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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.

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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

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ELECTROMAGNETIC WAVES

For waves traveling at the same velocity: the longer the

wavelength, the smaller the frequency.

ELECTROMAGNETIC SPECTRUM

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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.

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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?

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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

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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

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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.

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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.

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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

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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

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ATOMIC LINE SPECTRA

BRIGHT-LINE SPECTRA

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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)

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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

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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

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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.

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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 = ∞.

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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

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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

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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.

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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.

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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.

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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.

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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

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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.

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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.

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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

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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.

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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

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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

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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

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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

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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.

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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.

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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.

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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

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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

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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:

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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.

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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

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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

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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

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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.

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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

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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).

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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)

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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

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