7 Atomic Structure and Periodicity - Google Search

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Chapter 7

Atomic Structure and Periodicity

The Nature of Light: Classification of Electromagnetic Radiation

Characteristics of Waves, c = ln 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 108 m/s

Amplitude,

The depth or height of a wave

Speed of light in a vacuum:

c = 3.00´108 ms

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 interference is 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

Think About It Make sure your units cancel properly. A common error in this type of problem is neglecting to convert wavelength to nanometers.

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Electromagnetic Waves

For electromagnetic waves c = ln

n = cl

= 3.00 ´108 ms

4.4´10-6m= 6.8´1013 1

s = 6.8´1013Hz

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(NO3)2 and SrCO3 are heated.

Calculate the frequency of red light of wavelength

6.50 x 102 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

~2000K ~2000K

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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 NOT emit or absorb any quantity of energy. Energy is only emitted or absorbed in discrete quantities, like small

packages or bundles of energy. A quantum is 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 quanta of size hn.

E = hn

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 higher ν, 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

Photons and the Photoelectric Effect

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Dual Nature of Light: Exhibits Wave Properties AND Particulate Properties

Energy is quantized and occurs in discrete units called quanta.

Photons of Light

E = hn

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,

Ephoton = hn = 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 102 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 1010 cm

3. 0.714 cm

4. This can’t be solved unless you are given the frequency.

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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

1.54 105 J, what is the frequency of this useful energy?

1. 2.59 10–15 s–1

2. 3.86 1014 s–1

3. 1.56 1023 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

Rydberg Equation: used to calculate the wavelength of light emitted when an electron in a

hydrogen atom changes orbits

RH = Rydberg constant = 1.096776 x 107 m-1

n1 and n2 are positive integers

n2 > n1

1l

= RH( ) 1n1

2 -1n2

2

æ

è ç

ö

ø ÷

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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?

1l

= RH( ) 1n1

2 -1n2

2

æ

èç

ö

ø÷= 1.096776 ´107 m-1( ) 1

22 -142

æ

èç

ö

ø÷= 2056455 m-1

l = 12056455 m-1 = 4.86 ´10-7 m = 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 H2?

DE = hn = 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

En 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 state.

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As an electron gets closer to the nucleus, n decreases. En 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 = 3.00×108 m/s

2.18×10-18 J (6.63×10-34 J∙s )(3.00×108 m/s)

1 λ = 1

22 1 42 -

= 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 the equation gives a positive result.

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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 = h mu = 1.16×10-5 m 6.63×10-34 kg∙m2/s

(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 aluminum foil

Electron diffraction pattern of aluminum foil.

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The Heisenberg uncertainty principle states that it is impossible to know simultaneously both the momentum p and 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 reactive?

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Lithium Metal (Li) and H2O Sodium Metal (Na) and H2O

Potassium Metal and H2O 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.

n = 1

n = 2 n = 3

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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 higher in energy

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

Electrons in lower numbered shells are closer to the nucleus and are lower in 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 (1s,2s,3s,etc)

p orbitals are dumbbell-shaped and there are 3 p-orbitals for every energy level starting with n = 2 (2p, 3p, 4p, 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

d orbitals 3d, 4d, 5d, etc

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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 (ml): 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 ml value 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) ml = 0

1 ml value 1 s-orbital Spherical

l = 1 (p) ml = -1, 0, +1

ml values 3 p-orbitals

dumbbell

l = 2 (d) ml = -2, -1, 0, 1, 2

5 ml values 5 d-orbitals

l = 3 (f) ml = -3, -2, -1, 0, 1, 2, 3

7 ml values 7 f-orbitals

Quantum numbers designate shells, subshells, and orbitals.

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Quantum Numbers

Quantum numbers are required to describe the distribution of 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 ms = +1/2, -1/2: Two Allowed Spin States for the Electron

ms = +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

2px

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

2s angular momentum quantum number (l = 0) ml = 0; only 1 orientation possible

principal quantum number (n = 2)

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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

angular momentum quantum number, l = 2

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.)

Rule [1]

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There are n types of subshells in the nth energy level. There are n2 orbitals in the nth 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 [4]

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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 3rd 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 total of the superscripts in the electron configuration for Se

is 34, its atomic number.

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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-1s22s22p2

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

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

• Aluminum- 1s22s22p63s23p1

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

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

• Nickel- 1s22s22p63s23p64s23d8

– 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

– 1s22s22p63s23p64s23d3

– 1s22s22p63s23p64s23d104p65s24d9

– 1s22s22p63s23p6

• 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 1s22s22p63s23p64s23d4 (using the first 5 rules)

Cr 1s22s22p63s23p64s13d5 (using rule 6)

Abbreviated Electronic Configuration

He 1s2

Ne 1s22s22p6

Ar 1s22s22p63s23p6

Kr 1s22s22p63s23p64s23d104p6

Xe 1s22s22p63s23p64s23d104p65s24d105p6

Rn 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p6

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 1s22s22p4 [He] 2s22p4

Mn 1s22s22p63s23p64s23d5 [Ar] 4s23d5

Lu 1s22s22p63s23p64s23d104p65s24d105p66s24f145d1 [Xe] 6s24f145d1

Cr 1s22s22p63s23p64s13d5 [Ar] 4s13d5

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

He 1s2

C 1s22s22p2 [He]2s22p2 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 inner (or core) electron configuration is that of the noble gas at the end of the previous row (Ar); the row number gives the shell, and the location of the elements in the row containing As tells where the d-subshell and valence electrons are located: 4s23d104p3. Overall, the configuration is [Ar]4s23d104p3.

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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: 3s23p3. Overall, the configuration is [Ne]3s23p3.

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]4s13d5

Cu: 4s13d10

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 Orbital

Notation Electron

Configuration

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 Orbital

Notation Electron

Configuration

1s

2s 1s 2p

Ne (Z = 10) 10 electrons

1s22s22p6

2s 1s 2p

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Electron configuration, Noble Gas Notation, and Orbital Diagram for Ca

Ca 20 electrons

1s22s22p63s23p64s2

Element Orbital

Notation

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: 1s22s22p63s23p3

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 1s22s22p6, and so

on. Indicate how many electrons are unpaired in each case.

a. element number 37 1s22s22p63s23p64s23d104p65s1 1 unpaired e–

b. Si 1s22s22p63s23p2 2 unpaired e–

c. cobalt 1s22s22p63s23p64s23d7 3 unpaired e–

d. Ar 1s22s22p63s23p6 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]4s23d6

a) The outermost subshell is the 4s subshell.

b) The highest energy subshell is the 3d subshell.

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Electron Configurations

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

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

1s1

principal (n = 1)

angular momentum (l = 0)

number of electrons in the orbital or subshell

1s

2s 2p 2p 2p

Ener

gy

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

Ground state electron configuration of hydrogen


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

2s1

1s

2s 2p 2p 2p

Ener

gy

A possible excited state electron configuration of hydrogen


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.


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)


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

1s2

1s

2s

2p 2p 2p

Ener

gy

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


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

1s22s1

1s

2s

2p 2p 2p

Ener

gy

The ground state electron configuration of Li

The 1s orbital can only accommodate 2 electrons (Pauli exclusion principle)

The third electron must go in the next available orbital with the lowest possible energy.

Li has a total of 3 electrons

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1s

2s

2p 2p 2p

Ener

gy 1s22s2

The ground state electron configuration of Be

Be has a total of 4 electrons



1s

2s

2p 2p 2p

Ener

gy

The ground state electron configuration of B

1s22s22p1

B has a total of 5 electrons


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.

1s22s22p2

1s

2s

2p 2p 2p

Ener

gy

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



1s22s22p3

1s

2s

2p 2p 2p

Ener

gy

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



1s22s22p4

1s

2s

2p 2p 2p

Ener

gy

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.



1s22s22p5

1s

2s

2p 2p 2p

Ener

gy

The ground state electron configuration of F

F has a total of 9 electrons

When there are one or more unpaired electrons, as in the case of oxygen and fluorine, the atom is called paramagnetic.

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1s22s22p6

1s

2s

2p 2p 2p

Ener

gy

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 1s22s22p63s23p64s1. Because 1s22s22p63s23p6 is the electron configuration of argon, we can simplify potassium’s to [Ar]4s1.

1s22s22p63s23p64s1

The ground state electron configuration of K:

[Ar] [Ar]4s1

1s22s22p63s23p64s1

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]4s13d5 and not [Ar]4s23d4 as

expected. Copper (Z = 29) is [Ar]4s13d10 and not [Ar]4s23d9 as expected.

The reason for these anomalies is the slightly greater stability of d subshells that are either half-filled (d5) or completely filled (d10).

4s 3d 3d 3d 3d 3d [Ar] Cr

Greater stability with half-filled 3d subshell


There are several notable exceptions to the order of electron filling for some of the transition metals. Chromium (Z = 24) is [Ar]4s13d5 and not [Ar]4s23d4 as

expected. Copper (Z = 29) is [Ar]4s13d10 and not [Ar]4s23d9 as expected.

The reason for these anomalies is the slightly greater stability of d subshells that are either half-filled (d5) or completely filled (d10).


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

For Group 1A: [noble gas]ns1 For Group 2A: [noble gas]ns2

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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]ns2np5

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 1s22s2

Cl 1s22s22p63s23p5

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 (Zeff) is the actual magnitude of positive charge that is “experienced” by an electron in the

atom

Due to shielding, the value of Zeff increases steadily from left to right

because the core electrons remain the same but Z increases

Zeff = Z - SH

1.0 Li

1.3

Na 2.2 K

2.2 Rb 2.2

Cs 2.2

Be 1.95

B 2.60

C 3.25

N 3.90

O 4.55

F 5.20

Ne 5.85

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

Br radius = 114 pm

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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 Zeff 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

B 2.60

C 3.25

N 3.90

O 4.55

F 5.20

Ne 5.85

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 going down the size increases so Al is bigger than C.

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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 1st ionization energy increases

as you go left to right across a period as the Zeff increases.

Electrons in the filled 2s orbital provide some shielding for electrons in the 2p 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

1s 2s 2p

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The 1st 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 IE2, IE3, etc.

IE1(Na) = 496 kJ/mol

IE2(Na) = 4562 kJ/mol

Na(g) → Na+(g) + e−

Na+(g) → Na2+(g) + e−

It takes much more energy to remove core electrons than valence electrons

1s

2s

2p

3s

Mg 1s22s22p63s2

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 from a neutral atom.

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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- I1 = 580 kJ/mol

[Ne]3s23p1 [Ne]3s2

Al+(g) Al2+(g) + e- I2 = 1815 kJ/mol

[Ne]3s2 [Ne]3s1

Al2+(g) Al3+(g) + e- I3 = 2740 kJ/mol

[Ne]3s1 [Ne]

Al3+(g) Al4+(g) + e- I4 = 11,600 kJ/mol

[Ne] 1s22p5

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 Zeff 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

that already contains an electron.

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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− → O2−(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 Ar Al Cl Si S P

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 metallic element, Sn or Te?

Sn is a more metallic element than Te, according to the left-to-right metallic character trend in the periodic table.

The text links metallic character to the tendency to lose electrons in chemical reactions, and nonmetallic character to the tendency to gain electrons in chemical reactions. The metallic character trends therefore follow the ionization energy trends.

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Which is the more metallic element, Si or Sn?

Sn is a more metallic element than Si, according to the top-to-bottom metallic character trend in the periodic table.

Which is the more metallic element, Br or Te?

Te is a more metallic element than Br, according to both the top-to-bottom trend and the left-to-right metallic

character trend in the periodic table.

Which is the more metallic element, Se or I?

Here we can't tell which has the greater atomic size: I would be more metallic according to the top-to-bottom trend, but Se

would be more metallic according to the left-to-right metallic character trend. The effects tend to cancel.

How to write the electron configuration of an ion

formed by a main group element 1. Write the configuration for the atom.

2. Add or remove the appropriate number of electrons.

Na: 1s22s22p63s1 Na+: 1s22s22p6

Cl: 1s22s22p63s23p5 Cl− : 1s22s22p63s23p6

10 electrons total, isoelectronic with Ne

18 electrons total, isoelectronic with Ar

How to write the electron configuration of an ion formed by a d-block element

Ions of d – block elements are formed by removing electrons first from the shell with the highest value of n.

For Fe to form Fe2+, two electrons are lost from the 4s subshell not the 3d.

Fe can also form Fe3+, in which case the third electron is removed from the 3d subshell.

Fe: [Ar]4s23d 6 Fe2+: [Ar]3d 6

Fe: [Ar]4s23d 6 Fe3+: [Ar]3d 5

Trends in Ionic Radii

Cations are smaller than their parent

atoms.

Anions are larger than their parent atoms.

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In each of the following sets, which atom or ion has the smaller radius?

1. H or He

2. S2–, S, or S2+

3. Cu or Au

4. Rb or Ca

He

S2+

Cu

Ca

An isoelectronic series is a series of two or more species that have identical electronic

configurations, but different nuclear charges.

O2− : 1s22s22p6 F−: 1s22s22p6 Ne: 1s22s22p6

isoelectronic

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