Teacher Notes

Electronic and Vibrational Spectroscopy

Guided-Inquiry Learning Activity for AP® Chemistry

Additional Materials Required

Online computer access
Periodic table (one per student group)

Teacher Tips

  • “Electronic and Vibrational Spectroscopy” is one of the Guided-Inquiry Learning Activities for AP® Chemistry, Catalog No. AP7710, from Flinn Scientific. The other two guided-inquiry activities in the AP7710 learning activity package are “Photoelectron Spectroscopy”and “Mass Spectrometry.”
  • We recommend a collaborative, peer-learning environment for students to complete this guided-inquiry learning activity. Effective group sizes may include 3−4 students. To ensure active participation by all members of the group, it is helpful to assign rotating roles to the students and call on different students in each group to be the “reporter” for class discussion and explanation of key parts of the activity.
  • The organization of this activity into three parts provides convenient stopping points to discuss key questions and explain relevant learning goals.
  • Online computer access is recommended for students to answer questions concerning the structures of organic functional groups in Part B of this guided-inquiry activity.
  • The evidence and reasoning in this learning activity should be used to recall, review, and reinforce learning objectives related to quantization of electron energy levels. Part C also provides an excellent review of solution concentration calculations and stoichiometry. Students should also review the relative energy of all types of electromagnetic radiation, from gamma rays and x-rays to microwaves and radio waves.
  • Common misconceptions and conceptual hurdles or stumbling blocks in this unit on spectroscopy include:
    • What is the relationship between the observed (transmitted) color of a solution and the wavelength and color of light absorbed by a substance in solution?
    • How are absorbance and percent transmittance related?
    • What is the difference between atomic emission and absorption spectroscopy?
    • IR spectra are typically plotted as %T versus wavenumber. A “peak” is the “bottom” of the scale in %T.
  • A ball-and-spring model provides a great analogy for vibrational energy levels, especially stretching vibrations. As the mass of the ball increases, the ease of stretching increases and the frequency of vibration decreases. As the spring becomes “stiffer,” the force constant increases and the energy required to stretch the spring increases. The spring or force constant is analogous to the bond strength in a molecule.

Further Extensions

Alignment with the Curriculum Framework for AP® Chemistry

Essential Knowledge
1D3: The interaction of electromagnetic waves or light with matter is a powerful means to probe the structure of atoms and molecules, and to measure their concentration.

  1. The energy of a photon is related to the frequency of the electromagnetic wave through Planck’s equation (E = hv). When a photon is absorbed (or emitted) by a molecule, the energy of the molecule is increased (or decreased) by an amount equal to the energy of the photon.
  2. Different types of molecular motion lead to absorption or emission of photons in different spectral regions. Infrared radiation is associated with transitions in molecular vibrations and so can be used to detect the presence of different types of bonds. Ultraviolet/visible radiation is associated with transitions in electronic energy levels and so can be used to probe electronic structure.
  3. The amount of light absorbed by a solution can be used to determine the concentration of the absorbing molecules in that solution, via the Beer-Lambert Law.

Learning Objectives
1.15 The student can justify the selection of a particular type of spectroscopy to measure properties associated with vibrational or electronic motions of molecules.
1.16 The student can design and/or interpret the results of an experiment regarding the absorption of light to determine the concentration of an absorbing species in solution.

Correlation to Next Generation Science Standards (NGSS)

Science & Engineering Practices

Developing and using models
Analyzing and interpreting data
Constructing explanations and designing solutions
Obtaining, evaluation, and communicating information

Disciplinary Core Ideas

HS-PS1.A: Structure and Properties of Matter
HS-PS1.B: Chemical Reactions

Crosscutting Concepts

Cause and effect
Scale, proportion, and quantity
Structure and function

Performance Expectations

HS-PS1-1. Use the periodic table as a model to predict the relative properties of elements based on the patterns of electrons in the outermost energy level of atoms.
HS-PS1-2. Construct and revise an explanation for the outcome of a simple chemical reaction based on the outermost electron states of atoms, trends in the periodic table, and knowledge of the patterns of chemical properties.
HS-PS1-3. Plan and conduct an investigation to gather evidence to compare the structure of substances at the bulk scale to infer the strength of electrical forces between particles.
HS-PS2-6. Communicate scientific and technical information about why the molecular-level structure is important in the functioning of designed materials.

Answers to Questions

Part A. Electronic Energy Levels and UV/Vis Spectroscopy

  1. Molecules that absorb energy in the UV/Vis range tend to have broad absorbance bands, rather than discrete peaks, due to the possibility of different vibrational energy states associated with each electronic transition. Add arrows in Figure 4 to show at least three different energy transitions corresponding to promotion of an electron from E0 to E1.
    {13776_Answers_Figure_4_Electronic and vibrational transitions}
  2. UV spectra are typically measured from 200 to 400 nm. The ultraviolet spectrum of the amino acid tyrosine is shown in Figure 5. What are the wavelengths of maximum absorbance for the two major peaks in the UV spectrum of tyrosine?
    {13776_Answers_Figure_5_Ultraviolet spectrum of tyrosine}

    Tyrosine has two major peaks in its UV spectrum, at 230 nm and 275 nm, respectively.

  3. Using Equation 1 and Planck’s law (see the Background section), calculate the energy for the lower energy transition in the UV spectrum of tyrosine (see Figure 5). The speed of light c = 3.0 x 108 m/sec.

    The lower energy transition occurs at 275 nm. (Energy is inversely proportional to wavelength.)

  4. The UV spectra of organic molecules involve electrons in pi-molecular orbitals (double and triple bonds) or nonbonding orbitals (atoms with unshared pairs of electrons). Absorption of UV light results in promotion of these electrons to antibonding molecular orbitals. How will the wavelength of light absorbed by an organic molecule change as the energy gap between bonding and antibonding molecular orbitals decreases?

    Reducing the energy gap between bonding and antibonding orbitals will decrease the energy absorbed by an organic molecule for a corresponding electron transition. Lower energy is associated with longer wavelengths, thus the wavelength for the electron transition will increase.

  5. Indicators and other organic dyes contain a series of double bonds separated by single bonds. This pattern of bonding, called conjugation, reduces the separation between the ground state and the excited state of the pi-electrons. Use this fact to explain why organic dyes are highly colored.

    As noted in Question 4, reducing the energy gap for promotion of an electron to a higher energy level will shift the absorption to higher wavelength. This will shift the absorption maximum from the ultraviolet region to the visible region. Substances that absorb visible light are colored. (The transmitted color is complementary to the absorbed color or wavelength.)

  6. Transition metal ions have filled or partially filled d-orbitals. The presence of water molecules or other ligands surrounding a transition metal ion leads to energy differences among the d-orbitals. Depending on the metal ion involved, the energy difference may correspond to different wavelengths and energies of visible light. This property of transition metal ions gives many their characteristic—and beautiful—colors. Why are most zinc salts colorless?

    Zinc ions have a valence electron configuration of 3d10 after loss of their 4s2 electrons. Since Zn2+ ions have filled d-orbitals, electrons cannot be promoted or excited to a higher-energy, unfilled d-orbital.

  7. What does the transmitted color of a solution tell you about the color of light it absorbs?

    The color of light transmitted by a solution is complementary to the absorbed color or wavelength.

  8. Chromium(III) ions appear purple or violet in aqueous solution. Which wavelengths of light would you expect to be most strongly absorbed by Cr3+ ions?

    A solution that is violet should absorb the complementary color, yellow. The expected wavelength is 560–600 nm.

Part B. Vibrational Energy Levels and Infrared Spectroscopy
  1. Bond formation is due to a decrease in potential energy when two atoms approach each other, as shown below for HBr (see Figure 6). Label the point on the x-axis corresponding to the equilibrium H−Br bond length.
    {13776_Answers_Figure_6_Potential energy diagram for HBr}

    The equilibrium bond length occurs at the potential energy minimum in the diagram.

  2. Vibrational energy levels are shown by means of dashed lines in Figure 6. Explain how the dashed lines illustrate the relationship between the absorption of infrared radiation and bond stretching.

    The dashed lines indicate a range of internuclear separations (H−Br bond lengths) that have similar potential energy. This range of bond length distances is associated with normal vibrations or stretching of the bond. The dashed lines represent higher vibrational energy states due to bond stretching.

  3. IR spectra are typically plotted as shown (see Figure 7), with percent transmittance on the y-axis versus wavenumber on the x-axis.
    {13776_Answers_Figure_7_IR spectrum of acetone}
    1. What are the units of wavenumber? Describe the expected relationship between wavenumber and frequency.

      Wavenumber is given in units of inverse centimeters, cm−1, and is calculated as the reciprocal of the wavelength (1/λ) in cm. Wavenumber is proportional to frequency and is considered a frequency unit.
      v = c/λ and v/c = 1/λ; therefore wavenumber = v/c

    2. Is wavenumber directly or inversely proportional to energy? Label Figure 7 to show the direction of increasing vibrational energy along the x-axis.

      Wavenumber is a frequency unit and is thus directly proportional to energy (see Figure 7).

  4. Estimate the wavenumber for maximum absorbance observed between 1500 and 2000 cm−1 in the IR spectrum of acetone.

    Acetone shows a strong absorption band at about 1720 cm−1. Recall that transmittance and absorbance are inversely related. As transmittance decreases, absorbance increases.

  5. Predict the general relationship between bond strength, the energy required to stretch a bond, and the stretching frequency observed for the bond in the IR spectrum.

    As the bond strength increases, the energy required to stretch the bond also increases. In general, for atoms having the same or similar masses, the stretching frequency is greater for stronger bonds.

  6. The stretching frequency for a carbon−carbon bond occurs at approximately 1600 cm−1 for a C=C double bond and at 2100−2200 cm−1 for a C≡C triple bond. Explain the relative frequencies based on bond strength.

    A C≡C triple bond is stronger than a C=C double bond. The stretching frequency (wavenumber) increases as bond strength increases.

  7. The intensity of a band in the IR spectrum depends on the change in dipole moment that accompanies a vibrational transition in a molecule. Predict the relative intensity of IR bands due to C=C versus C=O stretching vibrations.

    Compounds containing C=O bonds are more polar and thus have more intense stretching frequencies than C=C double bonds.

Table 1 summarizes general infrared absorption frequencies for common structural units (functional groups) in organic molecules.
  1. The IR spectrum of aspirin (acetylsalicylic acid) is shown, along with its structure. Use the information in Table 1 to identify the peaks in the IR spectrum corresponding to stretching vibrations of the following bonds:
    1. O–H group

      The OH stretch gives rise to a broad band between 2700−3000 cm−1. This band overlaps the frequencies due to C−H stretches.

    2. C=O groups (there are two)

      1700 and 1750 cm−1

    3. C=C bonds in the benzene ring

      Sharp peak at 1600 cm−1

    {13776_Answers_Figure_8_IR spectrum and structure of aspirin}
  2. A portion of the IR spectrum of ethyl alcohol, CH3CH2OH, is shown. Note the broad band associated with the O−H group, as opposed to the sharp peak for the C−H stretching vibration. The width of the O−H band is usually attributed to hydrogen bonding. Why would hydrogen bonding lead to a broad band for the O–H stretching vibration?

    Hydrogen bonding leads to a variation in the O−H bond length as the hydrogen atom is attracted to unpaired electrons on adjacent atoms. The hydrogen bonds are fleeting or transient as molecules are able to move about freely in solution. The stretching frequency varies over a wide range as the bond strength and bond length vacillate.

Part C. Beer’s Law and Quantitative Analysis

Figure 9 shows a typical visible spectrum. The wavelength range for visible spectroscopy is 350−700 nm. In general, absorbance is proportional to concentration for each wavelength. The higher the concentration of a substance in solution, the more intense its color will be, and the greater its absorbance. The linear relationship between absorbance (A) and concentration (c) is expressed in Beer’s law, Equation 3, where b is the path length, in cm, of solution the light passes through, and a is a proportionality constant.
If c is given in units of molarity (M = moles/L), then a is known as the molar absorptivity coefficient, with units M−1 cm−1
{13776_Answers_Figure_9_Visible spectrum of nickel nitrate}
  1. The graph is for a 0.08 M solution of Ni2+ ions. At what wavelength is the value of molar absorptivity the highest?

    The absorption maximum occurs at about 390 nm.

  2. If the path length, b, is 1 cm, calculate the molar absorptivity (a) of the Ni2+ ion at this wavelength.

    A = 0.40 at 390 nm.
    0.40 = a x (1 cm)(0.08 M)
    a = 5 M−1 cm−1

  3. Predict the absorbance value for a 0.03 M Ni(NO3)2 solution if measured at this wavelength.

    If absorbance is directly proportional to concentration, the absorbance value for a 0.03 M solution should be equal to (0.03)/(0.08) x 0.40, or about 0.15. (Note that absorbance values have no units.)

  4. What concentration of Ni2+ ions would be expected to have an absorbance of 0.25 at this wavelength?

    0.25/0.40 = x/(0.08) x = approximately 0.05 M

A digital spectrophotometer measures both the percent transmittance of light and the absorbance. When light is absorbed, the radiant power (P) of the light beam decreases. Transmittance (T) is the fraction of incident light (P/Po) that passes through the sample. The relationships between transmittance and percent transmittance (%T) and between transmittance and absorbance (A) are given in Equations 4 and 5, respectively.
  1. What is the percent transmittance, %T, when A equals zero? When A equals 1.0?

    A = −log T
    When A = 0, T = 10o = 1 (100% transmittance = zero absorbance)
    When A = 1, T = 10–1 = 0.1 and %T = 10%. Only 10% of the radiant light passes through the solution when the absorbance is 1.0.

  2. For most substances, Beer’s law is linear for absorbance values below 1.000 and above 0.025. Based on the mathematical relationship between absorbance and transmittance in Question 5, explain why Beer’s law graphs of absorbance versus concentration may deviate from a straight line when A < 0.1 and when A > 1.

    Absorbance measurements are most accurate in the range 0.1–1.0. A spectrophotometer detects the difference or ratio between the power of the incident and transmitted light. At very low absorbance values, the percent transmittance is high (for A = 0.1, %T = 80%). Since almost all of the incident light goes through, there is not a great difference between the incident power and the transmitted light for the instrument to detect. At very high absorbance values the percent transmittance is very low. The instrument does not accurately detect low power of transmitted light.

  3. The Beer’s law graph shown below was generated by measuring absorbance values at 630 nm for a series of standard solutions containing known concentrations of Cu2+ ions. A sample of steel was treated with acid to dissolve the copper content, and the absorbance of the resulting solution at 630 nm was found to be 1.345, outside of the range of the Beer’s law plot. Which step (a, b or c) would you recommend for determining an accurate concentration of copper in this sample? Explain.
    1. Extrapolate the curve to include 1.345 A.
    2. Run the calibration curve using higher concentrations of copper.
    3. Dilute the sample to bring its absorbance between 0.200 A and 0.800 A.

      The best course of action is c. Dilute the sample using quantitative (analytical) transfer techniques. Extrapolating the Beer’s law plot to A > 1 will not give accurate concentration values. Similarly, a calibration curve obtained at higher concentration values may not give a linear graph.

    Stainless steel is an alloy of iron with other metals that are added to increase its corrosion resistance. One specific alloy of stainless steel contains 21% chromium. Quality control groups routinely test batches of this alloy by spectroscopy to determine the element content, including % chromium. The absorption spectra for both Fe3+ ions and Cr3+ ions are shown in Figure 10.
    {13776_Answers_Figure_10_Visible spectra of Fe3+ and Cr3+ solutions}
  4. A calibration curve (Beer’s law plot) was created using the wavelength of 405 nm. A series of Cr(NO3)3 standard solutions were tested to generate this calibration curve. The absorbance of the steel sample at 405 nm was determined. When the % Cr in the sample was calculated, the answer was a whopping 79%! What is the likely source of this very large error?

    Iron(III) ions also show a very high absorbance at 405 nm and thus interfere with the determination of chromium(III) ions at this wavelength. The absorbance of the dissolved steel sample at 405 nm is due to both Fe3+ and Cr3+ ions.

  5. A second calibration curve was generated for Cr3+ ions, this time using the absorbance at 575 nm (see Figure 11). When a 1.215-gram steel sample was dissolved in acid and diluted to 50.0 mL, the absorbance of the resulting solution at 575 nm was 0.520 A. Calculate the % Cr in the steel sample.
    {13776_Answers_Figure_11_Calibration curve for Cr3+ ions at 575 nm}

    When A = 0.520, [Cr3+] = 0.090 M in the sample solution.
    The amount of chromium in 50 mL of the solution is (0.09 moles/L) (0.050 L) (52 g/mole), or 0.234 g.
    Percent chromium = (0.234 g/1.215 g) x 100, or 19%.


AP® Chemistry Curriculum Framework: 2013−2014; The College Board, 2011.

Student Pages

Electronic and Vibrational Spectroscopy

Guided-Inquiry Learning Activity for AP® Chemistry


The absorption of electromagnetic radiation, whether in the ultraviolet, visible or infrared range, allows chemists to determine not only the structures of atoms and molecules but also, under certain conditions, their concentration in solution. Each type of electromagnetic radiation tells a different story about the composition of matter.


  • Electromagnetic radiation
  • Planck’s law
  • Absorption and emission
  • Electron energy levels
  • UV/Vis spectroscopy
  • Molecular vibrations
  • Infrared spectroscopy
  • Beer’s law
  • Quantitative analysis


Spectroscopy is a general term referring to methods of instrumental analysis based on the interaction of electromagnetic radiation with matter. Depending on the type or energy of electromagnetic radiation, the interaction of light with matter may lead to a variety of different types of transitions in atoms and molecules. The absorption of ultraviolet (UV) or visible (Vis) radiation results in electronic transitions, in which electrons are promoted or excited to higher energy levels, while the absorption of infrared (IR) radiation excites vibrational transitions in molecules. A spectrum is produced by measuring the intensity of radiant energy absorbed by a substance as a function of the wavelength of light from an external energy source. Electronic and vibrational spectra may be used to obtain information concerning the identity of elements and compounds, their atomic and molecular structures and the concentration of a substance in solution.

Electromagnetic (EM) radiation has a dual wave-particle nature, behaving and exhibiting properties of both particles and waves. The wave properties of EM radiation are characterized or defined by its velocity (c, the speed of light), wavelength (λ, lambda), and frequency (v, nu). See Equation 1. Different types of absorption spectroscopy are usually identified by the wavelength of light involved, namely, X-ray, UV, visible, infrared, microwave or radiofrequency.

The quantum theory or nature of light describes electromagnetic radiation in terms of mass-less particles, called photons, each possessing a quantum or packet of energy (E), where the energy is directly proportional to the frequency (v) of radiation (see Equation 2, known as Planck’s law). Planck’s law defines the relationship between EM radiation as a wave and as a photon of energy.
The proportionality constant h, known as Planck’s constant, is equal to 6.626 x 10−34 J • s. Combining Equation 1 with Planck’s law, we know that c is a constant (the speed of light), and therefore wavelength and frequency are inversely proportional. Thus, photons with higher energies have shorter wavelengths and photons with lower energies have longer wavelengths. The energy of EM radiation in UV spectroscopy is measured in wavelength, while IR spectra are typically plotted in terms of wavenumber, which is a frequency unit. (See Parts A and B.)

The quantum nature of light finds a direct parallel in the quantization of electron energy levels within an atom or molecule. The interaction of EM radiation with matter is best described as the increase in energy of an atom or molecule caused by the absorption of a photon. At the molecular or atomic level, these increases in energy are quantized. Electronic and vibrational spectroscopy are due to the following types of absorption.
  • Promotion of an electron in an atom or molecule (A) from a lower energy state to a higher energy, excited state, which is denoted A*. See Figure 1. The wavelength of radiation needed to cause electronic transitions is in the range of ultraviolet and visible light, 10 nm to 700 nm (1 nm = 1 nanometer = 1 x 10−9 m).
  • Change in vibrational energy levels associated with stretching and bending chemical bonds—see Figure 2. Like electronic energy levels, vibrational energy levels are also quantized. The wavelength of radiation needed to cause vibrational energy transitions occurs in the infrared region, 700−1000 nm.
The absorption of electromagnetic radiation may be represented using an energy level diagram (see Figure 3). E0 represents the ground state, or at-rest, energy level of an electron in a molecule. The possible excited energy levels for the electrons are quite numerous, since each electron energy level has multiple vibrational energy levels. (Note that there are also rotational energy levels embedded within the different vibrational levels.)
{13776_Background_Figure_3_Energy level diagram}

Experiment Overview

The purpose of this guided-inquiry learning activity is to investigate the principles and applications of electronic and vibrational spectroscopy. The activity is divided into three parts:

  1. Electronic energy levels and UV-Vis spectroscopy.
  2. Vibrational energy levels and IR spectroscopy.
  3. Beer’s law and quantitative analysis.

Student Worksheet PDF


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