Mr. Braswell's Chemistry Help

Learning Zone Presents:


Modern science identifies this "substance" through its physical properties; the most common current definition of matter is anything that has mass and occupies volume.

However, this definition has to be revised in light of quantum mechanics, where the concept of "having mass", and "occupying space" are not as well-defined as in everyday life.

A more general view is that bodies are made of several substances, and the properties of matter (including mass and volume) are determined not only by the substances themselves, but by how they interact.

In other words, matter is made up of interacting "building blocks", the so-called particulate theory of matter.

Matter is commonly said to exist in four states (or phases): solid, liquid, gas and plasma.

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                       The  Atom

Subatomic Particlesfree web site graphic


protons (p+)  

1.     positively charged subatomic particle located in nucleus of an atom

2.     mass = 1.673 x 10-24 g   or 1.0073 u

3.     always moving (atomic vibration)

free web site graphic




neutrons (n)

1.     subatomic particle located in nucleus that has no electrical charge

2.     mass = 1.675 x 10-24 g   or   1.0087 u

3.     a free neutron will decay into proton, electron, and electron anti-neutrino (ADVANCED)

4.     half-life of free neutron is about 890 seconds (ADVANCED)

5.     The amount of time that a neutron survives before it decays plays a role in determining which elements were created during the Big Bang.  It has a strong influence on the amount of Helium-4 in the universe.

free web site graphic




electron (e-)

1.     very small negatively charged subatomic particle found in orbit around nucleus

2.     mass = 1.1 x 10-29 g     or    0.000 549 u

3.     mass is 1/1836 that of proton

4.     moves very rapidly and is rather easily accelerated

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             History of atomic development

Year Milestone Scientist(s)
~400 BC The first coherent atomic theory Democritus
1804 First "modern" atomic theory J. Dalton
1869 First periodic table D. Mendeleev
1896-9 Radioactivity discovered and identified H. Becquerel, M. Curie, E. Rutherford
1897 Electron discovered J.J. Thomson
1909-11 Identification of an atomic nucleus E. Rutherford, H. Geiger
1913 Mass of electron determined R. A. Millikan
1913 Positive charge in the nucleus and naturally occuring isotopes discovered J.J. Thomson
1913 Atomic number determined, periodic table reorganized H. Moseley
1919 Proton existance confirmed, neutron proposed E. Rutherford
1931 Neutron Identified J. Chadwick
---- Summary and Symbology ---------

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                      PERIODIC TABLE


A group or family is a vertical column in the periodic table. Groups are considered the most important method of classifying the elements.

In some groups, the elements have very similar properties and exhibit a clear trend in properties down the group.

These groups tend to be given trivial (unsystematic) names, e.g., the alkali metals, alkaline earth metals, halogens, pnictogens, chalcogens, and noble gases.

Some other groups in the periodic table display fewer similarities and/or vertical trends (for example Group 14), and these have no trivial names and are referred to simply by their group numbers.


A period is a horizontal row in the periodic table.

Although groups are the most common way of classifying elements, there are some regions of the periodic table where the horizontal trends and similarities in properties are more significant than vertical group trends.

This can be true in the d-block (or "transition metals"), and especially for the f-block, where the lanthanides and actinides form two substantial horizontal series of elements.








Because of the importance of the outermost shell, the different regions of the periodic table are sometimes referred to as periodic table blocks, named according to the subshell in which the "last" electron resides.

 The s-block comprises the first two groups (alkali metals and alkaline earth metals) as well as hydrogen and helium.

The p-block comprises the last six groups (groups 13 through 18) and contains, among others, all of the semimetals. The d-block comprises groups 3 through 12 and contains all of the transition metals.

The f-block, usually offset below the rest of the periodic table, comprises the rare earth metals.


The chemical elements are also grouped together in other ways. Some of these groupings are often illustrated on the periodic table, such as transition metals, poor metals, and metalloids.

Other informal groupings exist, such as the platinum group and the noble metals.

Periodicity of chemical properties

The main value of the periodic table is the ability to predict the chemical properties of an element based on its location on the table.

It should be noted that the properties vary differently when moving vertically along the columns of the table than when moving horizontally along the rows.

Trends of groups

Modern quantum mechanical theories of atomic structure explain group trends by proposing that elements within the same group have the same electron configurations in their valence shell, which is the most important factor in accounting for their similar properties.

Elements in the same group also show patterns in their atomic radius, ionization energy, and electronegativity. From top to bottom in a group, the atomic radii of the elements increase.

Since there are more filled energy levels, valence electrons are found farther from the nucleus.

From the top, each successive element has a lower ionization energy because it is easier to remove an electron since the atoms are less tightly bound.

Similarly, a group will also see a top to bottom decrease in electronegativity due to an increasing distance between valence electrons and the nucleus.

Trends of periods

Periodic trend for ionization energy. Each period begins at a minimum for the alkali metals, and ends at a maximum for the noble gases.

Elements in the same period show trends in atomic radius, ionization energy, electron affinity, and electronegativity.

Moving left to right across a period, atomic radius usually decreases.

This occurs because each successive element has an added proton and electron which causes the electron to be drawn closer to the nucleus.

This decrease in atomic radius also causes the ionization energy to increase when moving from left to right across a period.

The more tightly bound an element is, the more energy is required to remove an electron.

Similarly, electronegativity will increase in the same manner as ionization energy because of the amount of pull that is exerted on the electrons by the nucleus.

Electron affinity also shows a slight trend across a period.

Metals (left side of a period) generally have a lower electron affinity than nonmetals (right side of a period) with the exception of the noble gases.


Although Lavoisier grouped the elements into gases, metals, non-metals, and earths, chemists spent the following century searching for a more precise classification scheme.

In 1829, Johann Wolfgang Döbereiner observed that many of the elements could be grouped into triads (groups of three) based on their chemical properties. Lithium, sodium, and potassium, for example, were grouped together as being soft, reactive metals.

Döbereiner also observed that, when arranged by atomic weight, the second member of each triad was roughly the average of the first and the third.[4]

This became known as the Law of triads. 

German chemist Leopold Gmelin worked with this system, and by 1843 he had identified ten triads, three groups of four, and one group of five.

Jean Baptiste Dumas published work in 1857 describing relationships between various groups of metals.

Although various chemists were able to identify relationships between small groups of elements, they had yet to build one scheme that encompassed them all.[4]

German chemist August Kekulé had observed in 1858 that carbon has a tendency to bond with other elements in a ratio of one to four.

Methane, for example, has one carbon atom and four hydrogen atoms. This concept eventually became known as valency.

In 1864, fellow German chemist Julius Lothar Meyer published a table of the 49 known elements arranged by valency.

The table revealed that elements with similar properties often shared the same valency.

English chemist John Newlands published a series of papers in 1864 and 1865 that described his attempt at classifying the elements: When listed in order of increasing atomic weight, similar physical and chemical properties recurred at intervals of eight, which he likened to the octaves of music. 

This law of octaves, however, was ridiculed by his contemporaries.

Russian chemistry professor Dmitri Ivanovich Mendeleev and Julius Lothar Meyer independently published their periodic tables in 1869 and 1870, respectively.

They both constructed their tables in a similar manner: by listing the elements in a row or column in order of atomic weight and starting a new row or column when the characteristics of the elements began to repeat.

The success of Mendeleev's table came from two decisions he made: The first was to leave gaps in the table when it seemed that the corresponding element had not yet been discovered.

Mendeleev was not the first chemist to do so, but he went a step further by using the trends in his periodic table to predict the properties of those missing elements, such as gallium and germanium.

The second decision was to occasionally ignore the order suggested by the atomic weights and switch adjacent elements, such as cobalt and nickel, to better classify them into chemical families.

With the development of theories of atomic structure, it became apparent that Mendeleev had inadvertently listed the elements in order of increasing atomic number.

With the development of modern quantum mechanical theories of electron configurations within atoms, it became apparent that each row (or period) in the table corresponded to the filling of a quantum shell of electrons.

In Mendeleev's original table, each period was the same length. However, because larger atoms have more electron sub-shells, modern tables have progressively longer periods further down the table.

In the years that followed after Mendeleev published his periodic table, the gaps he left were filled as chemists discovered more chemical elements.

The last naturally occurring element to be discovered was francium (referred to by Mendeleev as eka-caesium) in 1939

The periodic table has also grown with the addition of synthetic and transuranic elements.

The first transuranic element to be discovered was neptunium, which was formed by bombarding uranium with neutrons in a cyclotron in 1939.

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                       Nature of Light

Speed of light calculations video 

c = λν 

Each symbol in the equation is discussed below.

1) λ is the Greek letter lambda and it stands for the wavelength of light.

Wavelength is defined as the distance between two successive crests of a wave.

When studying light, the most common units used for wavelength are: meter, centimeter, nanometer, and Ångström.

Even though the offical unit used by SI is the meter, you will see explanations and problems which use the other three.

Less often, you will see other units used; picometer is the most common one among the less-often used wavelength units.

Ångström is a non-SI unit commonly included in discussions of SI units because of its wide usage.

Keep in mind these definitions:

one centimeter equals 10¯2 meter
one nanometer equals 10¯9 meter
one Ångström equals 10¯8 centimeter

The symbol for the Ångström is Å.

Most certainly, you will need to move easily from one unit to the other. For example, notice that1 Å = 10¯10 meter. This means that 10 Å = 1 nm. So, if you are given an Ångström value for wavelength and a nanometer value is required, divide the Ångström value by 10. If you can't make easy transitions between various metric units, you'd better go back to study and practice that area some more.

2) ν is the Greek letter nu.

It is NOT the letter v, it is the Greek letter nu!!! It stands for the frequency of the light wave.

 Frequency is defined as the number of wave cycles passing a fixed reference point in one second.

When studying light, the unit for frequency is called the Hertz (its symbol is Hz).

One Hertz is when one complete cycle passes the fixed point, so a million Hz is when the millionth cycle passes the fixed point.

There is an important point to make about the unit on Hz.

It is NOT commonly written as cycles per second (or cycles/sec), but only as sec¯1 (more correctly, it should be written as s¯1; you need to know both ways).

The "cycles" part is deleted, although you may see an occasional problem which uses it.

A brief mention of cycle: imagine a wave, frozen in time and space, where a wave crest is exactly lined up with our fixed reference point.

Now, allow the wave to move until the following crest is exactly lined up with the reference point, then freeze the wave in place.

This is one cycle of the wave and if all that took place in one second, then the frequencey of the wave is 1 Hz.

In any event, the only scientifically useful part of the unit is the denominator and so "per second" (remember, usually as s¯1) is what is used.

The numerator "cycles" is not needed and so its presence is simply understood and, if writing a fraction is necessary, a one would be used, as in 1/sec.

3) c is the symbol for the speed of light, the speed at which all electromagnetic radiation moves when in a perfect vacuum.

(Light travels slower when passing through objects such as water, but it never travels faster than when in a perfect vacuum.)

Both ways shown below are used to write the value. You need to be aware of both:

3.00 x 108 m/s
3.00 x 1010 cm/s

The actual value is just slightly less, but the above values are the one generally used in introductory classes.

(sometimes you'll see 2.9979 rather than 3.00.)

Be careful about using the exponent and unit combination.

Meters are longer than centimeters, so there are less of them used above.

Since there are two variables (λ and ν), we can have two types of calculations: (a) given wavelength, calcuate frequency and (b) given frequency, calculate wavelength.

By the way, these are the two types of problems teachers generally ask on the test.

Note that we can easily rearrange the main equation to fit these two types of problems:

(a) given the wavelength, calculate the frequency; use this equation: ν = c / λ

(b) given the frequency, calculate the wavelength; use this equation: λ = c / ν


Example problem #1: What is the frequency of red light having a wavelength of 7000 Å?

The solution below depends on converting Å into cm. This means you must remember that the conversion is 1 Å = 10¯8 cm. The solution:

(7000 x 10¯8 cm) (x) = 3.00 x 1010 cm/sec

Notice how I did not bother to convert 7000 x 10¯8 into scientific notation. If I had done so, the value would have been 7.000 x 10¯5.

Note also that I effectively consider 7000 to be 4 significant figures. I choose to do this because I know wavelength measurements are very accurate and that 6, 7, or even 8 sig figs are possible. At an introductory level, you will not know this, so that is why I am telling you here. Also, the value for the speed of light is known to nine significant figures, as in 299,792,458 m s¯1. However, 3.00 is good enough for introductory work.

The answer is 4.29 x 10141

Example problem #2: What is the frequency of violet light having a wavelength of 4000 Å?

The solution below depends on converting Å into cm. This means you must remember that the conversion is 1 Å = 10¯8 cm. The solution:

(4000 x 10¯8 cm) (x) = 3.00 x 1010 cm/sec

Notice how I did not bother to convert 4000 x 10¯8 into scientific notation. If I had done so, the value would have been 4.000 x 10¯5. Note also that I effectively consider 4000 to be 4 significant figures.

The correct answer is 7.50 x 10141

Be aware that the range of 4000 to 7000 Å is taken to be the range of visible light. Notice how the frequencies stay within more-or-less the middle area of 1014, ranging from 4.29 to 7.50, but always being 1014. If you are faced with this calculation and you know the wavelength is a visible one (say 555 nm), then you know the exponent on the frequency MUST be 1014. If it isn't, then YOU (not the teacher) have made a mistake.

Example problem #3: What is the frequency of EMR having a wavelength of 555 nm? (EMR is an abbreviation for electromagnetic radiation.)

First, let us convert nm into meters. Since one meter contains 109 nm, we have the following conversion:

555 nm x (1 m / 109 nm)

555 x 10¯9 m = 5.55 x 10¯7 m

Inserting into λν = c, gives:

(5.55 x 10¯7 m) (x) = 3.00 x 108 m s¯1

x = 5.40 x 10141

Example problem #4: What is the wavelength (in nm) of EMR with a frequency of 4.95 x 10141?

Substitute into λν = c, as follows:

(x) (4.95 x 10141) = 3.00 x 108 m s¯1

x = 6.06 x 10¯7 m

Now, we convert meters to nanometers:

6.06 x 10¯7 m x (109 nm / 1 m) = 606 nm

Example problem #5: What is the wavelength (in both cm and Å) of light with a frequency of 6.75 x 1014 Hz?

The fact that cm is asked for in the problem allows us to use the cm/s value for the speed of light:

(x) (6.75 x 10141) = 3.00 x 1010 cm s¯1

x = 4.44 x 10¯5 cm

Next, we convert to Å:

(4.44 x 10¯5 cm) x (1 Å / 10¯8 cm) = 4440 Å

Worksheet - Calculate frequency when given wavelength

Worksheet - Calculate wavelength when given frequency

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The hydrogen spectrum shows 4 bright lines: red, yellow (sometime rather faint), turquoise and purple.

Calculating the energy of a photon requires the light frequency (ν) and Planck’s constant (h).

h = 6.63 x 10-34 Js

Ephoton = hν

Problem: The spectroscopes give us data in terms of wavelength (l) in nanometers.

So, we need to use math to calculate the frequency from the wavelength.

c = ln, where c = the speed of light in a vacuum = 3.00 x 108 m/s.

Hey, what’s nu?  ν = c/λ

Problem: The spectroscope gives us wavelength in nm, but speed of light is in m.

Step 1: convert nm wavelengths to m.

Step 2: use ν = c/λ to convert wavelength to frequency.

Step 3: use Planck’s equation to calculate photon energy.

Example: Let’s say you observed a purple line at 4.5 in the spectroscope.  4.5 x 100 = 450 nm.

Step 1: 450 nm x (1m/ 1 x 109 nm)    = 4.5 x 10-7 m

Step 2:  ν = 3.00 x 108 m/s /4.5 x 10-7 m    =   6.7 x 1014/s  (or 6.7 x 1014 s-1)             

Step 3: Ephoton = hν = 6.63 x 10-34 Js (6.7 x 1014/s) = 4.4 x 10-19 J.

So, the energy of a single purple photon of 450 nm wavelength is

4.4 x 10-19 Joules.

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          Electronic Structure of Atoms

Bohr's model of the hydrogen atom

In 1913 Niels Bohr developed a theoretical explanation for a phenomenon known as line spectra.

Bohr's Model of the Hydrogen Atom

Line Spectra

Lasers emit radiation which is composed of a single wavelength. However, most common sources of emitted radiation (i.e. the sun, a lightbulb) produce radiation containing many different wavelengths.

When the different wavelengths of radiation are separated from such a source a spectrum is produced.

A rainbow represents the spectrum of wavelengths of light contained in the light emitted by the sun

  • Sun light passing through a prism (or raindrops) is separated into its component wavelengths
  • Sunlight is made up of a continuous spectrum of wavelengths (from red to violet) - there are no gaps

Not all radiation sources emit a continuous spectrum of wavelengths of light

  • When high voltage is applied to a glass tube containing various gasses under low pressure different colored light is emitted
    • Neon gas produces a red-orange glow
    • Sodium gas produces a yellow glow
  • When such light is passed through a prism only a few wavelengths are present in the resulting spectra
    • These appear as lines separated by dark areas, and thus are called line spectra

When the spectrum emitted by hydrogen gas was passed through a prism and separated into its constituent wavelengths four lines appeared at characteristic wavelengths

Balmer's Formula              

What was the formula that Balmer found?
Balmer examined the four visible lines in the spectrum of the hydrogen atom; their wavelengths are 410 nm, 434 nm, 486 nm, and 656 nm. He played around with these numbers and eventually figured out that all four wavelengths (symbolized by the Greek letter lambda) fit into the equation

R is the Rydberg constant, whose value is

The number n is just an integer; the above formula gives the longest wavelength, 656 nm, when n=3, and gives each of the shorter wavelengths as n increases up to 6.

656 nm is the red line in that picture, right?    Exactly; the shorter wavelengths correspond to the blue and violet lines you can see. (The 410 nm line is very faint, but it's there if you look closely.) This set of lines is called the Balmer series. Later, other researchers found that the series could be extended into ultraviolet wavelengths; the same formula still worked, with larger values of n.
Hmm...from Balmer's equation, it looks like when n gets bigger, the lines should start getting really close together.                                                                                                                      
That's exactly right; as n gets larger, 1 over n squared gets smaller, so there's less and less difference between the consecutive lines. You can see that the series has a limit-- that is, as n gets larger and larger, the wavelength gets closer and closer to one particular value. If n is infinity, then 1 over n squared is 0, and if you work out the numbers, you'll find that the wavelength is about 365 nm. That's just what experimentalists saw; around 365 nm, the lines became too close together to distinguis
h.  So do all of the emission lines from a hydrogen atom fit somewhere into the Balmer serie
s?  No, they don't. As scientists looked further into the non-visible parts of the spectrum, they found other series--which obeyed formulas hauntingly similar to Balmer's. For example, the Lyman series, which is entirely in the ultraviolet, fits the equation


 Those integers must have something to do with Bohr's energy levels. 
 Yes, they do; I'll show you how Bohr was able to deduce the electron's angular momentum from this formula..

Bohr's Model

  • Bohr began with the assumption that electrons were orbiting the nucleus, much like the earth orbits the sun.
  • From classical physics, a charge traveling in a circular path should lose energy by emitting electromagnetic radiation
  • If the "orbiting" electron loses energy, it should end up spiraling into the nucleus (which it does not). Therefore, classical physical laws either don't apply or are inadequate to explain the inner workings of the atom
  • Bohr borrowed the idea of quantized energy from Planck
    • He proposed that only orbits of certain radii, corresponding to defined energies, are "permitted"
    • An electron orbiting in one of these "allowed" orbits:
      • Has a defined energy state
      • Will not radiate energy
      • Will not spiral into the nucleus

If the orbits of the electron are restricted, the energies that the electron can possess are likewise restricted and are defined by the equation:

Where RH is a constant called the Rydberg constant and has the value

2.18 x 10-18 J

'n' is an integer, called the principle quantum number and corresponds to the different allowed orbits for the electron. Thus, an electron in the first allowed orbit (closest to the nucleus) has n=1, an electron in the next allowed orbit further from the nuclei has n=2, and so on.

Thus, the relative energies of these allowed orbits for the electrons can be diagrammed as follows:

All the relative energies are negative

  • The lower the energy, the more stable the atom
  • The lowest energy state (n=1) is called the ground state of the atom
  • When an electron is in a higher (less negative) energy orbit (i.e. n=2 or higher) the atom is said to be in an excited state
  • As n becomes larger, we reach a point at which the electron is completely separated from the nucleus
    • E = (-2.18 x 10-18 J)(1/infinity) = 0
    • Thus, the state in which the electron is separated from the nucleus is the reference or zero energy state (actually higher in energy than other states)

Bohr also assumed that the electron can change from one allowed orbit to another

  • Energy must be absorbed for an electron to move to a higher state (one with a higher n value)
  • Energy is emitted when the electron moves to an orbit of lower energy (one with a lower n value)
  • The overall change in energy associated with "orbit jumping" is the difference in energy levels between the ending (final) and initial orbits:
DE = Ef - Ei

Substituting in for the previously defined energy equation:

When an electron "falls" from a higher orbit to a lower one the energy difference is a defined amount and results in emitted electromagnetic radiation of a defined energy (DE)

  • Planck had deduced that the energy of the photons comprising EM radiation is a function of its frequency (E = h)
  • Therefore, if the emitted radiation from a falling electron had a defined energy, then it must have a correspondingly defined frequency
  • DE is positive when nf is greater than ni, this occurs when energy is absorbed and an electron moves up to a higher energy level (i.e. orbit).
  • When DE is negative, radiant energy is emitted and an electron has fallen down to a lower energy state

Revisiting Balmer's equation:

In 1885 a Swiss school teacher figured out that the frequencies of the light corresponding to these wavelengths fit a relatively simple mathematical formula:

where C = 3.29 x 1015 s-1 (not the 'c' used for the speed of light)

Since energy lost by the electrons is energy "gained" by the emitted EM energy, the EM energy from Bohr's equation would be:

Thus, Balmer's constant 'C' = (RH/h) (Rydberg constant divided by Planck's constant), and nf = 2. Thus, the only emitted energies which fall in the visible spectrum are from those electrons which fell down to the second quantum orbital. Those which fell down to the first orbital have a higher energy (frequency) than can be seen in the visible spectrum.

Calculate the wavelength of light that corresponds to the transition of the electron from the n=4 to the n=2 state of the hydrogen atom. Is the light absorbed or emitted by the atom?

Since the electron is "falling" from level 4 down to level 2, energy will be given up and manifested as emitted electromagnetic radiation:

DE = (2.18 x 10-18 J)((1/16)-(1/4)) = -4.09 x 10-19 J (light is emitted)
4.09 x 10-19 J = (6.63 x 10-34 Js) * (n)
6.17 x 1014 s-1 = n
l = (3.00 x 108 m s-1)/ (6.17 x 1014 s-1) = 4.87 x 10-7m = 487 nm

Bohr's model of the atom was important because it introduced quantized energy states for the electrons. However, as a model it was only useful for predicting the behavior of atoms with a single electron (H, He+, and Li2+ ions). Thus, a different model of the atom eventually replaced Bohr's model. However, we will retain the concept of quantized energy states

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                           de Broglie Waves

de Broglie Waves in the Bohr Atom

This is a dynamic situation: in the illustration below the wave is "waving," something that is hard to show in a static picture.

Moreover, if the electron momentum is not just right so that its de Broglie wavelength doesn't exactly wrap around the circumference, we get no standing wave; destructive interference results, and the electron cannot exist as part of a stable H atom if it has that momentum.

De Broglie's vision of Bohr's atom   

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            Electron Configurations


 Electron Configuration and the Table

The "periodic" nature of chemical properties that Mendeleev had discovered is related to the electron configuration of the atoms of the elements.

In other words, the way in which an atom's electrons are arranged around its nucleus affects the properties of the atom.

Bohr's theory of the atom tells us that electrons are not located randomly around an atom's nucleus, but they occur in specific electron shells.

 Each shell has a limited capacity for electrons.

As lower shells are filled, additional electrons reside in more-distant shells.

The capacity of the first electron shell is two electrons and for the second shell the capacity is eight.

Thus, oxygen, with eight protons and eight electrons, carries two electrons in its first shell and six in its second shell.

Fluorine, with nine electrons, carries two in its first shell and seven in the second.

Neon, with ten electrons, carries two in the first and eight in the second.

Because the number of electrons in the second shell increases, we can begin to imagine why the chemical properties gradually change as we move from oxygen to fluorine to neon.

Sodium has eleven electrons.

Two fit in its first shell, but remember that the second shell can only carry eight electrons.

Sodium's eleventh electron cannot fit into either its first or its second shell.

This electron takes up residence in yet another orbit, a third electron shell in sodium.

The reason that there is a dramatic shift in chemical properties when moving from neon to sodium is because there is a dramatic shift in electron configuration between the two elements.

 But why is sodium similar to lithium? Let's look at the electron configurations of these elements.

        Notice that the size (atomic radius) decreases from left to right.

As you can see in the following illustration, while sodium has three electron shells and lithium two, the characteristic they share in common is that they both have only one electron in their outermost electron shell.

           An atom's valence shell   "covers" inner electron shells               


These outer-shell electrons (called valence electrons) are important in determining the chemical properties of the elements.

An element's chemical properties are determined by the way in which its atoms interact with other atoms.
If we picture the outer (valence) electron shell of an atom as a sphere encompassing everything inside, then it is only the valence shell that can interact with other atoms - much the same way as it is only the paint on the exterior of your house that "interacts" with, and gets wet by, rain water.
The valence shell electrons in an atom determine the way it will interact with neighboring atoms, and therefore determine its chemical properties.
Since both sodium and lithium have one valence electron,  they share similar chemical properties.

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                   Dalton's Theory

Dalton's Atomic Theory

 1) All matter is made of atoms. Atoms are indivisible and indestructible.

2) All atoms of a given element are identical in mass and properties

3) Compounds are formed by a combination of two or more different kinds of atoms.

4) A chemical reaction is a rearrangement of atoms.

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                        Bohr Models

Bohr model animation



While the Rutherford model focused on describing the nucleus, Niels Bohr turned his attention to describing the electron. 

Prior to the Bohr Model, the accepted model was one which depicted the electron as an orbiting planet. 

The flaw with the planet-like model is that an electron particle moving in a circular path would be accelerating (see circular motion). 
An accelerating electron creates a changing magnetic field.

This changing magnetic field would carry energy away from the electron, eventually slowing it down and allowing it to be "captured" by the nucleus.

Bohr built upon spectroscopic observations of atoms.
Spectroscopists noticed that an atom can only absorb certain energies (colors) of light (the absorption spectrum) and once excited can only release certain energies (the emission spectrum) and these energies happen to be the same. 
Bohr used these observations to argue that the energy of a bound electron is "quantized." 


 absorption spectrum (4k)

                                                  Absorption Spectrum

Emission spectrum  (3k)

                                                   Emission Spectrum


Quantized is a fancy word meaning only certain quantities of energy are allowed. 

This explanation addresses the true origin of light.  

Since only certain energy levels are allowed it is actually possible to diagram the atom in terms of its energy levels. 
In the animation below you will see a model of a Hydrogen atom and to the right of it, a Bohr energy level diagram.

Bohr atom & energy levels (25k)
The Hydrogen Atom

In the animation you will notice that if the energy of the photon of light is just right, it will cause the electron to jump to a higher level. 

When the electron jumps back down, a photon is created for each jump down. 

A photon without the right amount of energy (the pink one) passes through the atom with no effect. 

Photons with too much energy will cause the electron to be ejected which ionizes the atom.

An ionized electron is said to be in the n=infinity energy level. 

Keep in mind that these rings are not actually orbits, but are levels that represent the location of an electron wave.  



The number n corresponds to the number of complete waves in the electron.

Ephoton = Einitial - Efinal

This formula can be used to determine the energy of the photon emitted (+) or absorbed(-).

Ephoton = hv

where h = 6.6 x 10-34 Js
or 4.14 x 10-15 eVs

This formula can be used to determine the energy of a photon if you know the frequency of it.  Planck's constant, h, can be used in terms of Joule(s) or eV(s).

Emission animation


The Millikan Oil Drop Experiment

The following section is for Honor's classes only. 





An experiment performed by Robert Millikan in 1909 determined the size of the charge on an electron.

He also determined that there was a smallest 'unit' charge, or that charge is 'quantized'. He received the Nobel Prize for his work.

We're going to explain that experiment here, and show how Millikan was able to determine the size of a charge on a single electron.

What Millikan did was to put a charge on a tiny drop of oil, and measure how strong an applied electric field had to be in order to stop the oil drop from falling.

Since he was able to work out the mass of the oil drop, and he could calculate the force of gravity on one drop, he could then determine the electric charge that the drop must have.

By varying the charge on different drops, he noticed that the charge was always a multiple of -1.6 x 10 -19 C, the charge on a single electron.

 This meant that it was electrons carrying this unit charge.

Here's how it worked. Have a look at the apparatus he used:

An atomizer sprayed a fine mist of oil droplets into the chamber. Some of these tiny droplets fell through a hole in the upper floor.

Millikan first let them fall until they reached terminal velocity.

Using the microscope, he measured their terminal velocity, and by use of a formula, calculated the mass of each oil drop.
Next, Millikan applied a charge to the falling drops by illuminating the bottom chamber with x-rays.

This caused the air to become ionized, and electrons to attach themselves to the oil drops.

By attaching a battery to the plates above and below this bottom chamber, he was able to apply an electric voltage.

The electric field produced in the bottom chamber by this voltage would act on the charged oil drops; if the voltage was just right, the electromagnetic force would just balance the force of gravity on a drop, and the drop would hang suspended in mid-air.

First, allow the drops to fall. Notice how they accelerate at first, due to gravity.

But quickly, air resistance causes them to reach terminal velocity.

Now focus on a single falling drop, and adjust the electric field upwards until the drop remains suspended in mid-air.

At that instant, for that drop, the electric force on it exactly equals the force of gravity on it.

Some drops have more electrons than others, so will require a higher voltage to stop.


 O.K., let's look at the calculation Millikan was now able to do.

When a drop is suspended, its weight m · g is exactly equal to the electric force applied q · E

The values of E, the applied electric field, m the mass of a drop, and g, the acceleration due to gravity, are all known values. So you can solve for q, the charge on the drop:

Millikan determined the charge on a drop.

Then he redid the experiment numerous times, each time varying the strength of the x-rays ionizing the air, so that differing numbers of electrons would jump onto the oil molecules each time.

He obtained various values for q.

The charge q on a drop was always a multiple of -1.6 x 10 -19 C, the charge on a single electron.

This number was the one Millikan was looking for, and it also showed that the value was quantized; the smallest unit of charge was this amount, and it was the charge on a single electron.


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                     Lewis Structures

Lewis Dot Structures: Lewis dot structures are a shorthand to represent the valence electrons of an atom.
The structures are written as the element symbol surrounded by dots that represent the valence electrons.
 The Lewis structures for the elements in the first two periods of the periodic table are shown below.


lewis_H Lewis Dot Structureslewis_He
lewis_Li lewis_Be lewis_B lewis_C lewis structure-nitrogen lewis_O lewis_F lewis_Ne


Lewis structures can also be used to show bonding between atoms.

The bonding electrons are placed between the atoms and can be represented by a pair of dots or a dash (each dash represents one pair of electrons, or one bond).

Lewis structures for H2 and O2 are shown below.

H2 H:Hor H-H



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             Thomson's Experiment

In 1897, J. J. Thomson dramatically changed the modern view of the atom with his discovery of the electron.

Thomson's work suggested that the atom was not an "indivisible" particle as John Dalton.

John Dalton had suggested but, a jigsaw puzzle made of smaller pieces.

Thomson's notion of the electron came from his work with a nineteenth century scientific curiosity: the cathode ray tube.

For years scientists had known that if an electric current was passed through a vacuum tube, a stream of glowing material could be seen; however, no one could explain why.

Thomson found that the mysterious glowing stream would bend toward a positively charged electric plate.

Thomson theorized, and was later proven correct, that the stream was in fact made up of small particles, pieces of atoms that carried a negative charge.

 These particles were later named electrons!

Thomson imagined that atoms looked like pieces of raisin bread, a structure in which clumps of small, negatively charged electrons (the "raisins") were scattered inside a smear of positive charges.


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            Rutherford's Experiment

In 1908, Ernest Rutherford, a former student of Thomson's, proved Thomson's raisin bread structure incorrect.

Rutherford performed a series of experiments with radioactive alpha particles

While it was unclear at the time what the alpha particle was, it was known to be very tiny. 

Rutherford fired tiny alpha particles at solid objects such as gold foil. 

He found that while most of the alpha particles passed right through the gold foil, a small number of alpha particles passed through at an angle (as if they had bumped up against something) and some bounced straight back like a tennis ball hitting a wall. 

Rutherford's experiments suggested that gold foil, and matter in general, had holes in it! 

These holes allowed most of the alpha particles to pass directly through, while a small number ricocheted off or bounced straight back because they hit a solid object.
In 1911, Rutherford proposed a revolutionary view of the atom.

He suggested that the atom consisted of a small, dense core of positively charged particles in the center (or nucleus) of the atom, surrounded by a swirling ring of electrons.

The nucleus was so dense that the alpha particles would bounce off of it, but the electrons were so tiny, and spread out at such great distances, that the alpha particles would pass right through this area of the atom.

Rutherford's atom resembled a tiny solar system with the positively charged nucleus always at the center and the electrons revolving around the nucleus.

He proved this by using alpha particles in his gold foil experiment, where most of the particles past through the foil.

This proved that the atom is mostly empty space.
The positively charged particles in the nucleus of the atom were called protons

Protons carry an equal, but opposite, charge to electrons, but protons are much larger and heavier than electrons.  



To enhance your understanding view the animations below:

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             Chadwick's Experiment

In 1930, James Chadwick discovered a third type of subatomic particle, which he named the neutron.

Neutrons help stabilize the protons in the atom's nucleus.

Because the nucleus is so tightly packed together, the positively charged protons would tend to repel each other normally.

Neutrons help to reduce the repulsion between protons and stabilize the atom's nucleus.

Neutrons always reside in the nucleus of atoms and they are about the same size as protons.

However, neutrons do not have any electrical charge; they are electrically neutral.

Atoms are electrically neutral because the number of protons (+ charges) is equal to the number of electrons (- charges) and thus the two cancel out. 

As the atom gets larger, the number of protons increases, and so does the number of electrons (in the neutral state of the atom). 

The illustration linked below compares the two simplest atoms, hydrogen and helium.

Simulated hydrogen and helium atoms

Atoms of different elements are distinguished from each other by their number of protons (the number of protons is constant for all atoms of a single element; the number of neutrons and electrons can vary under some circumstances).

To identify this important characteristic of atoms, the term atomic number (z) is used to describe the number of protons in an atom.

For example, z = 1 for hydrogen and z = 2 for helium.

Another important characteristic of an atom is its weight, or atomic mass.

The weight of an atom is roughly determined by the total number of protons and neutrons in the atom.

While protons and neutrons are about the same size, the electron is more that 1,800 times smaller than the two.

Thus the electrons' weight is inconsequential in determining the weight of an atom - it's like comparing the weight of a flea to the weight of an elephant.

Refer to the animation above to see how the number of protons plus neutrons in the hydrogen and helium atoms corresponds to the atomic mass.


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                  Electron cloud model

Erwin Schrödinger built upon the thoughts of Bohr yet took them in a new direction. 

He developed the probability function for the Hydrogen atom (and a few others). 

The probability function basically describes a cloud-like region where the electron is likely to be found. 

It can not say with any certainty, where the electron actually is at any point in time, yet can describe where it ought to be.  

Clarity through fuzziness, is one way to describe the idea. 

The model based on this probability equation can best be described as the cloud model.

The Cloud model (5k)The cloud model represents a sort of history of where the electron has probably been and where it is likely to be going. 

The red dot in the middle represents the nucleus while the red dot around the outside represents an instance of the electron. 


Imagine, as the electron moves it leaves a trace of where it was.  

This collection of traces quickly begins to resemble a cloud. 

The probable locations of the electron predicted by Schrödinger's equation happen to coincide with the locations specified in Bohr's model.



When the number of electrons changes in an atom, the electrical charge changes.

If an atom gains electrons, it picks up an imbalance of negatively charged particles and therefore becomes negative.

If an atom loses electrons, the balance between positive and negative charges is shifted in the opposite direction and the atom becomes positive.

In either case, the magnitude (+1, +2, -1, -2, etc.) of the electrical charge will correspond to the number of electrons gained or lost.

Atoms that carry electrical charges are called ions (regardless of whether they are positive or negative).

For example, the animation below shows a positive hydrogen ion (which has lost an electron) and a negative hydrogen ion (which has gained an extra electron).

The electrical charge on the ion is always written as a superscript after the atom's symbol, as seen in the animation.


Hydrogen Ion Simulation

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The number of neutrons in an atom can also vary.

Two atoms of the same element that contain different numbers of neutrons are called isotopes.

For example, normally hydrogen contains no neutrons.

An isotope of hydrogen does exist that contains one neutron (commonly called deuterium).

The atomic number (z) is the same in both isotopes; however the atomic mass increases by one in deuterium as the atom is made heavier by the extra neutron.

An element can exist in a number of forms, called isotopes. 

Isotopes are forms of the same atom that vary in mass.  



For example, there are two different types (isotopes) of copper atoms. 
One type of copper atoms weighs in at 62.93 amu, the other has a mass of 64.94 amu. 
The lighter isotope is more common with  69.09% of the naturally occurring copper having a mass of 62.93 amu per atom. 
The remainder of the atoms, 30.91 %, have a mass of 64.94 amu.  
To find the AVERAGE ATOMIC MASS of an atom, we take into account all of the isotopes that exist and the percentage of each type. 
The calculation of the average atomic mass is a WEIGHTED AVERAGE.
Average atomic mass = Σ (mass of isotope × relative abundance)
The bottom line is that to find the average atomic mass of copper, we insert the information about copper’s isotopes into the formula and solve. 
There are two isotopes, so we will be adding the contributions of 2 isotopes. 
(That’s where the Σ sign comes in. ) 
Average atomic mass of copper = (62.93 amu × 0.6909) + (64.94 amu  ×  0.3091)
                                               = 63.55
From the calculation, we know that an AVERAGE atom of copper has a mass of 63.55 amu. 
Notice that in this problem, we would predict that the average is closer to the weight of the lighter isotope. 
This is because the lighter form of copper is more abundant.

more examples:

Example #1: Carbon

mass number exact weight percent abundance
12 12.000000 98.90
13 13.003355 1.10

To calculate the average atomic weight, each exact atomic weight is multiplied by its percent abundance (expressed as a decimal). Then, add the results together and round off to an appropriate number of significant figures.

This is the solution for carbon:

(12.000000) (0.9890) + (13.003355) (0.0110) = 12.011

Example #2: Nitrogen

mass number exact weight percent abundance
14 14.003074 99.63
15 15.000108 0.37

This is the solution for nitrogen:

(14.003074) (0.9963) + (15.000108) (0.0037) = 14.007

Video: How to Calculate an Average Atomic Weight.

Example #3: Chlorine   Example #4: Silicon
mass number exact weight percent abundance   mass number exact weight percent abundance
35 34.968852 75.77   28 27.976927 92.23
37 36.965903 24.23   29 28.976495 4.67
        30 29.973770 3.10
The answer for chlorine: 35.453   The answer for silicon: 28.086

This type of calculation can be done in reverse, where the isotopic abundances can be calculated knowing the average atomic weight.

Go to tutorial on reverse direction.

Hydrogen Isotope Simulation

  The Periodic Table of Elements

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                 Line Spectra

line spectra

When matter is heated, it gives off light.

For example, turning on an ordinary light bulb causes an electric current to flow through a metal filament that heats the filament and produces light.

The electrical energy absorbed by the filament excites the atoms' electrons, causing them to "wiggle".

This absorbed energy is eventually released from the atoms in the form of light.

When normal white light, such as that from the sun, is passed through a prism, the light separates into a continuous spectrum of colors:


Bohr knew that when pure elements were excited by heat or electricity, they gave off distinct colors rather than white light.

This phenomenon is most commonly seen in modern-day neon lights, tubes filled with gaseous elements (most commonly neon).

When an electric current is passed through the gas, a distinct color (most commonly red) is given off by the element.

When light from an excited element is passed through a prism, only specific lines (or wavelengths) of light can be seen.

These lines of light are called line spectra. For example, when hydrogen is heated and the light is passed through a prism, the following line spectra can be seen:

To Bohr, the line spectra phenomenon showed that atoms could not emit energy continuously, but only in very precise quantities (he described the energy emitted as quantized).

Because the emitted light was due to the movement of electrons, Bohr suggested that electrons could not move continuously in the atom (as Rutherford had suggested) but only in precise steps.

Bohr hypothesized that electrons occupy specific energy levels.

When an atom is excited, such as during heating, electrons can jump to higher levels.

When the electrons fall back to lower energy levels, precise quanta of energy are released as specific wavelengths (lines) of light.

Under Bohr's theory, an electron's energy levels (also called electron shells) can be imagined as concentric circles around the nucleus.

Normally, electrons exist in the ground state, meaning they occupy the lowest energy level possible (the electron shell closest to the nucleus).

When an electron is excited by adding energy to an atom (for example, when it is heated), the electron will absorb energy, "jump" to a higher energy level, and spin in the higher energy level.

After a short time, this electron will spontaneously "fall" back to a lower energy level, giving off a quantum of light energy.

Key to Bohr's theory was the fact that the electron could only "jump" and "fall" to precise energy levels, thus emitting a limited spectrum of light.

The animation linked below simulates this process in a hydrogen atom.

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Bohr's Atom: Quantum Behavior in Hydrogen
Concept simulation - Reenacts electron's "jump" and "fall" to precise energy levels in a hydrogen atom.

Not only did Bohr predict that electrons would occupy specific energy levels, he also predicted that those levels had limits to the number of electrons each could hold.

Under Bohr's theory, the maximum capacity of the first (or innermost) electron shell is two electrons.

For any element with more than two electrons, the extra electrons will reside in additional electron shells.

For example, in the ground state configuration of lithium (which has three electrons) two electrons occupy the first shell and one electron occupies the second shell.

This is illustrated in the animation linked below.

The Lithium atom
For further details, the table linked below shows the electron configurations of the first eleven elements.
Atomic structure animation table 
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Here we see the electron configuration for the element helium.  This electron configuration provides us with the following information:


  • The large number "1" refers to the principle quantum number "n" which stands for the energy level.  It tells us that the electrons of helium occupy the first energy level of the atom.

  • The letter "s" stands for the angular momentum quantum number "l".  It tells us that the two electrons of the helium electron occupy an "s" or spherical orbital.

  • The exponent "2" refers to the total number of electrons in that orbital or sub-shell.  In this case, we know that there are two electrons in the spherical orbital at the first energy level.

Before we continue, we must review some information that you will need to fill   in electron configurations correctly.


I.  Principle Quantum Number (n) and Sublevels

The number of sublevels that an energy level can contain is equal to the principle quantum number of that level. 

So, for example, the second energy level would have two sublevels, and the third energy level would have three sublevels. 

The first sublevel is called an s sublevel.  The second sublevel is called a p sublevel. 

The third sublevel is called a d sublevel and the fourth sublevel is called an f sublevel. 

Although energy levels that are higher than 4 would contain additional sublevels, these sublevels have not been named because no known atom in its ground state would have electrons that occupy them.


II.  Sublevels and Orbitals

An orbital is a space that can be occupied by up to two electrons. 

Each type of sublevel holds a different number or orbitals, and therefore, a different number of electrons. 

s sublevels have one orbital, which can hold up to two electrons. 

p sublevels have three orbitals, each of which can hold 2 electrons, for a total of 6. 

d sublevels have 5 orbitals, for a possible total of 10 electrons. 

f sublevels, with 7 orbitals, can hold up to 14 electrons.  The information about the sublevels is summarized in the table below:

Orbital and Electron Capacity for the Four Named Sublevels

Sublevel# of orbitalsMaximum number of electrons


III.  Total Number of Orbital and Electrons per Energy Level

An easy way to calculate the number of orbitals found in an energy level is to use the formula n2

For example, the third energy level (n=3) has a total of 32, or nine orbitals. 

This makes sense because we know that the third energy level would have 3 sublevels; an s sublevel with one orbital, a p sublevel with 3 orbitals and a d sublevel with 5 orbitals.  1 + 3 + 5 = 9, so the formula n2 works!


IV.  Total Number of Electrons per Energy Level

An easy way to calculate the total number of electrons that can be held by a given energy level is to use the formula 2n2.  

For example, the fourth energy level (n=4) can hold 2(4)2 = 32 electrons. 

This makes sense because the fourth energy level would have four sublevels, one of each of the named types. 

The s sublevel hold 2 electrons, the p sublevel holds 6 electrons , the d sublevel holds 10 electrons and the f sublevel holds 14 electrons. 

2 + 6 + 10 + 14 = 32, so the formula 2n2 works! 

We can summarize this information in the table below:

Orbitals and Electron Capacity of the First Four Principle Energy Levels
Principle energy level (n)Type of sublevelNumber of orbitals per typeNumber of orbitals per level(n2)Maximum number of electrons (2n2)

V.  Order of Filling Sublevels with Electrons

The next thing that you need to recall is the fact that the energy sublevels are filled in a specific order that is shown by the arrow diagram seen below:

Remember to start at the beginning of each arrow, and then follow it all of the way to the end, filling in the sublevels that it passes through.  In other words, the order for filling in the sublevels becomes; 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d,7p.


                      Molecular orbitals

 Chemical bonding occurs when the net attractive forces between an electron and two nuclei exceeds the electrostatic repulsion between the two nuclei.

For this to happen, the electron must be in a region of space which we call the binding region.

 Conversely, if the electron is off to one side, in an anti-binding region, it actually adds to the repulsion between the two nuclei and helps push them away.

The easiest way of visualizing a molecular orbital is to start by picturing two isolated atoms and the electron orbitals that each would have separately.

 These are just the orbitals of the separate atoms, by themselves, which we already understand.

We will then try to predict the manner in which these atomic orbitals interact as we gradually move the two atoms closer together.

 Finally, we will reach some point where the internuclear distance corresponds to that of the molecule we are studying.

The corresponding orbitals will then be the molecular orbitals of our new molecule.


 The hydrogen molecule ion: the simplest molecule

To see how this works, we will consider the simplest possible molecule, H2+.

This is the hydrogen molecule ion, which consists of two nuclei of charge +1, and a single electron shared between them.

As two H nuclei move toward each other, the 1s atomic orbitals of the isolated atoms gradually merge into a new molecular orbital in which the greatest electron density falls between the two nuclei.

Since this is just the location in which electrons can exert the most attractive force on the two nuclei simultaneously, this arrangement constitutes a bonding molecular orbital 

Regarding it as a three- dimensional region of space, we see that it is symmetrical about the line of centers between the nuclei; in accord with our usual nomenclature, we refer to this as a σ (sigma) orbital. 


Bonding and antibonding molecular orbitals

Orbital conservation

There is one minor difficulty: we started with two orbitals (the 1s atomic orbitals), and ended up with only one orbital.

Now according to the rules of quantum mechanics, orbitals cannot simply appear and disappear at our convenience.

For one thing, this would raise the question of at just what internuclear distance do we suddenly change from having two orbitals, to having only one?

It turns out that when orbitals interact, they are free to change their forms, but there must always be the same number.

This is just another way of saying that there must always be the same number of possible allowed sets of electron quantum numbers.


In-phase and out-of-phase wave combinations

How can we find the missing orbital?

To answer this question, we must go back to the wave-like character of orbitals that we developed in our earlier treatment of the hydrogen atom.

You are probably aware that wave phenomena such as sound waves, light waves, or even ocean waves can combine or interact with one another in two ways: they can either reinforce each other, resulting in a stronger wave, or they can interfere with and partially destroy each other.

A roughly similar thing occurs when the “matter waves” corresponding to the two separate hydrogen 1s orbitals interact; both in-phase and out-of-phase combinations are possible, and both occur.

The in-phase, reinforcing interaction yields the bonding orbital that we just considered.

The other, corresponding to out-of-phase combination of the two orbitals, gives rise to a molecular orbital that has its greatest electron probability in what is clearly the antibonding region of space.

This second orbital is therefore called an antibonding orbital.

When the two 1s wave functions combine out-of-phase, the regions of high electron probability do not merge.

In fact, the orbitals act as if they actually repel each other.

Notice particularly that there is a region of space exactly equidistant between the nuclei at which the probability of finding the electron is zero.

This region is called a nodal surface, and is characteristic of antibonding orbitals.

It should be clear that any electrons that find themselves in an antibonding orbital cannot possibly contribute to bond formation; in fact, they will actively oppose it.

We see, then, that whenever two orbitals, originally on separate atoms, begin to interact as we push the two nuclei toward each other, these two atomic orbitals will gradually merge into a pair of molecular orbitals, one of which will have bonding character, while the other will be antibonding.

In a more advanced treatment, it would be fairly easy to show that this result follows quite naturally from the wave-like nature of the combining orbitals.

What is the difference between these two kinds of orbitals, as far as their potential energies are concerned?

More precisely, which kind of orbital would enable an electron to be at a lower potential energy?

Clearly, the potential energy decreases as the electron moves into a region that enables it to “see” the maximum amount of positive charge.

In a simple diatomic molecule, this will be in the internuclear region— where the electron can be simultaneously close to two nuclei.

The bonding orbital will therefore have the lower potential energy.

2  Molecular orbital diagrams

This scheme of bonding and antibonding orbitals is usually depicted by a molecular orbital diagram such as the one shown here for the dihydrogen ion H2+.

Atomic valence electrons (shown in boxes on the left and right) fill the lower-energy molecular orbitals before the higher ones, just as is the case for atomic orbitals.

Thus, the single electron in this simplest of all molecules goes into the bonding orbital, leaving the antibonding orbital empty.

Since any orbital can hold a maximum of two electrons, the bonding orbital in H2+ is only half-full.

This single electron is nevertheless enough to lower the potential energy of one mole of hydrogen nuclei pairs by 270 kJ— quite enough to make them stick together and behave like a distinct molecular species.

Although H2+ is stable in this energetic sense, it happens to be an extremely reactive molecule— so much so that it even reacts with itself, so these ions are not commonly encountered in everyday chemistry.


If one electron in the bonding orbital is conducive to bond formation, might two electrons be even better?

We can arrange this by combining two hydrogen atoms-- two nuclei, and two electrons.

Both electrons will enter the bonding orbital, as depicted in the Figure.

We recall that one electron lowered the potential energy of the two nuclei by 270 kJ/mole, so we might expect two electrons to produce twice this much stabilization, or 540 kJ/mole.


Bond order is defined as the difference between the number of electron pairs occupying bonding and nonbonding orbitals in the molecule.
A bond order of unity corresponds to a conventional "single bond".

Experimentally, one finds that it takes only 452 kJ to break apart a mole of hydrogen molecules.

The reason the potential energy was not lowered by the full amount is that the presence of two electrons in the same orbital gives rise to a repulsion that acts against the stabilization.

This is exactly the same effect we saw in comparing the ionization energies of the hydrogen and helium atoms.


With two electrons we are still ahead, so let’s try for three.

The dihelium positive ion is a three-electron molecule.

We can think of it as containing two helium nuclei and three electrons.

This molecule is stable, but not as stable as dihydrogen; the energy required to break He2+ is 301 kJ/mole.

The reason for this should be obvious; two electrons were accommodated in the bonding orbital, but the third electron must go into the next higher slot— which turns out to be the sigma antibonding orbital.

The presence of an electron in this orbital, as we have seen, gives rise to a repulsive component which acts against, and partially cancels out, the attractive effect of the filled bonding orbital.

Taking our building-up process one step further, we can look at the possibilities of combining to helium atoms to form dihelium.

You should now be able to predict that He2 cannot be a stable molecule; the reason, of course, is that we now have four electrons— two in the bonding orbital, and two in the antibonding orbital.

The one orbital almost exactly cancels out the effect of the other.

Experimentally, the bond energy of dihelium is only .084 kJ/mol; this is not enough to hold the two atoms together in the presence of random thermal motion at ordinary temperatures, so dihelium dissociates as quickly as it is formed, and is therefore not a distinct chemical species.

Diatomic molecules containing second-row atoms

The four simplest molecules we have examined so far involve molecular orbitals that derived from two 1s atomic orbitals.

If we wish to extend our model to larger atoms, we will have to contend with higher atomic orbitals as well.

One greatly simplifying principle here is that only the valence-shell orbitals need to be considered.

Inner atomic orbitals such as 1s are deep within the atom and well-shielded from the electric field of a neighboring nucleus, so that these orbitals largely retain their atomic character when bonds are formed.



For example, when lithium, whose configuration is 1s22s1, bonds with itself to form Li2,

we can forget about the 1s atomic orbitals and consider only the σ bonding and antibonding orbitals.

Since there are not enough electrons to populate the antibonding orbital, the attractive forces win out and we have a stable molecule.

The bond energy of dilithium is The 104.6 kJ/mole; notice that this value is far less than the 270 kJ bond energy in dihydrogen, which also has two electrons in a bonding orbital.

The reason, of course, is that the 2s orbital of Li is much farther from its nucleus than is the 1s orbital of H, and this is equally true for the corresponding molecular orbitals.

It is a general rule, then, that the larger the parent atom, the less stable will be the corresponding diatomic molecule.

Lithium hydride

All the molecules we have considered thus far are homonuclear; they are made up of one kind of atom.

As an example of a heteronuclear molecule, let’s take a look at a very simple example— lithium hydride.

Lithium hydride is a stable, though highly reactive molecule.

The diagram shows how the molecular orbitals in lithium hydride can be related to the atomic orbitals of the parent atoms.

One thing that makes this diagram look different from the ones we have seen previously is that the parent atomic orbitals have widely differing energies; the 1s orbital of lithium has a much lower energy than the 1s hydrogen orbital, owing to the greater nuclear charge of lithium.

On this diagram, the 2p orbitals of both parent atoms, and the 2s orbital of hydrogen, are shown.

These orbitals are not occupied in the ground states of the parent atoms, so we need not concern ourselves with them, although in a more thorough treatment their existence would need to be taken into account.

There are two occupied atomic orbitals on the lithium atom, and only one on the hydrogen.

With which of the lithium orbitals does the hydrogen 1s orbital interact?

The lithium 1s orbital is the lowest-energy orbital on the diagram.

Because this orbital is so small and retains its electrons so tightly, we can forget about it as far as bonding goes; we need consider only 2s orbital of lithium which combines with the 1s orbital of hydrogen to form the usual pair of sigma bonding and antibonding orbitals.

Of the four electrons in lithium and hydrogen, two are retained in the lithium 1s orbital, and the two remaining ones go into the 2s bonding orbital.

The resulting molecule is 243 kJ/mole more stable than the parent atoms.

As we might expect, the bond energy of the heteronuclear molecule is very close to the average of the energies of the corresponding homonuclear molecules.

Actually, it turns out that the correct way to make this comparison is to take the geometric mean, rather than the arithmetic mean, of the two bond energies.

The geometric mean is simply the square root of the product of the two energies.

The geometric mean of the H2 and Li2 bond energies is 213 kJ/mole, so it appears that the lithium hydride molecule is 30 kJ/mole more stable than it “is supposed” to be.

This is attributed to the fact that the electrons in the 2σ bonding orbital are not equally shared between the two nuclei; the orbital is skewed slightly so that the electrons are attracted somewhat more to the hydrogen atom.

This bond polarity, which we considered in some detail near the beginning of our study of covalent bonding, arises from the greater electron-attracting power of hydrogen— a consequence of the very small size of this atom.

The electrons can be at a lower potential energy if they are slightly closer to the hydrogen end of the lithium hydride molecule.

It is worth pointing out, however, that the electrons are, on the average, also closer to the lithium nucleus, compared to where they would be in the 2s orbital of the isolated lithium atom.

So it appears that everyone gains and no one loses here!

3  Sigma and pi orbitals

The molecules we have considered thus far are composed of atoms that have no more than four electrons each; our molecular orbitals have therefore been derived from s-type atomic orbitals only.

If we wish to apply our model to molecules involving larger atoms, we must take a close look at the way in which p-type orbitals interact as well.

Although two atomic p orbitals will be expected to split into bonding and antibonding orbitals just as before, it turns out that the extent of this splitting, and thus the relative energies of the resulting molecular orbitals, depend very much on the nature of the particular p orbital that is involved. 

The importance of direction

You will recall that there are three possible p orbitals for any value of the principal quantum number.

You should also recall that p orbitals are not spherical like s orbitals, but are elongated, and thus possess definite directional properties.

The three p orbitals correspond to the three directions of Cartesian space, and are frequently designated px, py, and pz, to indicate the axis along which the orbital is aligned.

Of course, in the free atom, where no coordinate system is defined, all directions are equivalent, and so are the p orbitals.

But when the atom is near another atom, the electric field due to that other atom acts as a point of reference that defines a set of directions.

The line of centers between the two nuclei is conventionally taken as the x axis.

If this direction is represented horizontally on a sheet of paper, then the y axis is in the vertical direction and the z axis would be normal to the page.

These directional differences lead to the formation of two different classes of molecular orbitals.

The above figure shows how two px atomic orbitals interact.

In many ways the resulting molecular orbitals are similar to what we got when s atomic orbitals combined; the bonding orbital has a large electron density in the region between the two nuclei, and thus corresponds to the lower potential energy.

In the out-of-phase combination, most of the electron density is away from the internuclear region, and as before, there is a surface exactly halfway between the nuclei that corresponds to zero electron density.

This is clearly an antibonding orbital— again, in general shape, very much like the kind we saw in hydrogen and similar molecules.

Like the ones derived from s-atomic orbitals, these molecular orbitals are σ (sigma) orbitals.

Sigma orbitals are cylindrically symmetric with respect to the line of centers of the nuclei; this means that if you could look down this line of centers, the electron density would be the same in all directions.

When we examine the results of the in- and out-of-phase combination of py and pz orbitals, we get the bonding and antibonding pairs that we would expect, but the resulting molecular orbitals have a different symmetry: rather than being rotationally symmetric about the line of centers, these orbitals extend in both perpendicular directions from this line of centers.

 Orbitals having this more complicated symmetry are called π (pi) orbitals.

There are two of them, πy and πz differing only in orientation, but otherwise completely equivalent.

The different geometric properties of the π and σ orbitals causes the latter orbitals to split more than the π orbitals, so that the σ* antibonding orbital always has the highest energy. The σ bonding orbital can be either higher or lower than the π bonding orbitals, depending on the particular atom.

4 Splitting patterns for the second-row diatomics

If we combine the splitting schemes for the 2s and 2p orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table.

Remember that only the valence orbitals of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character.

Notice that the relative energies of the 2p-derived σ and π bonding molecular orbitals are reversed in O2 and F2.

This is attributed to interactions between the 2s orbital each atom with the 2px orbital of the other, an effect similar to hybridizaton.

However, the order in which these two orbitals are filled has no effect on the predicted bond orders, so there is ordinarily no need to know which molecules follow which scheme.



Carbon has four outer-shell electrons, two 2s and two 2p.

For two carbon atoms, we therefore have a total of eight electrons, which can be accommodated in the first four molecular orbitals.

The lowest two are the 2s-derived bonding and antibonding pair, so the “first” four electrons make no net contribution to bonding.

The other four electrons go into the pair of pi bonding orbitals, and there are no more electrons for the antibonding orbitals— so we would expect the dicarbon molecule to be stable, and it is.

(But being extremely reactive, it is known only in the gas phase.)

You will recall that one pair of electrons shared between two atoms constitutes a “single” chemical bond; this is Lewis’ original definition of the covalent bond.

In C2 there are two paris of electrons in the π bonding orbitals, so we have what amounts to a double bond here; in other words, the bond order in dicarbon is two.


The potential energy of molecular oxygen is 494 kJ/mole below that of the parent atoms.

This is a smaller bond energy than nitrogen has — not surprising, considering that oxygen has two electrons in an antibonding orbital, compared to nitrogen’s one.

Although this paramagnetic ground-state form of O2 (also known as triplet oxygen) is the energetically favored, and therefore is the common form of this element, another variety, in which the two electrons are paired up in a single pi antibonding orbital, is also well known.

This singlet oxygen, as it is called, has a bond energy of only 402 kJ/mole.

The lower value reflects the action of electrostatic repulsion between the two electrons in the same orbital.

Singlet oxygen, being less stable, does not exist under normal conditions.

It can be formed by the action of light and in certain chemical reactions, and it has an interesting and unique chemistry of its own.

The two unpaired electrons of the dioxygen molecule give this substance an unusual and distinctive property: O2 is paramagnetic.

The paramagnetism of oxygen can readily be demonstrated by pouring liquid O2 between the poles of a strong permanent magnet; the liquid stream is trapped by the field and fills up the space between the poles.

Watch this video demo

Since molecular oxygen contains two electrons in an antibonding orbital, it might be possible to make the molecule more stable by removing one of these electrons, thus increasing the ratio of bonding to antibonding electrons in the molecule.

Just as we would expect, and in accord with our model, O2+ has a bond energy higher than that of neutral dioxygen; removing the one electron actually gives us a more stable molecule.

This constitutes a very good test of our model of bonding and antibonding orbitals. In the same way, adding an electron to O2 results in a weakening of the bond, as evidenced by the lower bond energy of O2.

The bond energy in this ion is not known, but the length of the bond is greater, and this is indicative of a lower bond energy.

These two dioxygen ions, by the way, are highly reactive and can be observed only in the gas phase.

What you should be able to do

Make sure you thoroughly understand the following essential ideas which have been presented above.

  • In what fundamental way does the molecular orbital model differ from the other models of chemical bonding that have been described in these lessons?
  • Explain how bonding and antibonding orbitals arise from atomic orbitals, and how they differ physically.
  • Describe the essential difference between a sigma and a pi molecular orbital.
  • Define bond order, and state its significance.
  • Construct a "molecular orbital diagram" of the kind shown in this lesson for a simple diatomic molecule, and indicate whether the molecule or its positive and negative ions should be stable.


Concept Map 


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            Molecular Orbital Theory (HONOR'S ONLY)

In valence bond theory, the electrons are thought of as still belonging to their individual atoms. 

However, in reality, the electrons in the bonds really are distributed in the “molecule”, and do not still belong to their individual atoms.

Also, there are problems with valence bond theory:  the O2 molecule should have a double bond between the two O atoms and each O atom should have 2 lone-pairs of electrons. 

This predicts that the O2 molecule will be diamagnetic; but, when we measure it, O2 is paramagnetic, having two unpaired electrons. 

So our model is not very satisfactory.


Molecular orbital theory (MO) allows us to think of the electrons in the molecule as a whole, rather than being associated only with specific atoms.

Bonding and Antibonding Molecular Orbitals

In MO theory, we think of the overlap of the two electrons in the two atomic orbitals in the H2 molecule as forming 2 molecular orbitals:  one is called the bonding MO orbital and the other is called the antibonding MO orbital.  

The bonding MO is always at lower energy (and therefore more stable) than either of the atomic orbitals from which it is made. 

The antibonding MO is at higher energy (and therefore less stable) than either of the atomic orbitals from which it is made. 

The names “bonding” and “antibonding” are very telling: 

if electrons are in bonding orbitals, then a bond is formed and is stable (and electron density is greatest between the nuclei of the 2 atoms), and if electrons are in antibonding orbitals, this is an unstable bond (and electron density is zero between the nuclei). 

The bonding orbitals (which are waves) interact in such a way that we have constructive interference in the resultant wave, whereas antibonding orbitals interact such that we have destructive interference.

σ Molecular Orbitals

In H2, the bonding MO orbital formed from two1s orbitals is called a sigma1s (σ1s) orbital, while the antibonding MO orbital is called σ*1s

Just as in atomic orbitals, each molecular orbital can hold 2 electrons, with opposite spins.

2s and 3s atomic orbitals come together to form σ2s and σ*2s  MO orbitals and σ3s and σ*3s  MO orbitals in the same way.

Show H2 and then the hypothetical He2 molecules.

Bond Order
For a diatomic molecule using MO theory, we can easily calculate the bond order. 

The magnitude of the bond order describes how stable a molecule is. 

The higher the bond order, the more stable the molecule. 
To calculate the bond order:  bond order = [# electrons in bonding MO - # electons in antibonding MO] / 2

Ex.:  for  H2, the 2 electrons are in the σ1s orbital and there are no electrons in the σ1s* orbital;
therefore the bond order = [2-0] / 2 = 1.

for  He2, there are 2 electrons in the σ1s orbital and two more electrons in the σ1s* orbital;
therefore the bond order = [2-2] / 2 = 0. 

MO theory predicts that a molecule with BO = 0 does not exist, and He2 does not.

If we do the same for Li2 and Be2, we can find and calculate that Li2 molecule has BO = 1 = stable molecule, and Be2 has

BO = 0 = this molecule does not exist.

π Molecular Orbitals                                                                                                              
For p orbitals, it is more complex:  two 2p orbitals can approach each and combine constructively to give bonding MOs that are lower in energy then the original atomic orbitals, and destrcutively, to give anti-bonding MO that are higher in energy than the original atomic orbitals. 

But the px, py and pz orbitals give rise to 2 different types of MOs:  when 2 px atomic orbitals come together to form a MO, they mix end-to-end to form both a sigma (σ) bonding and a sigma* (σ*) antibonding MO orbitals. 

 But, the two py and pz electrons can interact sideways to generate a pi (π) bonding and a pi (π*) antibonding MO orbitals. 

In a pi MO orbital (bonding or antibonding), the electron density is above and below the line joining the 2 nuclei of the bonding atoms.

A double bond is always 1 sigma bond and 1 pi bond; a triple bond is always 1sigma bond and 2 pi bonds. 

Molecular Orbital Diagrams

Rules for MO configuration & stability:  arrange the MOs in order of increasing energy:  (σ1s, σ*1s) = always there, and so therefore we can forget them;
then:   σ2s, σ*2s π2py = π2pz, σ2px, π*2py = π*2pz, σ*2px for Li2 --> N2       

and:   σ2s, σ*2s, σ2px, π2py = π2pzπ*2py = π*2pz, σ*2px for O2 --> Ne2

the number of MOs formed is always equal to the number of atomic orbitals you start with.

the more stable the bonding MO, the less stable is the corresponding antibonding MO orbital.

fill from low to high energies.  In a stable molecule, the number of electrons in bonding MOs is always greater than the number in antibonding MOs.

each MO can hold a maximum of 2 electrons with opposite spins (just as in atomic orbitals).

When electrons are added to MOs of the same energy, then they follow Hund’s rule => same spins (just as in atomic orbitals).

the total number of electrons in MOs is equal to the total of all the electrons on the bonding atoms.

Bond order = ½ [# electrons in bonding MOs  -  # electrons in antibonding MOs]

We will now predict the stability of the following ions and molecules:  H2+, H2, He2+ and He2. 

In H2+, there is a total of 1 electron shared by the two nuclei, and the bond order = ½.  Since a bond consists of two
electrons, and we only have one here, this ion may be stable.  The electron configuration is (σ1s)1.

In H2, there are 2 electrons and they both are in the bonding MO; bond order = 1 and therefore the molecule is stable, and we write  (σ1s)2

In He2+, there are 2 electrons in the bonding MO and 1 electron in the antibonding MO;

bond order =  ½ [2-1] = ½, and the stability is like that in H2+.    

We write  (σ1s)2(σ*1s)1

In He2, there are 2 electrons in the bonding MO and 2 electrons in the antibonding MO;

bond order =  ½ [2-2] = 0, and therefore no net stability.            

We write  (σ1s)2(σ*1s)2 

2nd period elements:  consider the following species and predict their relative stabilities:  Li2, Be2, B2, C2, N2, O2, F2, Ne2

For Li2, there are 6 electrons total to place: 2 electrons are in the bonding (σ1s)2 and 2 electrons in the antibonding (σ*1s)2; these essentially cancel each other out (as far as stability is concerned). 

There are then 2 more electrons in the (σ2s)2 bonding MO.  This forms the single covalent bond, with bond order = 1 = stable.    This molecule is diamagnetic.

When p orbitals are involved, this becomes much more involved:  2 p orbitals can come together to form either a sigma (σ) bond = head-to-head overlap, or pi (π) bonds = sideways overlap. 

We consider that the sigma bond is the overlap of  two 2px orbitals along the internuclear axis (one from each nucleus); these are called σ2px

The pi bonds are the sideways overlap of either 2py with 2py orbitals or 2pz with 2pz orbitals; these are called π2py  and π2pz.

The order of filling in the MOs is:

σ1s  <  σ*1s   <  σ2s < σ*2s  < π2py  =  π2pz  < σ2px  <  π*2py  =  π*2pz  <   σ*2px

This order is fine through N2, but the σ2px orbital lies lower than the π2py  =  π2pz  orbitals for O2 and F2.

For Be2, there are 8 electrons total to place: 2 electrons are in the bonding (σ1s)2 and 2 electrons in the antibonding (σ*1s)2; these essentially cancel each other out (as far as stability is concerned), leaving the 4 valence electrons to place.

Two electrons are placed in the (σ2s)2 bonding MO and 2 electrons are placed in the (σ*2s)2 antibonding orbital. 

The bond order = 0, and we predict that Be2 is not stable (and we find that this is correct experimentally).

For B2, there are 10 electrons total.  There are 6 valence electrons to place. 

Two electrons are placed in the (σ2s)2 bonding MO and 2 electrons are placed in the (σ*2s)2 antibonding orbital, canceling each other out as far as bond order is concerned.

We have to place the remaining 2 electrons in the π2py  =  π2pz  orbitals, one in each, with the same spin. 

Therefore the bond order = 1, and we predict that B2 is a stable compound (and we find that this is correct experimentally; bond length = 159 pm).  This molecule is paramagnetic.

For C2, there are 12 electrons total.  There are 8 valence electrons to place. 

Two electrons are placed in the (σ2s)2 bonding MO and 2 electrons are placed in the (σ*2s)2 antibonding orbital, canceling each other out as far as bond order is concerned.

We have to place the remaining 4 electrons in the π2py  =  π2pz  orbitals, 2 in each. 

Therefore, the bond order = 2, and we predict that C2 is a stable compound (and we find that this is correct experimentally, bond length = 131 pm; this is a double bond and is stronger than the bond in B2).   

This molecule is diagmagnetic.

For N2, there are 14 electrons total.  There are 10 valence electrons to place. 

Two electrons are placed in the (σ2s)2 bonding MO and 2 electrons are placed in the (σ*2s)2 antibonding orbital, canceling each other out as far as bond order is concerned. 

We have to place the remaining 6 electrons: 4 go into the π2py  =  π2pz  orbitals, 2 in each one, and the remaining 2 go into the (σ2p)2 orbital. 

The bond order = ½ [8 - 2] = 3, and we predict that N2 is a stable compound (and we know this is correct experimentally), and its bond length = 110 pm; this is a triple bond. 

This molecule is diamagnetic.


For O2, there are 16 electrons total; therefore, we have to place 12 valence electrons:

2s)2   (σ*2s)2p)2    2py)2pz)(π*2py)(π*2pz)1

The last 2 electrons go into antibonding orbitals according to Hund’s rule, and the bond order = 2 = double bond with a bond length = 121 pm. 

Also, note that this predicts that O2 molecule is paramagnetic (borne out by experiment).

In the valence bond model, all of the electrons seem to be paired, and therefore, the MO model is a better model for O2.


For F2, there are 18 electrons total; therefore, we have to place 14 valence electrons:

2s)2   (σ*2s)2p)2py)2pz)(π*2py)(π*2pz)2

The last 4 electrons go into antibonding orbitals, and the bond order = ½ [8 - 6] = 1 = single bond with a bond length = 143 pm (found by experiment).  This molecule is diamagnetic.


For Ne2, there are 20 electrons total; therefore, we have to place 16 valence electrons:

2s)2   (σ*2s)2p)2py)2pz)(π*2py)(π*2pz)(σ*2p)2

The bond order = ½ [8 - 8] = 0; therefore this molecule does not exist (found by experiment).

For exercises, predict whether the following ions are stable by calculating their bond order:  H2-, N2-, N2+, F2+, Ne2+

9.7 Bonding Theories and Descriptions of Molecules with Delocalized Bonding

The bonding theories we have seen have both strengths and weaknesses.  We will try to show these:

Strength:  we can make qualitative predictions about bond strengths and bond lengths.  Easy to draw and are used by chemists.

Weakness:  The structures are 2-dimensional, whereas real molecules are 3-D.  It also fails to account why bonds form.


Strength:  We can predict the shapes of many molecules and polyatomic ions.  Easy to draw and are used by chemists.

Weakness:   It also fails to account why bonds form (because it if based on Lewis theory of bonding).


Strength:  Describes the formation of covalent bonds as overlap of atomic orbitals. Bonds form because the resulting molecule has lower potential energy than the original, isolated atoms.

Weakness:   It fails to explain bonding in many molecules such as BeCl2, BF3 and CH4, where the central atom (in the groud state) does not have enough electrons to form the observed number of bonds.


Strength:  This is really only an extension of valence bond theory.  Using hybrid orbitals (sp, sp2, sp3, sp3d, sp3d2), we can understand the bonding and the geometry surrounding the central atom(s) of some molecules that we could not previously describe properly, such as:  BeCl2, BF3 and CH4

Weakness:   Valence bond theory and hybrid orbitals fail to predict some important properties of some molecules, such as the paramagnetism of O2.


Strength:  This theory allows us to predict accurately the magnetic and other properties of molecules and ions.

Weakness:   Pictures of molecular orbitals can be (and are) very complex.

Bottom line:  even though MO theory is the most powerful of all of the bonding models (gives us a more acurate picture of what molecules look like), we continue to use the othere modles when they can do a good job of explaining or predicting the properties of a molecule. 

Example:  if you need ato predict the 3-D shape of a AB5 molecule on an exam,you should draw its Lewis structure and apply VSEPR model.  This will give you much of the information you need.

If you try to draw the MO diagram, you will be helplessly bogged down becasue it very quickly gets too complicated.

If you are asked to predict whether a diatomic ion (ex., He2+) is stable or not, and whether it is paramagnetic or
diamagnetic, then you should draw a MO diagram.

It is best to use the simplest theory that can answer a particular question.


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Pure Substances and Mixtures Vocabulary


Pure substances:___________________________________________________________________

Saturated substances:______________________________________________________________

Unsaturated solutions:_____________________________________________________________

Concentrated substances:__________________________________________________________

Dilute substances:__________________________________________________________________

Homogenous substances:___________________________________________________________

Heterogeneous substances:_________________________________________________________

Fill in the blanks:
1. Anything that takes up space and has mass is called ____________.

2. There are four forms of matter;_________, _________, ___________ and _________.

3. _________ are hard objects and have definite shapes.

4. Matter is made up of tiny __________ with __________ between them.

5. The particles in a matter are always in ___________.

6. The more energy the particles have, the more _________ they move.

7. In a ________, the particles are close together.

8. In ________, the particles are able to slide past each other.

9. In a _________, the attractive forces between the particles are the weakest.

10. A _______________ contain only one kind of particles throughout.

11. A mixture contain two or more ___________.

12. A _________________ is a mechanical substance.

13. Air is a solution of _________.

14. __________ is frequently used to separate particles from a mechanical mixture.

15. Steel is a ____________ mixture.

16. Another name of Homogeneous mixture is ___________.

17. Homogeneous means____________.

18. Heterogeneous means____________.

19. ___________ is a substance that is dissolved in a solvent to form a solution.

20. ___________ is a substance that dissolves a solute to form a solution.

21. ___________ affects the dissolving rate of solutions.

22. Solutions with a low concentration of _________ are called __________ solutions.

23. Solutions with a high concentration of _________ are called __________ solutions.

24. ________ in the air allows fuel to burn.

                Atomic Structure Worksheet

Atomic Structure Worksheet

Fill in the blanks for the elements in this chart.  For the purposes of this chart, round all atomic masses to the nearest whole number.






Number of Protons

Number of Neutrons

Number of Electrons

Atomic Mass

Atomic Number


































































































Atomic Structure Worksheet – Solution Key

Fill in the blanks for the elements in this chart.  For the purposes of this chart, round all atomic masses to the nearest whole number.





Number of Protons

Number of Neutrons

Number of Electrons

Atomic Mass

Atomic Number

































































































































































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                  Atoms and Ions Practice Problems

Atomic Theory – Atoms and Ions




# of protons

# of electrons

# of electrons gained or lost

Net charge

Neon atom






Lithium ion




Lost 1



































Lost 1































Selenide ion



































Iodine atom


















Gained 2



Unnilseptium atom









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              Metals, Metalloids, and Nonmetals

Metals, Metalloids, and Nonmetals


1)    What’s the difference between a chemical and a physical property?  Give two examples of each and explain how they are different.


2)     Give four properties that are generally present in metals.


3)   If steel (a metal) is hard and granite (a nonmetal) is hard, why don’t we make automobile engines out of granite?


 4)       What are metalloids used for, and how does this affect modern technology?

Metals, Metalloids, and Nonmetals

Answer Key

1)   What’s the difference between a chemical and a physical property?  Give two examples of each and explain how they are different.

Chemical properties can only be tested by doing a chemical reaction, while physical properties don’t require making a chemical change.  

Examples of chemical properties are flammability, acidity, basicity, corrosion resistance, reactivity with water (or anything else), smell and taste (though these are debatable), and toxicity. 

Physical properties include color, shape, texture, melting and boiling point, density, mass, tensile strength, and brittleness.

2)   Give four properties that are generally present in metals.

          Metals generally have the following properties:

·       They are malleable (can be made into sheets)

·       They are ductile (can be made into wires)

·       They conduct electricity

·       They conduct heat and have low specific heat capacity

·       They are shiny

·       They react with acids and oxygen


3)   If steel (a metal) is hard and granite (a nonmetal) is hard, why don’t we make automobile engines out of granite?

Although the property of “hardness” is similar for steel and granite, other properties aren’t as desirable for manufacturing automobile engines. 

For example, the low heat conduction of rocks make them explode under high temperature. 

They are also brittle, making them poor for use in engine blocks.

4)   What are metalloids used for, and how does this affect modern technology?

Metalloids are used in the fabrication of computer chips. 

Because they only conduct electricity under some conditions, they are good for making electronic switches.


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 For each problem below, write the equation and show your work. Always use units and box in your final answer.


a. What are the similarities and differences between atomic orbitals and molecular orbitals?

b. Why is the bonding molecular orbital of H2 at lower energy than the electron in a hydrogen atom?

c. How many electrons can be placed into each MO of a molecule?


2. Consider the H2+ ion.

 a. Sketch the molecular orbitals of the ion, and draw its energy-level diagram.

 b. How many electrons are there in the H2+ ion? ______ Can you write a Lewis structure for the ion? _________ Explain

 c. Write the electron configuration of the ion in terms of its MOs.

 d. What is the bond order in H2+?

 e. Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state H2+ ion to fall apart? Explain.

 Molecular Orbitals

 For each problem below, write the equation and show your work. Always use units and box in your final answer.


a. What are the similarities and differences between atomic orbitals and molecular orbitals?

b. Why is the bonding molecular orbital of H2 at lower energy than the electron in a hydrogen atom?

c. How many electrons can be placed into each MO of a molecule?


2. Consider the H2+ ion.

a. Sketch the molecular orbitals of the ion, and draw its energy-level diagram.  

b. How many electrons are there in the H2+ ion? ______ Can you write a Lewis structure for the ion? _________ Explain

c. Write the electron configuration of the ion in terms of its MOs.

d. What is the bond order in H2+?

e. Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state H2+ ion to fall apart? Explain.  


a. Sketch the σ and σ* molecular orbitals that can result from the combination of two 2pz atomic orbitals.

b. Sketch the π and π* MOs that result from the combination of two 2px atomic orbitals.

c. Place the MOs from parts (a) and (b) in order of increasing energy, assuming no mixing of 2s and 2p orbitals.

4. Provide explanations for the following:

a. in Li2 the energy separation between the σ1s and σ*1s MOs is much less than that between the σ2s and σ*2s MOs.

b. The peroxide ion, O22-, has a longer bond than does the superoxide ion, O21-.

c. The π2p MOs of B2 are lower in energy than the σ2p MO.

5. Using Figures 9.34 and 9.39 as guides, give the molecular orbital electron configuration for each of the following cations:

a. B2+ b. Li2+ c. N2+ d. Ne22+

In each case indicate whether the addition of an electron would increase or decrease the bond order of the species.


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              Lewis Dot Diagram Worksheet


Lewis Dot Diagram- A notation of showing the valence electrons.

Directions: Show the Lewis dot diagrams for the following elements.

1) carbon






2) sodium






3) magnesium






4) aluminum






5) phosphorus






6) zinc






7) bromine






8) krypton








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                Periodic Table worksheet


Name ___________________________________________________________

 1. Define a family. _______________________________________________________ 

2. What is a period? ________________________________________________________ 

3. What is the symbol for the following elements. 

a. Magnesium _____________ b. Potassium ______________   c. Iron _______________ 

d. Copper ____________ 

4.  What are the names of the following elements. 

a. C __________________  b. Cl _________________  c. Au _________________  d. Sr _________________ 

5. What period are the following elements in?   a. He _______________  b. Ge _________________ 

c. Rb _______________  d. I __________________ 

6.  What group are the following elements?   a. Sulfur _______________  b. Ca _________________ 

c. Iodine _______________  d. Fe _________________  

7.  Write the name of an atom with the following characteristics. 

a. Halogen _________________

b. Nonmetal ________________ 

c. Alkali metal ______________

d. metalloid ________________ 

e. Lanthanide series __________

f. Alkaline Earth metal ________________ 

g. Transition metal ___________

h. Nobel gas ________________  

8. Write the electron configuration for  

a.  Li _______________________________________________________________ 

b. Na ______________________________________________________________ 

c.  K _______________________________________________________________ 


9.  How are each of the substances in problem 3 alike? 




10. Write the electron configuration for  

a. N ___________________________________________________________ 

b. P ___________________________________________________________ 

c. As __________________________________________________________  

11.  What are valence electrons? ______________________________________________________________________________________________


12.  How many valence electrons are in the following element?  

a. F ________  b. Cl ___________ c. Br ____________ d. I _____________  e. O ________ 

f. S ___________ g. Se ____________ h. Te ____________ 

13.  By looking at the valence electrons why are F, Cl, Br and Iodine in the same column of the periodic table?  _________________________________________________________________________________________________________



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1. What are cathode rays?

Why are they called cathode rays?

Do the cathode rays have a charge, if so what is the charge on each particle?

2. What are the differences in charge and mass among protons, neutrons, and electrons?

3. A particular atom of potassium contains 19 protons, 19 electrons, and 20 neutrons.

What is the atomic number of this atom?

What is the mass number?

Write the symbol for this potassium nucleus

4. How many electrons, neutrons, and protons are in atoms of chlorine with a mass number of 35  (neutral state)?

5. Yttrium was discovered in 1794. It is one of the elements used in superconductors.

How many electrons, protons, and neutrons are in an atom of yttrium-88 (neutral state)?

8. How many neutrons and protons are in each of the following nuclides?

a. carbon-14

# of protons  ___________________    # of neutrons  ______________________

b. phosphorus-32

# of protons  ___________________    # of neutrons  ______________________

c. nickel-63

# of protons  ___________________    # of neutrons  ______________________

d. iridium-192

# of protons  ___________________    # of neutrons  ______________________

e. iron-54

f. neptunium-235

# of protons  ___________________    # of neutrons  ______________________

10. Find the average atomic mass of the unknown element if the relative amounts are as follows:


Isotope Mass


103 amu


104 amu


105 amu


106 amu


107 amu


108 amu



11. What is the energy of a quantum of light with a frequency of 4.31 x 1014Hz?

12. What is the energy of light with a wavelength of 662 nm?

First find the frequency in hertz of this wavelength of light.


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                    Light Problems


Directions:  Solve the following problems.  Show proper set-up, work, and units for full credit. Box in your final answer. 


1.  A wave has a frequency of 22 Hz and a wavelength of 4.0 m. What is its velocity?



2.  What is the frequency of a wave if its wavelength is 3.6 x 10–9 m and its velocity is 3.0 x 108 m/s?



3.  As you move across the continuous spectrum from red to violet, what happens to…


a.  wavelength?


b.  frequency?


4.  A beam of microwaves has a frequency of 1.0 x 109 Hz. A radar beam has a frequency of 5 x 1011 Hz. Which type of radiation…


a.  has the longer wavelength?


b.  is nearer to visible light in the electromagnetic spectrum?


c.  is closer to X-rays in frequency value?


5.  A bright line spectrum contains a line with a wavelength of 518 nm. Determine…


a.  the wavelength, in meters. (Hint: 1 x 109 nm = 1 m)


b.  the frequency.


c.  the energy.


d.  the color of the line.


6.  A photon has an energy of 4.00 x 10–19 J. Find…


a.  the frequency of the radiation.


b.  the wavelength of the radiation.


c.  the region of the electromagnetic spectrum that this radiation represents.


7.  A photon of light has a wavelength of 3.20 x 105 m. Find…


a.  the frequency of the radiation.


b.  the energy of the photon.


c.  the region of the electromagnetic spectrum that this radiation represents.


8.  Determine the frequency of light with a wavelength of 4.257 x 10–7 cm.


9.  How many minutes would it take a radio wave with a frequency of  7.25 x 105 Hz to travel from Mars to Earth if the distance between the two planets is approximately 8.0 x 107 km?


10.  Cobalt-60 is an artificial radioisotope that is produced in a nuclear reactor for use as a gamma-ray source in the treatment of certain types of cancer. If the wavelength of the gamma radiation from a cobalt-60 source is 1.00 x 10–3 nm, calculate the energy of a photon of this radiation.


Selected  Answers:       

1.  88 m/s                                                 5a.  5.18 x 10–7 m                           5c.  3.84 x 10–19 J     

6b.  4.97 x 10–7 m                                  7 b.  6.21 x 10–31 J                            9.  4.4 minutes


2.  8.3 x 1016 Hz                                    5b.  5.79 x 1014 Hz                          6a.  6.03 x 1014 Hz    

7a.  938 Hz                                               8.  7.047 x 1016 Hz                        10.  1.99


 555 nm x (1 m / 109 nm)

555 x 10¯9 m = 5.55 x 10¯7 m

Inserting into λν = c, gives:

(5.55 x 10¯7 m) (x) = 3.00 x 108 m s¯1

x = 5.40 x 10141


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                  Electron Configurations

1.  What is electromagnetic radiation? 

How are frequency, wavelength, and speed of EMR related? 

Which of these forms of EMR has the shortest wavelength: IR, X-ray, microwaves, or blue light?

2.  What did the photoelectric effect prove?  How?

3.  What is a photon? 

How is the energy contained in a photon related to the frequency of the radiation? 

How is it related to the wavelength of the radiation?

4.  What happens inside an atom when it absorbs a photon of energy? 

What is this state of an atom called?

How are line spectra produced? 

5.  What does each line on a line spectrum correspond to?

What is a spectroscope used for?

6.  In which region of the electromagnetic spectrum are lines in the Paschen series of hydrogen found?

7.  How did each of these scientists contribute to our current knowledge of matter and energy: Max Planck, Niels Bohr, Louis de Broglie, Erwin Schrodinger, and Werner Heisenberg?

8.  What is the quantum theory?

9.  What is explained by the wave nature of electrons?

10.  What is an orbital, and how does it compare to Bohr’s orbits?

11.  Why can’t we know both the position and velocity of an electron?

12.  What are the four quantum numbers for an electron, and what does each one indicate?

13.  Draw, on three dimensional axes, an s orbital and the three p orbitals.

14.  Magnetism is a direct result of what phenomenon?

15.  List and explain diamagnetism and paramagnetism.

16.  What is the Aufbau principle?  How is it used?

17.  State and give an example of Hund’s rule.

18.  What does the Pauli exclusion principle state?

19.  What is an electron pair?

20.  What is significant about a noble gas configuration?  Which orbitals are filled in such a configuration?

Answer the following questions about arsenic (Z = 33), based on its position on the periodic table.  Do not write a configuration to help you answer the questions

What is its principle quantum number? 

How many electrons are in the outer shell? 

How many kernel electrons does it have? 

How many unpaired electrons does it have? 

How many full p orbitals does it have? 

Which energy level and sublevel are the last to be filled according to the Aufbau principle?

Write a complete and shorthand electron configuration for an atom of bismuth (Z = 83). 

Write the electron-dot notation for this atom.

Write an orbital notation for an atom of nickel (Z = 28).

Microwaves with a wavelength of 1.25 dm have what frequency? 

What amount of energy would one photon of this radiation contain? 

What amount of energy would a mole of these photons contain?

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            Electron Configuration Worksheet

1. Explain what each number and letter means in the following notation:   3p6


2. Determine the following for the 4th electron shell (4th main energy level) of an atom.

a)      The number of subshells it contains.


b)      The designation used to describe each of the first three subshells.


c)      The number of orbitals in each the first three subshells.


d)      The maximum number of electrons that can occupy the 4th electron shell.


e)      The maximum number of electrons that can occupy each of the first three subshells.


3. Fill in the numerical value(s) that correctly complete(s) each of the following statements.

a)      A 5f subshell holds a maximum of ______ electrons.


b)      A 4s orbital holds a maximum of ______ electrons.


c)      The maximum number of electrons in the third electron shell is ______.


d)      The forth shell contains ______ subshells, ______ orbitals, and a maximum of ______ electrons.


4. Give the maximum number of electrons that can occupy each of the following units.

a)  2p subshell              b)  5d orbital    c)   3s orbital     d)  5d subshell            e)  Fifth shell


5. Name the elements whose electron configurations are:

   a)      1s2 2s2 2p6 3s2 3p6 4s2 3d3


   b)      1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d9


6.  Write electron configurations for these elements:

    a)     Copper


    b)     Bromine

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1)         “3”= energy level, “p” = subshell designation, “6” = number of electrons

2a)       4

2b)       4s, 4p, 4d

2c)       1, 3, 5

2d)       32

2e)       2, 6, 10 

3a)       14

3b)       2

3c)       18

3d)       4, 16, 32

4a)       6

4b)       2

4c)       2

4d)       10

4e)       50

5a)       Vanadium

5b)       Silver

6a)       1s2 2s2 2p6 3s2 3p6 4s2 3d9

6b)       1s2 2 s2 2p6 3s2 3p6 4s2 3d10 4p5


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                             Atomic Structure Practice

1.   Calculate the wavelength of the spectral lines when an electron makes a transition from level 3 to level 2.

2.   Calculate the wavelength of a photon emitted when an electron in a hydrogen atom makes a transition from n=4 to n=2 using Rydberg's constant

3.  Calculate the energy of a photon emitted when an electron makes a transition from n=3 to n=2 and the wavelength from the result obtained

4.   The Vividh Bharati station of All India Radio, Delhi broadcasts on a frequency of 1,386kHz. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter

5.   The wavelength range of visible spectrum extends from violet (400nm) to red (750nm). Express the wavelengths in frequencies (Hz).

6.    The frequency of strong yellow line in the spectrum of sodium is 5.09 ´ 1014 per second. Calculate the wavelength of this light in nanometers

7.    Calculate the energy in SI units, the photons associated with the following wavelengths:
(a) 5000 Å     (b) 10-10cm   (c) 6000Å

8.    What is the color of the radiation when the electron of a hydrogen atom undergoes a transition from fourth energy level to second energy level?

9.   Calculate the wavelength of the limiting lines in the Balmer series (n=¥)

10.  How many moles of photons of lights having wavelength 400nm provide I.0J of energy?

11.   A photochemical reaction requires 9.6 X 10-16 Joule energy per molecule. Calculate the number of photons per molecule of light with wavelength 250nm that is just sufficient to initiate the reaction.

12.   Calculate the frequency and wave number of blue light (Use tables)

13.   If the energy difference between the electronic states of a hydrogen atom is 214.68kJ/mol, what will be the frequency of light emitted when the electron drops from higher energy level to lower energy level? (Planck's constant = 39.79 X 10-14 kJ sec /mol)

14.   A quantum of light having energy E has wavelength equal to 720 nm. Calculate the frequency of light that corresponds to energy equal to one quarter of E.

15.   Calculate the wavelength of the radiation that would cause the photochemical dissociation of a chlorine molecule. The bond dissociation energy of Cl-Cl bond is 245kJ/mole.

16.   Calculate the
(a) Number of electrons which together weigh one gram  

(b) Mass of one mole of electrons

(c) Charge of one mole of electrons

17.   One of the spectral lines of Cesium has a wavelength of 456nm.                                    Calculate frequency of  the line

18.   Calculate the number of each of the subatomic particles in the following neutral atoms:
                               He,        Cl,          N,            H,           Na ,            Mg,                    U

19.   The wavelength of first spectral line in the Balmer series is 656 nm. Calculate the wavelength of the second spectral line in the Balmer series.

20.   Calculate the energy of 1 mole of photons of radiations with frequency 5 ´ 1014 Hz.

21.   In the Balmer series of atomic spectra of hydrogen atom, a line corresponding to wavelength 656.4nm was obtained. Calculate the number of the higher orbit from which the electron drops to produce this line.


1.       Explain the electronic configurations of copper(Z=29) and chromium(Z=24) in ground states. Also list the 4 quantum numbers of the unpaired electron in the ground state.

2.       Compare the mass and charge of the 3 fundamental particles of an atom

3.       State Faraday's laws of electrolysis. Define radioactivity. What are the 3 principle radiations?

4.       How were the mass and charge of an electron determined?

5.       What were the observations made in Rutherford's experiment? How did this discard the first model of an atom? Explain Rutherford's nuclear model of an atom.

6.       What are isotopes? Give examples.

7.       What do you understand by electronic structure? Compare the Daltonian model with the quantum mechanical model of an atom. Explain the dual nature of light.

8.       The atomic spectra of elements are not continuous but consist of several lines. What assumptions did Bohr make to account for the line spectrum of an element? How did this explain the stability of the atom unlike the Rutherford model?

9.       The ionization energy of a hydrogen atom is +1312kJ/mol. Explain this with respect to Bohr's theory.

10.   State Heisenberg's uncertainty principle. What are atomic orbitals? What is the importance of the Schrodinger wave equation?

11.   What experimental evidences exist to explain the nature of electrons suggested by de Broglie? State the de Broglie relation.

12.   State Pauli's exclusion principle. How many electrons are there in a 5f-subshell?

13.   What sub shells are possible in the third energy level? How many orbitals can exist in this?

14.   Explain the significance of each of the four quantum numbers.

15.   State the AufBau principle and the n + l rule. Arrange the various orbitals from 1st energy to 5th energy level in increasing order of their energies

16.   How many electrons in Zinc (Z=30) have n + l =4?

17.   How many electrons in p-sub shell of argon have positive spins?

18.   An atom has 2 electrons in the K shell, 8 electrons in the L shell and 2 electrons in the M shell. Give the electronic configuration and find out the following:
(a) Atomic number                       (b) Total number of principal quantum numbers
(c) Total number of sub-levels     (d) Total number of s-orbitals and p-electrons

19.   Why is the 3d-subshell skipped in the building up of Calcium and Potassium atoms?

20.   How do you account for the following:       (a) the volume of the nucleus is very small as compared to the total volume of the atom   (b) e/m ratio of an electron is different from that of a proton                                                    (c) electron is a universal constituent of matter

21.   What is Hund's rule of maximum multiplicity? Explain with carbon and nitrogen as example

22.   Explain why half-filled and completely filled orbitals have extra stability.

23.   With the help of Pauli's exclusion principle and concept of atomic numbers for orbitals, show that M shell cannot have more than 18 electrons

24.   How are 2 electrons occupying the same orbital distinguished? What are the permitted values of the quantum numbers at the 5th energy level?

25.   Compare the shapes of 1s and 2s orbitals. What is the difference between the symbols l and L

26.   The expected electronic configuration of Cu is [Ar]3d94s2 but actually is [Ar]3d104s1. Give reasons and cite another example.

27.   What is meant by quantisation of energy? What is the shape of an s-orbital and a p-orbital?

28.   Write the electronic configuration for the ground state of B , Ne, Al, Cl, Cu, Zn, Sc, and Ca.

29.   (a) What are nucleons? Which shell would be the first to have a g-sub shell?
(b) Which series of hydrogen spectrum lies in the visible region?
(c) Which of the following sets of quantum numbers for orbitals in hydrogen atom has larger energy?                  1) n=3, l=2, m= -1, s =+1   

2) n=3, l=2, m= +1, s = - 1

30.   An atomic orbital has  n=3. List the possible values of l,m,and s.

31.   List the four quantum numbers of the unpaired electron in copper atom.

32.   Draw the shapes of 3s and 2px orbitals. Indicate the nodal plane when l=1?

33.   Write down the electronic configurations and quantum numbers of any electron in the outermost orbital:  copper, potassium, helium, phosphorus, scandium, chromium

34.   What is the maximum number of unpaired electrons in copper, bromide ion and K+ ion?

35.   In terms of the Bohr theory of the atom, why was it that the electrons do not spiral into the nucleus? Why was it abandoned?

36.   What is atomic spectrum? How is it different from an electromagnetic spectrum?

                        Electron Configuration Practice Worksheet

In the space below, write the unabbreviated electron configurations of the following elements:


1)         sodium           ________________________________________________


2)         iron                 ________________________________________________


3)         bromine         ________________________________________________


4)         barium           ________________________________________________


5)         neptunium    ________________________________________________


In the space below, write the abbreviated electron configurations of the following elements:


6)         cobalt             ________________________________________________


7)         silver               ________________________________________________


8)         tellurium        ________________________________________________


9)         radium           ________________________________________________


10)      lawrencium   ________________________________________________


Determine what elements are denoted by the following electron configurations:


11)         1s22s22p63s23p4  ____________________


12)         1s22s22p63s23p64s23d104p65s1  ____________________


13)         [Kr] 5s24d105p3  ____________________


14)         [Xe] 6s24f145d6  ____________________


15)         [Rn]  7s25f11  ____________________


Determine which of the following electron configurations are not valid:


16)         1s22s22p63s23p64s24d104p5 ____________________


17)         1s22s22p63s33d5  ____________________


18)         [Ra] 7s25f8  ____________________


19)         [Kr]  5s24d105p5 ____________________


20)         [Xe]  ____________________