AP#Chemistry12# #...

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~ Unit X: Spectroscopy ~ Topics to Review: Review the history of the atom (Dalton’s theory, the discovery of the electron and nucleus (Thomson model and Rutherford’s gold foil experiment), Bohr’s model of the hydrogen atom, and electron configuration). You will need to also review periodic trends, particularly ionization energy. Khan Academy has an excellent section on atomic structure, electron configuration, and Bohr’s model here: https://www.khanacademy.org/science/ap-‐chemistry/electronic-‐structure-‐of-‐atoms-‐ap

Background Information Light: Electromagnetic Waves (EM radiation) Electromagnetic radiation is one way that energy is transferred or propagated. Light, a form of EM radiation, exhibit wavelike properties, described by the diagram below.

Waves have the following characteristics: • Wavelength, λ (lambda) – distance from crest to crest • Frequency, ν (nu) – number of wave cycles per unit time • Amplitude – distance from midline to crest or midline to trough Wavelength is often measured in meters or nanometers (depending on the specific type of EM) (1 m = 109 nm) Frequency is usually measured in Hertz (Hz) which represent one cycle per second (1 Hz = 1 s-‐1). It represents how often the wave repeats. Wavelength and frequency are inversely proportional. As the wavelength gets longer, the shorter the frequency. Think about the two waves (on the diagram to the left) travelling from left to right. If they travel at the same speed, the wave with the longer wavelength will only go through about one cycle while the wave with the shorter wavelength will go through three cycles.

The speed of a wave can be found by multiplying wavelength and frequency. The speed of all EM in a vacuum is the speed of light, c, 3.00 x 108 m/s.

𝑐 = 𝜆 𝜈 Example 1: A particular wave of EM radiation has a frequency of 2.95 x 1012 Hz. What is the wavelength?

𝑐 = 𝜆 𝜈

𝜆 =𝑐𝜈=3.00 ∗ 10!𝑚𝑠2.95 ∗ 10!" 1𝑠

= 1.02 ∗ 10!! 𝑚

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The wavelength of the EM radiation is 1.02 x 10-‐4 m. This is approximately 0.1 mm, which corresponds to microwave radiation. EM radiation generally ranges from 10-‐16 m (gamma rays) to 108 (long radio waves). The

diagram on the next page shows the wavelength and frequency of different types of EM radiation. As we look through the spectrum, we may recognize specific types of radiation as harmful to humans. These potentially include gamma rays, X-‐rays, and UV radiation. All of these are ‘high energy’ radiation, and thus can cause damage to our DNA. Other forms of radiation, like visible light, microwaves, and radiowaves, surround us constantly. Their frequencies are so low that they are not harmful to us. How is frequency related to harmfulness? Dual Nature of Light: Quantization of Energy It has been known for centuries that light behaves like a wave. However, in 1900, Max Planck studied blackbody radiation—objects that glowed (emitted radiation) when heated. He discovered that when objects absorbed or emitted radiation, they could only do so in discrete amounts rather than continuous amounts. So instead of being able to transfer any amount of energy, Planck discovered that it was only possible to do so in individual packets, known as a quantum. We can make an analogy to this by thinking of money. When we make transactions, the lowest amount we can pay is in pennies (let’s assume we still have the Canadian penny). We can’t pay, for example, $0.5985 since we don’t have anything ‘units’ of money that is less than a cent. Energy works in a similar way. The smallest unit of energy is related to the Planck’s constant, h. This means that energy can only be transferred in multiples of ‘h’, just like how we can only transfer money in multiples of ‘pennies’.

𝑃𝑙𝑎𝑛𝑐𝑘!𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡: ℎ = 6.626 ∗ 10!!"𝐽 ∗ 𝑠 This discovery led to the idea that not only did light behave like a wave, it also displayed particle-‐like properties. This is known today as the wave-‐particle duality (or dual nature of light, wavelike and particle-‐like behavior). It turns out that even matter displays wave-‐particle duality… Think about this for a second. When energy is transferred from one object to another through the emission (or absorption) or light (think of our Sun’s rays of light), it occurs through the transfer of a photon, the particle of light, or a stream of photons. Each photon emitted carries a specific amount of energy, related to the frequency of the radiation (aha! Frequency is related to energy of light, which relates to its harmfulness to human health). The equation is given by Planck’s equation:

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𝐸 = ℎ𝑣 This equation tells us that the energy of a photon is equal to its frequency multiplied by Planck’s constant. Again, the amount of energy transfer must be a multiple of the constant. Example 2: A photon of red light at 650 nm is emitted by a laser pointer. What is the energy of this photon?

𝐸 = ℎ𝑣 𝑎𝑛𝑑 𝑐 = 𝜆𝑣 We can rearrange the speed of light equation for 𝑣 and substitute it in the energy equation.

𝑣 =𝑐𝜆 𝑎𝑛𝑑 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸 = ℎ

𝑐𝜆

𝐸 = 6.626 ∗ 10!!"𝐽 ∗ 𝑠 3.00 ∗ 10!𝑚𝑠650 ∗ 10!!𝑚

= 3.06 ∗ 10!!"𝐽

The energy of one photon of red light at 650 nm is 3.06 x 10-‐19 J. This is a TINY amount of energy, as expected, since this is just one particle of light. Example 3: Calculate the energy of one mole of photons of gamma radiation with wavelength of 1 x 10-‐14 m.

𝐸 = ℎ𝑐𝜆

𝐸 = 6.626 ∗ 10!!"𝐽 ∗ 𝑠 3.00 ∗ 10!𝑚𝑠1 ∗ 10!!"𝑚

= 2 ∗ 10!!!𝐽

Since this is only the energy of one photon, we need to multiply that by Avogadro’s number to get a mole.

𝐸 = 2 ∗ 10!!!𝐽

𝑝ℎ𝑜𝑡𝑜𝑛 ∗ 6.02 ∗ 10!"

𝑝ℎ𝑜𝑡𝑜𝑛𝑠𝑚𝑜𝑙𝑒

= 1 ∗ 10!"𝐽

Electron Transitions:

We mentioned earlier that matter can only absorb and emit in specific amounts of energy. Why is that? Let us look at the electronic structure of hydrogen. The hydrogen atom has one electron at the n = 1 shell (ground state). The shell level number is also called the principal quantum number. Recall that the electron configuration for hydrogen is 1s1. Since the electron cannot be in between energy levels, it must gain a specific amount of energy in order to be promoted to a higher energy level (excited state). For example, it must absorb exactly 1.634 * 10-‐18 J to jump to the n = 2 shell. This corresponds to UV light at 122 nm. This would mean that the electron at n=1 would need to absorb light at 122 nm for it to jump to n=2.

The energy of an individual level, n, is given by the equation 𝐸 = −2.178 ∗ 10!!" !!

!!𝐽𝑜𝑢𝑙𝑒𝑠. Z is the atomic

number, and in this case, it is 1 for hydrogen. Thus, the relationship between the energy of a transition between levels is calculated from this equation:

∆𝐸 = −2.178 ∗ 10!!" ∗ 𝑍!1

𝑛!"#$%! −1

𝑛!"!#!$%! 𝐽𝑜𝑢𝑙𝑒𝑠

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A negative energy change in this case means energy is emitted (electron relaxing from a higher to lower state) while a positive energy change would mean energy is absorbed (electron exciting from lower to higher state). Important: Even though we can use this equation for different atoms, it only works well for hydrogen and other 1-‐electron systems (i.e. Li2+ or O7+, which are generally unrealistic). Example 4: Calculating the energy transition when an n=4 electron relaxes to n=2 in a hydrogen atom. Determine the type of EM radiation this is.

∆𝐸 = −2.178 ∗ 10!!" ∗ 𝑍!1

𝑛!"#$%! −1

𝑛!"!#!$%! 𝐽𝑜𝑢𝑙𝑒𝑠

∆𝐸 = −2.178 ∗ 10!!" ∗ 1!12!−14!

𝐽𝑜𝑢𝑙𝑒𝑠

∆𝐸 = −2.178 ∗ 10!!" ∗14−116

𝐽𝑜𝑢𝑙𝑒𝑠

∆𝐸 = −4.084 ∗ 10!!"𝐽 The photon emitted has an energy of 4.084 x 10-‐19J. Since E = hv and c = λv

λ =cv=cEh=chE=

3.00 ∗ 10! 6.626 ∗ 10!!"

4.084 ∗ 10!!"= 4.87 ∗ 10!!𝑚 = 487 𝑛𝑚

EM radiation with 487 nm is visible light (greenish blue) Absorption Spectrum and Emission Spectrum: If we look at the various emissions (electron relaxing) possible in a hydrogen atom, we see that depending on where it starts and ends at, we can get EM radiation with various wavelengths. The Balmer series (where nfinal = 2) has transitions at just the right energy level such that the EM radiation emitted are in the visible light region (refer to Example 4). The Lyman series have transitions that are higher in energy (UV light emissions) while the Paschen series have transitions that are lower in energy (infrared emissions). How do the electrons get to the higher energy shells in the first place? If a group of hydrogen atoms are blasted with energy (electricity or radiation), electrons in the different hydrogen atoms will absorb whatever energy they can to transition to some other shell. For example, one electron might absorb a lot of energy and transition from n=1 to n=7 while another electron will absorb smaller amounts of energy and transition from n=1 to n=2. Yet, some other electron can make multiple jumps by first making the transition from n=1 to n=2, then n=2 to n=5. Because each transition corresponds to a particular wavelength, blasting radiation of the ‘wrong’ wavelength will not be absorbed.

To the left is the absorption spectrum of hydrogen from ~400 nm to ~700 nm. Notice that there are bands of black. These correspond to the wavelengths of light that were absorbed by

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the electrons in the hydrogen atoms. These wavelengths of light are no longer detected once the electrons absorb them to make an electronic transition. On the other hand, the colours of light that were not absorbed are sensed by the detector. These colours do not provide the electron with the correct amount of energy to make the necessary transitions. Finally, notice the dark band ~480 nm. According to Example 4, what electronic transition did the electron that absorbed this wavelength of light make? Now that the light has been absorbed and the electrons are all excited, the electrons need to emit light since an excited state is unstable. The electrons will generally all eventually head back to the ground state, but they can take detours along the way. For example, a n=7 electron can first transition to n=5, then to n=2, then to n=1. When these electronic transitions occur, they emit light with the wavelength corresponding to the appropriate energy level difference. For example, a transition from n =5 to n= 2 in a hydrogen atom corresponds to a wavelength of light of ~434 nm, blue light (calculate this for yourself!). The emission spectrum is shown on the right. The bands on colour that you see correspond to the wavelengths of light emitted from the hydrogen atom’s electrons relaxing from higher energy levels to lower energy levels. If we put the two spectra next to each other, something obvious should stand out. The absorption spectrum is the ‘opposite’ of the emission spectrum. This should make sense—Any wavelength of light that is absorbed to make a transition (e.g. n=2 to n= 3) must be the same light that is emitted when the electron relaxes (n=3 back to n=2). The absorbed light will show up as a ‘dark band’ in the absorption spectrum and the emitted light will show up as a ‘coloured band’ in the emission spectrum.

Likewise, if it takes ~434 nm light to promote an electron from n=2 to n=5, then 434nm light will be emitted when the electron relaxes from n=5 to n=2. Note: A more accurate view of these diagrams would be to treat an atom as a sphere, and the circular shells as energy levels. However, the lines all pointing inwards can seem confusing. Recall that the most inner shells are the ones that are closest to the nucleus, and thus are the most stabilized and have the lowest amount of energy. The energy diagrams in the previous pages flattened out the circles into lines to easily distinguish between n=1, n=2, n=3, etc. We finally now have the background knowledge to understand spectroscopy J

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Spectroscopy: Overview Spectroscopy is the set of laboratory techniques used to interact radiation (EM) with substances. For example, we have seen how certain ions absorb various colours of light in spectrophotometry or colourimetry. Depending on the radiation we use, we can focus on specific structures. For example, the carbon oxygen double bond in a carboxylic acid interacts with IR radiation, while the extended conjugated (alternating double and single bonds) structure in dyes interacts with visible light. Since different structures react differently to different types of EM radiation, you would choose the wavelength of light that is appropriate to what you want to study. Important note: Even though we have only looked at electronic transitions in atomic orbitals, specifically for the hydrogen atom, electronic transitions between molecular orbitals and between vibrational modes occur in a similar fashion—these transitions occur because electrons involved in the bond absorbs EM radiation at a specific wavelength. For example, an electron involved in the bonding between C=O in a ketone absorbs light with a frequency of 5 x 1013 Hz. This corresponds to the IR radiation region, and thus allows us to identify a molecule as having ketone group whenever they absorb at this frequency. Below is a table that generalizes the type of radiation a scientist would want to use in different studies. We will go into a little more detail for the techniques that are bolded.

Technique Frequency and Source Radiation Utility of Technique

Photoelectron Spectroscopy (PES)

X-‐Ray to High Energy Ultraviolet

1016 to 1019 Hz

• Excitation and ejection of core atomic electrons • Sample is a solid or in the gas phase • Photoelectron spectrum is based on energy levels of

individual orbitals (1s vs 4d) • Allows identification of atoms

UV-‐Vis Spectroscopy, Colourimetry,

Spectrophotometry)

Ultraviolet to Visible Light

1014 to 1016 Hz

• Excitation of molecular π-‐bond electrons, atomic valence electrons, or metal ion valence electrons

• Sample is in an aqueous solution and is either an organic molecule with extensive pi-‐bonding or a transition metal ion

• Absorbance spectrum is based on energy levels of absorbing structures

• Allows determination of concentrations using Beer’s Law • Allows identification of (large) structures such as benzene

rings or extended double bond structures in organic molecules

IR (Vibrational) Spectroscopy

Infrared

1012 to 1014 Hz

• Excitation of molecular vibrations in covalent bonds • Sample is in solution, a liquid or solid • IR spectrum is based on functional groups present (a

ketone C=O bond absorbs a different IR wavelength when compared to an alkene C=C bond)

• Allows determination of specific types bonds in molecules

Microwave (Rotational) Spectroscopy

Microwave

108 to 1012 Hz

• Excitation of molecular rotations in molecules • Sample is in gas phase • Allows determination of chemical composition of matter in

space using radio telescopes • Also helped us understand intermolecular forces

Nuclear Magnetic Resonance

Spectroscopy (NMR)

Radio wave

<108 Hz

• Excitation of nuclei in atoms and molecules • Sample is in solution • Allows determination of location of hydrogen atoms (and

other atoms) in organic molecules • Ubiquitous in the field of organic chemistry

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Example 5: For each of the following, pick the appropriate spectroscopic technique to use. a. Determine the concentration of a blue dye. b. Determine how much energy is required in total to remove all the electrons in an oxygen atom. c. Distinguish between a C=O bond and a C-‐O bond. d. Determine whether we have a sample of sodium atoms or magnesium atoms. e. Determine the absorptivity coefficient of Mn2+ ions.

Answers:

a. UV/Vis spectroscopy using Beer’s Law b. Photoelectron Spectroscopy (PES) c. IR Spectroscopy d. PES e. UV/Vis spectroscopy using Beer’s Law

Photoelectron Spectroscopy

Theory: The diagrams on the right shows the PES technique. A monochromatic beam of X-‐ray light (monochromatic means one wavelength only) is absorbed by the sample. Since the energy of X-‐rays is so high, electrons have enough energy not only to jump to a higher energy level, but they completely escape the surface of the sample. By analyzing the kinetic energy of these escaped/ejected electrons, we can calculate how strongly they were bound to the surface (binding energy) and thus analyze the sample for its identity. This technique is related to the Einstein’s discovery of the photoelectric effect which states that the amount of kinetic energy an electron has after ejection from a surface is equal to the amount of energy the light source has minus the binding energy (how tightly held the electrons are). Einstein won his Nobel Prize for this discovery! 𝑲𝑬𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 = 𝑬𝒑𝒉𝒐𝒕𝒐𝒏 − 𝑬𝒃𝒊𝒏𝒅𝒊𝒏𝒈 Let us look at the photoelectron spectrum (to the left) of an element with a low atomic number, and thus low

number of electrons. Which element is the PES for? The exact values of the binding energies are not important, only the relative values. Also, notice that the scale on the x-‐axis is not linear, but logarithmic, and that the scale itself is decreasing from left to right. If we look at the height of each peak, notice that the lowest-‐whole-‐number-‐ratio is 2:2:3. This would most

likely mean there are 7 electrons. The atom with 7 electrons is nitrogen. Let’s check to see if the rest of the information agrees. The electron configuration of nitrogen is 1s22s22p3. We know that core electrons are held the tightest since they are closest to the nucleus, and in nitrogen, these would be the 1s electrons. Nitrogen also has five valence electrons, and these would be the remaining 2s and 2p electrons. These electrons are held less tightly since they are further away from the nucleus. To summarize, the first peak corresponds to the two electrons in the 1s orbital, the second peak is the 2s orbitals, and the third peak is the 3s orbital.

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Example 6: Analyse the PES and determine the identity of the element. Once identified, assign each peak to an orbital. Answer: There are three peaks, and the relative height of each peak is 1:1:3. If we add all three numbers together, we get 5. The element with 5 electrons is boron.

If we analyse boron’s electron configuration, we get 1s22s22p1. This does not give us the 1:1:3 ratio, so the element cannot be boron. If we take the ratio and double all the numbers, we get 2:2:6. Adding up these numbers, we get 10 electrons. The element corresponding 10 electrons is neon. The electron configuration of neon is 1s22s22p6. This agrees with our ratio of 2:2:6, and thus we can conclude the element is neon, and the first peak is the 1s, second peak 2s, and third peak 2p. Notice once again that the specific binding energy is not important. Example 7: The PES peak for the 3s electrons are shown for Na and K.

a. The ionization energy of K is known to be lower than Na. Explain why. (This is a Chem 11 question!)

b. Why does the peak for K have a higher binding energy than Na?

c. Why is the intensity for potassium double that of sodium?

Answers: a. The ionization energy of K is lower than that of Na because the outer most electron in each atom is in different orbitals. The outermost electron of K

is the 4s electron and the outermost electron in Na is the 3s. • Since 4s electrons are much further away from the positively charged nucleus compared to 3s electrons, they are easier to remove due to

weaker electrostatic attraction (attractive) • Moreover, the 4s electrons are more shielded by core electrons, meaning they don’t feel the attraction of the nucleus as much [another

term for this is lower effective nuclear charge] (repelled by the core electrons: 1s22s22p63s23p6 = 18 electrons) while 3s electrons are less shielded (repelled by 1s22s22p6 = 10 electrons) (repulsive)

b. When we are removing the 3s electrons in each atom, we only need to consider the attractive forces to the nucleus since the amount of shielding will be the same (10 electrons). There are more protons in the nucleus of potassium than in sodium (19 vs 11). This means that the attraction is stronger. A stronger attraction results in more energy required in removing the electron (thus binding energy).

c. The intensity of K is doubled because there are two 3s electrons in K but only one 3s electron in Na.

Example 8: The complete PES of aluminum is shown on the right. On the same graph, predict how the PES of magnesium would look like. Explain briefly how you obtained your PES. Answer: First verify that the peaks correspond to aluminum. The peak ratios are 2:2:6:2:1. If we add these numbers together, we get 13 electrons. This is the number of electrons aluminum has. Moreover, these peaks correspond to 1s22s22p63s23p1 perfectly.

The number of electrons in magnesium is 12, resulting in the electron configuration 1s22s22p63s2. This means that magnesium will have 4 peaks that are similar to aluminum. However, magnesium has fewer protons (thus, a smaller nuclear charge). A smaller nuclear charge would mean that the attraction to the electrons is weaker. Thus, the binding energy would be lower for ALL orbitals.

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For Al, the 151 MJ/mol peak is the 1s2 peak. We can move this slightly to the right for magnesium since the binding energy is lower. We would do this for each of the other orbitals as well. However, we would NOT create a similar peak for Al’s 3p1 since Mg does not have a 3p electron. Note: The exact locations and values of the peaks is not important. What is important is the relative positions of the peaks (the same orbitals are all lower in Binding Energy). Example 9: The PES diagram for scandium is below.

a. Rationalize the location and height of the peaks and assign orbitals. Pay attention to the last two peaks. b. Explain why scandium commonly forms +3 charge ions (more common than +2 charge ions), while not

forming higher charged cations? Answers:

a. The peaks from left to right are as follows: 1s22s22p63s23p63d14s2. Although 4s electrons are filled before 3d electrons, 4s electrons are also removed first during ionization. This is because they are held less tightly by the nucleus (n=4 versus n=3).

b. The 4s2 and 3d1 electrons are very close in energy. Because of this, the amount of energy required to remove electrons from the 4s and 3d (<1MJ) is similar, and thus, they are usually removed together. Removing even more electrons requires significantly more energy (3+ MJ for the 3p electrons)

The PES for nitrogen and oxygen are shown to the right. We would expect oxygen to have a higher binding energy for all electrons since oxygen has a higher number of protons with no extra shielding. However, if we look at the 2p peak, we see that it has a lower binding energy for oxygen. To understand why, let’s write out the electron configuration of each atom and their orbital diagrams.

O: 1s2 2s2 2p4 N: 1s2 2s2 2p3

Notice that for oxygen, one electron in the p-‐orbital is paired (recall that there are three 2p orbitals, each can hold 2 electrons for a total of 6). The pairing of the electron causes repulsion, and thus the electrons are less tightly bound than expected. Less tightly bound would mean it is easier to remove, and has a lower binding energy. We expect this to be true also between phosphorus atoms and sulphur atoms.

151 12.1

7.9

1.09 0.58

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Real PES: Real photoelectron spectra are much more convoluted in the sense that they contain a lot of background ‘noise’ and interference. Peaks are also not well-‐defined, and many peaks overlap.

Overlapping peaks occur because PES is usually a solid-‐phase technique. With so many atoms nearby, ejected core electrons can leave holes that other valence electrons can drop into. The relaxation of these valence electrons into the now vacant spot causes an emission of low energy photons that can eject other electrons. This leads to electrons with low kinetic energy, and thus calculations would show that these electrons have high binding energy, when in fact, they do not.

We will not need to analyze these spectra, but be aware that the spectra we have been analyzing are idealized (and simulations, rather than real).

UV-‐Vis Spectroscopy/Colourimetry/Spectrophotometry We have already covered this in the kinetics chapter. Theory: Recall that dissolved, coloured compounds absorb colours of light that are complementary (‘opposite’) to their own colour. For example, a red dye will absorb most strongly green-‐blue light and a purple dye will absorb most strongly yellow light. This is because white light is a mixture of all colours of light, and any light that is absorbed would mean that it is not passed through the solution. A dye that absorbs green light most strongly would mean that all other light can pass through. This would mean you see a mixture of all colours except green, and we (aka ‘our brain’) interpret that as a red-‐orange colour.

The spectrum to the left shows which wavelengths of light the compound, indigo carmine, absorbs the most. The wavelengths of light it absorbs most is 200nm, then 290nm, then 250nm and 610nm. The lower (<400 nm) three values are in the UV-‐light region, and thus do not affect the colour of the compound itself, but absorbance at 610 nm means that indigo carmine absorbs most strongly yellow-‐orange light, leaving the compound a blue-‐indigo colour. A colourimetry or spectrophotometer allows

you to select specific wavelengths of light to illuminate a solution. It then measures the intensity of light entering, Io, and amount of light exiting the solution, I. Less light exiting would mean the solution has absorbed most of it. With indigo carmine, we would use the wavelength of 610 nm since that is the coloured light that it absorbs most. It is important that we select the most absorbing wavelength since this means that the instrument would be more sensitive to concentration changes. If a 0.100 M and a 0.0800 M both absorb at 0.001

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y = 2145x + 0.002

0

1

2

3

0 0.0002 0.0004 0.0006 0.0008 0.001

Absorban

ce

Concentra-on of Red Dye (M)

Absorbance of Red-‐Dye at 495 nm

absorbance units (wavelength of light used is lowly absorbed), then this experiment would be quite useless. It would be much better if the two solutions had significantly different absorbing values (wavelength of light used is highly absorbed). An absorbance value (no units, or ‘absorbance’ units) is calculated by the machine, and a reading is given.

𝐴 = − log𝐼𝐼!

Solutions that are coloured and more concentrated will absorb light more strongly. This is given by Beer’s Law:

𝐴 = ɛ𝑏𝑐 A is the absorbance of the solution at a particular wavelength. ɛ is the molar absorptivity coefficient (or molar extinction coefficient) and has units of M-‐1 cm-‐1 (or M-‐1 m-‐1). We can

think of the molar absorptivity coefficient as how absorbing a particular compound is. The higher it is, the stronger it will absorb.

b is the path length of the cuvette, a container (how far the light travels through the solution) c is the concentration of the solution. Because the absorptivity coefficient and path length can be fixed for any solution at a particular wavelength, we can make solutions of known concentrations and measure their absorbances to create a calibration curve. Calibration Curves: Determining the concentration of a substance can be done if we know its absorbance value and the relationship between concentration and absorbance. The absorbance values are easy to obtain as long as you have a spectrophotometer. However, to find the relationship between the concentration and absorbance, we first need a calibration curve that matches Beer’s Law, A = ɛbc. Let’s look at some data and see how we can construct a calibration curve: Several solutions at different known concentrations of red dye have their absorbances measured and recorded. The wavelength of light used is 495 nm and the cuvette length is 0.500 cm. Side note: The choice for the length of the cuvette is up to the experimenter. It is most common to use b = 1.00cm. The following data is collected and graphed (next page). We can see from the data itself that as the concentration increases, the absorbance increases linearly. This follows Beer’s Law.

The equation of the calibration curve is y = 2145x + 0.002. The y-‐intercept, 0.002, should be zero in the ideal case (since if concentration is zero, absorbance should also be zero), but due to experimental and/or random errors, it can fluctuate.

Concentration Absorbance

0.00100 M 2.147

0.00080 M 1.718

0.00060 M 1.289

0.00040 M 0.860

0.00020 M 0.431

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Since our y-‐axis is absorbance and our x-‐axis is concentration, we can write A = 2145c + 0.002. This matches with Beer’s Law (A = ɛbc), if we ignore the y-‐intercept. Since the slope of the graph is 2145 M-‐1, ɛb must also equal 2145 M-‐1. In this example then, since b = 0.500 cm, we can determine the absorptivity coefficient of this red dye at 495 nm.

ɛ!"#!" =2145𝑀!!

0.500𝑐𝑚 = 4290 𝑀!!𝑐𝑚!!

Generally, coloured compounds that have ɛ > 10000 M-‐1 cm-‐1 are considered strongly absorbing while compounds that have 10 M-‐1cm-‐1 < ɛ < 1000 M-‐1 cm-‐1 are considered weakly absorbing. Now that we have our calibration curve equation A = 2145c + 0.002, we can solve for any unknown concentration of red dye at 495 nm as long as we can measure its absorbance. Example 10: After creating the calibration curve (y=2145x + 0.002) of a red dye using the wavelength of light at 495 nm, a solution of red dye with unknown concentration was poured into a 0.500 cm cuvette and its absorbance is measured. The absorbance is measured to be 1.220. What is the concentration of the solution? Answer: You can answer this in two ways: Using the graph or using the equation. Using the graph is less accurate since it requires estimating, so let’s use the equation.

𝑦 = 2145𝑥 + 0.002 1.220 = 2145𝑥 + 0.002

𝑥 = 0.00057 𝑀 Since x is the concentration, [red dye] = 5.7 x 10-‐3 M. Example 11: A calibration curve is created for thymol blue (λ=300nm) and the linear equation obtained (with a 1.00 cm cuvette) is y = 12405x – 0.0012. As part of a kinetics experiment, multiple solutions of thymol blue are measured at different times. Calculate their concentrations. Time(s) 0.00 30.00 60.00 90.00 120.00 Absorbance 1.948 1.714 1.480 1.246 1.012 Concentration (M) Answer: This is done in the same way as in Example 10. The answers are below: Concentration (M) 1.571 x 10-‐4 1.383 x 10-‐4 1.194 x 10-‐4 1.005 x 10-‐4 8.168 x 10-‐5 Limitations of Spectrophotometry:

1) Spectrophotometry allows us to determine the concentrations of solutions of unknown concentrations by first measuring the absorbances of solutions of known concentrations (calibration curve). However, not all compounds can be measured this way.

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In general, many transition metals have various colours, so these can also be studied using a spectrophotometer. To the left is a list of some metal ions and their colours (metal ion colours can be affected by what they are coordinated to, so the list is not complete, and only gives you a rough idea of what colour they can be). Moreover, many organic compounds with extended conjugated structures (alternating single and double bonds) absorb colours. In fact, most of the dyes we used for food and for fabrics are all organic compounds. To the right is the molecular structure of Allure Red dye. Notice its conjugated structure.

Although certain compounds do not absorb coloured light, they may significantly absorb wavelengths of light corresponding to the UV region. Hence, this technique is called UV-‐Vis spectroscopy. The same principles apply to the UV region, except that no colours are visible. The solution will look colourless to us, but still absorbs at UV wavelengths. Organic molecules that absorb in this region of light are typically less conjugated molecules (smaller, with fewer alternating single and double bonds). Therefore, if we are working with molecules that do not contain coloured transition metal ions or organic molecules with conjugated structures, it is not possible to measure their concentration using UV-‐vis spectroscopy. 2) Notice that all the values we have worked with have relatively low concentrations. At high concentrations, dissolved molecules and ions start interacting with one another through intermolecular forces. The interactions interfere with how they absorb light and absorbance can significantly alter. Moreover, at high concentrations, the molecules may react with themselves (especially for chemical indicators that are pH sensitive). This results in a deviation from Beer’s Law, and the calibration curve becomes inaccurate. To the right is a theoretical graph of what could happen at high concentrations. Notice that the early parts of the graph show linearity, but at high concentrations, the graph curves and no longer shows a linear relationship (it can even start to decrease in absorbance!) Example 12: Which of the following molecules do we expect to be able to analyse with UV-‐vis spectroscopy?

a. Beta-‐carotene (found in carrots) b. Curcumin (turmeric powder)

c. Fructose

(sugar) d. Anthracene e. Salt, NaCl

f. Potassium permanganate, KMnO4

Transition Metal Ion

Colour

Co2+ Pink, Blue Cu2+ Blue-‐green Fe2+ Olive-‐green Ni2+ Bright green Fe3+ Brown to yellow CrO42-‐ Orange Cr2O72-‐ Yellow Ti3+ Purple Cr3+ Violet Mn2+ Pale pink MnO4-‐ Purple Zn2+/Ag+ Colourless

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Answers: a. Yes, extended conjugated structure. b. Yes, extended conjugated structure. c. No, only single bonds (the wedges and dashes are irrelevant here) d. Yes, but most likely only in the UV region since it is not conjugated extensively. e. No, Na+ is not a transition metal f. Yes, MnO4-‐ contains a transition metal

IR Spectroscopy/Vibrational Spectroscopy

Theory: Molecules contains bonds, and these bonds vibrate at a specific frequency. Depending on the structure of the molecule, and thus its bonds, they may absorb IR radiation. In general, molecules with vibrations that cause a change in dipole moment will be IR active (meaning it absorbs IR radiation and can be analysed with IR Spectroscopy). All polar molecules are IR active. For example, let’s consider a few molecules: O2, HCl, and CO2. O2: This molecule is symmetrical and is non-‐polar. No matter how you stretch or bend the bonds, it will still remain non-‐polar (no change in dipole moment). Thus, O2 is IR inactive. HCl: This molecule is polar and stretching the bond will change its polarity. Thus, HCl is IR active. CO2: This molecule is non-‐polar, however, bending due to vibrations of bonds can cause it to become polar. Consider the diagram to the right. Since CO2 can become polar, CO2 is IR active. Important: Since IR spectroscopy is generally used for organic compounds, and they are almost always complex molecules, we can assume that all of them are IR active. How is IR spectroscopy used? The set-‐up of IR spectroscopy is similar to that of UV-‐vis spectroscopy. IR radiation (of varying wavelengths) is given off by a lamp source and directed onto the sample to be analysed. The sample will absorb various wavelengths of IR and the rest is passed onto a detector. The computer processes the information and prints out a spectrum. IR spectroscopy is generally used as an qualitative analytical technique to identify functional groups of organic molecules. Quantitatively, it is less useful (for example, it cannot help determine concentration, like in UV-‐vis spectroscopy).

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Let’s see an example of an IR spectrum: This is the IR spectrum of vanillin. Notice a few things:

1) Instead of absorbance, transmittance is used instead (it just means the opposite, the lower the transmittance, the higher t he absorbance).

2) Frequency is not in the traditional units of s-‐1/Hz, but cm-‐1. This is just a different unit (wavenumber).

If we look at vanillin’s spectrum, we see that there are many ‘peaks’. Peaks in IR spectrums are upside down. For example, we would say there is a peak at 3500cm-‐1. There are also many peaks from 500-‐1700cm-‐1. All vanillin samples will have the same IR spectrum, and thus, if we had an unknown sample that gave us this IR spectrum, we could identify it as vanillin. As mentioned before, specific functional groups absorb at specific wavenumbers. Here is a diagram: This diagram tells us that a C=O bond absorbs strongly between 1500cm-‐1 and 1800cm-‐1. If we look back at our vanillin example, we indeed did have a C=O bond (the aldehyde at the top of the molecule). If we look at the spectrum, we do see a strong peak at 1700cm-‐1. This diagram also tells us that O-‐H bond absorbs strongly at 3500 cm-‐1. If we look back at our vanillin example, indeed, we do see a 3500 cm-‐1 peak and we also have an O-‐H in our molecule! IR spectroscopy allows us to identify functional groups in our molecules if we see specific peaks. However, there is generally a lot of extra information too. Look at all the peaks at ~3000 cm-‐1 in vanillin, and peaks from <1500 cm-‐1. All those peaks either correspond to C-‐H (which all organic molecules have, so it is not useful) or they overlap so much that we cannot tell what is what. Organic chemists generally use IR spectroscopy to identify O-‐H, C=O, N-‐H, and C≡C bonds in molecules. Example 13: Analyse the IR spectrum for hexanoic acid. Try to find the peaks corresponding to the C=O and O-‐H in hexanoic acid. You may need to refer to the diagram above that lists the IR absorption wavenumbers for functional groups.

These are considered ‘peaks’ although they are upside down.

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Answer: We are only interested in the O-‐H stretch and C=O stretch at ~3000 cm-‐1 and ~1700 cm-‐1 respectively.

Notice that earlier, the O-‐H stretch was at ~3500cm-‐1, but when bonded as COOH, the numbers can change slightly. It is not expected that you know this, but experienced chemists will use this information to tell apart a molecule that has C=O and O-‐H as a ketone + alcohol versus COOH as carboxylic acid.

Example 14: You produced an organic molecule with an oxygen group attached, and you know the empirical formula is C5H10O. Referring to the diagram on the left: Which of the molecules best fit the IR spectrum given?

Answer: A peak at ~3500cm-‐1 shows that there is an O-‐H group in the molecule. The only one that matches is d). Moreover, if the molecule was either a) or c), we would expect a large peak at ~1700 cm-‐1. But we do not see it in the spectrum, so we can rule out a) and c).

Example 15: An organic compound with the formula C6H11N has an IR spectrum peak at 2169 cm-‐1, no peak near 1600cm-‐1, and no peak from 3200cm-‐1 to 4000cm-‐1. Which of the following functional groups could be present in this molecule?

a. Alkene (C=C) b. Imine (C=N) c. Amine (C-‐N)

d. Alkyne (C≡C) e. Nitrile (C≡N) f. Ketone (C=O)

Answer: Ruling out wrong answers: A peak near 2169cm-‐1 means it could be a variety of things based on the diagram on the previous page, but since the molecule only has one N (no S, O), it could either be an alkyne (C≡C) or nitrile (C≡N), and we can immediately rule out f). No peak from 3200-‐4000cm-‐1 lets us rule out any amides, alcohols, or amine, so c) cannot be an answer. No peak at 1600cm-‐1 means it cannot be a) or b). Thus, the only answers can be d) or e).

Based on the formula of the molecule, C6H11N, it can only have one triple bond, and not any more double or triple bonds (try drawing out molecules with more than a triple bond: You would have left over hydrogens that can’t go anywhere). If the molecule had an alkyne, d), then that means the nitrogen would have to be singly bonded. As we know, the amine (C-‐N) option has been ruled out from before, so it cannot have an alkyne either.

Thus, the answer is only e). Note: The last part about comparing e) and d) is sophisticated. In your answer, if you circle both d) and e), I would accept that answer as well.

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Summary of Spectroscopy

• Electrons can transition from one energy level to another by absorbing electromagnetic radiation. The type of EM radiation they absorb depend on which energy level they jump from and jump to. These transitions can happen in atoms (atomic orbitals) or in molecules (electrons in bonds)

• Photoelectron spectroscopy involves electrons ejected from their atomic orbitals using extremely high energy EM radiation. The ejected electrons have their kinetic energy measured and that indirectly measures the binding energy of the electrons. We can match binding energy to specific orbitals from where the electrons come from

• UV-‐vis spectroscopy involves compounds that absorb strongly in the UV light region or visible light region. Substances that absorb these wavelengths can be analysed for their concentration since their absorbance is related to concentration according to Beer’s Law, A=ɛbc.

• IR spectroscopy involves the vibrations of bonds in molecules. Since specific bonds absorb at different frequencies of IR radiation, we can analyse a spectrum and determine what bonds are present. This is used extensively in organic chemistry to identify functional groups that are present.

• There are many other types of spectroscopic techniques used in chemistry and other fields of science. Each of them take advantage of a different aspect (rotation of molecules, magnetic behavior of nucleus, vibration of bonds, electron transitions, etc)

Example 16 (from AP exam): The following spectrum is taken for a pure substance. Out of the following type of transitions (molecular rotations, molecular vibrations, nuclei spin transition, electronic transition), which type of transitions is observed for each region in the absorption spectrum?

Answer: Region X corresponds to electronic transitions in bonds; region Y corresponds to molecular vibrations; and region Z corresponds to molecular rotations.

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