07 Atomic Structure and Periodicity 6.pdf

Chapter 7

Atomic Structure and Periodicity

The Nature of Light: Classification of

Electromagnetic Radiation

Characteristics of Waves,

c

=

l

n

Wavelength, l, m

distance between

identical points on

successive waves

Frequency, n

,

1/s or Hz

number of waves that

pass a point per second

Speed, c

2.9979 x 10

8

m/s

Amplitude, The depth or height of a wave

Speed of light in a vacuum:

c

=

3.00

´

10

8_m

s

Electromagnetic Radiation

 A beam of light consists of two mutually

perpendicular oscillating fields: an oscillating

electric field and an oscillating magnetic field.

 The direction of propagation of the beam is also perpendicular to the electric and magnetic fields.

 Light is a wave.

When light passes through two closely spaced slits, an interference pattern is produced.

The Double-Slit Experiment

Constructive interferenceis a result of adding waves that are in phase.

Destructive interference is a result of adding waves that are out of phase.

This type of interference is typical of waves and demonstrates the wave nature of light.

A laser commonly used in the treatment of vascular skin lesions has a wavelength of 532 nm. What is the frequency of this radiation?

Conversion Between Wavelength and Frequency

Strategy We must convert the wavelength to meters and solve for frequency using c = λν.

Solution λ = 532 nm×

ν =

Setup Rearranging the equation to solve for frequency gives ν = . The speed of light, c, is 3.00×108 _m/s.

c λ

1×10-9 _m

1 nm = 5.32×10-7 m

3.00×108 _m/s

5.32×10-7 _{m = 5.64}×1014 s-1

Electromagnetic Waves

For electromagnetic waves

c

=

ln

n

=

c

l

=

3.00

´

10

8m s

4.4

´

10

-6

m

=

6.8

´

10

13 1s

=

6.8

´

10

13

Hz

Determine the frequency of light with a wavelength of 4.4 mm.

Calculate Frequency From Wavelength

c =

ln

,

n

=

c

/

l

The brilliant

red

colors

seen in fireworks are due

to the emission of light

with wavelengths around

650 nm

when

strontium

salts

such as

Sr(NO

3

)

2

and

SrCO

3

are heated.

Calculate the

frequency

of

red light

of wavelength

6.50 x 10

2

nm

.

Characteristics of Waves:

Wavelength, Frequency, and Energy

Our understanding of electromagnetic radiation (EM)

played a critical role in our understanding of atomic

structure. Compare red light, blue light, and x-rays and

arrange them in order so that the longest wavelength is

first, highest frequency is second, and finally the highest

energy is at the end of the list.

1. X-ray; x-rays; x-rays

2. Red; x-rays; blue

3. Blue; x-rays; red

4. Red; x-rays; x-rays

Energy of EM Radiation

While sitting in a dentist

’

s chair you are hit with several

forms of electromagnetic radiation. The bright light is

sending energy in the visible region, the body heat of the

dentist is emanating energy in the infrared region; perhaps

a radio is playing and some energy in the radio waves

region approach your ears; and you are about to receive an

x-ray. Which arrangement places the EM waves in proper

order with respect to increasing energy per photon?

1. Visible; infrared, radio; x-rays 2. X-rays; visible; radio; infrared 3. Infrared; radio; visible; x-rays 4. Radio; infrared; visible; x-rays

Quantum Theory



Matter (particles)



Has mass



Position in space can be specified



Energy (waves)



No mass



Position in space can not be specified

1900: Matter and energy are distinctly different

Blackbody

Radiation

Light given off by a hot blackbody The amount of energy given off at a certain

temperature depends on the wavelength.

~1000K

~1500K

 Classical physics failed to explain Blackbody radiation

 Assumed that matter could emit or absorb any quantity of energy.

 Plank’s explanation for Blackbody radiation

 Matter could NOTemit or absorb any quantity of energy.  Energy is only emitted or absorbed in discrete quantities, like small

packages or bundles of energy.

 A quantumis the smallest quantity (bundle) of energy that can be absorbed or emitted as EM radiation.

 Energy of atoms is gained or lost only in whole-number multiples of the quantity hn, i.e., energy is absorbed or released by atoms in discrete “chunks” called quantaof sizehn.

E =

h

n

h = 6.626 x 10-34 _{Js (Plank’s constant)}

n

= frequency of the em radiation

Blackbody Radiation:

Electrified Pickle Emits Yellow

Light Due to Excited Na

+

Ions

Albert Einstein used Planck’s theory to explain the photoelectric effect.

Electrons are ejected from the surface of a metal exposed to light of a certain minimum frequency, called the

threshold frequency.

The number of electrons ejected is proportional to the intensity.

Below the threshold frequency no electrons were ejected, no matter how bright (or intense) the light.

1905: Photons and the Photoelectric Effect

Einstein proposed that the beam of light is really a stream of particles.

These particles of light are now called

photons.

Each photon (of the incident light) must posses the energy given by the equation:

Photons and the Photoelectric Effect

Shining light onto a metal surface can be thought of as shooting a beam of particles – photons – at the metal atoms.

If the ν of the photons equals the energy the binds the electrons in the metal, then the light will have enough energy to knock the electrons loose.

If we use light of a higherν, then not only will the electrons be knocked loose, but they will also acquire some kinetic energy.

Photons and the Photoelectric Effect

This is summarized by the equation

KE is the kinetic energy of the ejected electron

W is the binding energy of the electron

Dual Nature of Light: Exhibits Wave

Properties AND Particulate Properties

Energy is quantized and occurs in discrete units called quanta.

Photons of Light

E

=

h

n

nl

=

c

on rearranging,

n

=

c

l

on substituting,

E

=

hc

l

Energy of Photons

Red Orange Yellow Green Blue Indigo Violet

E

=

hc

l

750 nm

400 nm

l

Energy

Whose Lightsaber is More Deadly?

Mace Windu

4.47 x 10

–19

J/photon

Obi-Wan Kenobi

4.18 x 10

–19

J/photon

Darth Vader

3.06 x 10

–19

J/photon

Yoda

3.90 x 10

–19

J/photon

Calculate

the

Energy

of a Photon

Given

the

Wavelength

of EM Radiation,

E

photon

=

h

n

=

hc

/

l

The

blue

color in fireworks is often achieved by

heating

copper (I) chloride (CuCl)

to about 1200

°

C.

The compound emits blue light having a

wavelength

of 450 nm

.

What is the increment of energy

(the quantum) that is emitted

at

4.50 x 10

2

nm by CuCl

?

Calculate

Wavelength

of a Photon With

a

Given

Energy

,

E

=

hc

/

l

so

l

=

hc/

E

Suppose that a microwave oven uses photons with an

energy

of 1.42



10

–23

joules

to provide you with a

cooked popcorn snack. You have less than two minutes

before the corn pops to determine the

wavelength

of the

microwaves.

(Note: use Planck

’

s constant as 6.626



10

–34

Js.)

1. 1.40 cm

2. 2.14



10

10

cm

3. 0.714 cm

Calculate

Frequency

of a Photon

Given

the

Energy

,

E

=

hn

so

n

=

E

/h

The wavelength of the laser light that allows you to listen

to your favorite tunes on a CD player lies in the red area

of the visible spectra. If one mole of the photons delivers

1.54



10

5

J

, what is the frequency of this useful energy?

1. 2.59



10

–15

s

–1

2. 3.86



10

14

s

–1

3. 1.56



10

23

s

–1

4. I don

’

t know (I only listen to vinyl).

Bohr’s Theory of the Hydrogen Atom

When white light is passed through a prism, a

continuous spectrum of colors results which

contains all of the wavelengths of visible light

Visible Light

Spectrum

A Continuous

Spectrum

Contains all of the

wavelengths of

visible light.

The

Hydrogen

Line

Emission

Spectrum

When a sample of hydrogen gas is excited by

electricity and passed through a prism, only a few

lines are seen, each of which corresponds to a discrete

wavelength.

Every Element Has Its Own

Unique Emission Line Spectrum

wavelength of light emitted when an electron in a

Rydberg Equation:

used to calculate the

hydrogen atom changes orbits

RH

= Rydberg constant = 1.096776 x 10

7

m

-1

n

1

and

n

2

are positive integers

n

2

>

n

1

1

l

=

( )

R

H

1

n

₁2

-1

n

₂2

æ

è

ç

ö

Calculate the wavelength of light emitted when

an electron in a hydrogen atom changes orbits

What is the wavelength of light emitted when an

electron falls from the

n

= 4 shell to the

n

= 2 shell?

1

l

=

( )

RH 1

n₁2 -1

n₂2 æ è

ç ö

ø

÷=

(

1.096776´107_m-1

)

1 22

-1 42 æ è

ç ö_ø÷=2056455 m-1

l

= 1

2056455 m-1=4.86´10-7m=486 nm

Bohr Attributed the Emission of Radiation to the

Electron Dropping From a Higher Energy Orbit

to a Lower One

1.Electrons orbit the nucleus in circular orbits.

2.Only orbits of certain radii are permitted.

3.An electron in a permitted orbit has a specific energy.

4.Energy is emitted or absorbed as a photon when the electron changes its orbit.

What color of light is emitted when an excited electron in the hydrogen atom falls from:

a) n =5 to n = 2 b) n = 4 to n = 2 c)n = 3 to n = 2

What is the significance of the line spectrum of H

2

?

D

E = h

n

= hc/

l

Only certain energies are possible;

that is, the electron energy levels are quantized

The Energies That an Electron in a

Hydrogen Atom Can Possess

E

_n

is the energy

n

is a positive integer

The energy of a free electron is arbitrarily assigned a value of zero.

Electron Energies

Calculate the energy required to remove the electron from a hydrogen atom in its ground state (i.e., calculate the ionization

energy).

n = 

When n = , the electron has been removed from the atom. This is called the zero-energy

As an electron gets closer to the nucleus, n decreases.

E

n becomes larger in absolute value (but more negative) as n gets

smaller.

En is most negative when n = 1.

Called the ground state, the lowest energy state of the atom

For hydrogen, this is the most stable state

The stability of the electron decreases as n increases.

For H, every energy state in which n > 1 is called an excited state.

The Energies That an Electron in a Hydrogen

Atom Can Possess

Summary: The Bohr Model Explains The Line Spectrum of Hydrogen

Radiant energy absorbed by the atom causes the electron to move from the ground state (n = 1) to an excited state (n > 1).

Conversely, radiant energy is emitted when the electron moves from a higher-energy state to a lower-energy excited state or the ground state.

The quantized movement of the electron from one energy state to another is analogous to a ball moving and down steps.

nf is the final state

ni is the initial state

The Transition of an Electron from a Higher to a Lower State Results in the Emission of a Photon of Energy E = hn

Suppose an electron is initially in an excited state, ni.

During emission, the electron drops to a lower energy state, nf.

The energy difference between the initial and final states is

Bohr’s Theory of the Hydrogen Atom

To calculate wavelength, substitute c/λ for ν and rearrange:

Bohr’s Theory of the Hydrogen Atom

nf is the final state

ni is the initial state

Calculate the wavelength (in nm) of the photon emitted when an electron transitions from the n = 4 state to the n = 2 state in a hydrogen atom.

Calculate the Wavelength of a Photon Emitted when an Electron Transitions from a Higher to Lower State

Solution Setup

h = 6.63×10-34 _{J∙s and c = 3.00×10}8 _m/s

2.18×10-18 _J (6.63×10-34 _{J∙s )(3.00}×108 _m/s) 1

λ = 21 2 - ₄1 2

= 2.055×106 _m-1

λ = 4.87×10-7 _m 1 nm

1×10-9 _{m = 487 nm} ×

Think About It Look again at the line spectrum of hydrogen and make sure your result matches one of them. Note that for an emission, ni, is always greater than nf, and

Wave Properties of Matter

Louis de Broglie reasoned that if light can behave like a stream of particles (photons), then electrons could exhibit wavelike properties. According to de

Broglie, electrons behave like standing waves.

Only certain wavelengths are allowed.

At a node, the amplitude of the wave is zero.

e

-

Bound to the Nucleus

Similar to a Standing Wave

A string attached at both ends

vibrates to produce a musical tone.

“

Standing

”

waves because they

are stationary and maintain a

constant amplitude.

There must be a whole number of

half-wavelengths in any of the

allowed motions of a standing

wave.

The Hydrogen e

-

Visualized as a

Standing Wave Around the Nucleus

Only certain circular orbits have a

circumference into which a whole number

of wavelengths of the standing electron

wave will

“

fit

”

.

Circular orbits with any other

circumference produce destructive

interference of the standing electron wave

and are not allowed.

Waves Can Behave Like Particles and Particles (Electrons) Can Behave Like Waves

De Broglie deduced that the particle and wave properties are related by the following expression:

λ is the wavelength associated with the particle

m is the mass (in kg)

u is the velocity (in m/s)

The wavelength calculated from this equation is known as the de Broglie wavelength.

Calculate the de Broglie wavelength of the “particle” in the following two cases: (a) a 25-g bullet traveling at 612 m/s and (b) an electron (m = 9.109×10-31 _kg) moving at 63.0 m/s.

Calculate the de Broglie wavelength of a Particle

Solution (a)25 g ×

(b) Setup

h = 6.63×10-34 _{J∙s, or 6.63}×10-34 _kg∙m2_{/s; Remember m must be expressed in} kg.

= 0.025 kg

λ = 1 kg 1000 g h mu 6.63

×10-34 _kg∙m2_/s (0.025 kg)(612 m/s)

= = 4.3×10-35 _m

λ = _mu h 6.63×10-34 kg∙m2/s = 1.16×10-5 _m (9.109×10-31 _{kg)(63.0 m/s)}

=

Think About It While you are new at solving these problems, always write out the units of Planck’s constant (J∙s) as kg∙m2_{/s. This will enable you to check your unit}

cancellations and detect common errors such as expressing mass in grams rather than kilograms. Note that the calculated wavelength of a macroscopic object, even one as small as a bullet, is extremely small. An object must be at least as small as a subatomic particle in order for its wavelength to be large enough for us to observe.

Electron Diffraction Experiments Demonstrated that Electrons Exhibit the Wave Property of Interference

Experiments have shown that electrons do indeed possess wavelike properties:

X-ray diffraction pattern of

The Heisenberg uncertainty principlestates that it is impossible to know simultaneously both the momentum pand the position x of a particle with certainty.

Δx is the uncertainty in position in meters

Δp is the uncertainty in momentum

Δu is the uncertainty in velocity in m/s

m is the mass in kg

Quantum Mechanics

If the position of a particle is known more

precisely, then it’s velocity measurement

must become less precise

In the picture, we

know the exact location of the cars, but we have

no idea how fast they are moving.

If the velocity of a particle is measured more

precisely then the position must become

correspondingly less precise

In the

picture, we

know the

speed of the

cars, but we

have no idea

exactly

where they

are.

How to Use the Heisenberg Uncertainty Principle

Strategy The uncertainty in the velocity, 1 percent of 5×106 _{m/s, is Δu.} Calculate Δx and compare it with the diameter of they hydrogen atom. An electron in a hydrogen atom is known to have a velocity of 5×106 _{m/s + 1} percent. Using the uncertainty principle, calculate the minimum uncertainty in the position of the electron and, given that the diameter of the hydrogen atom is less than 1 angstrom (Å), comment on the magnitude of this uncertainty compared to the size of the atom.

Setup The mass of an electron is 9.11×10-31 _{kg. Planck’s constant, h, is} 6.63×10-34 _kg∙m2_/s.

How to Use the Heisenberg Uncertainty Principle

Solution

Δu = 0.01 × 5×106 _{m/s = 5}×104 _m/s

Δx =

Δx = h 4π ∙ mΔu

6.63×10-34 _kg∙m2_/s

4π(9.11×10-31 _kg)(5×104 _{m/s) > 1}×10-9 m An electron in a hydrogen atom is known to have a velocity of 5×106 _{m/s + 1} percent. Using the uncertainty principle, calculate the minimum uncertainty in the position of the electron and, given that the diameter of the hydrogen atom is less than 1 angstrom (Å), comment on the magnitude of this uncertainty compared to the size of the atom.

The minimum uncertainty in the position x is 1×10-9 _{m = 10Å. The uncertainty} is 10 times larger than the atom!

Think About It A common error is expressing the mass of the particle in grams instead of kilograms, but you should discover this inconsistency if you check your unit cancellation carefully. Remember that if one uncertainty is small, the other must be large. The uncertainty principle applies in a practical way only to submicroscopic particles. In the case of a macroscopic object, where the mass is much larger than that of an electron, small uncertainties, relative to the size of the object, are possible for both position and velocity.

Why are the noble gases

unreactive and the

alkali metals are very

Lithium Metal (Li) and H

2

O

Sodium Metal (Na) and H

2

O

Potassium Metal and H

2

O

Why are the noble gases unreactive

and the alkali metals are very

reactive?

Because of the different

arrangements of

electrons

in

the different elements!

The Modern Model of the Atom

The precise paths of electrons cannot be

determined accurately. Instead, the

PROBABILITY

of finding electrons

in a specific location can be

determined.

The location and energy of electrons can be

specified using three terms:

Shell (aka level)

Subshell (aka sublevel)

Orbital

An additional fourth term,

spin number

, indicates

whether the electron is spinning clockwise or

counterclockwise.

The

shell

number is indicated by

assigning a number, n.

The lowest n number is 1, the next higher number is 2, etc.

(The n numbers correlate to the rows of the periodic table, 1-7)

Higher n numbers correspond to higher energies and greater

distances from the nucleus.

Principal Quantum Number

(

n

) (aka shell):

Size and Energy

n = 1, 2, 3, 4, 5, 6, 7 (correspond to periods)



As

n

increases, the orbital becomes larger

and the electron spends more time farther from the nucleus.



An increase in

n

also means

higher energy

, because the electron is less tightly bound to the nucleus, and the energy is less negative.

As

n

increases, the size of

the orbitals increases

 Electrons in higher numbered shells are further

from the nucleus and are higherin energy

 Shells with larger numbers (n) are farther from the nucleus (larger) and can hold more electrons.

 Electrons in lowernumbered shells are closerto the nucleus and are lowerin energy

The Angular Momentum Quantum Number (

l

)

(aka subshell):

Shape

l

= 0 to

n

– 1 for each value of

n

The value of l for a particular sublevel is assigned a letter.

The shell and subshell are used together to identify the subshell clearly (e.g. 2p subshell, or 3s subshell). All the electrons in the same subshell have the same energy.

The

Subshell

is indicated by assigning a letter,

l

Different values of

l correspond to different letter designations:

l

=

0,

1,

…

,

n

-

1

The subshell (broadly) indicates the shape (e.g. all s orbitals are spherical in shape)

The shell and subshell are used together to identify the subshell clearly (e.g. 2p subshell, or 3s subshell).

All the electrons in the same subshell have the same energy.

l

=

0

s subshell

l

=

1

p subshell

l

=

2

d subshell

l

=

3

f subshell

An

orbital

is a volume of space

in which electrons are found.



s

orbitals are

spherical

and there is 1

s

-orbital for

every energy level starting with

n

= 1 (1

s

,2

s,3

s,etc)



p

orbitals are

dumbbell-shaped

and there are 3

p

-orbitals for every energy level starting with

n

= 2 (2

p

,

3

p

, 4

p

, etc)

s

orbital

p

orbitals

1s orbital

2py orbital

2pz orbital

2px orbital

d

-orbitals are shaped like clovers or double

dumbbells and there are

5 d-orbitals for every energy level starting

with

n

= 3

1s orbital

f-orbitals are shaped like flowers or triple

dumbbells and there are 7 f-orbitals for

every energy level starting with

n

= 4

1s orbital

f

orbitals

4f, 5f, etc

Magnetic Quantum Number (

m

_l

):

Orientation of the Orbital in Space

ml = –

l

to +

l

l= 0 (s) l = 1 (p) l = 2 (d)

ml = 0 ml= -1, 0, +1 ml= -2, -1, 0, 1, 2

1 mlvalue 3 ml values 5 ml values

1 s-orbital 3 p-orbitals 5 d-orbitals

The orbital is a volume of space in the which the electrons are found.

l

= 0 (

s

)

m

l

= 0

1

m

l

value

1

s

-orbital

Spherical

l

= 1 (

p

)

m

l

= -1, 0, +1

m

l

values

3

p

-orbitals

dumbbell

l

= 2 (

d

)

m

l

= -2, -1, 0, 1, 2

5

m

l

values

5

d

-orbitals

l

= 3 (

f

)

m

l

= -3, -2, -1, 0, 1, 2, 3

7

m

l

values

7

f

-orbitals

Quantum Numbers

Quantum numbersare required to describe the distributionof electron density in an atom.

There are three quantum numbers necessary to describe an atomic orbital.

The principal quantum number (n) – designates size and energy

The angular moment quantum number (l) – describes shape

The magnetic quantum number (ml) – specifies orientation

Practice with the Allowed Values of Quantum Numbers

Strategy Recall that the possible values of ml depend on the value of l, not on the value of n.

What are the possible values for the magnetic quantum number (ml) when the principal quantum number (n) is 3 and the angular quantum number (l) is 1?

Solution The possible values of ml are -1, 0, and +1. Setup The possible values of ml are – l,…0,…+l.

Think About It Consult Table 3.2 to make sure your answer is correct. Table 3.2 confirms that it is the value of l, not the value of n, that determines the possible values of ml.

Spin Quantum Number

m

s

= +1/2, -1/2:

Two Allowed Spin States for the Electron

m

s

= +1/2, -1/2

The electron can spin in

one of two opposite

directions.

Pauli Exclusion Principle

 In a given atom, no two electrons can have the same

four quantum numbers (n, l, ml, and ms).

 n, l, ml : describe the atomic orbital

 ms : describes the electron spin

 Since only two values of ms are allowed, an orbital can hold only two electrons, and they must have opposite spins.

A Summary of Quantum Numbers

To summarize quantum numbers: principal (n) – size angular (l) – shape magnetic (ml) – orientation

electron spin (ms) direction of spin

Required to describe an atomic orbital

Required to describe an electron in an atomic orbital

2

p

x

principal (n = 2)

angular momentum (l = 1)

related to the magnetic quantum number (ml )

All

s orbitals

are spherical in shape but differ in size:

1s < 2s < 3s

2

s

angular momentum quantum number (l = 0)

ml= 0; only 1 orientation

All

p orbitals

are shaped like dumb-bells:

Three orientations:

l = 1 (as required for a p orbital)

ml = –1, 0, +1

Four

d orbitals

are shaped like double dumb-bells,

the fifth is a dumb-bell with a doughnut waist:

The d orbitals:

Five orientations:

l= 2 (as required for a d orbital)

ml = –2, –1, 0, +1, +2

Energies of Orbitals

The energies of orbitals in the hydrogen atom depend only on the principal quantum number.

2nd _{shell (n = 2)}

3d subshell (n = 3; l = 2)

2p subshell (n = 2; l = 1)

3rd _{shell (}n = 3)

2s subshell (n = 2; l = 0)

3p subshell (n = 3; l = 1) 3s subshell (n = 3; l = 0)

How to Label Orbitals with Quantum Numbers

Strategy Consider the significance of the number and the letter in the 4d designation and determine the values of n and l. There are multiple values for ml, which will have to be deduced from the value of l.

List the values of n, l, and ml for each of the orbitals in a 4d subshell.

Solution 4d

Possible ml are -2, -1, 0, +1, +2.

Setup The integer at the beginning of the orbital designation is the principal quantum number (n). The letter in an orbital designation gives the value of the angular momentum quantum number (l). The magnetic quantum number (ml) can have integral values of – l,…0,…+l.

principal quantum

number, n = 4 quantum number, l = 2 angular momentum

Think About It Consult the following figure to verify your answers.

The

Electronic Configuration

shows how the

electrons are arranged in an atom’s orbitals

The ground state is the lowest energy arrangement. By following a set of rules, we can predict which orbitals are

filled and how many electrons each contains.

Rules to Determine the Ground State

Electron Configuration of an Atom

• Electrons are placed in the lowest energy

orbital beginning with the 1s orbital.

• Orbitals are then filled in order of increasing

energy.

• Within an atom, the lower the value of n,

the more stable (lower in energy) will be

the orbital (e.g. the 1s orbital is lower in

energy than the 2s, which is lower than the

3s, etc.)

There are

n

types of subshells in the

n

th energy level.

There are

n

2

orbitals in the

n

th energy level.

Rule [2]

There is one

s

, three

p

, five

d

, and seven

f

orbitals.

Rule [3]

How many

subshells

are in each

shell

?

How many

orbitals

are in each

subshell

?

How many

orbitals

are in each

shell

?

Subshell Blocks of the Periodic Table

Each orbital can contain a maximum of

2 electrons, and the electrons must be

spinning in opposite directions.

Subshell

# of Orbitals

Max # of e

-

s

1

2

p

3

6

d

5

10

f

7

14

Rule [5]

Order of Filling Orbitals

Electrons in an atom



fill orbitals in sublevels of the same type with one

electron each until all sublevels are half full



then pair up in the orbitals using opposite spins

Reading Electron Configuration

Directly from the Periodic Table

•

Think of each element box on the periodic table as an

electron.

•

Count and name the electrons on the table reading from

left to right.

•

Make 3 adjustments:

–For this purpose, we pretend that helium is directly next to

hydrogen, putting it in the s-sublevel, not the p-sublevel.

–Because of energy overlaps, the 1st _{d-sublevel is 3d not 4d.}

–Because of energy overlaps, the 1st _{f-sublevel is 4f, not 6f.}

•

To check:

Look at the last part of the electron

configuration, find this square on the periodic

table-it should be the element represented by the

electron configuration.

Reading Electron Configuration

Directly from the Periodic Table

Chlorine's Electron Configuration

•

The large numbers represent the energy level (

n

).

•

The letters represent the sublevel. (

l

=

s

and

p

)

•

The superscripts indicate the number of electrons in the

sublevel.

•

Think of each square on the periodic table as an electron.

Counting the number of squares will give you the number

of electrons in each energy level and sublevel.

To check an electron configuration: The last notation

in the electron configuration represents the location

of the element on the periodic table.

Silicon:

the 3p2_{in the configuration for Si indicates its location as the} 2nd _{square in the p sublevel on the 3}rd _row

of the periodic table.

To check an electron configuration: The total of the

superscripts in an electron configuration equals the atomic

number of the element (the total # of electrons!).

Selenium:

The superscripts add up to the

total # of electrons in the atom.

The

last electron added

gives

the location of the element

in the periodic table.

•

Carbon-1

s

2

2

s

2

2

p

2

–Carbon’s atomic # (# of electrons) is 6.

–C is the 2nd _{square in the p-sublevel of the 2}nd _{energy level.}

•

Aluminum- 1

s

2

2

s

2

2

p

6

3

s

2

3

p

1

–Aluminum’s atomic # (# of electrons) is 13.

–Al is the 1st _{square in the p-sublevel of the 3}rd _{energy level.}

•

Nickel- 1

s

2

2

s

2

2

p

6

3

s

2

3

p

6

4

s

2

3

d

8

–Nickel’s atomic # (# of electrons) is 28.

–Ni is the 8th square in the d-sublevel of the 3rd _{energy level.}

Electron Configuration Practice Problems

•

Name the elements whose electron configurations

are

–

1

s

2

2

s

2

2

p

6

3

s

2

3

p

6

4

s

2

3

d

3

–

1

s

2

2

s

2

2

p

6

3

s

2

3

p

6

4

s

2

3

d

10

4

p

6

5

s

2

4

d

9

–

1

s

2

2

s

2

2

p

6

3

s

2

3

p

6

•

Write electron configurations for these elements:

–

Potassium

–

Lanthanum

–

Copper

–

Bromine

Rule [6]

Half-filled and

filled subshells have

special stability so the

electronic configuration

of transition metals

Cr

and

Cu

show an

exception to rule 5

.

3d 4s

3d 4s

Cr

1

s

2

2

s

2

2

p

6

3

s

2

3

p

6

4

s

2

3

d

4

(using the first 5 rules)

Cr

1

s

2

2

s

2

2

p

6

3

s

2

3

p

6

4

s

1

3

d

5

(using rule 6)

Abbreviated Electronic Configuration

He 1s2

Ne 1s2_2s2_2p6

Ar 1s2_2s2_2p6_3s2_3p6

Kr 1s2_2s2_2p6_3s2_3p6_4s2_3d10_4p6

Xe 1s2_2s2_2p6_3s2_3p6_4s2_3d10_4p6_5s2_4d10_5p6

Rn 1s2_2s2_2p6_3s2_3p6_4s2_3d10_4p6_5s2_4d10_5p6_6s2_4f14_5d10_6p6

The

noble gases

on the far right side of the periodic

table have

completely filled subshells

. Thus, they

are

chemically very stable

and do not react readily

with other substances.

The electron configuration can be

shortened

by using

Noble Gas Notation

.

Examples:

O 1s2_2s2_2p4 _{[He] 2s}2_2p4

Mn 1s2_2s2_2p6_3s2_3p6_4s2_3d5 _{[Ar] 4s}2_3d5

Lu 1s2_2s2_2p6_3s2_3p6_4s2_3d10_4p6_5s2_4d10_5p6_6s2_4f14_5d1 _{[Xe] 6s}2_4f14_5d1

Write the Symbol of the previous Noble Gas, then add the electronic configuration of the additional electrons.

He 1s2

C 1s2_2s2_2p2 _[He]2s2_2p2

element: nearest noble gas:

Noble Gas Notation Electron

Configuration

The periodic table gives the

electron configuration for Arsenic

• The electron configuration can be determined by the periodic table because the shell and subshell information is embedded in the table.

The periodic table gives the

electron configuration for Phosphorus

• The inner electron configuration is that of the noble gas at the end of

the previous row (Ne); the row number gives the shell, and the location of the elements in the row containing P tells where the

valence electrons are located: 3s2_3p3_{. Overall, the configuration is}

[Ne]3s2_3p3_.

1. Identify the element in group 18 (8A) closest to,

but not greater than, your element.

2. Put that element in brackets.

3. The

“

leftover stuff

”

always starts with an s orbital.

4. The

number before the s orbital

is the

row number

of your element

.

5. Calculate how many electrons are in your

“

leftover stuff.

”

6. Continue filling in orbitals using your guide until

your account for all the electrons in the

“

leftover

stuff.

”

Abbreviated Electronic Configuration

Half-filled and Filled Subshells

Have Special Stability:

Exceptions Include Cr and Cu

Ar

Cr: [Ar]4

s

1

3

d

5

Cu: 4

s

1

3

d

10

An Orbital Diagram uses a

box

to

represent

each

orbital

and

arrows

to

represent

electrons

.

•

. Boxes are labeled with the principal quantum

number, n, and the sublevel letter.

•

Two electrons must have paired spins (opposite

directions) to fit into the same orbital.

•

Possible spins are clockwise and counterclockwise,

represented by up and down arrows.

an orbital a single,

unpaired electron

an electron pair

Orbital diagram for H-atom and He-atom

The arrow represents the electron; the electron is in

the ground state because it is in the lowest-energy

shell and subshell possible.

H (Z = 1)

1 electron 1s 1s1

He (Z = 2)

2 electrons 1s 1s2

Element Notation Orbital Configuration Electron

Order of Filling Orbitals

Electrons in an atom

 fill orbitals in sublevels of the same type with one electron each until all sublevels are half full  then pair up in the orbitals using opposite spins

Li (Z = 3)

3 electrons 2s 1s22s1

C (Z = 6)

6 electrons 1s22s22p2 Element Notation Orbital Configuration Electron

1s

2s

1s 2p

Ne (Z = 10)

Electron configuration

,

Noble Gas

Notation

, and

Orbital Diagram

for Ca

Ca 20 electrons

1s2_2s2_2p6_3s2_3p6_4s2

Element Notation Orbital

Electron

Configuration Noble Gas Notation

2s

1s 2p 3s 3p 4s 4s is lower in energy;

it is filled before 3d.

[Ar]4s2

Determine the Number of Unpaired

Electrons in a Phosphorus Atom.

P: 1

s

2

2

s

2

2

p

6

3

s

2

3

p

3

Determine the Number of Unpaired Electrons in a

Carbon

Atom.

Determine the Number of Unpaired Electrons in a

Silicon

Atom

(

same as Carbon because in the same group

)

What is the maximum number of electrons that

can be contained in each of the following?

a. A 2p orbital Answer: 2 Each orbital can hold 2 e– max

b. A 2p subshell Answer: 6 There are three 2p orbitals c. The second shell Answer: 8 one 2s + three 2p orbitals

2s 2p

Write an electronic configuration for each of the

following elements, using the form 1

s

2

2

s

2

2

p

6

, and so

on. Indicate how many electrons are unpaired in each

case.

a. element number 37 1s22s22p63s23p64s23d104p65s1 1 unpaired e–

b. Si 1s2_2s2_2p6_3s2_3p2 _{2 unpaired e}–

c. cobalt 1s2_2s2_2p6_3s2_3p6_4s2_3d7 _{3 unpaired e}–

d. Ar 1s2_2s2_2p6_3s2_3p6 _{0 unpaired e}–

Identify the outermost subshell occupied by

electrons in iron atoms.

Identify the highest energy subshell occupied

by electrons in iron atoms.

Iron

[Ar]4s

2

3d

6

a)

The outermost subshell is the 4s subshell.

Electron Configurations

The electron configurationdescribes how the electrons are distributed in the various atomic orbitals.

In a ground state hydrogen atom, the electron is found in the 1s

orbital.

1

s

1

principal (n = 1)

angular momentum (l = 0)

number of electrons in the orbital or subshell

1s

2s 2p 2p 2p

Ene

rgy

The use of an up arrow indicates an electron with ms = + ½

Ground state electron configuration of hydrogen

Electron Configurations

If hydrogen’s electron is found in a higher energy orbital, the atom is in an excited state.

2

s

1

1s

2s 2p 2p 2p

Ene

rgy

A possible excited state electron configuration of hydrogen

Electron Configurations

The helium emission spectrum is more complex than the hydrogen spectrum.

There are more possible energy transitions in a helium atom because helium has two electrons.

Electron Configurations

In a multi-electron atoms, the energies of the atomic orbitals are split.

Splitting of energy levels refers to the splitting of a shell (n=3) into subshells of different energies (3s, 3p, 3d)

Electron Configurations

According to the Pauli exclusion principle, no two electrons in an atom can have the same four quantum numbers.

1

s

2

1s

2s

2p 2p 2p

Ene

rgy

The ground state electron configuration of helium

Quantum number

Principal (n) Angular moment (l) Magnetic (ml) Electron spin (ms)

1 0 0

+ ½ 1 0 0

‒ ½

describes the 1s orbital

describes the electrons in the 1s orbital

Electron Configurations

The Aufbau principlestates that electrons are added to the lowest energy orbitals first before moving to higher energy orbitals.

1

s

2

2

s

1

1s

2s

2p 2p 2p

Ene

rgy

The ground state electron configuration of Li

Electron Configurations

The Aufbau principlestates that electrons are added to the lowest energy orbitals first before moving to higher energy orbitals.

1s

2s

2p 2p 2p

Ene

rgy

1

s

2

2

s

2

The ground state electron configuration of Be

Be has a total of 4 electrons

Electron Configurations

The Aufbau principlestates that electrons are added to the lowest energy orbitals first before moving to higher energy orbitals.

1s

2s

2p 2p 2p

Ene

rgy

The ground state electron configuration of B

1

s

2

2

s

2

2

p

1

B has a total of 5 electrons

Electron Configurations

According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized.

1

s

2

2

s

2

2

p

2

1s

2s

2p 2p 2p

Ene

rgy

The ground state electron configuration of C

The 2p orbitals are of equal energy, or degenerate.

Put 1 electron in each before pairing (Hund’s rule).

C has a total of 6 electrons

Electron Configurations

According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized.

1

s

2

2

s

2

2

p

3

1s

2s

2p 2p 2p

Ene

rgy

The ground state electron configuration of N

The 2p orbitals are of equal energy, or degenerate.

Put 1 electron in each before pairing (Hund’s rule).

N has a total of 7 electrons

Electron Configurations

According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized.

1

s

2

2

s

2

2

p

4

1s

2s

2p 2p 2p

Ene

rgy

The ground state electron configuration of O

O has a total of 8 electrons

Once all the 2p orbitals are singly occupied, additional electrons will have to pair with those already in the orbitals.

Electron Configurations

According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized.

1

s

2

2

s

2

2

p

5

1s

2s

2p 2p 2p

Ene

rgy

The ground state electron configuration of F

F has a total of 9 electrons

Electron Configurations

According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized.

1

s

2

2

s

2

2

p

6

1s

2s

2p 2p 2p

Ene

rgy

The ground state electron configuration of Ne

Ne has a total of 10 electrons

When all of the electrons in an atom are paired, as in neon, it is called diamagnetic.

Electron Configurations and the Periodic Table

The electron configurations of all elements except hydrogen and helium can be represented using a noble gas core.

The electron configuration of potassium (Z = 19) is 1s2_2s2_2p6_3s2_3p6_4s1_.

Because 1s2_2s2_2p6_3s2_3p6 _{is the electron configuration of argon, we}

can simplify potassium’s to [Ar]4s1_.

1

s

2

2

s

2

2

p

6

3

s

2

3

p

6

4

s

1

The ground state electron configuration of K:

[Ar]

[Ar]4

s

1

1

s

2

2

s

2

2

p

6

3

s

2

3

p

6

4

s

1

Electron Configurations and the Periodic Table

There are several notable exceptions to the order of electron filling for some of the transition metals.

Chromium (Z = 24) is [Ar]4s1_3d5 _{and not [Ar]4s}2_3d4 _as

expected.

Copper (Z = 29) is [Ar]4s1_3d10 _{and not [Ar]4s}2_3d9 _{as expected.}

The reason for these anomalies is the slightly greater stability of d

subshells that are either half-filled (d5_{) or completely filled (d}10_).

4s 3d 3d 3d 3d 3d

[Ar]

Cr

Greater stability with half-filled 3d subshell

Electron Configurations and the Periodic Table

There are several notable exceptions to the order of electron filling for some of the transition metals.

Chromium (Z = 24) is [Ar]4s1_3d5 _{and not [Ar]4s}2_3d4 _as

expected.

Copper (Z = 29) is [Ar]4s1_3d10 _{and not [Ar]4s}2_3d9 _{as expected.}

The reason for these anomalies is the slightly greater stability of d

subshells that are either half-filled (d5_{) or completely filled (d}10_).

Electron Configurations and the Periodic Table

4s 3d 3d 3d 3d 3d

[Ar]

Cu

Greater stability with filled 3d

subshell

There is a distinct pattern to the

electron configurations of the elements

in a particular group

The Outermost Electrons of an Atom

are called the Valence Electrons

For Group 1A: [noble gas]ns1

valence core

For Group 2A: [noble gas]ns2

valence core

For Group 7A: [noble gas]ns2_np5

valence core

The

chemical properties

of an element depend

on the number of electrons in the

valence shell

.

The

valence shell

is the outermost shell

(

highest value of n

)

Be

1

s

2

2

s

2

Cl

1

s

2

2

s

2

2

p

6

3

s

2

3

p

5

valence shell:

n

= 2

# of

valence electrons = 2

valence shell:

n

= 3

# of

valence electrons = 7

Elements in the same group have the same

number of valence electrons

Group # 1A – 8A = # valence electrons (except He = 2)

Ex: All elements in Group 2A have 2 valence electrons.

Effective nuclear charge (

Z

eff

)

is the actual magnitude of positive charge

that is “experienced” by an electron in the

atom

Due to

shielding

, the value of

Z

eff

increases

steadily from

left to right

because the core electrons remain the

same but

Z

increases

Z

eff

=

Z

-

S

H 1.0 Li 1.3 Na 2.2 K 2.2 Rb 2.2 Cs

Be

1.95 2.60 B 3.25 C 3.90 N 4.55 O 5.20 F 5.85 Ne

Moving left to right across period 2, the nuclear charge increases by one with

each new element, but the effective nuclear charge increases only by an

average of 0.64

The Radius of an Atom (r) is Defined as Half

the Distance Between the Nuclei in a Molecule

Consisting of Identical Atoms

228 pm

Atomic Size Increases Going Down a

Group and Decreases Across a Period

Atomic

Size

Increases

Down

a

Group

because

the Outermost Electrons in Higher Energy Levels

are Farther from the Nucleus

Simplified explanation using Bohr atom n = 1

n = 2 n = 3

n = 4

Atomic

Size

Decreases

Across

a

Period

Because the Increasing

Z

eff

Pulls the

Valence Electrons Closer to the Nucleus

n = 1 n = 2

n = 3 n = 4

Zeff H

1.0 Li 1.3 Na 2.2 K 2.2 Rb 2.2 Cs 2.2

Be

1.95 2.60 B 3.25 C 3.90 N 4.55 O 5.20 F 5.85 Ne

Which has a larger size, C or O?

C is bigger than O.

C has 6 protons in the nucleus for a +6 charge at the

nucleus, while O has 8 protons in the nucleus, for

a +8 charge at the nucleus.

Which has a larger size, Li or K?

K is bigger than Li.

K has a valence electron in energy level 4 while Li has a valence electron in level 2. This means the K atom is larger than the Li atom because the valence electron is farther away from the

nucleus.

Which has a larger size, C or Al?

Both trends indicate that Al is larger than C. Going across the size decreases so C is smaller than Al, and

Which has a larger size, Se or I?

I is bigger than Se.

The increase in size going down a group is bigger than

the decrease in size going across a group.

Ionization Energy (IE) is the Energy Required to

Remove an Electron from an Atom

 The result is an ion, a chemical species with a net charge.

 Sodium has an ionization energy of 495.8 kJ/mol.

 Specifically, 495.8 kJ/mol is the first ionization energy of sodium, IE1(Na), which corresponds to the removal of the most loosely held electron.

Na(g) → Na+_{(g) + e}−

The first Ionization energy is the energy needed

to remove an electron from an atom.

Na + energy Na+ + e–

•Ionization energies decrease down a

column as the valence e− _get farther away from

the positively charged nucleus.

•Ionization energies increase across a row as the number of protons in the nucleus increases.

The 1

st

ionization energy

increases

as you go

left to right across a

period

as the Z

eff

increases

.

Electrons in the filled 2

s

orbital provide some

shielding for electrons in the 2

p

orbital from the

nuclear charge.

B 1s 2s 2p

Be _{1s 2s 2p}

Within a given shell, electrons with a higher value of

l

are higher in energy and thus, easier to remove.

Removing a paired electron is easier

because of the repulsive forces between

two electrons in the same orbital.

N

O

1s 2s 2p

The 1

st

ionization energy

decreases

as you go

down a group

as the

distance between the nucleus and

valance electrons increases

.

Which has the higher ionization energy, As or Sb?

As has a higher ionization energy than Sb, according to the top-to-bottom ionization trend in the periodic table.

Which has the higher ionization energy, N or Si?

N has a higher ionization energy than Si, according to both the top-to-bottom and the left-to-right ionization trends in the

periodic table.

Which has the higher ionization energy, O or Cl?

Here we can't tell which has the higher ionization energy: O

would be higher according to the top-to-bottom trend, but Cl would be higher according to the left-to-right ionization

trend. The effects tend to cancel.

It is possible to remove additional electrons in

subsequent ionizations, giving

IE

2

,

IE

3

, etc.

IE1(Na) = 496 kJ/mol

IE2(Na) = 4562 kJ/mol

Na(g) → Na+_{(g) + e}−

Na+_{(g) → Na}2+_{(g) + e}−

It takes much more energy to remove core

electrons than valence electrons

1s 2s

2p 3s

Mg

1

s

2

2

s

2

2

p

6

3

s

2

738 kJ/mol 1451 kJ/mol

7733 kJ/mol Core electrons experience greater Zeff because of fewer filled shells shielding them from

the nucleus. It is harder to remove an electron from a positive ion then

Ionization Energy: X(

g

)



X

+

(

g

) + e

-



The energy required to remove an electron from a

gaseous atom or ion.



Al(

g

)



Al

+

(

g

) + e

-

I

1

= 580 kJ/mol

[Ne]3

s

2

3

p

1



[Ne]3

s

2



Al

+

(

g

)



Al

2+

(

g

) + e

-

I

₂

= 1815 kJ/mol

[Ne]

3

s

2



[Ne]3

s

1



Al

2+

(

g

)



Al

3+

(

g

) + e

-

I

3

= 2740 kJ/mol

[Ne]

3

s

1



[Ne]



Al

3+

(

g

)



Al

4+

(

g

) + e

-

I

4

= 11,600 kJ/mol

[Ne]



1

s

2

2

p

5

Sodium and aluminum both have one unpaired electron in

their neutral ground state atoms. Why is the first ionization

energy of Na lower than Al, but the second ionization of

energy of Al lower than the second ionization energy of Na?

1. The smaller size of Al makes it difficult to ionize at first, but after losing one e–_{, the other atoms can expand more and make it}

easier to ionize a second e–_.

2. The 3s electrons of Na does a better job of shielding than the 3p electron of Al.

3. The second e– _{taken from Na must be taken from a new lower}

level; the second e– _{from Al is in same level as before.}

4. The first e– _{ionized from both is single, but the second one in}

Na is paired, while the second in Al is still single

.

Electron Affinity (

EA

) is the energy

released when an atom in the gas

phase accepts an

e

–

.

Cl(g) + e−_{→ Cl}−_(g)

349.0 kJ/mol of energy is released.

A positive electron affinity indicates a

process that is energetically favorable.

Like ionization energy,

electron affinity increases from

left to right

across a period as

Z

eff increases.

Easier to add an electron as the positive charge of the nucleus increases.

It is easier to add on electron to an

s

-orbital than to add one to a

p

-orbital with the

same principle quantum number.

Within a

p

-subshell, it is easier to add an electron

to an empty orbital then to add one to an orbital

While many first electronic affinities

are positive, subsequent electron

affinities are always negative.

Considerable energy is required to overcome the

repulsive forces between the electron and the

negatively charged ion.

Process Electron Affinity

O(g) + e− _{→ O}−_(g)

O− _{(g) + e}− _{→ O}2−_(g)

EA1 = 141 kJ/mol

EA2 = −741 kJ/mol

Properties of Metals

 Metals tend to

Be shiny, lustrous, malleable (can be pounded into a sheet), and ductile (can be drawn into a wire)

Be good conductors of heat and electricity

Have low ionization energies (commonly form positive cations)

Properties of Nonmetals

 Nonmetals tend to

Very in color and are not shiny

Be brittle, rather than malleable

Be poor conductors of heat and electricity

Have high electron affinities (commonly form negative anions)



Metals

/ On the L



Give up one or more e

-to form a positive ion

X



X

+

+ e

-

low IE



Nonmetals

/ On the R



Ability to gain one or

more e

-

to form an anion

when reacting with a

metal

X + e

-



X

-

large IE and the most

negative EA

Metalocity: how metallic the element is.

Trends Within the Periodic Table

Metals Metalloids Nonmetals

C

Si

Ge

Sn Pb

–

The elements get more metallic as you

go down a group.

Na Mg Al Si P S Cl Ar

The elements get less metallic as you

go left to right.

Using these generalizations, francium is the most metallic element while helium is the least metallic.

Which is the more meta