electrons would be 2 in the 1s subshell, 2 in the 2s
subshell, 6 in the 2p subshell, 2 in the 3s subshell, and 1 in the 3p subshell.
Thus, the electron
configuration of aluminum is
Al: 1s22s22p63s23p1.
Electron configuration can also be written for ions.
(3 less than the neutral atom).
For example, Al3+ has only has 10 electrons
Its electron configuration would be
Al3+: 1s22s22p6.
Electron configuration can also be written using a short form by utilizing the noble gas electron
configuration. For example, the short of aluminum electron configuration is
Al: [Ne] 3s23p1, where [Ne] has electron configuration of 1s22s22p6.
This is a convenient way to write electron configuration for elements or ions having many
electron.
In general, the electron configuration of the noble gas element prior is used instead.
Electron configuration can also be drawn as a diagram.
For example, the diagram for
aluminum is
Al:
↑↓
↑↓
1s2
2s2
↑↓
↑↓
↑↓
2p6
↑↓
3s2

3p1
When placing electrons in subshells having more than one orbitals, Hund’s rule is applied,
where each orbital must first occupied by one electron of the same spin before a second electron
is added into the orbital.
For example, nitrogen atom has 7 electrons.
Its electron
configuration diagram is
N:
↑↓
↑↓
1s2
2s2



2p3
For elements having electrons in the d or f subshell, the Half-filled rule also applied, where halffilled subshell is more stable (has lower energy state) than partially filled subshell (has higher
energy state). For example, the atom of chromium has 24 electrons.
The short form of the
electron configuration is
Cr: [Ar]4s13d5,

4s1




3d5
(5)

(lower energy state)
Experiment 9 – Electron Configurations
instead of [Ar]4s23d4.
↑↓


4s2


(higher energy state)
3d4
Pre-Laboratory Assignment
1) The aqua color line of the Balmer series occurs at 486.1 nm,
a) Determine the frequency associated with the photon associated with this wavelength.
b) Determine the amount of energy released by this photon.
c) The Balmer series are the results of excited electron at higher level relaxing to the electronic
level of principle quantum number of nf = 2. Determine the excited electronic level, ni, of
this electron.
d) If this excited electron has the same wavelength as the emitted photon, determine its speed
during the transition.
2) Given the quantum number, n = 4.
a) How many subshells is in this shell?
Write all the quantum numbers (l) associated with the
subshells.
b) What is the maximum electron occupancy in this shell?
c) For each of the subshells, write the name of the subshells associate with each quantum
number l, the number of orbitals in the subshell, and the quantum numbers (ml) of the
orbitals.
3) Write all the possible sets of four quantum numbers for the subshell 3p.
4) Write the electron configuration of an element that has 28 electrons in both long and short forms.
5) What is Pauli Exclusion Principle?
6) What is Hund’s rule?
7) What is the Half-filled rule?
(6)
Experiment 9 – Electron Configurations
Procedure
1) Energy levels by subshells
Use the Aufbau Principle to list the subshells in the order of increasing energy in your lab
notebook.
2) Writing electron configurations
For each of the following specie, use the Aubau Principle to write the electron configuration in
both long and short form. Based on the short form, construct the atomic orbital diagram and
place the electrons into the orbitals.
a) Elements: 15P, 19K, 26Fe, 29Cu, 35Br, 38Sr, 42Mo, and 48Cd.
b) Ions: 9F−, 13Al3+, 16S2−, 20Ca2+, 26Fe2+, and26Fe3+.
3) Constructing a generalize electron configuration scheme
Use the Periodic Table template on the “Data” page, fill-in the each box with the correct electron
configuration omitting the Noble Gas configuration.
Post Laboratory Questions
1) The last two visible line observed in the Balmer series are of the wavelength 434.0 nm and 410.2
nm. For each of the wavelength,
a) Determine the frequency associated with the photon associated with this wavelength.
b) Determine the amount of energy released by this photon.
c) The Balmer series are the results of excited electron at higher level relaxing to the electronic
level of principle quantum number of nf = 2. Determine the excited electronic level, ni, of
this electron.
d) If this excited electron has the same wavelength as the emitted photon, determine its speed
during the transition.
2) How many electrons can have a set of four quantum numbers with n = 3 and ml = −2?
Write
the set of the four quantum numbers of each of these electrons.
3) How many electrons can be in the shell with the principle quantum number of 7?
4) How many orbital(s) are in each of the following subshells: s, p, d, and f?
5) For each of the following sets of four quantum numbers, determine if it is possible and explain
your reason.
a) (2, 2, 0, +½)
b) (4, 3, −2, −½)
c) (3, 1, −4, +½)
d) (−3, 0, 0, −½)
e) (6, 4, −2, +1)
(7)
Experiment 9 – Electron Configurations
1A
2A
3B
4B
5B
6B
7B
┌── 8B ──┐
1B
2B
3A
4A
5A
6A
7A
8A
Data
(8)

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