Physics/Chemistry concepts

Question 1

How does stability and reactivity correlate in a chemistry context?

Answer 1

• A system is considered thermodynamically stable when it has reached a state of minimum total energy, this means that any small perturbation will not lead to a significant change in the system’s state. Reactivity indicates how readily a substance will undergo change
• Generally, less stable reactants are often more reactive because they seek to achieve stability through reaction
• When it comes to transitional states, their stability also influences reactivity. A more stable transition state typically leads to faster reactions since it requires less energy to proceed through that state

Kinetic Stability: This refers to how quickly a system can react or change state. A kinetically stable system may not be at its lowest energy state but is resistant to change due to high activation energy barriers. For example, diamond is thermodynamically less stable than graphite but remains kinetically stable because the conversion from diamond to graphite occurs very slowly under normal conditions.

Question 2

Give a summary of possible molecular orbital symmetries in polyatomic molecules.

Answer 2

Symmetry labels
A: Symmetric with respect to the principal axis.
B: Antisymmetric with respect to the principal axis.
E: Doubly degenerate orbitals, meaning there are two orbitals with the same energy level but different orientations.
T: Triply degenerate orbitals, indicating three orbitals of the same energy.

Prime or double prime notation
‘: Symmetric under reflection in a horizontal plane (σh).
‘’: Antisymmetric under reflection in a horizontal plane (σh).

Bonding characteristics
Bonding Orbitals: Typically labeled with lower energy symmetry types (like a₁ or e).
Antibonding Orbitals: Often indicated by an asterisk (*) and higher energy states.
Non-bonding Orbitals: May have symmetry types that do not favor bonding interactions.
a orbitals typically favors bonding through constructive overlap and symmetry while b orbitals may exhibit antisymmetry or non-bonding characteristics.

Symmetry with respect to inversion center
g (gerade) and u (ungerade) indicate whether an orbital is symmetric or antisymmetric with respect to inversion through the center of symmetry of the molecule.

Numerical subscript might indicate multiple orbitals with same symmetry, e.g., a₁: symmetric with respect to the principal axis and a₂: another symmetric orbital.

Question 3

Define the Pauli repulsion

Answer 3

• Pauli repulsion is a consequence of the exclusion principle. How can the exclusion principle be violated? It can’t but it can be masked to some degree using the principle of uncertainty. You can place two identical fermions at the same place if you can’t be certain exactly where they are. Putting them closer means having lower uncertainty which requires higher energy. Since that energy increases with decreasing distance that gradient is a repulsive force.
• While there isn’t a single, explicit equation for Pauli repulsion, in a simplified model, Pauli repulsion can be related to the overlap of electron densities:
  E_Pauli ∝ ∫ ρ_A(r) ρ_B(r) dr
  where ρ_A(r) and ρ_B(r) are the electron densities of atoms A and B, respectively.
  In some computational models, Pauli repulsion is approximated using an effective potential:
  V_Pauli(r) = K r⁻ⁿ
  where K is a constant, r is the interatomic distance, and n is typically a large even number (e.g., 12).
• Pauli repulsion contributes to the Exchange energy

Question 4

How does Pauli repulsion influence bond formation in late transition metals?

Answer 4

• As the d-orbitals of late transition metals are nearly filled, there are fewer empty orbitals available for bonding
• The high electron density in the valence shell leads to stronger Pauli repulsion between the metal’s electrons and those of potential ligands
• The strong Pauli repulsion often results in weaker metal-ligand bonds compared to earlier transition metals
• Late transition metals tend to form stronger bonds with “soft” ligands that can better accommodate the Pauli repulsion
• The increased Pauli repulsion often leads to lower coordination numbers in late transition metal complexes

Question 5

why spin-orbit coupling is stronger in heavier elements?

Answer 5

• Increased Nuclear Charge: Heavier atoms have a larger nuclear charge, which creates a stronger electric field experienced by the electrons.
• Relativistic Effects: As atomic number increases, the speed of inner electrons approaches a significant fraction of the speed of light. This relativistic effect enhances the magnetic field generated by the electron’s orbital motion, amplifying the spin-orbit interaction
• Larger Orbital Radii: Heavier atoms have larger atomic radii, which results in electrons having larger orbital paths. This increased path length contributes to a stronger magnetic field generated by the electron’s motion, enhancing the SOC effect
• Reduced Shielding for Valence Electrons: Despite increased electron shielding in heavier atoms, the valence electrons still experience a stronger effective nuclear charge. This leads to a more pronounced SOC effect for the outer electrons, which are often most relevant for chemical and physical properties
• Quantum Mechanical Considerations: The spin-orbit coupling strength (ξ) is proportional to Z^4, where Z is the atomic number. This strong dependence on atomic number explains why SOC increases dramatically for heavier elements

Question 6

What is the conversion factor of the energy units kcal/mol, kJ/mol and eV?

Answer 6

1 eV = 96.5 kJ/mol
1 kcal/mol = 4.18 kJ/mol

Question 7

Define electronegativity.

Answer 7

• Electronegativity, symbolized as χ, is the tendency for an atom of a given chemical element to attract shared electrons (or electron density) when forming a chemical bond
• Pauling first proposed the concept of electronegativity in 1932 to explain why the covalent bond between two different atoms (A–B) is stronger than the average of the A–A and the B–B bonds. To calculate Pauling electronegativity for an element, it is necessary to have data on the dissociation energies, thus it is a semi-empirical method.
  |χ_A - χ_B| = 0.208 √∆
  ∆ = average bond dissociation energy
  the result is domensionless
• As only differences in electronegativity are defined, it is necessary to choose an arbitrary reference point in order to construct a scale. Hydrogen was chosen as the reference, as it forms covalent bonds with a large variety of elements: its electronegativity was fixed at 2.20. It is also necessary to decide which of the two elements is the more electronegative. This is usually done using “chemical intuition”: e.g., hydrogen bromide dissolves in water to form H+ and Br− ions, so it may be assumed that bromine is more electronegative than hydrogen.
• General rule: → Atomic radius decreases → Ionization energy increases → Electronegativity increases
• Less popular is the Mulliken electronegativity. Robert S. Mulliken proposed that the arithmetic mean of the first ionization energy (E_i) and the electron affinity (E_ea) should be a measure of the tendency of an atom to attract electrons:
  χ = (E_i-E_ea)/2
  As this definition is not dependent on an arbitrary relative scale, it has also been termed absolute electronegativity, with the units of kilojoules per mole or electronvolts. The values are similar to the Pauling scale.

The electron affinity (Eea) of an atom or molecule is defined as the amount of energy released when an electron attaches to a neutral atom or molecule in the gaseous state to form an anion.

X(g) + e− → X−(g) + energy

Question 8

What is the dissociation energy of a C-H bond in methane? How does it compare to C-H bond dissociation in other alkanes? What about an O-H bon in water?

Answer 8

• The energy of a C-H bond in methane is approximately 412-416 kJ/mol
• The C-H bond in methane is slightly stronger than those in larger alkanes due to the absence of other carbon atoms, which can cause slight electron density shifts in larger molecules
  Ethane (C2H6): 410 kJ/mol
  Propane (C3H8): ~400 kJ/mol for secondary C-H bonds
  Butane (C4H10): ~390 kJ/mol for tertiary C-H bonds
• The energy of an O-H bond in water molecules is approximately 464 kJ/mol

Question 9

Describe the types of production of ions

Answer 9

Question 10

Explain how the cooling of the ions in the trap occurs

Answer 10

• For this, we use helium as a buffer gas, which thermalizes the ions with the trap walls, which is cooled with liquid helium
• Mechanism: the cooling process occurs when molecules collide with the colder particles. During these collisions, energy is transferred from the hotter molecules to the colder collision partners. This energy transfer can involve various degrees of freedom:
  Translational cooling: Reduction in the overall kinetic energy of the molecules.
  Rotational cooling: Decrease in the rotational energy states of the molecules.
  Vibrational cooling: Lowering of the vibrational energy levels of the molecules.

Question 11

What is a non-innocent ligand?

Answer 11

• In chemistry, a (redox) non-innocent ligand is a ligand in a metal complex where the oxidation state is not clear. Typically, complexes containing non-innocent ligands are redox active at mild potentials.
• Dioxygen can be non-innocent, since it exists in two oxidation states, superoxide (O₂⁻) and peroxide (O₂²⁻).

Question 12

Give a summary of atomic quantum numbers.

Answer 12

Question 13

Briefly describe the aufbau principle.

Answer 13

1. Electrons are placed in orbitals to give the lowest total electronic energy to the atom. This means that the lowest values of n and l are filled first. Because the orbitals within each subshell (p, d, etc.) have the same energy, the orders for values of ml and ms are indeterminate.
2. The Pauli exclusion principle requires that each electron in an atom have a unique set of quantum numbers. At least one quantum number must be different from those of every other electron. This principle does not come from the Schrödinger equation, but from experimental determination of electronic structures.
3. Hund’s rule of maximum multiplicity requires that electrons be placed in orbitals to give the maximum total spin possible (the maximum number of parallel spins). Two electrons in the same orbital have a higher energy than two electrons in different orbitals because of electrostatic repulsion (see below); electrons in the same orbital repel each other more than electrons in separate orbitals. Therefore, this rule is a consequence of the lowest possible energy rule (Rule 1).

Further Hund’s rules:
- For a given multiplicity, the term with the largest value of the total orbital angular momentum quantum number L has the lowest energy.
- For a given term, in an atom with outermost subshell half-filled or less, the level with the lowest value of the total angular momentum quantum number J (J = L + S) lies lowest in energy. If the outermost shell is more than half-filled, the level with the highest value of J is lowest in energy.

Question 14

Why is it easier to observe high oxidation states in 4d and 5d transition metals, compared to the 3d row?

Answer 14

• 3d orbitals show small radial extension, in comparison to the 4d and 5d orbitals. E.g., atomic radii of the outer d orbitals: Fe ⁓ 0.36 Å, Ru ⁓ 0.62 Å, Os ⁓ 0.70 Å.
• The orbitals become more diffuse as we go down in the group and increase the number of radial nodes and the screening that the outer orbitals experience due to the core electrons.
• Such small extent of the 3d orbitals, results in increased Pauli repulsion
• If the ligands are particularly large, reaching higher coordination in order to promote higher oxidation states of the metal center is challenging
• On the other hand, ligands such as oxides and fluorides are of small size, but their 2p orbitals are also of small extent, and overlap with the 3d shell is poor, weakening the metal–ligand bonds
• Those effects are more critical for the 3d metals, while for the 4d metals, the presence of a node in the 4d orbitals diminishes this issue, and for the 5d metals, the relativistic expansion of the 5d orbitals improve the overlap between d and ligand orbitals.

a₀ ∝ 1/m