Unformatted text preview: Calculations
In Chemistry
*****
Modules 23 and 24
Light, Spectra, and Electron Configuration *****
Module 23 – Light and Spectra ....................................................................................631
Lesson 23A:
Lesson 23B:
Lesson 23C:
Lesson 23D:
Lesson 23E:
Lesson 23F:
Lesson 23G: Waves ................................................................................................................. 631
Waves and Consistent Units ............................................................................ 636
Planck's Law ...................................................................................................... 641
DeBroglie’s Wavelength .................................................................................. 645
The Hydrogen Atom Spectrum ....................................................................... 650
The Wave Equation Model .............................................................................. 656
Quantum Numbers .......................................................................................... 658 Module 24 – Electron Configuration...........................................................................662
Lesson 24A:
Lesson 24B:
Lesson 24C:
Lesson 24D:
Lesson 24E: The MultiElectron Atom.................................................................................. 662
Shorthand Electron Configurations ................................................................ 666
Abbreviated Electron Configurations............................................................. 669
The Periodic Table and Electron Configuration ........................................... 673
Electron Configurations: Exceptions and Ions ............................................ 678
For additional modules, visit www.ChemReview.Net Module 23 — Light and Spectra Module 23 — Light and Spectra
Timing: Begin this module when wavelength and frequency calculations are assigned.
Pretests: If you believe that you know the material in a lesson, try two problems at the end
of the lesson. If you can do those calculations, you may skip this lesson.
***** Lesson 23A: Waves
Waves and Chemistry
Electromagnetic energy includes gamma rays, xrays, ultraviolet, visible, and infrared
light, microwaves, and radio waves. Each of these types of energy occupies a different
region of the electromagnetic spectrum.
Chemical particles can both absorb and release electromagnetic energy. This absorption
and release of energy can be a powerful tool in identifying chemical particles. Exposure to
certain types of electromagnetic energy can also cause chemical particles to change and
react.
In some cases, the behavior of electromagnetic energy is best predicted by assuming that
the energy is a particle, but in other cases, energy is best understood as a wave. Let us
begin by investigating the properties of waves. Wave Terminology
Crest 1 Wavelength (λ) 0 90 180 270 360 450 540 630 720 810 900 990 Trough
The following are some of the components of a wave that are important in chemistry.
1. Wavelength is the distance between the crests of a wave, which is equal to the distance
between the troughs of a wave.
a. The symbol for wavelength is λ (the lowercase Greek letter lambda).
b. Since a wave length is a distance, the units of wavelength are distance units, such as
meters, centimeters, or nanometers. © 2011 ChemReview.Net v.2m Page 631 Module 23 — Light and Spectra 2. Frequency is a number of events per unit of time. The unit for frequency is 1/time.
For waves, frequency is the number of wave crests that pass a point per unit of time.
a. In wave equations, the symbol for frequency is υ (the lowercase Greek letter nu).
b. The SI unit for time, a fundamental quantity, is seconds. Because frequency is a
derived quantity that is 1/time, the SI unit must be 1/seconds (s―1). The unit
second―1 is also called a hertz (Hz). During calculations, it is best to write hertz as
s―1 . Hertz and s―1 are equivalent and can cancel.
c. When wave frequency is expressed as cycles per second,” wave cycles are the
entity being measured, and 1/seconds is the unit. When writing wave units, the
term “wave cycle” or “cycle” is often included as a helpful label in conversion
calculations, but is usually omitted as understood in equation calculations.
3. The speed of a wave is equal to its frequency times its wavelength.
wave speed = λ υ = (lambda)(nu). Memorize the equation for wave speed in words, symbols, and names for the symbols. Wave Calculations
Because wave relationships are often defined by multiterm equations, wave calculations
are generally solved using equations rather than conversions. We will start with a simple
problem that can be solved using both methods, but to practice with the equations that will
be required for more complex calculations, solve the problem below using the equation
method (for method review, see Lessons 17D or 21B).
Q. If ocean waves are traveling at 200. meters/minute and the crests pass a fixed point
at a rate of 15.0 waves per minute, what is the wavelength, in meters? * * * * * (When you see * * * * , cover below, solve, and then check below.)
Write the one equation learned so far for waves.
Wave speed = λ υ
List those three terms in a data table. After each term, write the data in the problem that
corresponds to the term. Add a ? and the desired unit after the WANTED symbol.
*****
Wave speed = 200 m/min. λ = ? meters
υ = 15.0 wave cycles/min. = 15.0 min.―1 (speed units are distance over time)
(the length of a wave is a distance)
(frequency units are 1/time) When solving wave frequency calculations using equations, “wave cycles” is usually
omitted as understood to be the object being measured.
Solve the equation in symbols for the WANTED symbol, then substitute the DATA.
Include the consistent units and check the unit cancellation.
***** © 2011 ChemReview.Net v.2m Page 632 Module 23 — Light and Spectra SOLVE: Since Wave speed = λ υ λ in meters = speed = speed • 1 = 200. m •
υ υ min. 1 = 13.3 meters 15.0 min.―1 Note in the unit cancellation in the denominator:
min.• min.―1 = min.1• min.―1 = min.0 = 1 . Anything to the zero power equals one. Practice A: Check your answers at the end of this lesson. 1. Write the SI units for
a. Wavelength b. Frequency c. Energy d. Speed 2. Street lights containing sodium vapor lamps emit an intense yellow light at two close
wavelengths. The more intense wave has a frequency of 5.09 x 1014 Hz. If light travels
at the speed of 3.00 x 108 m • s―1 , what is the wavelength of this intense yellow wave
in meters? (Use the equation method to solve.) Electromagnetic Waves
The movement of electric charge creates electromagnetic waves. The waves propagate:
they travel outward from the moved charge. The energy that was added to move the
charge is carried outward by the waves.
In a vacuum, all electromagnetic waves travel at the speed of light:
3.00 x 108 meters/second. The speed of light is the “speed limit of the universe:” the
fastest speed possible for energy or matter. In wave calculations, the speed of light is given
the symbol c.
Electromagnetic waves slow when they travel through a medium that is denser than a
vacuum, but when passing through air or other gases at normal atmospheric pressures, the
speed of light does not slow sufficiently to affect most calculations in chemistry.
For electromagnetic waves, this relationship will be true (and must be memorized):
Speed of Light = c = λ υ = 3.00 x 108 m/s in vacuum or air Since c is a constant, υ and λ are inversely proportional. As wavelength goes up, frequency
must go down. If υ goes up, λ must go down.
Further, as long as we work in consistent units and in air or vacuum, since c is constant, a
specific value for the frequency of an electromagnetic wave will always correlate to a
specific value for its wavelength. © 2011 ChemReview.Net v.2m Page 633 Module 23 — Light and Spectra The Regions of the Electromagnetic Spectrum
The electromagnetic spectrum goes from waves with very high frequencies (and very low
wavelengths) to waves with very low frequencies (and very high wavelengths. The regions
of the spectrum are assigned different names to help in predicting the types of interactions
that the energy will display. However, all of these forms of energy are electromagnetic
waves. The difference among the divisions of the spectrum is the length (or corresponding
frequency) of the waves.
The following table (no need to memorize) summarizes some of the general divisions of the
electromagnetic spectrum.
Frequency (s―1) Wavelength (m) Type of Electromagnetic Wave 1024 3 x 10─16 Gamma Rays 1021 3 x 10─13 1018 3 x 10─10 Xrays 1015 3 x 10─7 Ultraviolet, Visible, Infrared Light 1012 3 x 10─4 Microwaves 109 3 x 10─1 UHF Television Waves 106 300 Radio Waves Units For Frequency and Wavelength
Measurements of wavelengths and frequencies often
involve very large and very small numbers. Values are
often expressed using SI prefixes such as gigahertz
(GHz) or nanometers (nm). Prefixes needed most often
are those for powers of three. Engineering Notation
Scientific notation expresses a value as a significand
between 1 and 10 times a power of 10.
Engineering notation expresses values as a significand
between 1 and 1,000 times a power of 10 that is
divisible by 3. In wave calculations, answers are often
preferred in engineering rather than scientific notation
to ease conversion to the metric prefixes based on
powers of three. Prefix Symbol Means tera T x 1012 giga G x 109 mega M x 106 kilo k x 103 milli m x 10―3 micro μ x 10―6 nano n x 10―9 pico p x 10―12 femto f x 10―15 Examples: Converting scientific to engineering notation,
5.35 x 10―4 m = 535 x 10―6 m in engineering notation ( = 535 micrometers = 535 μm)
9.23 x 1010 Hz = 92.3 x 109 Hz in engineering notation ( = 92.3 GHz ) © 2011 ChemReview.Net v.2m Page 634 Module 23 — Light and Spectra To convert any exponential notation to engineering notation, adjust the exponent and
decimal position until the exponent is divisible by 3 and the significand is between 1
and 1,000.
(To review moving the decimal, see Lesson 1A). Try this example.
Q. Convert 5.27 x 10―11 m to engineering notation, then convert the exponential to a
metricprefix.
*****
A. 5.27 x 10―11 m = 52.7 x 10―12 m in engineering notation = 52.7 picometers or 52.7 pm
52.7 x 10―12 is the only way to write the given quantity that results in both an
exponent divisible by 3 and a significand between 1 and 1,000.
Engineering notation, like scientific notation, results in one unique expression for
each numeric value, and this makes answers easy to compare and check.
During electromagnetic wave calculations, you should work in general exponential
notation, then, at the end, convert your answers to either scientific or engineering notation,
depending on the system preferred for wave calculations in your course. Practice B: Do every other question. Complete the rest during your next study session. 1. By inspection, convert these to units in engineering notation, without prefixes.
a. 5.4 GHz b. 720 nm c. 96.3 MHz 2. Convert these first to engineering notation, then to measurements with metric prefixes
in place of the exponential terms.
a. 47 x 10―7 m b. 347 x 104 Hz c. 1.92 x 10―8 m d. 14,920 x 10―1 Hz e. 0.25 x 1011 Hz f. 7,320 m ANSWERS
Practice A
1. a. Wavelength is a distance, and the SI unit for distance is the meter (m).
b. Frequency is defined as 1/time, and the SI unit for time is the second, so the SI unit of υ is 1/s or
s―1, which is called a Hertz.
c. Energy The SI unit for energy is the joule (J) or kg • m2 • s―1 .
d. Speed is defined as distance over time, so the SI units are meters/second (m · s―1)
2. Wave speed = λ υ
Speed = 3.00 x 108 m • s―1 © 2011 ChemReview.Net v.2m Page 635 Module 23 — Light and Spectra λ in meters = ?
υ = 5.09 x 1014 Hz s―1 ( During calculations, write Hz as s―1 ) ? = λ = speed = 3.00 x 108 m • s―1 = 0.589 x 10―6 m = 5.89 x 10―7 m
υ
5.09 x 1014 s―1
Practice B
1a. 5.4 GHz = 5.4 x 109 Hz or s―1 1b. 720 nm = 720 x 10―9 m 1c. 96.3 MHz = 96.3 x 106 s―1
2a. 47 x 10―7 m = 4.7 x 10―6 m = 4.7 μm (if exponent is made larger, make significand smaller) 2b. 347 x 104 Hz = 3.47 x 106 Hz = 3.47 MHz 2c. 1.92 x 10―8 m = 19.2 x 10―9 m = 19.2 nm 2d . 14,920 x 10―1 Hz = 1.492 x 103 Hz = 1.492 kHz
2e. 0.25 x 1011 Hz = 25 x 109 Hz = 25 GHz (significand must be between 1 and 1,000)
2f. 7,320 m = 7.32 x 103 m = 7.32 km ***** Lesson 23B: Wave Calculations and Consistent Units
When using equations to solve science calculations, the units of measurements must be
consistent: for each quantity in the equation, all measurements of that quantity must be
converted to the same unit.
How do you decide which units to choose as the consistent units in a problem? When
choosing consistent units in equations we have studied previously, we have used two rules.
• When using a specified value and units for the gas constant R, our rule was:
convert the DATA to the units used in the constant. • When using specific heat capacities (c), our rule was: convert to the unit of the most
complex term in the DATA. In wave calculations, we will follow both of these rules. Specific constants will be required,
but those constants will generally contain the most complex units in the problem.
The following rules will help in choosing and converting to consistent units.
1. Write the equation needed for the problem.
2. If the equation has constants, in the DATA table, first write each constant’s symbol,
value and units, then below write the symbols for the variables.
3. In the DATA table, after each variable symbol, write its chosen consistent unit.
Example: DATA: λ in m = (a wavelength is a distance) To choose the consistent units to write after each variable symbol, apply these steps.
a. If the equation has constants, after each variable write a unit that is appropriate for
the symbol and is consistent with the units in the constants. © 2011 ChemReview.Net v.2m Page 636 Module 23 — Light and Spectra Example: If c= λυ is the equation that is needed, list DATA: c = 3.00 x 108 m · s―1 λ
υ in m =
in s―1 = (list constants in the equation first) (wavelength is a distance; the distance unit in c is meters)
(υ = 1/time, the time unit used in the constant c is seconds) b. If there are no constants in the equation, label each variable with an appropriate
unit matching the units used in the WANTED unit.
4. If the unit that is WANTED is not specified,
a. pick a WANTED unit to match the units in the constants of the equation.
b. If no constants are used, write after the WANTED unit the SI unit for the quantity.
5. In the DATA, write the data supplied for each symbol, then convert that DATA to the
consistent units if needed.
6. First solve for the WANTED symbol in the consistent unit, then convert to a different
WANTED unit if specified.
The problem below will help you to understand and remember the rules above.
Q. When neon gas at low pressure is subjected to high voltage electricity, it emits
waves of light. One of the more intense waves in the visible spectrum has a
wavelength of 640. nm, perceived by the eye as red light. What is the frequency of
this light in terahertz (THz)? Solve using the steps above.
*****
Answer
This problem involves a frequency (υ) and a wavelength (λ) for light. We know that
light travels at the speed of light (c), a constant. So far, we know only one equation that
relates those three symbols. So, to start, your paper should look like this:
c= λυ
DATA: c = 3.00 x 108 m · s―1
λ=
υ= For the speed of light (c), in conversion calculations the unit m/s must be used as a ratio
and written in the top/bottom format, but in equations, it will simplify unit cancellation
if the units are written in the “on one line” format: 3.00 x 108 m · s―1 .
Now assign a consistent unit to each variable. Since this equation has a constant (c),
after each variable symbol write a unit that both measures the variable and matches one
of the units used in the constant.
Do that step, then check below. © 2011 ChemReview.Net v.2m Page 637 Module 23 — Light and Spectra *****
Your DATA should look like the Rule 2a Example above, minus the (comments).
Next, after the = sign for each variable, write the data for that variable that is supplied
in the problem. Then, in the DATA table,
• convert the supplied units to the consistent units if needed. • For the WANTED variable, after the assigned consistent unit, write the unit
WANTED in the problem if it is not the consistent unit. Do those steps, and then check your answer below.
*****
Your paper should look like this.
c= λυ
DATA: c = 3.00 x 108 m • s―1
λ in m = 640. nm = 640. x 10―9 m
υ in s―1 ; find s―1 then THz WANTED For λ , meters is the chosen consistent unit, so you must convert nm to m. The easy way
is to substitute what the prefix means.
To SOLVE,
• First solve the equation for the WANTED symbol in symbols. • Substitute the DATA into the solved equation using the consistent units, and solve
including the units. • If needed, convert to the unit WANTED in the problem. Modify your work if needed and finish.
*****
SOLVE: ? = υ in s―1 then THz = c λ = 3.00 x 108 m • s―1 = 4.69 x 1014 s―1
640. x 10―9 m That solves in the consistent unit. To finish, convert to the WANTED unit.
*****
4.69 x 1014 s―1 Hz • 1 THz = 469 THz
1012 Hz Done!
This method of choosing to solve in the units of the constants, is arbitrary. You can use any
consistent units to solve. In physics, it is usually preferred (and simplifying) to solve all
problems in SI units. In many chemistry calculations, constants will be stated in SI units,
matching the physics practice.
However, “solving in the units of the constants” will save a few steps if data is provided in
mL, grams, or other nonSI base units, as will be the case in some science calculations. © 2011 ChemReview.Net v.2m Page 638 Module 23 — Light and Spectra Try Q2. How many wavelengths of the 640. nm red neon light above would fit into
one centimeter? (one wave cycle = 1 wave = 1 wavelength) *****
If you are not sure how to proceed, list the data, try to assign symbols, and see if the
symbols fit a known equation.
*****
The wanted unit is waves/cm, which is the inverse of wavelength, not wavelength.
Plus, none of the data has a frequency, so the data does not match the one equation we
know so far.
Note that the data includes an equality, and that all of the data can be listed as
equalities or ratios. That’s a hint: try conversions to solve.
*****
WANT:
DATA: ? waves
cm (you want the waves per one cm, a ratio unit) one wave = 1 wavelength (2 measures of the same object) one wavelength = 640. nanometers
Though “waves” or “wave cycles” is usually left out of wave equation calculations as
understood, including “waves” may help when using conversions. If needed, adjust
your work and then finish the problem.
*****
One way of several to SOLVE is
? waves =
1 wave
•
cm
1 wavelength 1 wavelength • 1 nm • 10―2 m = 15,600 waves
1 cm
cm
640. nm
10―9 m How many waves of red light fit into a centimeter? Quite a few. Summary: Frequency and Wavelength
1. Wavelength: the distance between the crests of a wave. The symbol is λ (lambda).
T The units are distance units: the base unit meters, or nanometers, etc.
2. Frequency: the number of times a wave crest passes a point per unit of time.
The symbol is υ (nu). The units are 1/time .
SI frequency units = wave cycles per second = 1/seconds = s―1 = hertz (Hz).
In calculations, write hertz as s―1 so that units will cancel properly.
3. The speed of a wave is equal to its frequency times its wavelength.
Wave speed = λ υ © 2011 ChemReview.Net v.2m = (lambda)(nu). Page 639 Module 23 — Light and Spectra 4. Electromagnetic waves travel at the speed of light (symbol c).
For all electromagnetic waves, c = λ υ = 3.00 x 108 m · s―1 in vacuum or air. 5. To simplify solving equations that include a constant that has units,
a. In the DATA table,
• list the constants first. • After each symbol for a variable, write a unit to convert DATA to. Choose if
available a unit that matches the constants of the equation. If there are no
constants, convert to the WANTED unit or to an SI unit. b. First solve in the consistent unit, then convert to the WANTED unit if needed. Practice (Additional practice with λ and υ will be provided in lessons that follow.) 1. If an AM radio station broadcasts a signal with a wavelength of 0.390 km, what is the
frequency of the signal on a radio tuner, in kHz?
2. If there are 225 waves per centimeter, what is the wavelength of the waves in meters? ANSWERS
1. (The data is a λ and wanted is a υ. The equation that relates those two variables is:)
c= λυ
DATA: c = 3.00 x 108 m • s―1 λ in m = 0.390 km = 0.390 x103
υ in s―1 = ? , then convert to kHz (list constants used in the equation first)
m = 390 m 1 kHz = 103 Hz = 103 s―1 ( convert to units used in the constants)
(listing metric conversions is optional) ? = υ in s―1, then kHz = c = 3.00 x 108 m • s―1 = 0.77 x 106 s―1 • 1 kHz = 770 kHz
λ
390 m
103 s―1
(Hertz and s―1 are equivalent and can cancel.)
2. (If you are not sure how to proceed, list the data, assign symbols, then see if the symbols fit a known
equation.)
WANTED: λ DATA: 225 waves = 1 cm in m = ? (or meters/wave) (If no equation seems to fit the DATA, try conversions to solve. If needed, use that hint and finish.
***** © 2011 ChemReview.Net v.2m Page 640 Module 23 — Light and Spectra If the WANTED unit is rewritten as meters/wave, a ratio is wanted, and a ratio is in the data to start from.
Arrange the given so that one unit is where is WANTED (Lesson 11B), but any order for these two
conversions works.)
SOLVE: ? meters =
1 cm •
wave
225 waves 10―2 m
1 cm = 4.44 x 10―5 m
wave The wavelength is 4.44 x 10―5 m.
***** Lesson 23C: Planck’s Law
Energy and Frequency
In 1900, the German physicist Max Planck, studying the blackbody radiation emitted by
objects at high temperature, discovered that energy is absorbed by or emitted from atoms in
bundles of a small but constant size, or in multiples of that constant size.
Building on Planck’s work, in 1905 Albert Einstein proposed an explanation for the
photoelectric effect: the observation that when UV light shines on a metal, the metal emits
electrons.
Einstein postulated that light can be considered to be made of small particles called photons.
In Einstein’s formulation, electromagnetic energy has characteristics of both a wave and a
particle, with photons acting as particles of energy that travel at the speed of light. These
particles he called quanta. A single particle is a quantum. The energy of each photon is
correlated to its frequency as a wave.
For photons, the equation that relates frequency and energy is called Planck’s law.
Ephoton = h υ
In this equation, h is a number with units called Planck’s constant.
Planck’s constant = h = 6.63 x 10―34 joule · seconds
Because frequency must be positive, and Planck’s constant is small but positive, the
equation E = hυ means that as the energy of a wave increases, its frequency must increase,
and that higher frequency waves have higher energy.
Since calculations using Planck’s constant involve electromagnetic waves, we can
• use our previous equation for the speed of those waves, • solve that equation for υ: υ = c / λ , then • write the photon energy equation as E = h υ c=λυ , or, substituting for υ , E= h· c
λ These two general forms of Planck’s law are equivalent. The first solves for energy in terms
of frequency, the second in terms of wavelength. The first should be memorized, and the
second either memorized (“for wavelength, the two constants are on top”) or derived as
needed. © 2011 ChemReview.Net v.2m Page 641 Module 23 — Light and Spectra Together, these equations mean that in the case of electromagnetic waves, for the three
variables energy, frequency, and wavelength, if you know any one, you can calculate both
of the other two.
Further, it will always be true that as photon energies go up, the corresponding frequencies
go up and wavelengths go down. Waves with higher energy have higher frequency and
shorter wavelength. Calculations Using Planck’s Law
In solving the problems in these lessons, it is expected that you will be able to write from
memory the equations for waves and the value of the speed of light (which is used quite
often in science), but you will be supplied the value for Planck’s Constant (h) when it is
needed in a problem.
When using Planck’s law, to simplify problemsolving we will use the same rules as other
equations.
• Identify and write the needed equation. • In the DATA table, list the constants in the equation and their values first. • After each symbol for a variable, write a consistent unit. Choose the units of the
constants if the equation has constants. If not, choose the WANTED unit if it is
specified, or choose the SI unit for each quantity. • In the DATA table, convert all DATA to the consistent units. • Solve for the WANTED symbol in the consistent unit, then convert to other
WANTED units if needed. Try those steps on this problem.
Q. Cosmic rays are highenergy radiation that enters the earth’s atmosphere from
space. The energy of a single cosmic ray photon can be as high as 50. joules. What
would be the frequency of this radiation in Hz? ( h = 6.63 x 10―34 J · s ) *****
To decide which equation is needed to solve a problem, try this method: as you read the
problem, write the symbol that fits the unit for each item of WANTED and DATA you
encounter. Simply listing the symbols as you read will often quickly identify the
equation that you need to solve.
*****
50 J = E, ? = υ . The fundamental equation that relates E and υ is ?
***** E = hυ Write a data table and solve. *****
DATA: h = 6.63 x 10―34 J · s (list constants first, use their units) E in J = 50 J υ in s―1 = ? © 2011 ChemReview.Net v.2m then convert to Hz WANTED Page 642 Module 23 — Light and Spectra ? = υ ( in Hz ) = E = h 50. J
6.63 x 10―34 J · s = 7.5 x 1034 s―1 Hz Photons with this is extremely high energy and frequency are produced by nuclear
processes in stars. Let’s try a problem with a more commonly encountered energy.
Q2. A microwave oven warms food by producing radiation with a typical wavelength
of about 12 cm. What is the energy of this wave? ( h = 6.63 x 10―34 J · s )
*****
Answer
As you read, you encounter a λ and a E. The equation that uses λ and E is ?
***** E= h · c
λ
DATA: c = 3.00 x 108 m · s―1
h = 6.63 x 10―34 J · s (list the two constants first, use their units) E in J = ? λ in m = 12 cm = 12 x 10―2 m = 0.12 m ( h uses joules )
( c uses m, c = x 10―2) (In the DATA table, convert DATA to the consistent unit.)
***** E= h · c
λ = ( 6.63 x 10―34 J · s ) ( 3.00 x 108 m · s―1 ) = 1.7 x 10―24 J
0.12 m Practice
1. Based on E = h υ and the rules for unit cancellation, what must the SI unit for
Planck’s constant be? Why?
2. The human eye can generally see energy waves in the range of 400 to 700 nm. When
hydrogen gas at low pressure is subjected to high voltage, it emits four waves of light in
the visible region of the spectrum: one red, one bluegreen, one blueviolet, and one
violet. ( h = 6.63 x 10―34 J • s )
a. A photon of red light from the hydrogen spectrum has an energy of 3.03 x 10―19 J.
What is the wavelength of this light in nanometers?
b. The bluegreen line consists of waves with a frequency of 615 THz. What is the
energy of these waves?
c. The blueviolet line has a wavelength of 434 nm. What is the frequency of these
waves in Hz? © 2011 ChemReview.Net v.2m Page 643 Module 23 — Light and Spectra ANSWERS
1. Since the SI unit for E is joules and for υ is s―1 ., the units of E = h υ must be J = (? Units) • s―1.
For unit cancellation to work, the units of h must equal J • s (see Lesson 21D).
(In SI base units, since Energy = work = mad = (mass)(acceleration)(distance) = J units = kg · m2 · s―2 ,
the units of h = ? = J · s also = kg · m2 · s―1 ), but base units are not needed for most calculations).
2a. (Part (a) involves E and λ . The equation that relates those variables is) E= h · c
λ
h = 6.63 x 10―34 J · s DATA: c = 3.00 x 108 m · s―1 (list the two constants, convert DATA to those units) E in J = 3.03 x 10―19 J λ in m = ? then convert to nm
1 nm = 10―9 m ( c uses meters ) (listing fundamental metric conversions is optional in DATA ) ***** λ (in m) = h · c = ( 6.63 x 10―34 J · s ) ( 3.00 x 108 m · s―1 ) = 6.56 x 10―7 m
E
3.03 x 10―19 J
λ (in nm) = 6.56 x 10―7 m = 656 x 10―9 m (engineering notation) = 656 nm
or λ (in nm) = 6.56 x 10―7 m •
2b. 1 nm = 6.56 x 102 nm = 656 nm
10―9 m (Part (b) involves E and υ . The equation that relates those symbols is) E=hυ
DATA: h = 6.63 x 10―34 J • s (list constants first. Use their units for WANTED and DATA) E in J = WANTED υ
SOLVE: in s―1 = 615 THz = 615 x 1012 Hz = 615 x 1012 s―1 ( h uses seconds ) E (in J) = h υ = ( 6.63 x 10―34 J • s ) ( 615 x 1012 s―1 ) = 4.08 x 10―19 J 2c. (The problem has λ and υ. The equation that relates λ and υ for electromagnetic waves is)
c= λυ
DATA: c = 3.0 x 108 m · s―1 (list constants used in the equation first) λ in m = 434 nm = 434 x 10―9 m
υ in s―1 = ? , then convert to Hz © 2011 ChemReview.Net v.2m Page 644 Module 23 — Light and Spectra SOLVE:
? = υ in s―1 , then Hz = c = 3.00 x 108 m · s―1 = 0.00691 x 1017 s―1 = 6.91 x 1014 Hz
λ
434 x 10―9 m
***** Lesson 23D: De Broglie’s Wavelength
Timing: Do this section if you are asked to solve calculations using the De Broglie
equation, and/or if you plan to take physics.)
***** De Broglie’s Wavelength Equation
In 1923, the French physicist Louis De Broglie proposed that, just as energy has particlelike
properties, particles can have wavelike properties. In the equation De Broglie derived to
predict the characteristics of these “matter waves,” the length of the wave associated with a
particle depends on the mass of the particle and its speed (or velocity). The equation is
λ= h
mass · speed Though De Broglie’s equation can be applied to any moving particles, it is most often
applied to small particles such as electrons.
Apply De Broglie’s wavelength equation to the following problem. If you need a hint, read
a part of the answer below, then try again.
Q. Calculate the wavelength of an electron (mass of electron = 9.11 x 10―28 grams) that
is traveling at onetenth the speed of light.
*****
The problem involves wavelength (λ), mass, and speed. The equation that relates those
three variables is the De Broglie wavelength equation.
λ= h
mass · speed DATA: h = 6.63 x 10―34 J · s
mass = 9.11 x 10―28 g (list the two constants first) speed = 0.100 x 3.00 x 108 m · s―1 = 3.00 x 107 m · s―1
λ= ?
SOLVE: ? = λ = h
=
6.63 x 10―34 J · s
mass · speed
(9.11 x 10―28 g)( 3.00 x 107 m · s―1) =??? We have a problem. The units do not cancel. You could ignore that, but if you do, you will
do a lot of work to arrive at an answer that is not correct. © 2011 ChemReview.Net v.2m Page 645 Module 23 — Light and Spectra When units do not cancel properly, something is likely wrong. Let us see if we can fix this
problem.
Joules is an abbreviation for a combination of SI base units used to measure distance, mass,
and time. In many calculations in chemistry, joules can be left “as is” in an answer unit, as
a measure of energy. The De Broglie wavelength equation, however, a case where, for the
units to cancel, joules must be converted to the base units that joules is an abbreviation for.
To convert joules to its base units, you have two choices. You can memorize the base units
that are equivalent to joules, or you can derive the base units from simple relationships if
you know a bit of physics. Let’s review the process for converting unit abbreviations such
as newtons, pascals, watts, and volts to the fundamental SI base units that those named units
are abbreviating. This conversion process will be helpful to know, especially if you take
future courses in chemistry, physics, or engineering. Converting Joules to Base Units
What is a joule of energy?
Energy is the capacity to do work. There are many forms of energy: including potential,
kinetic, thermal, and electrical. There are many units that can be used to measure energy,
including calories, ergs, BTUs, and kilowatthours. The SI unit used to measure energy is
the joule.
Energy is a derived quantity. It can be defined in terms of the fundamental quantities of
distance, mass, and time. The easiest way to do so is to define energy in terms of what in
physics is termed “mechanical work.”
Work is defined as the product of a force acting over a distance. In equation form:
Energy = work = force times distance = F • d
Since force is equal to mass times acceleration (F = ma), we can write
Energy = work = F • d = (mass times acceleration) times distance = m·a·d , or
Energy = work = m·a·d (This relationship can be remembered as “work is mad!”) The joule, the derived SI unit measuring energy, is defined in terms of the SI base units that
measure the fundamental quantities. The SI baseunits are derived from what was
historically known as the mks system.
• Distance is measured in the SIbase unit of meters (m). • The SI base unit used to measure mass is the kilogram (kg), not the gram.
(In the SI system, base units for all other fundamental quantities do not include a
metric prefix. The kilogram used for mass is the exception.) • The base unit for time is seconds (s), and acceleration is defined as distance/time2,
so acceleration is measured in base units of meters/second2 ( = m · s―2 ). Based on Energy = work = m·a·d , the SI unit measuring energy (one joule) is defined as
the energy needed to accelerate one mass base unit (one kg) by one acceleration base unit
(1 m · s―2) for one distance base unit (1 m). By substituting into the definition of energy
Energy = m · a · d © 2011 ChemReview.Net v.2m the base units used to measure each of those four terms: Page 646 Module 23 — Light and Spectra 1 Joule = 1 kg · (1 m · s―2 ) · 1 m , this produces a definition of
Joule = kg · m2 · s―2 This key equality relates joules to its SI base units. In the unit for h, if we substitute those base units in place of joules, and measure the other
variables in the data in those base units, the units will cancel properly in De Broglie
wavelength and other calculations.
To summarize: To solve the De Broglie equation, for the units of h, substitute for joules the
base units that joules are equivalent to. Either memorize the base units for joules, or
remember that “work is mad!” and substitute the SI base units (mks) that measure m, a,
and d.
In general, SI derived units that are abbreviations for combinations of base units (such as
joules, newtons, watts, and pascals) can be converted to SI base units using these steps:
To convert an SI combined unit to its base units
1. Write the equation that defines the derived quantity that the unit measures using
quantities for which the SI base units are known; then
2. Substitute into the equation the base units used to measure those quantities.
(For more on converting unit abbreviations such as Joules to SI base units, see Lesson 19C).
*****
Now let’s return to our De Broglie wavelength calculation.
In your DATA table, substitute for joules the base units of joules, adjust your DATA units to
match these new units for the constant, solve, and then check your answer below.
*****
To arrange for the units to cancel, begin by substituting the base units for joules into h:
DATA: h = 6.63 x 10―34 J ( kg · m2 · s―2 ) · (s) (list constants first) Write after the mass variable in the DATA table the units for mass used in the constant h,
then convert the mass DATA to those units.
mass in kg = 9.11 x 10―28 g • 1 kg
103 g = 9.11 x 10―31 kg Mass, to have consistent units that will cancel in the equation, must be kg and not g.
Speed (distance over time) is supplied in the problem in the base units used in h.
speed (in m/s) = 0.100 x 3.00 x 108 m · s―1 = 3.00 x 107 m · s―1
Since the unit for the WANTED wavelength is not specified, pick a unit to attach to the
symbol that either uses the units in the constant or is an SI unit. In this and in most
problems, both choices will result in the same unit. Since wavelength is a distance, solve
for λ in the distance SI base unit: meters.
SOLVE:
? = λ (in m) = =
6.63 x 10―34 ( kg · m2 · s―2 ) · s = 2.43 x 10―11 m
h
mass · speed
(9.11 x 10―31 kg) (3.00 x 107 m · s―1) © 2011 ChemReview.Net v.2m Page 647 Module 23 — Light and Spectra Mark the unit cancellation on your paper carefully. If the units cancel to give the WANTED
unit, then it is likely that the numbers were put in the right place to get the right answer.
When should you substitute base units for units such as joules, pascals, volts, and newtons?
Do so only when it is necessary in order for units to cancel to obtain a WANTED unit. If
the WANTED unit and/or other DATA units include one of those complex derived units,
you will probably not need to convert to base units to solve. Summary: Equations and Constants For Electromagnetic Waves
Be sure that these 7 relationships can be recalled from memory before doing the
following practice. Treat practice as a practice test.
1. Wave speed = λ υ 2. For electromagnetic waves, speed of light =
3. Planck’s law: E = h υ and c = λ υ = 3.00 x 108 m · s―1 E=h·c
λ 4. Planck’s constant = h = 6.63 x 10―34 J · s
5. The De Broglie wavelength equation:
6. Joule = kg · m2 · s―2 λ= h
mass · speed (memorize or derive as needed) 7. When working in problems that mix derived and fundamental SI units, convert all
DATA to mks: distance in meters, mass in kg, time in seconds. Practice
1. If F = m · a , and the SI unit for force (the newton) is defined using the SI base units for
mass and acceleration, what base units are equivalent to the newton?
2. If the wavelength of a moving electron is measured to be 36.3 picometers, what is the
speed of the electron? (mass of electron = 9.11 x 10―28 grams) The Heisenberg Uncertainty Principle
In 1927, the German physicist Werner Heisenberg postulated the uncertainty principle:
that for very light particles such as electrons, it is not possible to be certain of both position
and velocity at the same time. Mathematically, his equationtion is
(uncertainty in position) x (uncertainty in velocity) ≥ h/(4π · mass)
Since the mass of a particle is constant, all of the terms on the right side of this equation are
constant, and the two terms on the left are variables. The equation is therefore in the form
xy = c and is an inverse proportion. As the uncertainty of one variable on the left goes
down, the uncertainty of the other goes up. © 2011 ChemReview.Net v.2m Page 648 Module 23 — Light and Spectra This result is that if we know where an electron is, we cannot say precisely its velocity (its
speed and direction, which tells us where it will be next). If we know the velocity of the
electron, we cannot say precisely where it is. This means that the location of electrons must
be stated in probabilities rather than certainties.
If you need to solve calculations using the uncertainty equation, simply memorize the
equation and apply the rules for solving equations discussed above.
***** ANSWERS
1. Since F = m · a , one unit of force (one Newton) equals one base unit of mass (1 kg) times one base unit
of acceleration (one meter per second2 = 1 m · s―2).
newton = kg · m · s―2
2. (The problem involves λ, speed, and mass. Use the De Broglie equation.)
λ= h
mass · speed (So that units cancel in the De Broglie equation, substitute the base units for joules into h:)
DATA: h = 6.63 x 10―34 J · s ( kg · m2 · s―2 ) · s (list constants first) For consistent units that will cancel, mass must be in kg and not g.
mass in kg = 9.11 x 10―28 g • 1 kg = 9.11 x 10―31 kg
103 g For the WANTED speed, the units are not specified, so pick units for speed (distance per unit
of time) that are used in the constant (h): meters and seconds.
speed (in m/s) = ? λ (in m) = 36.3 picometers •
To SOLVE:
= ? = speed (in m/s) = 10―12 m
1 pm = 36.3 x 10―12 m = 6.63 x 10―34 ( kg · m2 · s―2 ) · s
h
mass · λ
(9.11 x 10―31 kg) (36.3 x 10―12 m) = 6.63
x 10―34+31+12 ( kg · m2 · s―2 ) · s = 0.0200 x 109 m · s―1
(9.11 x 36.3)
kg · m = 2.00 x 107 m/s
***** © 2011 ChemReview.Net v.2m Page 649 Module 23 — Light and Spectra Lesson 23E: The Hydrogen Atom Spectrum
Bohr and Atomic Spectra
When atoms are heated to high temperatures, or vaporized and electrified, they glow: they
emit light. Viewed through a prism or a diffraction grating, this light separates into thin
lines of color in the order of the colors of the rainbow: energy waves at sharply defined
wavelengths. Additional waves may be emitted in regions of the spectrum not visible to
the eye. Together, these energy waves are called the emission spectrum or line spectrum
of the atom.
Each atom has a characteristic line spectrum. Astronomers can analyze the light from
distant stars to identify the atoms that are present in the stars.
In 1913, the Danish physicist Neils Bohr proposed that the light of an atom’s emission
spectrum results from electrons falling from higher to lower fixed energy levels inside an
atom, like marbles falling down a staircase (but one in which each step has a different rise).
Based on a simple mathematical equation, En = ― 2.18 x 10―18 J/atom
n2 Bohr’s model predicted exactly the observed wavelengths of the hydrogen atom spectrum.
Bohr’s model did not correctly explain other properties of the hydrogen electron, nor did it
correctly predict electron behavior in other atoms. However, his proposal that electrons
must occupy energy levels in the atom, and his equation to calculate the energy levels in
hydrogen, remain a key part of our current model for atomic structure. Bohr’s Model For Hydrogen Energy Levels
A neutral hydrogen atom consists of one proton at the center of the atom and one electron
outside the nucleus. (A hydrogen nucleus may also include one or two neutrons, but these
neutral particles do not affect the key electrical interaction between the positive proton and
the negative electron.) Nearly all of the volume of the atom is due to the space occupied by
the electron’s movement around the nucleus.
The hydrogen electron can be described as existing at certain fixed levels of energy that are
similar to a staircase. The Hatom electron must be found on one of the energy levels of the
staircase: it cannot rest between steps.
The steps of the Hatom staircase are uneven, but they follow a consistent pattern. The first
step up from the bottom step is large, but the rise for each subsequent step is smaller.
The staircase has an infinite number of steps, numbered from n = 1 to n = ∞ . Step n = 1 is
the bottom step and n= ∞ is at the top of the staircase.
In nature, systems tend to go to their lowest potential energy. The Hatom electron is
therefore normally found at the bottom step n = 1, where it is said to be in its ground state.
However, if energy is added to the Hatom, such as from heat, energy waves, or high
voltage, the electron can be promoted up the staircase. If the Hatom electron is above the
bottom step, it is unstable and said to be in an “excited state.” © 2011 ChemReview.Net v.2m Page 650 Module 23 — Light and Spectra An Hatom electron promoted to an upper step is unstable because it is not at its lowest
possible potential energy. The electron will therefore tend to fall back down the staircase.
It falls like a marble, either hitting every step or skipping some steps, until it reaches the
bottom step n = 1. The electron may “pause” on any step, but the electron cannot pause
between steps.
The electron falling down the staircase must lose energy. To do so, it emits energy waves
(photons). The energy waves it emits are seen as the lines of the Hatom spectrum. To
calculate the energy of the lines in the spectrum, let’s add energy values to this Hatom
model. The H Energy Levels
Using this equation: En = ― 21.8 x 10―19 J
n2 (per each atom) calculate and fillin each missing En value for the steps of the Hatom listed below.
To add and subtract numbers in exponential notation by hand, the numbers must have the
same exponential term (see Lesson 1B). To simplify upcoming calculations, we will write
all of the En values as “ x 10―19 J “ as in the above equation. If you need help, check the
sample calculation below.
n = ∞ ________________ E∞ = 0 Joules
…
n = 6 ________________ E6 = ― 0.606 x 10―19 J
n = 5 ________________ E5 = ↓ Calculate the remaining En values n = 4 ________________ E4 =
n = 3 ________________ E3 =
n = 2 ________________ E2 = n = 1 ________________ E1 =
*****
For the energy level n = 6, : E6 = ― 21.8 x 10―19 J = ―0.606 x 10―19 J 62
Finish calculating the remaining energies if needed.
***** © 2011 ChemReview.Net v.2m Page 651 Module 23 — Light and Spectra Your answers should match
those at the right. (The
numbers are accurate, but
the spacing between the
steps is not drawn to scale.)
The movement of electric
charge creates
electromagnetic waves.
When the Hatom electron
falls from one energy level to
a lower level, it must lose
energy. To do so, it releases
an energy wave with an
energy equal to the difference
in energy between those two
levels. The Hatom Energy Levels
n = ∞ _____________ E∞ = 0 J
… n = 6 _____________ E6 = ― 0.606 x 10―19 J
n = 5 _____________ E5 = ― 0.872 x 10―19 J
n = 4 _____________ E4 = ― 1.36 x 10―19 J n = 3 _____________ E3 = ― 2.42 x 10―19 J n = 2 _____________ E2 = ― 5.45 x 10―19 J n = 1 _____________ E1 = ― 21.8 x 10―19 J In your notebook, calculate
the difference in energy
between energy levels
E3 and E2.
*****
E3 minus E2 = ― 2.42 x 10―19 J ― (― 5.45 x 10―19 J) = + 3.03 x 10―19 J
From the electron’s perspective, this difference is assigned a negative sign because it
represents the amount of energy that the electron must lose in falling from E3 to the lower
energy level E2: the electron will be minus this much energy. However, from the
perspective of the light wave emitted, this energy is positive: the wave must contain the
amount of energy that the electron loses.
In Lesson 23C, Problem 2a, you calculated the wavelength of a wave with this energy.
What was the value of that wavelength? According to Problem 1a, what color would your
eye perceive this wave to be?
*****
656 nm. When the Hatom electron falls from level n= 3 to n = 2, it produces an
energy wave at a wavelength of 656 nm that your eye sees as red.
When the electron falls between other energy levels, it emits light with other colors (plus
waves in other regions of the electromagnetic spectrum that the eye cannot see). © 2011 ChemReview.Net v.2m Page 652 Module 23 — Light and Spectra Practice: The HAtom Spectrum
Do Problems 1 and 3 below. Do more if you need more practice.
In Lesson 23C, Practice 2, and in the lesson above, we calculated these values for the HAtom Visible Spectrum
Color Wavelength (nm) Frequency (Hz) red 656 bluegreen 486 6.15 x 1014 blueviolet 434 6.91 x 1014 violet Energy (J/atom) 410 3.03 x 10―19 Transition
Step 3 2 4.08 x 10―19 In this table, the last column represents the movement of an electron to lower levels in the
H atom. The color, wavelength, frequency and energy in this table are properties of the
energy wave emitted as the electron falls.
Use the table as data for the problems below. Add your answers to the table as you
complete the following calculations. ( h = 6.63 x 10―34 J • s )
1. As the Hatom electron falls from level n = 5 to n = 2, based on the values for the energy
levels (En) that you calculated in this lesson,
a. what would be the energy of the wave emitted?
b. What would be the wavelength of the wave in nm?
c. What would be the color of the wave?
d. What information do these answers allow you to add to the table above?
2 Calculate the energy of the violet line of the Hatom visible spectrum
a. using Planck’s law.
b. Use the energylevel diagram for the Hatom to determine which transition
produces a photon with the energy of the violet line. 3. If sufficient energy is added to the Hatom in its ground state (with the electron at
n = 1), the electron can be ionized: it can be taken an essentially infinite distance away
from the proton that is attracting the electron. The energy needed to remove the
electron is the energy needed to promote the electron from level n = 1 to n = ∞. This
value is termed the ionization energy of the atom.
a. Based on the energy level diagram for H, what is the minimum amount of energy
that is needed to take away (ionize) an Hatom electron?
b. If an ionized electron falls from level n = ∞ down to level n = 1 in one transition,
what will be the energy of the wave emitted by the electron?
c. How does this energy value compare to the energy values of the lines in the visible
spectrum listed in the chart above?
d. What will be the wavelength of this energy wave in nm? © 2011 ChemReview.Net v.2m Page 653 Module 23 — Light and Spectra e. If the human eye can generally see energy waves in the range of 400 to 700 nm, will
you be able to see this wave when looking at an Hatom spectrum?
4. One of the lines in the ultraviolet region of the spectrum of hydrogen has a frequency of
2.46 x 1015 Hz. Which transition does this represent?
5. Using alternate units, the ionization energy of hydrogen is 13.6 electron volts (eV) per
atom. Convert this ionization energy to kilojoules per mole (kJ/mol).
(Use 1 eV = 1.60 x 10―19 J)
6. If the ionization energy of a hydrogen atom is ― 21.8 x 10―19 J, what is the energy of
level n = 3 ? ANSWERS
1. a. E5 ― E2 = ― 0.872 x 10―19 J ― (― 5.45 x 10―19 J) = + 4.58 x 10―19 J
b. This problem involves E (from part a) and λ . The equation that relates those variables is
E= h ·c
λ
DATA: h = 6.63 x 10―34 J · s c = 3.00 x 108 m · s―1 (list the two constants, convert DATA to their units) E in J = 4.58 x 10―19 J λ in m then convert to nm WANTED ( c uses meters ) nano means x 10―9 SOLVE
λ (in m) = h · c = ( 6.63 x 10―34 J · s ) ( 3.00 x 108 m · s―1 ) = 4.34 x 10―7 m
E
4.58 x 10―19 J
= 434 x 10―9 m = 434 nm
c. Blueviolet. See values in the table above.
d. For the blueviolet wave, E = 4.58 x 10―19 J , and the electron is falling from step 5 2. 2. a. Only λ is known in the table; E is WANTED. The form of Planck’s law that relates λ and E is E= h ·c
λ
DATA: c = 3.00 x 108 m · s―1
h = 6.63 x 10―34 J · s (list constants first, use their units to solve) E in J = WANTED λ in m = 410 nm = 410 x 10―9 m ( h uses joules )
(convert to the meters used in c ) E = h · c = ( 6.63 x 10―34 J · s ) ( 3.00 x 108 m · s―1 ) = 4.85 x 10―19 J
λ
410 x 10―9 m © 2011 ChemReview.Net v.2m Page 654 Module 23 — Light and Spectra b. Based on the pattern in the spectrum data table, the violet line should represent the 6 2 transition. Check it: E6 ― E2 = ― 0.606 x 10―19 J ― (― 5.45 x 10―19 J) = + 4.84 x 10―19 J
Allowing for rounding, this agrees with the 2a answer.
3. a. Enough energy must be added to promote the electron from n = 1 to n = ∞,
E∞ ― E1 = 0 J ― (― 21.8 x 10―19 J) = + 21.8 x 10―19 J
b. From n = ∞ down to n = 1, the energy difference is the same: E∞ ― E1 = + 21.8 x 10―19 J
The energy put in to pull away (ionize) the electron, starting from n = 1, must equal the total energy the
electron loses by emitting an energy wave when the electron returns to n = 1.
c. + 21.8 x 10―19 J is a larger energy than those for the visible waves listed in the table.
d. Part (d) involves E (from part c) and λ . The equation that relates those variables is
E= h·c
λ
DATA: h = 6.63 x 10―34 J · s c = 3.00 x 108 m · s―1 (list the two constants, convert DATA to those units) E in J = 21.8 x 10―19 J λ in m = ? then convert to nm
nano means x 10―9 (for units consistent with c, solve in m first ) SOLVE:
λ (in m) = h · c = ( 6.63 x 10―34 J · s ) ( 3.00 x 108 m · s―1 ) = 0.912 x 10―7 m
21.8 x 10―19 J
= 91.2 x 10―9 m = 91.2 nm
E
e. No. This wave has a shorter wavelength (and a higher frequency and energy) than the eye can see.
91.2 nm is in the ultraviolet (UV) region of the spectrum.
The highenergy ultraviolet lines of the Hatom spectrum can be imaged by special films.
Prolonged exposure to UV radiation is dangerous to the eyes and skin. The sun produces high
amounts of UV radiation, but most is absorbed by the ozone layer in the earth’s upper atmosphere
before it can reach the earth’s surface. If the ozone layer decays, the incidence of skin cancer on
earth would likely increase, among other harmful effects.
4. To find the transition using the energylevel diagram, the energy of the line is needed.
The equation that relates E and υ is
DATA: h = 6.63 x 10―34 J · s E=hυ
(list constants first, use their units for WANTED and DATA) E in J = ? υ in s―1 = 2.46 x 1015 Hz s―1
SOLVE: ( h uses seconds ) E (in J) = h υ = ( 6.63 x 10―34 J · s ) (2.46 x 1015 s―1 ) = 1.63 x 10―18 J Which transition has this energy?
***** © 2011 ChemReview.Net v.2m Page 655 Module 23 — Light and Spectra Convert this E to 16.3 x 10―19 J for easier comparison to the numbers in the energylevel diagram.
*****
E2 ― E1 = ―5.45 x 10―19 J ― (― 21.8 x 10―19 J) = + 16.4 x 10―19 J
The energy lost by an electron in falling from E2 to E1 matches the energy value of the wave emitted,
as calculated from the frequency above , allowing for rounding.
For all of the transitions where the electron falls to n = 1, the Hatom will emit UV radiation.
5. WANTED:
DATA: ? kJ
mol 13.6 eV = 1 atom
1 eV = 1.60 x 10―19 J
6.02 x 1023 atoms = 1 mole atoms (mixes atoms with moles of atoms) A ratio is WANTED. The data is two ratios/equalities/conversions. Try conversions (Lesson 11B).
*****
? kJ = 1.60 x 10―19 J • 1 kJ •
mol
1 eV
103 J 13.6 eV • 6.02 x 1023 atoms = 1.31 x 103 kJ
1 atom
1 mole
mol 6. The ionization energy is the energy need to promote the electron from n = 1 to n = ∞.
That energy is also the value in the energylevel equation En = ― 21.8 x 10―19 J
n2
E3 = ― 21.8 x 10―19 J = ― 2.42 x 10―19 J
For energy level n = 3, (per atom) 32
***** Lesson 23F: The Wave Equation Model
Schrődinger’s Wave Equation
Though the Bohr model explained the spectrum of hydrogen, other aspects of the model
were less successful at explaining the behavior of the hydrogen electron and the behavior of
electrons in other atoms.
In 1926, the German physicist Erwin Schrődinger developed equations which described the
electron as if it were a wave, similar to the “standing waves” created by stringed
instruments. Solutions to these equations are termed the quantum mechanical (wave)
model for the atom. This model remains today our best explanation for the behavior of the
hydrogen electron, and it is the basis for predicting the behavior of electrons in all other
atoms as well.
Solutions to the wave equation generally involve complex mathematics, and one question
may have multiple solutions. In many respects, however, the wave equation produces a
model for the hydrogen atom based on mathematical patterns that are elegant in their
simplicity.
The following points are part of the description of the hydrogen atom based on quantum
mechanics. © 2011 ChemReview.Net v.2m Page 656 Module 23 — Light and Spectra Predictions of the Wave Equation For Hydrogen
1. Inside the hydrogen atom are levels described as n = 1, 2, 3, …., ∞ called principal
quantum numbers. Higher quantum numbers represent higher energy levels. In the
hydrogen atom, these energy levels have the En values calculated by the En equation in
the Bohr model.
2. Around the Hatom nucleus are orbitals that describe the space where an electron is
likely to be found. Each orbital has a shape that is described in terms of probability. The
shape describes where an electron in the orbital will be 90% of the time. An electron in
an orbital has an energy equal to one of the values of Bohr’s En equation.
3. At each principal quantum number n, there are n2 total orbitals, and n different types of
orbitals.
a. At level n = 1, there is one orbital, the 1s orbital. An s orbital has a spherical
symmetry around the nucleus: in an s orbital, at a given distance from the nucleus
in all directions, there is an equal chance of finding an electron.
b. At level n = 2, there are two types of orbitals and four total orbitals: one 2s orbital
with spherical symmetry, and three 2p orbitals. The three p orbitals are
perpendicular to each other; they can be described as falling on x, y, and z axes
around the nucleus.
c. At level n = 3, there are three types of orbitals and nine total orbitals: one spherical
3s orbital, three perpendicular 3p orbitals, and five 3d orbitals. Most (but not all) of
the d orbitals are diagonal to the p orbitals.
d. At level n = 4, there are four types of orbitals and 16 total orbitals: one 4s, three 4p ,
five 4d, and seven 4f orbitals.
(It may help to remember the spdf order of the orbitals as “stupid pirates die fighting.”)
4. The Hatom electron in its ground state is in the 1s orbital. If sufficient energy is added
to an Hatom, its electron can be promoted into one of the higher energy orbitals.
The above points can be summarized by a diagram. Below is the model for the
Hatom predicted by the wave equation for the first four principal quantum numbers. Each
line ( ___ ) represents an orbital.
The Hydrogen Atom: Orbitals For the First Four Energy Levels
4s __ 4p __ __ __ 4d __ __ __ __ __ 4f __ __ __ __ __ __ __ ( = 16 orbitals) 3s __ 3p __ __ __ 3d __ __ __ __ __ ( = 9 orbitals) 2s __ 2p __ __ __ 1s __ © 2011 ChemReview.Net v.2m ( = 4 orbitals )
( = 1 orbital ) Page 657 Module 23 — Light and Spectra Though the above diagram applies only to hydrogen, it is the basis for the orbitals found in
all of the other atoms. Every atom has the same number and type of orbitals that are found
in hydrogen. The difference will be that in other atoms, the orbitals will have different (but
generally predictable) relative energies. Practice
Commit the diagram above to memory, then do the problems below.
1. At level n = 4 of the hydrogen atom are
a. how many types of orbitals?
b. How many total orbitals?
c. Write the number and letter used to identify the types of orbitals, and list the
number of orbitals there will be of each type, at n = 4.
2. At level n = 5, the H atom has orbitals designated 5s, 5p, 5d, 5f, and 5g. How many
orbitals are there of each type? ANSWERS
1. a. 4 types of orbitals b. 16 total orbitals c. c. One 4s, three 4p, five 4d, and seven 4f orbitals. 2. At n = 4, there are n2 = 16 total orbitals: 1 s, 3 p, 5 d, and 7 f .
At level n = 5, there must be n2 = 25 total orbitals. 25 – 16 = 9 additional orbitals at n = 5.
At level n = 5, there must be one 5s, three 5p, five 5d, seven 5f, and nine 5g orbitals.
***** Lesson 23G: Quantum Numbers
Quantum Numbers For Hydrogen
In the wave equation for the Hatom, each electron in an orbital can be identified by a series
of quantum numbers that predict the characteristics of the orbital and of an electron in that
orbital.
1. Principal quantum numbers are the n values: the integers 1, 2, 3, … The value of n
will predict the size and energy of the orbital. For higher n values, the orbitals occupy
more volume and are at higher potential energy. An electron in orbitals with a higher n
will on average be further from the nucleus than electrons in orbitals with a lower n in
the same atom. © 2011 ChemReview.Net v.2m Page 658 Module 23 — Light and Spectra 2. Angular momentum quantum numbers (symbol l – a lowercase script L ) at each n are
numbered from 0 to n─1. Each l value correlates with one of the types (s, p, d, or f) of
orbitals.
The s orbital is l = 0, p orbitals are l = 1, d orbitals are l = 2, and f ‘s are l = 3.
3. Magnetic quantum numbers (symbol ml ) have values from ─ l to 0 to +l . These
numbers identify the multiple p, d, and f orbitals at each n.
4. Electron spin quantum numbers (symbol ms ) identify the spin of an electron in an
orbital. An electron must have a spin of either +½ or ─½ . In these lessons, we will
represent an electron that has a positive spin as ↑ and one with a negative spin as ↓ .
A way to remember these rules for quantum numbers is to memorize the Hatom orbital
diagram above plus the quantum number diagram below for n = 4. Note the patterns of
the numbers going from the bottom up.
Relating Quantum Numbers and Orbitals
4s 4p ml = 0 l= 0 n= 4d
1 0 2 1 0 1 2 3 2 2 ↑
1 4 1 1 4f 0 1 2 3 3 The electron shown above in the 4f level would be described as having the quantum
numbers n = 4, l = 3, ml = ─1, and ms = +½ .
The diagram for level n = 3 is similar to n = 4 above, except that it will lack the f orbitals.
The diagram for level n = 2 will have only s and p orbitals. The diagram for level n = 1 will
have only the single s orbital. Practice
Commit the diagram in this lesson to memory, then do the problems below.
1. In a manner similar to the “Relating Quantum Numbers and Orbitals” diagram above,
add numbers below the orbitals to complete the following chart.
3s 3p 3d ml = l=
n=
2. At level n = 5, what quantum numbers are permitted for a. l ? © 2011 ChemReview.Net v.2m b. m l ? c. m s ? Page 659 Module 23 — Light and Spectra 3. Write the diagram for the first four (the lowest four) energy levels of the hydrogen
atom, then add to the diagram one electron that has quantum numbers
a. n = 3, l = 1, ml = 1, and ms = ―½ .
b. n = 4, l = 2, ml = ―2, and ms = +½ .
4. Write symbols and values for the four quantum numbers that characterize the
hydrogen atom electron when it is in the following orbitals.
a. 2s ↓ 2p b. 3s 3p 3d c. 4s 4p 4d ↓
4f ↑ ANSWERS
1. 3s 3p ml = 0 l= 3d 0
3 n= 1 1 2 1 1 2. a. l : 0 1 2 3 4 b. ml : a. 4s 4p 3s 3p 2s __ 3. 0 0 2 2
4  3 2 1 0 1 2 3 4 2p __ __ __ 4d
↓ 1 4f 3d c. ms : Always +½ and ―½ .
( = 16 total )
( = 9 total )
( = 4 total ) 1s __ ( = 1 total ) b. 4s 4p 4d ↑ 3s 3p 3d 2s __ 2p __ __ __ 1s __ 4f ( = 16 total )
( = 9 total )
( = 4 total )
( = 1 total ) 4. a. n = 2, l = 0, ml = 0, and ms = ― 1/2 .
b. n = 3, l = 2, ml = ―1, and ms = ― 1/2 .
c. n = 4, l = 3, ml = ―3, and ms = + 1/2 .
***** © 2011 ChemReview.Net v.2m Page 660 Module 23 — Light and Spectra Summary: Light and Spectra
1. Wavelength: the distance between the crests of a wave. The symbol is λ (lambda).
The units are distance units: either the base unit meters, or nanometers, etc.
2. Frequency is a number of events per unit of time. The units are 1/time. For waves,
frequency is the number of wave crests that pass a point per unit of time.
In wave equations, the symbol for frequency is υ (nu).
SI frequency units: In equations, use
In conversions, use 1/seconds = s―1 = hertz (Hz).
wave cycles or wave lengths/second In equation calculations, write hertz as s―1 so that units will cancel properly.
3. The speed of a wave is equal to its frequency times its wavelength.
Wave speed = λ υ = (lambda)(nu). 4. Electromagnetic waves travel at the speed of light (symbol c).
For all electromagnetic waves, c = λ υ = 3.00 x 108 m · s―1 in vacuum or air. 5. To simplify solving wave calculations using equations,
a. In the DATA table,
• list the constants first. • Convert the DATA to consistent units: those of the constant of the equation if
there is one, or those used in the WANTED unit, or an SI unit, in that order. b. SOLVE for the consistent unit, then convert to a different WANTED unit if needed.
6. Planck’s law that relates frequency and electromagnetic energy is E=hυ or E= h· c
λ where h = Planck’s constant = 6.63 x 10―34 J · s Higher frequency waves have higher energy and lower wavelength.
7. The energy levels inside a hydrogen atom can be calculated by
En = ― 21.8 x 10―19 J (per each atom)
n2
The spectrum of the hydrogen atom can be explained by assuming the hydrogen atom
electron moves between these energy levels.
8. The Hydrogen Atom: Orbitals For the First Four Energy Levels
4s __ 4p __ __ __ 4d __ __ __ __ __ 4f __ __ __ __ __ __ __ ( = 16 orbitals) 3s __ 3p __ __ __ 3d __ __ __ __ __ ( = 9 orbitals) 2s __ 2p __ __ __ ( = 4 orbitals ) 1s __ ( = 1 orbital )
##### © 2011 ChemReview.Net v.2m Page 661 Module 24 — Electron Configuration Module 24 — Electron Configuration
Pretests: In this module, if you believe that you know the material in a lesson, try two
problems at the end of the lesson. If you can do those calculations, you may skip the
lesson.
***** Lesson 24A: The MultiElectron Atom
Orbitals For the Other Atoms
Schrődinger’s wave equation predicts mathematically the observed behaviors of the
hydrogen atom. However, for atoms with more than one electron, the wave equation
provides less exact predictions of how an atom will behave.
Why? In the case of hydrogen, one proton and one electron attract. Mathematics is able to
precisely model the forces in this “two body” problem, but if a second electron is added to
the atom, the situation is more complex.
Because the protons are tightly packed into the nucleus, nuclei with more than one proton
behave as a single point of positive charge. Multiple electrons will be attracted to those
protons, but unlike the case of hydrogen with one electron, two or more electrons also repel
each other. How much will they repel? It depends in part on the types of orbitals that the
electrons occupy.
At n = 2, an electron in the 2s orbital will on average be closer to the nucleus than an
electron in a 2p orbital. This closer 2s electron, by repelling the 2p electron, will act to shield
the 2p electron slightly from the attraction of the protons in the nucleus. This means that
the 2p orbital will be slightly higher in energy than the 2s orbital.
The result of these and other factors is
• the wave equation predicts qualitatively, but not exactly, how electrons in atoms
other than hydrogen will behave, and • the orbital diagram for multielectron atoms will be different from that of hydrogen. In neutral atoms other than hydrogen, the orbitals have different energy values in each
atom, but the orbitals are generally arranged in the same order. This means that for the
atoms in the periodic table, there will be only two types of energy level diagrams to learn:
one for hydrogen, and one that works in most cases for all of the other atoms. The Orbital Diagram for MultiElectron Atoms
For neutral atoms other than hydrogen, these rules apply.
1. All atoms have the number and kind of orbitals predicted by the wave equation for the
hydrogen atom, but the orbitals are arranged in a different order based on their relative
energies.
2. The energy level diagram has clusters of orbitals. Large energy gaps separate these
clusters, but within a cluster, energy levels are close. © 2011 ChemReview.Net v.2m Page 662 Module 24 — Electron Configuration 3. The lowest energy level in a cluster is always an s orbital. The energy of this s orbital is
relatively high compared to the energy of the orbitals below it, but the s orbital is
relatively close in energy to the orbitals above it in the cluster.
4. The d orbitals are shielded by both the s and p orbitals, so much so that in neutral
atoms, the d orbitals rise to between the energy of the s and p orbitals with one higher
principal quantum number.
For the five lowest energy clusters, the sublevels in order of increasing energy are
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 5. The f orbitals are more shielded than the d orbitals, so much so that the 4f orbitals for
most neutral atoms have energy slightly above the 6s orbital, but below the 5d orbital.
For most neutral atoms, the 6th and 7th clusters have sublevels in this order:
6s 4f 5d 6p 7s 5f 6d 7p . These rules result in the
EnergyLevel Diagram for a MultiElectron Atom
7p 6d 6p 5d 7s 6s 5s 4s 3s 2s 5p 4p 5f 32 4f 32 4d 18 3d 18 3p 8 2p 8 2 1s © 2011 ChemReview.Net v.2m Page 663 Module 24 — Electron Configuration The numbers at the right side of the diagram show the number of electrons that the cluster
will hold.
When a series of orbitals is at the same energy, such as in the three p orbitals or five d
orbitals, the orbitals are said to be in a sublevel, and the orbitals are termed degenerate.
At this point, to understand why we write the sublevels in the order that we do, you need
to be able to draw the bottom four clusters of the above diagram from memory. At a later
point, you will need to draw the complete diagram, but as we will see, patterns that relate
the orbital diagram to the periodic table will make this task much easier. The Orbital Electron Configuration
A key step in understanding the chemical behavior of atoms is to write the electron
configuration of the atom: showing which orbitals are filled, halffilled, and empty. This
electron configuration can be determined by filling the orbitalenergylevel diagram with
the atom’s electrons.
Rules for Filling the Orbital Energy Level Diagram
To write the orbital electron configuration for a neutral atom, use these steps.
1. Find the number of electrons in the atom. The atomic number is the number of protons
in an atom. In a neutral atom, this is also the number of electrons.
2. Put the electrons into the orbitals one at a time. Each electron will fall to the unfilled
orbital with the lowest energy. This rule, that the orbitals fill from the lowest energy
up, is called the aufbau principle.
3. An orbital becomes filled when it has two electrons; it cannot have more than two. The
electrons must have opposite spins.
This rule is termed the Pauli exclusion principle, which says that no two electrons in
an atom can have the same four quantum numbers. An electron must have one of two
possible spins, described as spin up and spin down.. In these lessons, we will
represent an electron with a spin up as ↑ and a down spin as ↓ .
If an orbital has two electrons, the electrons are said to be paired and the orbital is
filled. If an orbital has only one electron, it is termed is halffilled and has an
unpaired electron.
4. Each electron has a spin with a value of + 1/2 or —1/2 . In these lessons, we will
represent an electron with a positive spin as ↑ and a negative spin as ↓ .
If a series of orbitals is at the same energy, fill each orbital with one electron, and give
all the electrons in this sublevel the same spin. By convention, positive spins ( ↑ ) are
assigned first), before you start to pair electrons. (This is Hund’s rule.)
Example: The neutral atom carbon has 6 electrons. Fill the orbital diagram with the
six electrons from the bottom up. Each s orbital is filled with two electrons
that have opposite spins.
For the 2p orbitals, since the three p orbitals are at the same energy, put one
electron with an up spin in each orbital before you start to pair electrons. © 2011 ChemReview.Net v.2m Page 664 Module 24 — Electron Configuration The resulting carbon electron configuration is
2s ↑↓ 2p ↑ ↑ 1s ↑↓
Carbon has two filled orbitals, two halffilled orbitals, and two unpaired electrons.
For some neutral atoms, the actual experimentally determined electron configurations
differ slightly from those predicted by the rules above. However, these rules predict the
actual configurations for over 90% of the neutral atoms in the periodic table. Practice
Practice writing the bottom 4 clusters (from the 1s to the 4p orbitals) of the orbital energylevel diagram until you can do so from memory, then do the problems below. Use a
periodic table.
1. The following are electron configurations for neutral atoms. Name the atoms.
a. 2s ↑↓ 2p ↑ b. 2s ↑↓ 1s ↑↓ 2p ↑↓ ↑↓ ↑ 1s ↑↓ 2. Draw the orbital electron configuration for
a. Nitrogen b. Neon c. Phosphorous d. Nickel 3. How many unpaired electrons are found in neutral atoms of
a. Nitrogen (2a above) b. Nickel (2d above) ANSWERS
1. a. Contains 5 electrons = Boron (B) b. 9 electrons = Fluorine (F) 2.
a. 2s ↑↓ 2p ↑ ↑ ↑ b. 1s ↑↓
c. 3s ↑↓
2s ↑↓ 2s ↑↓ 2p ↑↓ ↑↓ ↑↓ 1s ↑↓
3p ↑ ↑ ↑ 2p ↑↓ ↑↓ ↑↓ 1s ↑↓ d. 4s ↑↓
3s ↑ ↓
2s ↑↓ 4p 3d ↑↓ ↑↓ ↑↓ ↑ 3p ↑↓ ↑↓ ↑↓
2p ↑↓ ↑↓ ↑↓ 1s ↑↓ © 2011 ChemReview.Net v.2m Page 665 ↑ Module 24 — Electron Configuration 3. a. Nitrogen: 3 unpaired electrons b. Nickel: Two unpaired electrons ***** Lesson 24B: Listing the Sublevels
The Order of the Sublevels
A method that will show the electron configuration of an atom without drawing the orbital
diagram is to list the sublevels in order of increasing energy, and to include the number of
electrons in each sublevel.
Example: The configuration for silicon (Si) can be written as: 1s2 2s2 2p6 3s2 3p2 This method does not show as much information as writing the orbital diagram, but this
method is quicker, and from it we can determine which orbitals are totally filled, partially
filled, and unfilled. That is the information we will need most often to answer questions
about electron configuration.
To write the electron configuration for a neutral atom by listing the sublevels, follow these
steps.
1. Before you begin to write the configuration for specific atoms, it is helpful to write the
sublevels for the multielectron atom in order of increasing energy (from the bottom
up). This list is
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p … As you write the sublevel numbers and letters, leave a gap in front of each of the s
orbitals in the series. This gap indicates the large gap in energy between the clusters in
the orbital diagram, a factor that is important in t in determining how much energy
must be added to remove electrons from an atom or ion.
In listing the sublevels and clusters in order, it may help to note that each of the clusters
begins with the number of the cluster followed by an s, and ends (except for the first
cluster) with the number of the cluster followed by a p.
Another way to remember the order of 7s 7p the sublevels in the list above is to draw 6s 6p 6d 6f the memory device at the right. 5s 5p 5d 5f 5g To create this diagram, first 4s 4p 4d 4f draw the order for the orbitals of the 3s 3p 3d hydrogen atom (as shown), and then add 2s 2p the diagonal arrows pointed left and up. 1s 2. To write the electron configuration for a specific atom by listing the sublevels, first
find the number of electrons in the atom (which is equal to its atomic number). © 2011 ChemReview.Net v.2m Page 666 Module 24 — Electron Configuration 3. Then write the sublevels in order of increasing energy, as listed above. As you go, fill
each sublevel full of electrons. Add superscripts to indicate the number of electrons that
fill each sublevel.
• An s orbital is filled when it has 2 electrons. • A p orbital sublevel is full when it has 6 electrons. • A d orbital sublevel can hold 10 electrons. • An f orbital sublevel can hold 14 electrons. A full 1s orbital is written as 1s2 (read as “one s two”).
A full 3d sublevel is written: 3d10 (read as “three d ten”).
4. When not enough electrons remain to fill a sublevel, write the number of remaining
electrons as a superscript, then stop.
Examples:
a. Sodium is atomic number 11; the neutral Na atom has 11 electrons.
To write the Na electron configuration, write the list of sublevels, filling
them with electrons as you go, until you run out of electrons.
The Na shorthand configuration is written Na: 1s2 2s2 2p6 3s1 The sum of the superscripts must equal the number of electrons:
2 + 2 + 6 +1 = 11
Note that when the sublevels are written in order of increasing energy, all of
the sublevels before the last sublevel must be filled.
b. Iridium (Ir) has 77 electrons. Its shorthand configuration is:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d7 To check your answer, add up the superscripts. The total must be the
number of electrons in the atom. Practice A
Use a periodic table. Check your answers after each part.
1. Write the memory device that lists the orbitals in order of increasing energy, then write
the orbitals on one line, in order of increasing energy, from 1s to 7p.
2. Write the electron configuration by listing the sublevels for
a. Oxygen b. Sulfur 3. How many unpaired electrons are in
a. Oxygen b. Sulfur 5. A shortcut: note that for each atom, its highest cluster starts with the number of the row
in which the atom is found in the periodic table, followed by an s. © 2011 ChemReview.Net v.2m Page 667 Module 24 — Electron Configuration When listing the sublevels, find the number of the row that the atom is in, then write
full sublevels until you reach the s orbital with that number. That s orbital will start the
highest cluster.
Examples:
a. Sodium is in the third row of the periodic table (rows go across).
Its electron configuration is: 1s2 2s2 2p6 3s1 The highest cluster in which sodium has electrons begins with 3s .
b. Iridium is in the sixth row of the table. Its configuration is:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d7 6. To find the number of unpaired electrons in a neutral atom, draw the orbital diagram, but
do so only for the sublevel that is unfilled in the highest cluster.
Examples:
In sodium above, all of the levels below 3s1 are filled. They will contain no unpaired
electrons. The final term is 3s1, which is shorthand for the orbital configuration
3s ↑ . Sodium therefore has one unpaired electron.
In iridium above, all of the orbitals below 5d7 are filled. In the highest occupied
cluster, the one sublevel with unfilled orbitals has the electron configuration 5d7
which i represents the orbital configuration 5d ↑↓ ↑↓ ↑ ↑ ↑ . Neutral Iridium
has 3 unpaired electrons. Practice B
Use a periodic table. Check your answers after each part.
1. Listing the sublevels, write the electron configuration for
a. Fe b. Br c. Kr 2. How many unpaired electrons are in
a. Fe b. Br c. Kr 3. a. Strontium is in which row of the periodic table?
b. Listing the sublevels, write the electron configuration for strontium.
c. Into which orbital did strontium’s last (highest energy) electron go? © 2011 ChemReview.Net v.2m Page 668 Module 24 — Electron Configuration ANSWERS
Practice A
1. 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 2. a. Oxygen: 1s2 2s2 2p4 6s 4f 5d 6p 7s 5f 6d 7p. b. Sulfur: 1s2 2s2 2p6 3s2 3p4 3. To find the number of unpaired electrons, look only at the highest sublevel with electrons.
a. Oxygen: 2p4 = 2p ↑↓ ↑ ↑ = 2 unpaired e─ b. Sulfur: 3p4 = 3p ↑↓ ↑ ↑ = 2 unpaired e─ Practice B
1. a. Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6 Kr: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 c. b. Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 3. To find the number of unpaired electrons, look only at the highest sublevel with electrons.
a. Fe: 3d6 = 3d ↑↓ ↑ ↑ ↑ ↑
b. Br: 4p5 = 4p ↑↓ ↑↓ ↑ c. Kr: 4p6 = 4p ↑↓ ↑↓ ↑↓
b. Sr: 1s2 3. a. Row 5 = 4 unpaired e─ = 1 unpaired e─
= 0 unpaired e─
2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 c. 5s ***** Lesson 20C: Abbreviated Electron Configurations
The electron configurations written by listing the sublevels can be abbreviated by using the
symbol for the appropriate noble gas to represent totally filled lowerenergy clusters.
Example:
Lead, atomic number 82, has an electron configuration
Pb: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p2 but this configuration can be abbreviated as
Pb: [Xe] 6s2 4f14 5d10 6p2
The symbol [Xe] represents the five filled lower clusters.
This form for the electron configuration is known by a variety of names. In these lessons
we will call this the abbreviated electron configuration, but you may also see this referred to
as the shorthand or core electron or noble gas shortcut electron configuration.
Because this form for electron configurations is the fastest of the three types of
configurations to write, it is the one written most often. © 2011 ChemReview.Net v.2m Page 669 Module 24 — Electron Configuration Configurations written using noble gas symbols are also helpful because they separate the
core electrons (in the filled lower clusters) from the noncore electrons. The core electrons
around an atom do not change when the atom bonds or participates in chemical reactions.
The noncore electrons are those that are in the highest cluster that contains electrons: the
electrons written to the right of the noble gas symbol in the abbreviated electron
configuration.
In chemistry, the focus of our attention will be the electrons written to the right of the noble
gas symbol. Those noncore electrons are the electrons that are gained or lost in chemical
reactions and are shared in covalent bonds. Writing Abbreviated Electron Configurations For Neutral Atoms
1. Use the symbol for the appropriate noble gas to represent totally filled lowerenergy
clusters. The electrons in these filled lower clusters are termed the core electrons.
[He] = 1s2 = 2 electrons
[Ne] = 1s2 2s2 2p6 = 10 e─
[Ar] = 1s2 2s2 2p6 3s2 3p6 = 18 e─
[Kr] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 = 36 e─
[Xe] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 = 54 e─
[Rn] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 = 86 e─ Note that except for helium, all of the noble gas electron configurations end in p6, and
the number in front of the p6 is the row of the periodic table that the noble gas completes.
Example:
Aluminum is atomic number 13. The shorthand electron configuration for a neutral
Al atom is
Al: 1s2 2s2 2p6 3s2 3p1
To write the abbreviated configuration, note that Al has 13 electrons. The noble gas in
the chart above that can be used to represent the filled lower clusters is neon, which
has 10 electrons. The abbreviated electron configuration for Al is
Al: [Ne] 3s2 3p1 (10 + 2 + 1 = 13) The symbol [Ne] represents two filled lower clusters containing 10 electrons.
2. For a simplified way to write abbreviated configurations for neutral atoms, use these
steps.
a. Find the atom in the periodic table. Note its number of electrons.
b. Write in brackets [ ] the symbol of the noble gas at the end of the row above the atom.
c. After the [noble gas], write the number of the row in the periodic table that the atom
is in, followed by an s. This will get you started on writing the orbitals in the highest
unfilled cluster. © 2011 ChemReview.Net v.2m Page 670 Module 24 — Electron Configuration Example: Find vanadium (V) in the 4th row of the table. Ar is at the end of the
row above V.
Start vanadium’s abbreviated configuration by writing V: [Ar] 4s
d. Starting from the number of electrons in the noble gas, add electrons to fill the
sublevels until you run out.
e. To check your configuration, add the number of electrons in the noble gas (equal to
its atomic number) plus the superscripts to the right of the noble gas symbol. This
total must equal the number of electrons in the neutral atom (its atomic number).
Try the steps in Rule 2 on this question.
Q. Write the abbreviated electron configuration for tungsten (W), atomic number 74. *****
Tungsten is in the 6th row of the periodic table. The noble gas at the end of the 5th row
is xenon (Xe). Start the electron configuration by writing
W: [Xe] 6s
Xe has 54 electrons and W has 74. Fill the unfilled cluster until you run out of
electrons.
W: [Xe] 6s2 4f14 5d4 is the abbreviated configuration. Check: 54 + 2 + 14 + 4 = 74
*****
3. The valence electrons establish many of the properties of a neutral atom.
For the main group atoms (in the tall columns of the periodic table), the valence
electrons are defined as all of the s and p electrons in the highest unfilled cluster (the s’s
and p’s to the right of the noble gas symbol).
Examples:
Al: [Ne] 3s2 3p1 has two s and one p electron to the right of the noble gas symbol.
Aluminum therefore has the 10 core electrons represented by [Ne], plus
3 valence electrons.
I: [Kr] 5s2 4d10 5p5 has two s and five p electron to the right of the noble gas
symbol, so iodine has 7 valence electrons. For main group atoms, the d and f
electrons are not considered to be valence electrons.
The valence electrons are the electrons most likely to be involved in chemical reactions
and covalent bonding. Compared to the electrons in other orbitals, the valence
electrons are “loosely bound.”
For transition metals and the two rows below the main table, the electrons considered
valence electrons in some cases may also include the d or f electrons. In reactions of
atoms whose highest energy electron is in a d or f orbital, the d and f electrons are not as
loosely bound as the s electrons, but the d and f electrons are more loosely bound than
the core electrons, and they may be involved in chemical reactions and processes.
Apply those rules to this Q. How many valence electrons are in lead (Pb)? © 2011 ChemReview.Net v.2m Page 671 Module 24 — Electron Configuration *****
Pb: [Xe] 6s2 4f14 5d10 6p2 has two s and two p electron to the right of the noble
gas symbol, so lead has 4 valence electrons. For main group atoms, the valence
electrons are limited to the s and p electrons in the highest cluster that contains
electrons. Practice B: Use a periodic table. Check your answers after each part. 1. List in order the sublevels in the 4th cluster of the orbital energylevel diagram.
2. What noble gas symbol would be used to represent five full orbital clusters?
3. When writing an abbreviated electron configuration, what would be the first sublevel
that would be written after these noble gas symbols?
a. [He] b. [Kr] c. [Rn] 4. Write the abbreviated electron configuration for these neutral atoms.
a. Li (3) b. Cl (17) c. Iodine (53) d. Yttrium (39) e. Polonium (84) 5. Write the number of valence electrons for these Problem 4 atoms.
a. Li (3) b. Cl (17) c. Iodine (53) 6. Write the number of core electrons in e. Polonium (84) 4a. Li 4b. Cl 7. What is the highest number of valence electrons that a main group neutral atom can
have?
8. For the 2nd neutral atom in the 4th row of the periodic table,
a. what is its abbreviated electron configuration?
b. What ion does this atom tend to form?
c. What would be the electron configuration for the ion that this atom tends to form? ANSWERS
Practice A
1. 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p.
2. a. Oxygen: 1s2 2s2 2p4 b. Sulfur: 1s2 Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6 e. Kr: 1s2 2s2 2p6 3s2 3p6 2s2 2p6 3s2 3p4 4s2 3d10 4p6 c. d. Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 3. To find the number of unpaired electrons, look only at the highest sublevel with electrons.
a. Oxygen: 2p4 = 2p ↑↓ ↑ ↑ = 2 unpaired e─ b. Sulfur: 3p4 = 3p ↑↓ ↑ ↑ = 2 unpaired e─ © 2011 ChemReview.Net v.2m Page 672 Module 24 — Electron Configuration c. Fe: 3d6 = 3d ↑↓ ↑ ↑ ↑ ↑
d. Br: 4p5 = 4p ↑↓ ↑↓ ↑ e. Kr: 4p6 = 4p ↑↓ ↑↓ ↑↓ 4. a. Sr: 1s2 2s2 2p6 = 4 unpaired e─ = 1 unpaired e─
= 0 unpaired e─ 3s2 3p6 4s2 3d10 4p6 5s2 b. Row 5 c. 5s Practice B
1. 4s 3d 4p 2. [Xe] at the end of the 5th row. 3. a. [He] 2s The number of the s orbital is the row number that the atom is in. b. [Kr] 5s
4. a. Li (3): [He] 2s1 b. Cl (17): [Ne] 3s2 3p5 d. Yttrium (39): [Kr] 5s2 4d1 c. [Rn] 7s c. Iodine (53): [Kr] 5s2 4d10 5p5 e. Polonium (84): [Xe] 6s2 4f14 5d10 6p4 5. Valence electrons: a. Li (3) one b. Cl (17) seven c. Iodine (53) seven e. Polonium (84) six
6. a. Lithium has 2 core electrons represented by [He] b. Cl has 10 core electrons represented by [Ne] 7. Eight. The highest number of s and p electrons possible in the highest cluster is 2 in the s and 6 in the p.
8. a. Ca: [Ar] 4s2 b. Ca2+ c. [Ar] . To form the 2+ ion, calcium loses its two valence electrons. ***** Lesson 24D: The Periodic Table and Electron Configuration
The Shape of the Table
Atoms that are in the same column of the periodic table have similar behavior because those
atoms have similar configurations for their highest energy electrons.
The correlations between a standard periodic table and electron configuration includes:
1. The periodic table starts with hydrogen, which has one proton and one electron. One
proton and one electron are then added for each atom moving to the right across the
rows of the periodic table.
When a cluster in the orbital energy level diagram is filled, the result is the electron
configuration of a noble gas.
After each noble gas, a new row of the table is started. This convention places all of the
noble gases in the last column. All noble gases have totally filled clusters.
2. Comparing the orbital energy level diagram and the periodic table:
• The orbital energy level diagram has 7 clusters of energy levels and the periodic
table has 7 rows. • If the highest energy electron for an atom is added into the 4th cluster, that atom
is in the 4th row of the periodic table. • If an atom is in the 5th row of the table, its highest energy electron is found in the
5th cluster: the one that start with the 5s orbital. © 2011 ChemReview.Net v.2m Page 673 Module 24 — Electron Configuration 3. The number of electrons a cluster can hold equals the number of atoms in each row of the
periodic table.
• The lowest cluster of the orbital diagram can hold two electrons. The first row of
the periodic table has two atoms: H and He. • The second and third clusters in the orbital energy diagram each hold 8 electrons.
The second and third rows of the periodic table each have 8 atoms. Across these
rows, 8 electrons are added to the atoms, one electron at a time.
• The fourth cluster has the orbitals 4s 3d 4p . These three sublevels can hold 18
electrons. The fourth row of the table has how many atoms? *****
18. Count them to be sure.
• The sixth cluster has the orbitals 6s 4f 5d 6p . These four sublevels can hold 32
electrons. The sixth row of the periodic table has 32 atoms (because it includes
the first row of atoms in the two rows listed below the chart). Practice A: Use a periodic table. If needed, check your answers after each part. 1. What is the abbreviated electron configuration (using the noble gas symbol) for the first
neutral atom in the 7th row of the periodic table?
2. Which p orbitals are filling in the 5th row of the periodic table?
3. Which d orbitals are filling in the 4th row of the table?
4. Which f orbitals are filling in the 7th row of the table?
•
4. The shape of the periodic table is determined by the order in which the orbitals fill in the
orbital energylevel diagram. s
block p block
d
block f block © 2011 ChemReview.Net v.2m Page 674 Module 24 — Electron Configuration • All neutral atoms that are in the first two columns of the periodic table are in the s
block (see diagram above). For the atoms in this block, the s orbitals are filling:
these atoms always have their highest energy electron going into an s orbital. For
neutral atoms in the first column, all electron configurations end in s1. In the second
column, all end in s2. • All of the neutral atoms in the six tall columns at the right in the table are in the p
block: they have their highest energy electron going into a p orbital (except for
helium). In Group 3A (the boron family), all electron configurations end in p1. In
the noble gas column, all end in p6 (except for helium, which ends in s2 to fill the
first cluster). • All transition metals are in the d block: they have their highest energy electron
going into a d orbital. The transition metals are placed after the s tall columns but
before the p tall columns because the d orbitals fill the orbital diagram after the s, but
before the p orbitals.
In the each row of the periodic table, as in each cluster of the orbital energy level
diagram, the d orbitals that are filling have a principal quantum number that is one
less than the quantum number of the s orbital that fills immediately before them.
o After the 4s orbitals fill, the 3d orbitals fill. After the 5s fill, the 4d fill. Examples:
The first transition metal, scandium (Sc), has an electron configuration that ends
in 3d1. The second, titanium (Ti), ends in 3d2. The last transition metal in that
row, zinc (Zn) has an electron configuration that ends in 3d10.
Mercury (Hg), two rows below zinc in the table, has an electron configuration
that ends in 5d10. As you might expect, cadmium (Cd), in the row between zinc
and mercury, has an electron configuration that ends in ….?
*****
4d10
• The atoms in the two rows below the table are in the f block: they have their
highest energy electrons going into f orbitals. Since there is room for 14 electrons in
the seven f orbitals, there are 14 atoms in each of the two rows placed below the
table. In the first row below, called the lanthanides because the row begins with
lanthanum, the 4f orbitals are filling. In the second row, the actinides, the 5f
orbitals are filling.
In the each row of the periodic table, as in each cluster of the orbital energy level
diagram, the f orbitals that are filling have a principal quantum number that is two
less than the s orbital that fills immediately before them:
o After the 6s orbitals fill, the 4f orbitals fill. After the 7s fill, the 5f fill. © 2011 ChemReview.Net v.2m Page 675 Module 24 — Electron Configuration Practice B: Use a periodic table. If needed, check your answers after each part. 1. Which orbitals are being filled in the 6th row of the periodic table, in order?
2. Where in the periodic table are the neutral atoms that have electron configuration
ending in
a. s2 b . p6 c. p2 3. Name the four neutral atoms whose electron configuration ends in d2 . The Columns of the Table
4. Atoms in the same column of the table have similar electron configurations.
Examples:
• All of the atoms in column one have an electron configuration ending in s1.
H ends in 1s1, Li in the 2nd row ends in 2s1, Fr in the 7th row ends in 7s1. • All halogens (Group 7A) have electron configurations ending in p5. All halogens
have seven valence electrons: two s and five p electrons in their highest cluster. • All neutral atoms in the carbon family (Group 4A) have an electron configuration
ending in p2. All have four valence electrons: two s and two p electrons. • The number of valence electrons for an atom is the number of the main (A) group
(using the AB group notation) for its column of the periodic table.
Example: All atoms in Group 7A (the halogens) have seven valence electrons. The patterns in the periodic table will help you to write all of the various types of electron
configurations.
For example, the table below shows the highest energy sublevel for atoms in the last
column of each block of the table. Note the patterns 1s2
2s2 2p6 3s2 3p6 s
4s2
block 5s2
6s2 3d10 d
block 7s2 4p6 4d10 5p6 5d10 6p6 6d10 f block © 2011 ChemReview.Net v.2m p block 4f14
5f14 Page 676 Module 24 — Electron Configuration On this table, from the top down, reading across each row, write the filled sublevel
configurations shown for the first 4 rows of the table. Leave a little extra room before you
start a new row.
*****
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6…. Note that this is the order used when listing the sublevels to write an electron configuration..
This diagram will also help in quickly identifying the configuration of the highest energy
sublevel of an atom.
Example: Tellurium (Te) has 52 electrons. Find Te in the periodic table. Then mark
where it will be in the above table. Decide from the position in the above table the
configuration of the highest energy sublevel.
Based on its position, Te’s highest energy orbital will have the configuration: _______
*****
5p4 . For the shorthand configuration, simply write all of the bold orbitals above, in
order from the top and across the rows, until you get to Te’s 5p4 , then stop. Try it.
*****
Te: 1s2 2 s 2 2 p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p4 To write the abbreviated configuration, write the [noble gas] at the end of the row above Te,
then write the filled orbitals shown in the row that Te is in, with the last sublevel being the
highest energy sublevel for Te. Try it.
*****
Te: [Kr] 5s2 4d10 5p4
As a check for both types of configurations, count the electrons represented by each.
*****
You should get 52, the number of electrons in a neutral Te atom.
***** Mendeleev’s Periodic Table
In 1872, the Russian chemist Dmitri Mendeleev proposed a periodic table as a way to
predict chemical behavior. The existence of protons and electrons would not be discovered
for decades. In fact, at the time, many of the atoms currently in the periodic table had not
yet been discovered. Mendeleev bravely predicted that atoms would be discovered to fill
the holes he left in his table. Based on the patterns in his table, he described what the
characteristics of these undiscovered atoms would be. His predictions soon proved correct,
and his periodic table became the central framework for organizing chemical behavior .
Mendeleev’s 1872 table is remarkably similar to the modern periodic table.
We know today what Mendeleev did not: why the atoms are organized the way they are in
his table. The shape of the table is determined by the order of the electron orbitals that are
predicted to exist by Schrődinger’s wave equation. The atoms in columns have similar
behavior because they have similar electron configurations in their highest occupied
cluster. © 2011 ChemReview.Net v.2m Page 677 Module 24 — Electron Configuration Practice C: Use a periodic table. On problems with parts,
part. Save the rest for your next practice session. and complete every other 1. What do the rare earth atoms have in common with regard to their electron
configuration?
2. Why do the rare earth atoms fit into the periodic table before the transition metals
begin, rather than after?
3. Write the orbital configuration of just the highest unfilled sublevel for these neutral
atoms.
a. Iodine (53) b. Cobalt (27) c. Rubidium (37) 4. Write the shorthand electron notation (1s2 2s2…) for these atoms.
a. Gallium (31) b. Zirconium (40) 5. Write the predicted abbreviated electron configuration (using noble gas symbols) for
a. Magnesium (12)
d. Potassium (19)
g. Mercury (80)
j. Fluorine (9) b. Osmium (76) c. Aluminum (13) e. Phosphorous (15) f. Polonium (84) h. Plutonium (94)
k. Iron (26) i. Rutherfordium (104)
l. Beryllium (4) 6. Where are the atoms located that have
a. One valence electron? b. 7 valence electrons? ANSWERS
1. [Rn] 7s1
2. 5p . The principal quantum number of the filling p orbitals is the row number in the table.
3. 3d . The principal quantum number of the filling d orbitals is one less than the row number.
4. 5f . The principal quantum number of the filling f orbitals is two less than the row number.
Practice B
1. 6s 4f 5d 6p 2a. In Group 2A (the 2nd tall column). 2b . Noble gases below He. 2c. In the 2nd tall column to the right of the transition metals (the carbon family).
3. Titanium (Ti), Zirconium (Zr) Hafnium (Hf), and Rutherfordium (Rf).
Practice C
1. Using our rules for predicting electron configuration, the highest energy electron is going into an f orbital.
2. Using our rules, the f orbitals are at lower energy than the d orbital with one higher principal quantum
number; the f orbitals fill before the d’s.
3. a. Iodine 5p ↑↓ ↑↓ ↑ © 2011 ChemReview.Net v.2m b. Cobalt 3d ↑↓ ↑↓ ↑ ↑ ↑ c. Rubidium 5s ↑ Page 678 Module 24 — Electron Configuration 4. a. Ga: 1s2 b. Zr: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5. a. Mg: [Ne] 3s2 5s2 4d2 b. Os: [Xe] 6s2 4f14 5d6 c. Al: [Ne] 3s2 3p1 d. K: [Ar] 4s1 e. P: [Ne] 3s2 3p3 f. Po: [Xe] 6s2 4f14 5d10 6p4 g. Hg: [Xe] 6s2 4f14 5d10 h. Pu: [Rn] 7s2 5f6 i. Rf: [Rn] 7s2 5f14 6d2 j. F: [He] 2s2 2p5 k. Fe: [Ar] 4s2 3d6 l. Be: [He] 2s2 6. a. In the first column of the periodic table. b. In the halogen family: tall column 7. ***** Lesson 24E: Electron Configurations: Exceptions and Ions
Exceptions to the Standard Predictions
Not all of the actual electron configurations follow the rules above, but many of the
exceptions can be predicted. In predicting exceptions, the rules are
1. A series of orbitals at a sublevel (orbitals with the same energy) often have lower
potential energy if they are empty, halffilled or totally filled. The expected electron
configurations predicted by the standard rules often rearrange to form these energy
configurations that gain them special stability.
Example:
Write the standard predicted electron configuration for chromium (24): [ How many unpaired electrons would chromium have?
*****
By the standard rules, Cr = [Ar] 4s2 3d4
4p
↑
3d ↑ ↑ ↑
4s ↑↓
In this orbital diagram, chromium has four unpaired electrons, but chromium has an
exceptional configuration. Based on the rule above, predict what it might be.
The highest unfilled cluster would be *****
If one of the chromium 4s electrons is promoted slightly, up to the 3d orbitals, both the
4s and the 3d orbitals become halffilled.
This configuration is 4s ↑ 4p 3d ↑ ↑ ↑ ↑ ↑ = [Ar] 4s1 3d5 If a system can go to lower potential energy, it strongly tends to do so. Measurements
of the behavior of neutral chromium atoms find that they have the six unpaired
electrons predicted by this exception rule.
Other atoms below chromium in the same column may have similar behavior, but the
exceptions do not always occur. Experimental measurement of the number of unpaired © 2011 ChemReview.Net v.2m Page 679 Module 24 — Electron Configuration electrons is often required to determine the actual electron configuration of a neutral atom
or ion when exceptions are possible.
2. The electron configurations of the rareearth atoms contain many exceptions, especially
in the early columns. The reason is that the d and the f orbitals are very close in energy.
Many of the exceptions have a nonstandardprediction d1 electron, such as
Ce (58): [Rn] 6s2 4f1 5d1 and U (92): [Rn] 7s2 5f3 6d1 But, by a slim majority, most rare earth atoms have the s2 f* d0 configuration predicted
by the standard rules for filling the orbital energylevel diagram.
Some periodic tables list La and Ac in the 14 columns below the table. Others place La
and Ac in the transition metals, and instead list Lu and Lr at the end of the 14 lower
columns. Arguments can be made for both conventions. Practice A: Use a periodic table. If needed, check your answers after each part. 1. Write the abbreviated electron configuration (using noble gas symbols) that would be
predicted, using the standard rules, for
a. Nickel (28) b. Palladium (46) After each configuration above, write the predicted number of unpaired electrons.
2. By experiment, palladium is found to have no unpaired electrons. Write what this
exceptional electron configuration is likely to be, and explain why it would be stable.
3. Find the three coinage metals (copper, silver, and gold) in the periodic table. These
atoms have been treated as a special group since ancient times. All are metals that
retain their shine and resist corrosion when compared to most metals. Their
exceptional stability arises in part from their exceptional electron configurations.
a. Write the standard prediction of the abbreviated configuration for gold (79).
b. What would be the exceptional configuration for gold, and why would it be stable?
c. In its exceptional electron configuration, how many unpaired electrons does gold
have? How many valence electrons? Electron Configurations For Ions
Atoms tend to form ions that have the same number of valence electrons as the nearest
noble gas. The noble gases have eight valence electrons (except for helium that has 2).
To write electron configurations for monatomic ions, follow these steps.
1. First write the neutral electron configuration.
2. To form negative ions, add electrons in the standard order.
3. To form positive ions, take away electrons from the highest cluster. Take the electrons
out in this order: take first the p’s then the s’s, then the d’s, then the f’s. © 2011 ChemReview.Net v.2m Page 680 Module 24 — Electron Configuration Note that in making positive ions, the electrons are taken out of a neutral atom in a
different order than they go in. Electrons are lost first from the orbitals with the highest
principal quantum number. This means that for main group atoms, the valence p then s
electrons are taken away first.
Why? When an electron is taken away, the repulsion of all the electrons is reduced, and the
shielding of the d and f orbitals is reduced. In postitive ions, this drops the d and f orbitals
to a lower relative energy than they have in neutral atoms.
For example, the 3d orbitals, which in a neutral atom are higher in energy than the 4s
orbitals, in 4th row positive ions drop to an energy lower than the 4s orbitals.
Apply the three rules above to these two examples, then check your answers below.
Q1. Write the abbreviated electron configuration for the chloride ion, Cl─.
Q2. Write the abbreviated electron configuration for the tin (IV) ion, Sn4+.
*****
A1. Neutral chlorine is [Ne] 3s2 3p5, chloride ion has one more electron: [Ne] 3s2 3p6
This gives chloride the same electron configuration as argon, the nearest noble gas.
When two atoms have the same electron configuration, they are said to be
isoelectronic. Iso is a prefix from ancient Greek meaning equal. A2. Neutral tin has a configuration of [Kr] 5s2 4d10 5p2, and Sn4+ is [Kr] 4d10 . The 4+
ion has lost four electrons: the two 5p electrons, then the two 5s electrons. This gives
the tin ion zero valence electrons. Ions tend to have either zero or 8 valence
electrons, giving them the same valence configuration as the nearest noble gas. It is difficult to add an electron to, or to remove an electron from, a noble gas electron
configuration. Atoms with electron configurations close to that of a noble gas are very
reactive, usually undergoing reactions that result in their attaining a noble gas electron
configuration. This tendency is a driving factor in a wide range of chemical reactions. Practice B: Use a periodic table. If needed, check your answers after each part. 1. Add a charge to these symbol to show the monatomic ion that the following atoms tend
to form (review Lesson 7B if needed).
a. Cs b. S c. At d. Al e. Mg f. F 2. Write the electron configurations for each ion in Problem 1.
3. Which ion in Problem 1 is isoelectronic with the noble gas xenon?
4. Silver (atomic number 47) forms a 1+ ion.
a. Write the abbreviated electron configuration for Ag+.
b. How many unpaired electrons are in Ag+? 5. Iron (26) forms two cations: Fe2+ and Fe3+.
a. Write the abbreviated electron configuration for both ions.
b. What is the number of unpaired electrons in each ion? © 2011 ChemReview.Net v.2m Page 681 Module 24 — Electron Configuration c. Which ion lost more than its valence electrons?
d. Why might that ion lose more than its valence electrons?
6. Write the electron configuration for the two ions formed by copper. Which ion has a
configuration that is not predicted by the standard rules for filling the energy level
diagram and forming ions, and why might this ion form? ANSWERS
Practice A
1a. Ni: [Ar] 4s2 3d8 ; 2 unpaired electrons:
b. Pd: [Kr] 5s2 4d8 2 unpaired electrons: 4s ↑↓
5s ↑↓ 4p
5p 3d ↑↓ ↑↓ ↑↓ ↑ ↑
4d ↑↓ ↑↓ ↑↓ ↑ ↑ 2. Pd: has an actual configuration of [Kr] 5s0 4d10 This configuration has empty s and full d orbitals.
Orbitals at the same energy that are empty, halffilled, or totally filled have lowered potential energy.
3. a. Au: [Xe] 6s2 4f14 5d9
b. Au: [Xe] 6s1 4f14 5d10 This gives halffilled s , filled f , and filled d orbitals. Orbitals at the same
energy that are empty, halffilled, or totally filled have special stability.
c. The single 6s electron is the one unpaired and the one valence electron. Practice B
1. a. Cs+ b. S2─ 2. a. [Xe] b. [Ne] 3s2 3p6 or [Ar] d. [Ne] c. At─ e. [Ne] d. Al3+ e. Mg2+ f. F─ c. [Xe] 6s2 4f14 5d10 6p6 or [Rn] f. [Ne] 3. Only 1a: Cs+ . The abbreviated configuration for At─ can be written using [Xe] to start, but At─ is
isoelectronic with radon (Rn) (see 2c answer).
4. a. Neutral Ag would have a standard predicted configuration of [Kr] 5s2 4d9 ,
but if a coinage metal has a neutral configuration of s1 d10 , this results in halffilled s and totally
filled d orbitals. The actual configuration for neutral silver is Ag: [Kr] 5s1 4d10
a configuration that has only once valence electron. Since silver’s most common ion is one plus, this
fits the prediction that it only has one valence electron as a neutral atom.
To make Ag+, take away the one valence electron. Ag+ = [Kr] 4d10
b. Ag+ = [Kr] 4d10 has no unpaired electrons. All of the orbital sublevels are filled.
5. a. Write the neutral atom configuration first. Fe: [Ar] 4s2 3d6
Fe2+ has lost two its two valence electrons. Fe2+: [Ar] 3d6
Fe3+ has lost one more electron than Fe2+: Fe3+: [Ar] 3d5 © 2011 ChemReview.Net v.2m Page 682 Module 24 — Electron Configuration b. Fe2+ has four unpaired electrons. Fe2+: [Ar] 3d6 = 3d ↑↓ ↑
Fe3+ has five unpaired electrons. Fe3+: [Ar] 3d5
c. = 3d ↑ ↑↑ ↑ ↑↑↑ ↑ Fe3+ has lost its valence electrons, plus one more. d. By losing one extra electron, Fe3+ has gained a 3d5 configuration, which has degenerate d orbitals
that are halffilled and therefore have special stability.
6. Copper’s two ions are copper (I) and copper (II).
The neutral copper electron configuration predicted by the standard rules is [Ar] 4s2 3d9 . However, the
actual, measured copper electron configuration is [Ar] 4s1 3d10 . In this exceptional configuration, by
promoting one s electron to the d orbitals, Cu gains a halffilled s and a totally filled series of d orbitals
which have extra stability.
In some cases, copper behaves as if it has one valence electron, and in other cases as if it has two.
The Cu2+ configuration is [Ar] 3d9 . In this configuration, the neutral copper atom has formed an ion by
losing two electrons, and it has no valence electrons.
The Cu+ configuration is [Ar] 3d10 . In this exceptional configuration, Cu+ has zero valence electrons plus
a totally filled series of d orbitals. Both factors support extra stability.
##### © 2011 ChemReview.Net v.2m Page 683 ...
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