Spectroscopy notes for uni: rotational/vibrational

I am only publishing these as convenient access points for me. 

Rotational spectroscopy

Source: storage.googleapis.com/difyxmkmjwyase/rotational-spectroscopy-molecules-examples.html

Note: wavenumbers are a unit of energy. As E = hc/λ, the wavenumber ū = 1/λ = E/hc

• Focuses on rotational molecular energies, which correspond to the microwave region of the EM spectrum.
• Requires the molecule to have a permanent dipole moment.
• This is because it means the molecule is microwave active.
• The simplest model used in rotational spectroscopy is the rigid rotor.
• Consists of two masses a fixed distance apart from each other
• Moments of inertia
• Inertia is a property of objects in motion to stay in motion.
• A molecule can have up to three moments of inertia passing through the centre of mass, along three different axes (x, y, z).
• Linear and spherical rotors only have one moment of inertia
• In linear rotors, I₁=I₂, and I₃=0. Examples include linear molecules like HCl or CO₂.
• In spherical rotors, I₁=I₂=I₃. Examples include tetrahedral molecules with equal dipole moments, like CH₄.
• Symmetric rotors have two moments of inertia
• I₁=I₂≠I₃, and I₃≠0.
• Examples include tetrahedral molecules with different dipole moments, like CHCl₃ (the C-Cl dipole is a different magnitude to the C-H dipole, due to the effects of electronegativity).
• You can have oblate symmetric rotors, where I₁=I₂>I₃, and prolate symmetric rotors, where I₁=I₂<I₃. Benzene is an example of the former, and allenes are an example of the latter.
• Assymetric rotors have three different moments of inertia
• I₁≠I₂≠I₃, where none of them equal zero.
• An example is H₂O.
• To calculate the moment of inertia (I), you take the product of the mass and the interatomic distance.
• But this requires you to know where the centre of mass is.
• For a rigid diatomic - a rigid rotor with two atoms - you can use the reduced mass principle to use as the mass in the moment of inertia.
• Reduced mass = product of masses/sum of masses.
• I = μr₀², where μ is the reduced mass, and r₀ is the interatomic distance. 
• Rotational energies (in the rigid rotor)
• The energy for a given energy level defined by the quantum number J corresponds to:
• E_J = BJ(J+1), where B is the rotational constant, equal to h/8π²Ic. 
• Rotational energy transitions can only occur between directly successive energy levels, when ∆J = +1 (like in an excitation) or -1 (as in a relaxation).
• The energy gap between successive energy levels, ∆E:
• B(J+1)(J+2) - BJ(J+1) = B(J+1)(J+2-J) = 2B(J+1)
• But the gap between energy gaps, or ∆∆E (yes, really):
• 2B(J+1) - 2B(J) = 2B(J+1-J) = 2B
• In an absorption spectrum (which you'd obtain after doing rotational spectroscopy), you'd therefore get equally spaced lines.
• The distance between them would equal 2B. 
• For a given J, the number of degenerate energy levels equals 2J+1.
• Rotational energies (in the non-rigid rotor)
• To this point, we've assumed the interatomic distance is constant. In reality, this distance can change since bonds are able to distort. This is due to the centrifugal force acting on the atoms as they rotate.
• To account for this, we add a correction constant:
• E_J = BJ(J+1) - DJ²(J+1)², where D is the centrifugal distortion constant.
• D = h³/32π⁴I²r²kc, where k is the force constant. The reason why k is involved is due to Hooke's law coming into play, as we assume the bond is elastic. But I don't think I need to know more than this for the exam.
• With this correction constant, the energy gap will thus shrink for greater J values, since the DJ²(J+1)² term increases faster than the BJ(J+1) term.
• Therefore, in an absorption spectrum for a non-rigid rotor, we'd expect the spectrum lines to converge.

Vibrational spectroscopy

IR spectrum. Note: this doesn't look like the spectra for the an/harmonic oscillators.

• Focuses on vibrational molecular energies, which occur in the infrared region of the EM spectrum
• Vibrational transitions focus on bonds bending or stretching.
• For vibrational transitions to occur, there must be an induced change in dipole moment.
• Harmonic oscillator
• A simple model is to assume the bond will vibrate under simple harmonic motion, and that the bond is elastic.
• Using Hooke's law, we can derive the vibrational energy as equal to:
• E = ks²/2, where k is the force constant and s is the change in distance between the atoms during a vibration.
• We therefore get a parabola!
• The energy levels in vibrational spectroscopy depend now on the quantum number n. n can equal 0 or any greater positive integer.
• For a given energy level, E_n:
• E_n = (n+1/2)hv, where v is the vibrational frequency.
• Note that this means the minimum energy of a rigid rotor isn't equal to 0. This energy, for E₀, is known as the zero point energy.
• v is further equal to (1/2π)√(k/m_eff). Worth noting that for diatomics, the effective mass, or m_eff , = the reduced mass.
• This form is related to the angular velocity ω = 2πv. I could get into centrifugal forces and angular velocities, but that's more physics than chemistry, and I don't think I need to know more beyond this.
• Vibrational transitions only occur between directly successive energy levels. 
• Therefore, the energy gap ΔE:
• ((n+1)+1/2)hv - (n+1/2)hv = (n+3/2-n-1/2)hv = hv
• So ΔE is a constant value.
• This also means there will be only one line in the subsequent absorption spectrum.
•  Anharmonic oscillator
• For small changes in distance, the system will generally be harmonic. Yet for larger changes in distance, due to the effects of nuclear repulsion, we'd end up with the bond breaking.
• To manage this, we introduce a correction constant that explains the effects of anharmonicity:
• E_n = (n+1/2)hv - (n+1/2)²hv.x_a, where x_a is the anharmonicity constant.
• As with the non-rigid rotor, as n increases, (n+1/2)²hv.x_a increases faster than (n+1/2)hv, so we'd expect the energy gap to no longer be constant, and to instead converge over time.
• This means the bond can dissociate; we can use the values obtained from the spectral lines to graph out a Birge-Spooner plot, which enables us to calculate the dissociation energy when this would happen. 
• The dissociation also means we can begin to have transitions over more than one quantum number - these are known as overtones.  
• The transition from n=0 to n=1 is known as the fundamental
• the transition from n=0 to n=2 is the first overtone of the fundamental
• the transition from n=0 to n=3 is the second overtone of the fundamental
• and so on
• We can also have transitions from n>0, which can be seen at higher temperatures, where more electrons will be present in higher energy levels.
• The transition from n=1 to n=2 is known as the first hot band
• The transition from n=2 to n=3 is known as the second hot band
• We can also have overtones here, using the same naming conventions as before
• Identifying vibrations
• For simplicity's sake, we have linear and non-linear molecules. Linear molecules have bond angles of 180°, non-linear molecules don't.
• An atom can translate across the x, y and z axes, so has three degrees of freedom.
• In a molecule, this translates to there being 3N degrees of freedom overall, since there are N atoms and each atom can perform three translations.
• Some of these degrees of freedom will be rotations and vibrations, such as bends and stretches.
• A linear molecule will have 3N-5 vibrations
• A non-linear molecule will have 3N-6 vibrations 
• Not all of these vibrations will result in transitions, though, since the vibrations must result in a change in magnetic dipole.

All of this results in infrared spectroscopy - IR - which is used to analyse key functional groups present in a molecule. This is one of the most fundamental spectroscopic methods used, indeed it's the first one I learnt at A Level, and compared to simple rotational or vibrational methods, it's extremely useful alongside NMR in identifying what a molecule actually is.

Rotational-vibrational

Source: https://dpotoyan.github.io/Chem324/ch04/note04B.html

• Obviously, we can have both of these types of transitions occurring at the same time
• The Born-Oppenheimer approximation states that the total energy of a molecule is equivalent to the energy of various transitions, including rotational, vibrational, and electronic.
• After all, these transitions all exist somewhere along the EM spectrum, with rotational and vibrational being the lowest energy and electronic being the highest. 
• In theory, a photon could bring about all transitions so long as all the rules from earlier apply, and we also have coupling between molecular motions.
• Coupling essentially means "one motion will cause another motion to occur", so for example a rotation might bring about a vibration.
• And coupling can occur between rotations and vibrations since they both depend on the elasticity and length of the bond.
• We can thus combine both rotations and vibrations into one spectrum (electronic transitions are much higher energy). We can assume the rigid rotor and harmonic oscillators apply.
• The spectrum consists of a band origin in the middle (this is the Q branch), corresponding to a fundamental frequency, along with two branches:
• For relaxations, where ∆J = -1, we have the P branch on the left. A transition denoted as, say, P₃, corresponds to a relaxation from J'' = 3 to J' = 2. 
• For excitations, where ∆J = +1, we have the R branch on the right. A transition denoted as, say, R₂, corresponds to an excitation from J'' = 2 to J' = 3.
• J'' is the initial rotational level, J' is the final rotational level.
• Under the rigid rotor assumption, the energy spacing is equal to 2B. 
• We can also use the band origin and its fundamental frequency to work out k or the effective mass.
•  The shape of this spectrum is also determined by factors such as the temperature, which will affect the population of certain energy levels, and thus affect how the spectrum is shifted. 

Now I've completed all my notes, it's time to actually do some questions...