١/١٣/١٤٣٧
١
CHAPTER 6 – ELECTRONIC
STRUCTURE OF ATOMS
Prof. Dr. Nizam M. El-Ashgar
THE WAVE NATURE OF LIGHT
�The electronic structure of an atom: refers to the number
of electrons in an atom as well as distribution of the
electrons around the nucleus and their energies.
�To understand the electronic structure of atoms, one
must understand the nature of electromagnetic
radiation.
�Visible light is an example of electromagnetic radiation.
�All types of electromagnetic radiation move through a
vacuum at a speed of 3.00 x 108 m/s, the speed of light.
١/١٣/١٤٣٧
٢
LIGHT
Red Orange Yellow Green Blue Ultraviolet
WAVES
The wavelength λ: The distance between two adjacent peaks
(or two troughs) . Measured in nanometers (1 nm = 10-9 m)
The frequency of the wave ν: The number of complete
wavelengths that pass a given point each second.
Units: 1/s or Hz 1kHz = 103 Hz
1 S
If 108 cycles pass in
one second, the
frequency is 108/s or
108Hz
١/١٣/١٤٣٧
٣
ELECTROMAGNETIC WAVES
For waves traveling at the same velocity: the longer the
wavelength, the smaller the frequency.
ELECTROMAGNETIC SPECTRUM
١/١٣/١٤٣٧
٤
FORMULA
c = λν
c = speed of light (3.00 x 108 m/s)
λ = wavelength (m)
ν = frequency (Hz or s-1)
SAMPLE EXERCISE 6.1
(a) Which wave has the higher frequency? (b) If one
wave represents visible light and the other represents
infrared radiation, which wave is which?
(a) The lower wave has a longerwavelength (greater distancebetween peaks). The longer thewavelength, the lower thefrequency (v = c/λ).
Thus, the wave (1) has the lowerfrequency, and wave (2) has thehigher frequency.
(b) According to electromagneticWave (1) would be the infraredradiation and Wave (2) Visible.
١/١٣/١٤٣٧
٥
PRACTICE EXERCISE
If one of the waves represents blue light and the other
red light, which is which?
� Answer: The expanded visible-light portion of
Figure 6.4 tells you that red light has a longer
wavelength than blue light.
� Wave (1) has the longer wavelength (lower
frequency) and would be the red light and wave
(2) is the blue light.
SAMPLE EXERCISE 6.2
The yellow light given off by a sodium vapor lamp used for
public lighting has a wavelength of 589nm. What is the
frequency of this radiation?
______________________________________________________
١/١٣/١٤٣٧
٦
PRACTICE EXERCISE
(a) A laser used in eye surgery to fuse detached retinas
produces radiation with a wavelength of 640.0 nm. Calculate
the frequency of this radiation.
Answers: 4.688 1014 s–1
(b) An FM radio station broadcasts electromagnetic radiation
at a frequency of 103.4 MHz (megahertz; MHz = 106s–1).
Calculate the wavelength of this radiation. The speed of light
is 2.998x108 m/s to four significant digits
Answer:2.901 m
QUANTIZED ENERGY AND PHOTONS
There are three main phenomena that cannot be explained
by the wave theory of light:
1. Blackbody radiation
2. Photoelectric effect
3. Emission spectra
١/١٣/١٤٣٧
٧
BLACKBODY RADIATION
� When solids are heated, they emit radiation. As seen in
the red glow of heated metal.
� The wavelength emitted depends on the temperature of
the object.
� The wave nature of light does not explain how an object
can glow when its temperature increases.
� Planck assumed that energy could only be released or
absorbed in certain fixed quantities that he called a
quantum.
BLACKBODY RADIATION
١/١٣/١٤٣٧
٨
FORMULA
E = hν� E = energy (J)
� h = Planck’s constant (6.626 x 10-34
J.s)
� v = frequency (Hz or s-1)
Plank Theory:
� matter is allowed to emit or absorb
energy only in whole-number ratios
of hv .
� Because energy can be released or
absorbed only in specific amounts,
we say that energy is quantized.
Max Planck (1858 - 1947)
THE NATURE OF ENERGY
Max Planck explained it by assuming that
energy comes in packets called quanta.
١/١٣/١٤٣٧
٩
PHOTOELECTRIC EFFECT
� The photoelectric effect occurs when light of a certain
frequency shining on the surface of a metal causes
electrons to be emitted from the metal.
� In this case, light behaves like a stream of tiny energy
packets called photons.
SAMPLE EXERCISE 6.3
Calculate the energy of one photon or yellow light with
a wavelength of 589nm.
١/١٣/١٤٣٧
١٠
PRACTICE EXERCISE
A laser emits light with a frequency of 4.69 x 1014 s-1. What is
the energy of one photon of the radiation from this laser?
Answers: 3.11x10–19J.
PRACTICE EXERCISE
If the laser emits a pulse of energy containing 5.0 x 1017
photons of this radiation, what is the total energy of that
pulse?
Answer: 0.16 J.
If the laser emits 1.3 x 10-2J of energy during a pulse, how
many photons are emitted during this pulse?
Answer: 4.2 x1016 photons
١/١٣/١٤٣٧
١١
WAVE VS. PARTICLE
We must consider that light possesses both wavelike and
particle-like characteristics.
Phenomenon Can be explained Can be explained
by waves. by particles.
Reflection
Refraction
Interference
Diffraction
Polarization
Photoelectric
effect
LINE SPECTRA AND THE BOHR MODEL
� A spectrum is produced when radiation from a source isseparated into its different wavelengths.
� Continuous spectrum (a): spectrums from lightcontaining all wavelengths .
� Line spectrum (b): spectrum containing radiation ofonly specific wavelengths. Often referred to as“fingerprints” of the element
١/١٣/١٤٣٧
١٢
ATOMIC LINE SPECTRA
BRIGHT-LINE SPECTRA
١/١٣/١٤٣٧
١٣
LINE SPECTRA VS. CONTINUOUS EMISSION SPECTRA
The fact that the emission spectra of H2 gas and other
molecules is a line rather than a continuous emission
spectra tells us that electrons are in quantized energy
levels rather than anywhere about nucleus of atom.
26
LINE SPECTRA OF HYDROGEN
The Rydberg equation:
Balmer (initially) and Rydberg (later) developed theequation to calculate all spectral lines in hydrogen
RH = Rydberg constant (1.096776 x 107 m-1)
n1 and n2 are positive integers that represent energy
levels
1
λ= (RH )(
1
n12−
1
n22)
١/١٣/١٤٣٧
١٤
BOHR’S MODEL
� Bohr based his model on 3 postulates:
1. Only orbits of certain radii (definite energies) are
permitted for the electron.
2. An electron in an orbit has a specific energy and will not
spiral into the nucleus.
3. Energy emitted or absorbed by the electron can only
change in multiples of E = hv.Ground state: when an electron occupies
the lowest possible energy level
Excited state: when an electron absorbs
energy and is promoted, temporarily to
an energy level higher than its ground
state
Niels Bohr was the first to offer an explanation for line spectra
(1885-1962)
electron orbits
Bohr’s Model of the Hydrogen Atom
n = 1
n = 2
n = 3
n = 4
n = 5
n = 6
nucleus
BOHR’S MODEL OF THE HYDROGEN ATOM
١/١٣/١٤٣٧
١٥
Bohr’s Model of the Hydrogen Atom
n = 6n = 5n = 4
n = 3
n = 2
n = 1
En
erg
y
Ground State
nucleus
absorption of a photon
e
e
Bohr’s Model of the Hydrogen Atom
n = 6n = 5n = 4
n = 3
n = 2
n = 1
En
erg
y
Ground State
nucleus
e
e emission of a photon
١/١٣/١٤٣٧
١٦
ENERGY STATES OF HYDROGEN
� Bohr calculated the specific energy levels using the
equation below.
E = energy.
h = Planck’s constant.
c = speed of light.
RH = Rydberg constant,
n = positive integer that represents an energy level.
×−= −
218 1
J102.18n
E n
( )
−=2
1
nRchE Hn
ENERGY STATES OF HYDROGEN
� The energies of the electron of the hydrogen atom are negative
for all values of n.
� The more negative the energy is, the more stable the atom will
be.
� The energy is most negative for n = 1.
� As n gets larger, the energy becomes less negative.
� When n = ∞, the energy approaches zero.
� Thus, the electron is removed from the nucleus of the hydrogen
atom.
١/١٣/١٤٣٧
١٧
TRANSITIONS BETWEEN ENERGY LEVELS
An electron can change energy levels by absorbing energy
to move to a higher energy level or by emitting energy to
move to a lower energy level.
For a hydrogen electron, the energy change is given by:
−−=
2
i
2
f
H
11∆
nnREif∆ EEE −=
RH = 2.179 × 10−18 J, Rydberg constant
constant sPlanck'
∆ electron
=
ν==
h
hEE photon
−−==
2
i
2
f
H
11
nnRh λν /hc
�Light is absorbed by an atom when:
The electron transition is from lower n to higher n
nf > ni
In this case, ∆E will be positive.
�Light is emitted from an atom when:
The electron transition is from higher n to lower n.
nf < ni
In this case, ∆E will be negative.
An electron is ejected when: nf = ∞.
١/١٣/١٤٣٧
١٨
SAMPLE EXERCISE 6.4
Using Figure 6.14 in your book (p. 220), predict which of the
following electronic transitions produces the spectral line having
the longest wavelength: n = 2 to n =1, n = 3 to n = 2, or n = 4
to n = 3.
Solution:
�
Hence the longest wavelength willbe associated with the lowestfrequency.
From the Figure: The shortestvertical line represents the smallestenergy change.Thus, the n= 4 to n= 3 transitionproduces the longest wavelength(lowest frequency) line.
PRACTICE EXERCISE
Indicate whether each of the following electronic
transitions emits energy or requires the absorption of
energy:
a) n = 3 to n = 1
Answers: emits energy.
b) n = 2 to n = 4
Answe: requires absorption of energy
١/١٣/١٤٣٧
١٩
Example:Example:
n = 6
n = 5
n = 4
n = 3
n = 2
n = 1
For an electron moving from n = 4 to n = 2:
( )
−−=∆ 22
11
initialfinal
Hnn
RchE
( )
−−=∆
22 4
1
2
1HRchE
( )
−×−=∆ −
16
1
4
11018.2 18 JE
1875.01018.2 18 ××−=∆ − JE
∆E = - 4.09 x 10-19 J
E = 4.09 x 10-19 J
What wavelength (in nm) does this
energy correspond to?
λch
E =E
ch=λ
J
msJs19
1834
1009.4
1031063.6−
−−
××××
=λ
The energy of the photon emitted is:
λ = 486 x 10-9 m
= 486 nm
n = 6
n = 5
n = 4
n = 3
n = 2
n = 1
١/١٣/١٤٣٧
٢٠
LIMITATIONS OF THE BOHR MODEL
Bohr’s model can only explain the line spectrum of
hydrogen, not other elements.
Important ideas from Bohr’s model:
1. Electrons exist in specific energy levels.
2.Energy is involved when electrons jump to different
energy levels.
THE WAVE BEHAVIOR OF MATTER
� De Broglie suggested the as the electron moves around
the nucleus, it is associated with a particular
wavelength.
� De Broglie used the term matter waves to describe the
wave characteristics of material particles.
١/١٣/١٤٣٧
٢١
DE BROGLIE
He proposed that the wavelength of any particle depends
on its mass and velocity.
λ = wavelength.
h = Planck’s constant.
m = mass.
v = velocity.
Based on his formula, the wavelength of an object of
normal size would be so tiny that it could not be
observed.
λ =h
mv
SAMPLE EXERCISE 6.5
What is the wavelength of an electron moving with a speed of
5.97 x 106 m/s? The mass of the electron is 9.11 x 10-31 kg.
Comment: By comparing this value with the wavelengths of
electromagnetic radiation we see that the wavelength of this
electron is about the same as that of X-rays.
١/١٣/١٤٣٧
٢٢
PRACTICE EXERCISE
Calculate the velocity of a neutron whose de Broglie wavelength
is 500 pm. The mass of a neutron is 1.67 x 10-27 kg.
Answer: 7.92 102 m/s
DIFFRACTION
� A few years after de Broglie published his theory, it was
found that electrons (like x-rays) will diffract when
passed through a crystal.
� Diffraction is normally associated with waves.
١/١٣/١٤٣٧
٢٣
THE UNCERTAINTY PRINCIPLE
Heisenberg’s uncertainty principle states that when an
object has an extremely small mass (electron): it is
impossible for us to simultaneously know both the exact
momentum and location.
If the object is normal sized, then the uncertainty would
be too small to matter.
The uncertainty of mv becomes more significant as
the mass of the particle becomes smaller.
(∆x) (∆mv) ≥ h4π
QUANTUM MECHANICS AND ATOMIC ORBITALS
The Bohr model: assumes that the electron is in a circular
orbit of a particular radius.
In the quantum mechanical model, the electron’s location
is based on the probability that the electron will be in a
certain region of space.
١/١٣/١٤٣٧
٢٤
4 ATOMIC MODELS
Dalton model lum Pudding Model Rutherford’ sModel
Bohr’s Model Quantum Mechanical Model
THE QUANTUM MECHANICAL MODEL
� Energy is quantized - It comes in chunks.
� A quantum is the amount of energy needed to
move from one energy level to another.
� Since the energy of an atom is never “in between”
there must be a quantum leap in energy.
� In 1926, Erwin Schrodinger derived an equation
that described the energy and position of the
electrons in an atom
١/١٣/١٤٣٧
٢٥
SCHRODINGER’S WAVE EQUATION
22
2 2
8dh EV
m dx
ψ ψ ψπ
− + =
Equation for the
probabilityprobability of a single
electron being found
along a single axis (x-axis)Erwin Schrodinger
©
2012
Pea
rson
Edu
cati
on,
Inc.
QUANTUM MECHANICS
� Erwin Schrödinger developed a
mathematical treatment into
which both the wave and particle
nature of matter could be
incorporated.
� It is known as quantum
mechanics.
� The wave equation is designated
with a lowercase Greek psi (ψ).
No direct physical meaning.
� The square of the wave equation,
ψ2, gives a probability density
map of where an electron has a
certain statistical likelihood of
being at any given instant in
time.
١/١٣/١٤٣٧
٢٦
THE ELECTRONIC ORBITAL
Since the quantum mechanical model is based on the probability
of finding an electron, then the orbitals are normally shaded with a
fuzzy edge.
An atomic orbital is a region of space in which there is a high
probability of finding an electron
Within the border ofan atomic orbital,there is a 90% chanceof finding an electron.The darker theshading of the orbital,the higher the chanceof finding an electron.
QUANTUM NUMBERS
�Solving the wave equation gives a set of wave
functions, or orbitals, and their corresponding
energies.
�Each orbital describes a spatial distribution of
electron density.
�An orbital is described by a set of three quantum
numbers.
١/١٣/١٤٣٧
٢٧
1) Principal quantum number, n (for main shells or
main levels): This is the same as Bohr’s model.
�describes the energy level on which the orbital
resides.
�The values of n are integers ≥ 1.
As n becomes larger, the atom becomes larger and the
electron is further from the nucleus.
energy of electron in a given orbital: ( )
−=
2
1
nRchEH
n is always a positive integer: 1, 2, 3, 4 , 5, 6, 7,QQQ.
K, L, M, N, O, P, Q, QQQ
2) Angular momentum quantum number, l: (for
subshells or sublevels).
This quantum number defines the shape of the orbital.
This quantum number depends on the value of n.
Allowed values of l :
For any n: l is an integers ranging from 0 to n − 1.
Value of l 0 1 2 3
Type of orbital s P d f
١/١٣/١٤٣٧
٢٨
3) Magnetic Quantum Number, ml : (for orbitals).
Describes the three-dimensional orientation of the orbital.
� Values are integers ranging from -l to l:
−l ≤ ml ≤ l.
Number of ml (orbitals) = 2l +1
Therefore, on any given energy level, there can be up to:
One s orbital. (ml = 0)
Three p orbitals. (ml = -1, 0, 1)
Five d orbitals. (ml = -2,-1,0,1,2)
Seven f orbitals. (ml = -3,-2,-1,0,1,2,3)
And so on.
Important note: Total number of orbitals in the
main shell = n2
MAGNETIC QUANTUM NUMBER, ML
� Orbitals with the same value of n form a shell.
� Different orbital types within a shell are subshells.
١/١٣/١٤٣٧
٢٩
Level n 1 2 3
Sublevel l
Orbital ml
Spin ms
0 0
0 0 1 0 -1 0 1 0 -1 2 1 0 -1 -2
2101
+1/2
-1/2
Allowed Sets of Quantum Numbers for Electrons in Atoms
QUANTUM NUMBERS
quantum number symbol representspossible
numbers
principle quantum
numbern
energy
level
1, 2, 3, 4,
5, 6, 7
angular momentum
quantum numberl sublevel 0, 1, 2, 3
magnetic quantum
numberml orbital
-3, -2, -1,
0, 1, 2, 3
spin magnetic
quantum numberms spin
+1/2 or
-1/2
١/١٣/١٤٣٧
٣٠
SAMPLE PROBLEM
0
3, 1, 0, 1/2
2, 1, -1, -1/2
3, 2, -2, 1/2
Write the 4 quantum numbers (n, l, ml, ms) for the following
electrons:
PRACTICE PROBLEMS
Write the quantum numbers for the following
electrons:
The last electron in Mn:
The 6th electron
3d5 = __ __ __ __ __
3d
3, 2, 2, 1/2
__ __ __ __ __
1s 2s 2p 2, 1, 0, 1/2
١/١٣/١٤٣٧
٣١
SAMPLE EXERCISE 6.6
a) Predict the number of subshells in the fourth shell, n = 4.
n = 4 so l = 0, 1, 2, 3
b) Give the label for each of these subshells.
These subshells are labeled 4s, 4p, 4d, and 4f.
c) How many orbitals are in each of these subshells?
4s = 1 orbital value: 0
4p = 3 orbitals values : 1, 0, –1
4d = 5 orbitals values: 2, 1, 0, –1, –2
4f = 7 orbitals values: 3, 2, 1, 0, –1, –2, –3
PRACTICE EXERCISE
a) What is the designation for the subshell with n = 5 and l = 1?
b) How many orbitals are in this subshell?
c) ndicate the values of ml for each of these orbitals.
Answers: (a) 5p; b) 3; (c) 1, 0, –1
١/١٣/١٤٣٧
٣٢
REPRESENTATIONS OF ORBITALS
� Wave function provides information about an electron
probable location at 90 % in space.
S- orbital:
� Has spherical symmetry.
� As n increases, the distance from the nucleus increases for s
orbitals.
� An intermediate point at which a probability function goes to
zero is called a node.
� We see that s orbitals possess n−1 nodes, or regions where
there is 0 probability of finding an electron.
No nodes One node
Two nodes
١/١٣/١٤٣٧
٣٣
P-ORBITALS
� There are 3p orbitals which all have the same energy but
of different orientations.
� The electron density for p orbitals occurs in two regions
on either side of the nucleus, separated by a node.
� The two regions are called lobes.
� Each p orbital has the same energy.
D-ORBITALS
� There are 5 d orbitals which all have the same energy.
� More variations of lobes.
١/١٣/١٤٣٧
٣٤
F-ORBITALS
� There are 7 f orbitals which all have the same energy.
� Of complex shapes.
ENERGIES OF ORBITALS
� For a one-electron hydrogen atom, orbitals on the same energy level
have the same energy.
� That is, they are degenerate.
١/١٣/١٤٣٧
٣٥
ENERGIES OF ORBITALS
� As the number of electrons increases, though, so does therepulsion between them.
� Therefore, in many-electron atoms, orbitals on the sameenergy level are no longer degenerate.
4) SPIN QUANTUM NUMBER, MS
� In the 1920s, it was discovered that two electrons in the
same orbital do not have exactly the same energy.
� The “spin” of an electron describes its magnetic field,
which affects its energy.
١/١٣/١٤٣٧
٣٦
ms can only take two values, +1/2 or -1/2
PAULI EXCLUSION PRINCIPLE
�No two electrons in the
same atom can have
exactly the same energy.
�For example, no two
electrons in the same
atom can have identical
sets of quantum numbers.
2p2s
n=2, l=0, ml =0, ms = +1/2
n=2, l=0, ml =0, ms = -1/2
١/١٣/١٤٣٧
٣٧
HUND’S RULE
“For degenerate orbitals, the lowest energy is attained
when the number of electrons with the same spin is
maximized.”
Hund’s rule is based on the fact that electrons repel one
another.
THE DIAGONAL RULE
This results in the following order:
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6,
6s2, 4f14, 5d10, 6p6, 7s2, 5f14, 6d10, 7p6
١/١٣/١٤٣٧
٣٨
ELECTRON CONFIGURATION
Electron configuration : how electrons are
distributed in the various atomic orbitals.
Aufbau’s principle: states that electrons fill the
orbitals in order of increasing energy.
Ground state: electrons in lowest energy state.
Excited state: electrons in a higher energy
orbital.
Methods of electronic configuration:
١/١٣/١٤٣٧
٣٩
ELECTRON CONFIGURATIONS
ELECTRON PAIRING
Paired Electrons: Electrons having opposite spins when
they are in the same orbital.
Unpaired electron: is one that is not accompanied by a
partner of opposite spin.
١/١٣/١٤٣٧
٤٠
SAMPLE EXERCISE 6.7
a) Draw the orbital diagram for the electron configuration of
oxygen. b) How many unpaired electrons does an oxygen atom
possess?
O, Z =16 (orbital diagram)
1s22s2p4 (Symbol diagram)
The atom has two unpaired electrons._________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------_______________________________________________________________________________________________________
Practice Exercise.(a) Write the electron configuration for phosphorus, element
15. (b) How many unpaired electrons does a phosphorus
atom possess?
Answers: (a) 122s22p63s23p3, (b) three
PRACTICE ELECTRON CONFIGURATION AND ORBITAL NOTATION
Write the electron configuration and orbital notation for each atoms with:
Z = 20, Z = 35, Z = 26
Z = 20 1s2 2s2 2p6 3s2 3p6 4s2 (Ca)
Z = 35 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 (Br)
Z = 26 1s2 2s2 2p6 3s2 3p6 4s2 3d6 (Fe)
1s 2s 2p 3s 3p 4s
1s 2s 2p 3s 3p 4s 3d 4p
1s 2s 2p 3s 3p 4s 3d
١/١٣/١٤٣٧
٤١
NOBLE GAS CONFIGURATION
� In writing the condensed/noble gas electron
configuration, the nearest noble gas of lower atomic
number is represented by its symbol in brackets [ ].
Noble gas core configuration: can be used to represent
all elements but H and He
Example:
Z = 15 [1s22s22p6]3s23p3 (P)
[Ne] 3s23p3 (P)
� Valence electrons are electrons in the outer shell.
• These electrons are gained and lost in reactions.
� Core electrons are electrons in the inner shells.
• These are generally not involved in bonding.
ELECTRON CONFIGURATIONS AND THE PERIODIC TABLE
The periodic table is structured so that the elements with the
same valence electron configuration are arranged in columns.
s and p numbers = period number = n
d number = n-1 f number = n-2
١/١٣/١٤٣٧
٤٢
SAMPLE EXERCISE 6.8
What is the characteristic valence electron configuration of
the halogens?
The characteristic valence electron configuration of a
halogen is :
ns2np5
where n ranges from 2 in the case of fluorine to 6 in the case
of astatine.______________________________________________________________________________________________________________________________________________
Practice Exercise:Which family of elements is characterized by an ns2np2 electron
configuration in the outermost occupied shell?
Answer: group 4A
١/١٣/١٤٣٧
٤٣
SAMPLE EXERCISE 6.9
From PT: (a) Write the electron configuration for bismuth,
element number 83.
(b) Write the condensed electron configuration for this
element.
[Xe]6s24f145d106p3 or [Xe]4f145d106s26p3
(c) How many unpaired electrons does each atom of bismuth
possess?
Three unpaired electrons.
PRACTICE EXERCISE
Use the periodic table to write the condensed electron configurations
for (a) Co (atomic number 27)
(b) Te (atomic number 52).
Answers:
(a) [Ar]4s23d7 or [Ar]3d74s2.
(b) [Kr]5s24d105p4 or [Kr]4d105s25p4.
١/١٣/١٤٣٧
٤٤
ODD ELECTRON CONFIGURATIONS
Sublevels are the most stable when ½ full or completely
full.
It is more acceptable for a large sublevel to be stable.
Most odd electron configurations such as chromium and
copper are based on this fact.
� Not all elements follow the “order” of the diagonal rule
� Notable exceptions: Cu (Z = 29) and Cr (Z = 24)
Cu = [Ar]4s13d10
Cr = [Ar]4s13d5
Reason: slightly greater stability associated with filled and
half-filled d subshells
� This occurs because the 4s and 3d orbitals are very close
in energy.
� These anomalies occur in f-block atoms, as well.
SOME ANOMALIES
١/١٣/١٤٣٧
٤٥
Some Anomalies
SAMPLE INTEGRATIVE EXERCISE
a)Boron occurs naturally as two isotopes, 10B and 11B, with natural
abundances of 19.9% and 80.1%. In what ways do the two isotopes
differ from one another? Does the electron configuration of 10B differ
from that of 11B?
Solution:
Boron: Z = 5
The have same Z i.e. same atomic number (5 protons) but differ in
number of neutrons (10B have 5 neutrons and 11B, have 6 neutrons).
The two isotopes of boron have identical electron configurations,
1s22s22p1, because each has five electrons.
b) Draw the orbital diagram for an atom of 11B. Which electrons are
the valence electrons?
The valence electrons are the outermost occupied shell:
( 2s22p1 electrons).
١/١٣/١٤٣٧
٤٦
c) Indicate 3 major ways in which 1s electrons in boron differ
from its 2s electrons.
The 1s and 2s orbitals are both spherical.
They differ in three important respects:
1- The 1s orbital is lower in energy than the 2s orbital.
2- The 1s orbital is smaller than the 2s.
3- The 2s orbital has one node, whereas the 1s orbital has no nodes.
d. Elemental boron reacts with fluorine to form BF3, a gas. Write a
balanced chemical equation for the reaction of solid boron with
fluorine gas.
2 B(s) + 3 F2(g) →2 BF3(g)
e) ∆Hfo for BF3(g) is -1135.6kJ/mol. Calculate the standard
enthalpy change in the reaction of boron with fluorine.
2 B(s) + 3 F2(g) →2 BF3(g)
∆H= 2(–1135.6) –[0 + 0] = –2271.2 kJ.
The reaction is strongly exothermic.
f) When BCl3, also a gas at room temperature, comes into
contact with water, the two react to form hydrochloric
acid and boric acid, H3BO3, a very weak acid in water.
Write a balanced net ionic equation for this reaction.
BCl3(g) + 3 H2O(l) →H3BO3(aq) + 3 H+(aq) + 3 Cl–(aq)
١/١٣/١٤٣٧
٤٧
HW EXERCISES:
1, 6, 8, 13, 16, 18, 23,
27, 33, ,35, 39, 40, 45, 48,
52, 55, 57, 60, 65, 72, 74,
75, 77, 85, 91, 94, 98, 100
� ا�� وا����ة ��� .ا�0�1/ (.-�ظ � * (� ) ذ&� ��: $#"ل � � �� أ�� ط��� ر�� هللا ����
.ا�41/ داء (�?<، $>6; �7 ا�0�1/ أ:89 (� 67 � ���415"د
� -#7 �BC�?�D در ا�41/ (� أ�/�� �/أ * * *
*��6E �5 هللاCي ر/�G 854ا� HI ا� "���ت BG; أن $?; : $#"ل ا�>#�I أ"#� M5N /0�1ا� ;?$:4C/ه إ�* ا�415"د
-�Qأو� :TU#�$ ( VW. -�QI&�Xو :�QI � 8YZ$ ( [BI?).
-�Q9��Xو :�QI � /51$ ( [)\). -�Q6�.0_^ ا�8ب: ورا
-�Q4)�Nو :`I7"-ب ا��� ��� ` .$.
* * *
.: �5 ارT<E اf&�4ن، C d<X�eE"�� ا�.I"م وا�15�: $#"ل رو0"
* * * :$#"ل B� h��i/ ا�#/وس
�) ( VE�jا� "Q7 اخ �) V-j��� إ&�5 ا� "م � * (� أ� m5 ش-) m5� 8B_$ك mQYا"$ V�إ$�ك وا�41/إ$�ك وا�41/
e0 " 7� ا�e0 " 7 ( mB G /41� ا�mB G /41 ) ......وط� &>m4 � * ا��U6ء، وا87ح �>8ح ا8Nq$� واC\ر أن 41E/ أC/ا وط� &>m4 � * ا��U6ء، وا87ح �>8ح ا8Nq$� واC\ر أن 41E/ أC/ا mB G �7 � � GmB G �7/ر هللا 47-8ى ا��56] � * ا���س &mI � [5# وuC VQC87&� وھ5 ..G/ر هللا 47-8ى ا��56] � * ا���س &mI � [5# وuC VQC87&� وھ5