• Electrons display both particle properties and wave properties.
• Electrons were discovered by JJ Thompson
• Thompson also measured the charge/mass ratio
• Milikan was able to determine the charge on an electron
• Davisson and Germer discovered the wave nature of an electron at Bell Labs in 1927

Different models of atoms
• Thompson developed the plumb pudding model of an atom 1898
• Rutherford suggested the planetary model (i.e. electrons orbit the nucleus)
• Bohr applied concepts of quantization to
Rutherford’s model to develop the Bohr model
• Bohr model lasted 10 years, and was replaced by a wave model, called the quantum mechanical model, based on the wave nature of electrons.

Energy has properties of waves , for example electromagnetic energy
Wavelength (m)
Amplitude (m)
Speed 3.0X10
8 m/s
Energy (j)
Frequency (1/s, Hz)

• Wavelength (

) is the distance from one wave crest to another in meters.
• The frequency (

) of a wave expresses the number times a wave passes a given point in some unit of time in 1/s.
• Amplitude of a wave is the height of the crest or depth of the trough with respect to the center line of the wave in meters.

10 -12 10 -10 10 -8
Gamma rays X- rays Uv-rays
10 -7
Visible rays
10 -4
Infrared rays
10 -2
Micro
Waves
(Radar)
10 0 10 2
Radio and Television waves
Increasing wave length in meters
Increasing energy

Our major source of
EM comes from our sun.
EM travels at the speed of light
3.0x10
8 m/s
Has both wave properties and particle properties
Photons are the particles possessed by EM

• Waves refract or bend when they pass from one medium to another with different densities.
• Diffraction is the bending of electromagnetic radiation as it passes around the edge of an object or through narrow openings.
• Interference is the interaction of waves that results in either reinforcing their amplitudes or canceling them out.

White Light
The shortest wave lengths bend longer ones, thus violet is the shortest
Davisson and Germer discovered the wave nature of an electron at Bell Labs in 1927 by observing electron diffraction.
R
O
Y
G
B
I
V

The red-orange light from hydrogen gas passes through a prism to form a line spectra. Each different colored light has its own unique energy.

• Atomic emission spectra consist of bright lines on a dark background.
• Atomic absorption spectra consist of characteristic series of dark lines produced when free gaseous atoms are illuminated by external sources of radiation.

• Max Planck proposed that light can have both wavelike and particle-like properties.
• A quantum is the smallest discrete quantity of a particular form of energy.
• Particles of radiant energy are known as quanta .
• Quantum theory is based on the idea that energy is absorbed and emitted in discrete quanta .

• Something that is quantized has values that are restricted to whole-number multiples of a specific base value.
• The energy of a quantum of radiation is:
 E = h
 where h is Planck’s constant
 h = 6.6260755 x 10 -34 J•s
 Or E = hc/


• Each packet of electromagnetic radiation energy is called a quantum .
• Einstein called the packets photons .

• The photoelectric effect is the release of electrons from a metal as a result of electromagnetic radiation.
• The photoelectric effect can be explained if electromagnetic radiation consists of tiny particles called photons.


• Johannes Rydberg revised Balmer’s equation to describe the complete hydrogen spectrum.
1

= (1.0097 x 10 -2 (nm) -1 )
 n
1
2
1
- n
1
2
2


N
1 is a whole number that remains fixed for a series of calculations in which n
2 number with values of n
1
+1, n
1 is also a whole
+2,… for successive line in the spectrum.

What is the wavelength of the line in the visible spectrum corresponding to n and n
2
= 4?
1
= 2

• The electron in a hydrogen atom occupies a discrete energy level and may exist only in the available energy levels.
• The electron may move between energy levels by either absorbing or emitting specific amounts of energy.
• Each energy level is designated by a specific value for n, called the principal quantum number.

• Neils Bohr derived the following formula for the possible energy differences (

E) be any pair of energy levels with values n
1 and n
2
.

E =
2
 2 me
4 h
2


1 n
1
2
-
1 n
2
2

 m and e is the mass and charge of the electron.


• An energy level is an allowed state that an electron can occupy in an atom.
• Movements of electrons between energy levels are called electron transitions.

• The lowest energy level available to an electron in an atom is its ground state .
• An excited state of an electron in an atom is any energy state above the ground state.

In terms of the Bohr model absorption and emission looks like this.

Electrons move between energy levels by absorbing and emitting energy in the form of light.
We call the lowest energy level the ground state . The higher energy level is called the excited state .

• The Bohr model applies only to one electron atoms.
• The Bohr model doesn’t account for the observed spectra of multielectron elements or ions.
• The movement of electrons in atoms is much less clearly defined than Bohr allowed.

• If electromagnetic radiation behaves as a particle, de Broglie reasoned, why couldn’t a particle in motion, such as an electron, behave as a wave?
• de Broglie’s Equation
 
= h/mu (m in kg and u in m/s)

• De Broglie reasoned that an electron in a hydrogen atom could behave as a circular wave oscillating around the nucleus.
• If electrons are moving around the nucleus in a continuous manor, the state of the electron must be described by a quantum number, n.

http://www.youtube.com/watch?v=P0Fi1Vcb pAI

• Quantum mechanics allows us to predict the probabilities of where we can find an electron.
• We cannot map out on the path an electron travels.
 The Heisenberg’s uncertainty principle says that you cannot determine the position and momentum of an electron at the same time.

• The description of the behavior of particles as waves is called wave mechanics or quantum mechanics.
• The mathematical description of an electron wave is called the wave equation.
• Wave functions,

, are mathematical descriptions of the motion of electron waves as they vary with location and with time.

• The principle quantum number, n , is a positive integer that indicates the shell and relative size of orbital(s).
• The angular momentum quantum number, l , is an integer from zero to n-1. It defines the shape of the orbital and subshell.
Value of l
Letter identifier
0 1 2 3 4 s p d f g

• The magnetic quantum number, m l
, is an integer with a value from l to + l . It defines the orientation of an orbital in the space around the nucleus of an atom.
• The spin magnetic quantum number, m s
, is to account for the two possible spin orientations.
The values for m s are +1/2 and -1/2.

It takes a total of 4 quantum numbers to identify an electron in a particular atom. Like it’s student ID no.
4p y
+1/2 spin QN; m s
=1/2 (clockwise or counterclockwise magnetic QN; m l
=0 (shape orientation) angular momentum QN; l =1 (volume shape) principal QN; n =4 (size and energy)

4 n
1
2
3
l
0
1
2
3
0
1
2
0
0
1 m l
0
0
-1,0,+1
0
-1,0,+1
-2.-1,0,+1,+2
0
-1,0,+1
-2.-1,0,+1,+2
-3,-2.-1,0,+1,+2,+3
# orbitals
1
1
3
5
1
3
1
3
5
7

What are the letter designations of all the subshells in the n = 5 energy level or shell? What is total number of orbitals in the n = 5 shell?

• Psi squared,

2 , defines the space, called an orbital, in atom where the probability of finding an electron is high.
• A radial distribution plot is a graphical representation of the probability of finding an electron in a thin spherical layer near the nucleus of an atom.

1s Orbital

• Pauli’s exclusion principle - no two electrons in an atom may have the same set of four quantum numbers.
 An orbital can only hold two electrons and they must have opposite spins.

Write the set of quantum numbers which describe each electron in the three 2p orbitals.

Which of the following combinations of quantum number are allowed?
1.
n = 1, l = 1, m l
= 0
2.
n = 3, l = 0, m l
= 0
3.
n = 1, l = 0, m l
= -1
4.
n = 2, l = 1, m l
= 2

Hydrogen Atom
3s 3p 3d
E 2s 2p
1s

• They do not follow the diagram for the hydrogen atom.
• As l changes the energy of the orbital changes
 The lower the value of l the lower in energy the subshell

Beyond the 3p subshell the orbitals don’t fill in an obvious way.
For example the 4s level lies lower in energy than the 3d .

• Orbitals that have the exact same energy level are degenerate.
• Core electrons are those in the filled, inner shells in an atom and are not involved in chemical reactions.
• Valence electrons are those in the outermost shell of an atom and have the most influence on the atom’s chemical behavior.

The way in which electrons are organized into shells , subshells and orbitals in an atom is called the electronic configuration .
The electronic configuration of an atom can be determined using the “Aufbau rule” also known as the “building up principle” .
Aufbau comes from the German meaning construction although it was the Danish physicist Neils Bohr who came up with the idea !!

The Aufbau Principle states that:
“The orbitals of lower energy are filled in first with the electrons and only then the orbitals of high energy are filled .”
What is the lowest energy orbital of an atom?
What is the third lowest energy orbital of an atom?

The Aufbau Principle states that:
“The orbitals of lower energy are filled in first with the electrons and only then the orbitals of high energy are filled .”
What is the lowest energy orbital of an atom?
1s orbital
What is the third lowest energy orbital of an atom?

The Aufbau Principle states that:
“The orbitals of lower energy are filled in first with the electrons and only then the orbitals of high energy are filled .”
What is the lowest energy orbital of an atom?
1s orbital
What is the third lowest energy orbital of an atom?
2p orbital

How would we use our rules to “build up” the electron configuration of a Li atom?
Li has Z = 3 so has 3 e .

How would we use our rules to “build up” the electron configuration of a Li atom?
Li has Z = 3 so has 3 e .
1s subshell

How would we use our rules to “build up” the electron configuration of a Li atom?
Li has Z = 3 so has 3 e .
1s subshell

How would we use our rules to “build up” the electron configuration of a Li atom?
Li has Z = 3 so has 3 e .
1s subshell

How would we use our rules to “build up” the electron configuration of a Li atom?
Li has Z = 3 so has 3 e .
2s subshell
1s subshell

How would we use our rules to “build up” the electron configuration of a Li atom?
Li has Z = 3 so has 3 e .
2s subshell
1s subshell

How would we use our rules to “build up” the electron configuration of a Li atom?
Li has Z = 3 so has 3 e .
2s subshell
1s subshell
We can write this in shorthand as 1s 2 2s 1

If there are multiple orbitals with the same energy how do we decide which orbital to put an electron?
We use Hund’s rule which states:
“Electrons will not join other electrons in an orbital if an unoccupied orbital of the same energy is available”

As we have seen previously for p, d and f subshells there are multiple orbitals with the same energy .
In particular:
• p subshells have three orbitals with the same energy
• d subshells have five orbitals with the same energy
• f subshells have seven orbitals with the same energy
Each of these orbitals may accommodate a maximum of two electrons .

Using Hund’s rule how would we put three electrons in a p subshell ?
p subshell
 p x p y p z p x p y p z p subshell


When we do put two electrons in one orbital then they obey the
Pauli exclusion principle .
“only electrons with opposite spin can occupy the same orbital ” p subshell
 p x p y p z p subshell
 p x p y p z

How would we use our rules to “build up” the electron configuration of a N atom?
N has Z = 7 so has 7 e .
1s subshell

How would we use our rules to “build up” the electron configuration of a N atom?
N has Z = 7 so has 7 e .
1s subshell
We can write this in shorthand as 1s 2 2s 2 2p 3

How would we use our rules to “build up” the electron configuration of a N atom?
N has Z = 7 so has 7 e .
1s subshell

How would we use our rules to “build up” the electron configuration of a N atom?
N has Z = 7 so has 7 e .
2s subshell
1s subshell

How would we use our rules to “build up” the electron configuration of a N atom?
N has Z = 7 so has 7 e .
2s subshell
1s subshell

How would we use our rules to “build up” the electron configuration of a N atom?
N has Z = 7 so has 7 e .
2p subshell
2s subshell
1s subshell

How would we use our rules to “build up” the electron configuration of a N atom?
N has Z = 7 so has 7 e .
2p subshell
2s subshell
1s subshell

How would we use our rules to “build up” the electron configuration of a N atom?
N has Z = 7 so has 7 e .
2p subshell
2s subshell
1s subshell

How would we use our rules to “build up” the electron configuration of a N atom?
N has Z = 7 so has 7 e .
2p subshell
2s subshell
1s subshell
We can write this in shorthand as 1s 2 2s 2 2p 3

1s 2s 2p
H: 1s 1
He: 1s 2
Li: 1s 2 2s 1
Be: 1s 2 2s 2
B: 1s 2 2s 2 2p 1

1s 2s 2p
C: 1s 2 2s 2 2p 2
 or
C: 1s 2 2s 2 2p 2

Hund’s Rule tells us which configuration is correct .

Electron Configuration 1s 2s 2p
C: 1s 2 2s 2 2p 2
N: 1s 2 2s 2 2p 3
O: 1s 2 2s 2 2p 4
F: 1s 2 2s 2 2p 5
Ne: 1s 2 2s 2 2p 6

Fourth Period Elements
K 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1
Ca 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 or [Ar]4s 1 or [Ar]4s 2
Sc 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 1 or [Ar]4s 2 3d 1
Ti 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 or [Ar]4s 2 3d 2
V 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 3 or [Ar]4s 2 3d 3
Cr 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 or [Ar]4s 1 3d 5
Mn 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 5 or [Ar]4s 2 3d 5
•
•
•
Cu 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 or [Ar]4s 1 3d 10

• Chromium and Copper do not follow the pattern of the other elements.
 You should remember these two families, because other elements in these families exhibit the same types of configurations
• You can use the Periodic Table to guide you in writing electron configurations.

There is an easy way to remember the sequence of the energies of the subshells .

Write the electron configuration for Pt.
Write the electron configuration for I .
Write the electron configuration for Rh 2+ .

• Start with the configuration for the neutral atom, then add or remove electrons from the valence shells to make the desired ion.
• Atoms or ions that are isoelectronic with each other have identical numbers and configurations of electrons.

Orbital penetration occurs when an electron in an outer orbital has some probability of being close to the nucleus
• Penetration ability follows this order: s > p > d > f.

Effective nuclear charge (Z eff
) is the attractive force toward the nucleus experienced by an electron in an atom, usually the outer ( valence electrons) .
• Electrons in between the nucleus and the designated electron shield the designated electron from the nucleus.
• Electrons between the nucleus and the designated electron also repelled the designated electron
• Shielded electrons are further away from the nucleus than they would be if not shielded

The quantity of energy required to remove 1 mole of electrons from 1 mole of the gaseous atom or ion.
X(g) ---> X + (g) + e (g)

• First ionization energy: increases from left to right across a period ; decreases going down a group .

Successive Ionization Energies (kJ/mol)
Elements IE
1
H
Li
IE
2
1312
IE
3
IE
4
He 2372 5249
520 7296 12040
IE
5
IE
6
IE
1 is the first ionization energy, which is the energy to remove a valence electron from an atom to produce a cation.
Be 897 1758 15050 21070
B
C
801 2426 3660 24682 32508
1087 2348 4617 6201 37926 46956
IE
2 is the second ionization energy and is the energy required to
N
O
1402 2860 4581 7465 9391 52976
1314 3383 5298 7465 10956 13304 remove the second electron
Note: IE
2
>IE from a cation.
1
, since more energy is required to remove an electron from a positive ion compared to a neutral atom.

Elements IE
1
IE
2
IE
3
IE
4
IE
5
IE
6
H 1312
He 2372 5249
Li
Be
B
C
N
O
520 7296 12040
897 1758 15050 21070
801 2426 3660 24682 32508
1087 2348 4617 6201 37926 46956
1402 2860 4581 7465 9391 52976
1314 3383 5298 7465 10956 13304
Note: The large jump in ionization energy to the right of the red line is due to destroying the noble gas configuration, which we know to be very stable. Also notice that ionization energy increases as we remove an electron from a more positive cations, going left to right

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This ChemTour explores the relationship of frequency, wavelength, and energy using animations, interactive graphs, and equations. The quantitative exercises include graph reading and calculations using Planck’s constant and the speed of light.

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This animation recreates Thomas Young’s double-slit experiment and demonstrates how constructive and destructive interference occur.

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A boat moving with or against the direction of wave movement demonstrates the motion-induced shifts in wavelengths and frequency that are examples of the
Doppler effect.

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This ChemTour examines the emission and absorption spectra for sodium and hydrogen and relates them to energy level transitions.

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This ChemTour explores the idea that energies of electrons surrounding atomic nuclei are quantized. In Practice
Exercises, students learn to calculate the energies of specific states of hydrogen, and the energies involved in electronic transitions.

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In this ChemTour, students learn to apply the de Broglie equation to calculate the wavelength of moving objects ranging from baseballs to electrons. Includes Practice
Exercises.

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In this ChemTour, students explore the rules for designating quantum numbers. Includes Practice Exercises.

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This ChemTour explains how electrons are distributed within atomic orbitals. Students learn how to determine an element’s electron configuration and learn how to complete an orbital box diagram. Includes practice exercises.

Suppose two photons combine in a crystal to form a single photon of green light or "green photon."
Which of the following could be the colors of the two combining photons?
A) Green & green B) Blue & yellow C) Infrared & infrared
Combining Two Photons

Please consider the following arguments for each answer and vote again:
A. A green photon can only be produced by the combination of two other green photons of the same wavelength.
B. The color green is the result of combining the colors blue and yellow, just as a green photon will result from the combination of blue and yellow photons.
C. Only two infrared photons have the proper total energy needed to form a green photon.
Combining Two Photons

An electron in the ground state absorbs a single photon of light and then relaxes back to the ground state by emitting an infrared photon (1200 nm) followed by an orange photon (600 nm).
What is the wavelength of the absorbed photon?
A) 400 nm B) 600 nm
Absorption and Fluorescence of Light
C) 1800 nm

Please consider the following arguments for each answer and vote again:
A. The wavelength is inversely proportional to the energy, so for energy to be conserved the absorbed photon must have a wavelength of 400 nm.
B. The wavelength of the absorbed photon is the difference of the wavelength of the two emitted photons, which is 600 nm.
C. For the energy to be conserved, the sum of the wavelengths must be conserved. So the wavelength of the absorbed photon is 1800 nm.
Absorption and Fluorescence of Light

The diagram to the left depicts the interference pattern that results from the constructive and destructive interference of light waves that are diffracted as they pass through two slits. If the pattern is the result of green light passing through two slits, which of the following patterns would be the result of blue light passing through the same two slits?
A) B) C)
Two-Slit Diffraction and Interferometry

Please consider the following arguments for each answer and vote again:
A. The wavelength of blue light is shorter than that of green light, so constructive and destructive interference occurs at smaller intervals.
B. The interference pattern is dependent only on the width of and distance between the two slits. Therefore, the interference pattern should not change.
C. Blue light is higher in energy than green light and therefore would be less affected by the two slits.
Two-Slit Diffraction and Interferometry

When a photon of red light hits metal X, an electron is ejected. Will an electron be ejected if a photon of yellow light hits metal X?
A) Yes B) No C) Can't tell
Photoelectric Effect: Red and Yellow Light

Please consider the following arguments for each answer and vote again:
A. Photons of yellow light possess more energy than photons of red light, so a yellow photon also must eject an electron.
B. Each metal has a specific wavelength of light that will cause electrons to be ejected. If red light has the correct wavelength, yellow cannot.
C. Whether a yellow photon will eject an electron from the metal will depend on how tightly the electron is bound to the metal.
Photoelectric effect: Red and Yellow Light

When a photon of blue light hits metal X, an electron is ejected. Will an electron be ejected if a photon of green light hits metal X?
A) Yes B) No C) Can't tell
Photoelectric Effect: Blue and Green Light

Please consider the following arguments for each answer and vote again:
A. So long as enough photons of light hit the metal, an electron will always be ejected, regardless of the wavelength of the light.
B. The energy of a blue photon is higher than the energy of a green photon so an electron removed with blue light will not be removed with green light.
C. Whether a green photon will eject an electron from the metal will depend on how tightly the electron is bound to the metal.
Photoelectric effect: Blue and Green Light

A 300-nm photon can eject an electron from a metal surface with a certain kinetic energy.
What photon wavelength would be required to eject an electron from the same metal surface with twice the kinetic energy?
A) 150 nm B) 200 nm C) 600 nm
Photoelectric Effect: Kinetic Energies of Electron

Please consider the following arguments for each answer and vote again:
A. To eject an electron with twice the kinetic energy, twice the energy must be provided by the photon, so the photon wavelength must be halved.
B. A photon with a wavelength of 200 nm will overcome the work function and provide twice the kinetic energy.
C. To double the kinetic energy of the ejected electron, the wavelength of the impacting photon also must be doubled.
Photoelectric Effect: Kinetic Energies of Electrons

Suppose a hydrogen molecule (1H
2
) is traveling at 800 m/s and a deuterium molecule (2H
2
) is traveling at 400 m/s.
What can be said of the de Broglie wavelengths of the two molecules?
A) λH > lD B) λH < lD C) λH = lD
De Broglie Wavelengths of H
2
0 Molecules

Please consider the following arguments for each answer and vote again:
A. The kinetic energy of the deuterium molecule is twice that of the hydrogen molecule.
Therefore, the deuterium molecule will have a shorter de Broglie wavelength.
B. Because the speed of the hydrogen molecule is greater than the speed of the deuterium molecule, the de
Broglie wavelength of the hydrogen molecule will be shorter.
C. The hydrogen molecule and the deuterium molecule have the same momentum and therefore will have the same de Broglie wavelength.
De Broglie Wavelengths of H
2
O Molecules

One method for decreasing the temperature of atoms, known as laser cooling, involves bombarding an atom with photons of light, decreasing its overall momentum and thus its kinetic energy (just like one could slow a fastmoving car by colliding it with another car).
A sodium atom at a temperature of 60 K has a de Broglie wavelength of 66 pm (6.6x10-11 m). Approximately how many photons of red light (at λ = 660 nm) would it take to stop a sodium atom at 60 K?
A) ~1
Laser Cooling of Sodium Atoms
B) ~102 C) ~104

Please consider the following arguments for each answer and vote again:
A. A photon travels ~105 times faster than a sodium atom.
Therefore, only one photon is required.
B. The kinetic energy of a sodium atom is ~100 times less than the kinetic energy of a red photon.
C. The de Broglie wavelength of a sodium atom at 60 K is
~104 times shorter than the wavelength of a red photon, so it will take 104 photons to stop a single sodium atom.
Laser Cooling of Sodium Atoms

What color will a yellow object appear when it is seen through a filter with the absorption spectrum shown to the left?
A) Yellow B) Blue C) Black
Transmission of Light through a Color Filter

Please consider the following arguments for each answer and vote again:
A. The filter absorbs no yellow light, so the object will appear yellow.
B. Blue light is absorbed by the filter, so an object seen through the filter will appear blue.
C. No yellow light is absorbed by the filter, so the object will appear black.
Transmission of Light through a Color Filter

A)
Photon emission from a system possessing the energy level diagram to the left would produce which of the following spectra ?
B) C)
Emission Spectra

Consider the following arguments for each answer and vote again:
A. The photon wavelength depends only on the energy of the lowest state, so only 1 wavelength is possible.
B. There are 2 possible transitions—one from each of the
2 upper levels.
Thus, 2 wavelengths of light are emitted.
C. The 3 energy levels lead to 2 high-energy transitions and 1 low-energy transition.
Therefore, 3 different photon wavelengths are possible.
Emission Spectra

A)
Emission from which of the following energy level diagrams would produce the spectrum shown to the left?
B) C)
Energy Levels

Consider the following arguments for each answer and vote again:
A. The arrangement of the energy levels reflects the arrangement of the lines in the emission spectrum.
B. This energy level diagram allows only 1 low-energy transition, consistent with the emission spectrum.
C. Only this energy level diagram allows 3 high-energy transitions and 1 low-energy transition.
Energy Levels

The diagram to the left shows the spacing of the first five energy levels for a hydrogen atom.
Which of the following transitions in He+ has the same wavelength as the 4→2 transition in H?
A) 4→2 B) 8→4 C) 16→8
Transition in H and He +

Consider the following arguments for each answer and vote again:
A. He+ has the same electron configuration as H; therefore, the energy level diagram will be the same.
B. The atomic number of He+ is twice that of H.
Therefore, to produce the same energy splitting, the energy levels must be twice that of H.
C. The energy of the electron is proportional to Z2, which is 4 for He+. Therefore, the two levels, 4 and 2, must be increased by a factor of 4 to 16 and 8, respectively.
Transition in H and He +

Periodic Table
Which atom or ion can have the electron configuration 1s22s22p1?
A) Li B) BeC) B+
Electron Configurations

Consider the following arguments for each answer and vote again:
A. The answer must be lithium because it is the first element in row 2 to possess only one unpaired electron.
B. Beryllium in its ground state has the electron configuration 1s22s2, so Be- in its ground state will have the configuration 1s22s22p1.
C. In its ground state, boron has the electron configuration 1s22s22p1, so B+ must also have this configuration.
Electron Configurations

Which of the following has the lowest ionization energy?
A) H(1s1) B) He(1s13p1) C) He+(4p1)
Ionization Energies

Consider the following arguments for each answer and vote again:
A. Hydrogen has a lower nuclear charge than helium, so it always has a lower ionization energy than any helium atom or ion.
B. He(1s13p1) has almost the same ionization energy as
H(3p1), which has a lower ionization energy than either H(1s1) or He+(4p1).
C. Because the electron in He+(4p1) is in the fourth shell, the ionization energy of He+(4p1) is the lowest.
Ionization Energies

How does the ionization energy of
He(1s2) compare to the ionization energies of H(1s1) and He+(1s1)?
A) Higher B) Lower C) In-between
Ionization Energies of He(1s 2 )

Consider the following arguments for each answer and vote again:
A. It is harder to remove an electron from a doubly occupied orbital than from a singly occupied orbital.
B. Each electron offsets the charge of one of the protons, giving an effective nuclear charge of zero.
C. Each electron partially shields the other, leading to an effective nuclear charge that is between 1 and 2.
Ionization Energies of He(1s 2 )

Which of the following atoms
(or ions) has the smallest radius?
A) K+ B) Ar C) Cl-
Atomic and Ionic Radii

Consider the following arguments for each answer and vote again:
A. K+ has the highest nuclear charge and so has the smallest atomic radius.
B. Because it is a noble gas, Ar has the smallest atomic radius.
C. Cl- has the nucleus with the lowest mass, so it has the smallest atomic radius.
Atomic and Ionic Radii

Suppose an electron is transferred from a potassium atom to an unknown halogen atom.
For which of the following halogen atoms would this process require the least amount of energy?
A) Cl B) Br C) I
Electron Affinity of Halogen Atoms

Consider the following arguments for each answer and vote again:
A. Chlorine has the greatest affinity for electrons and so would release the most energy when an electron is added.
B. Electron donation is most favorable energetically when it occurs between atoms on the same row of the periodic table.
C. Because of its massive nuclear charge and large electron cloud, an iodine atom can most easily accept an additional electron.
Electron Affinity of Halogen Atoms