Chemical bonding lecture 3

Question 1

what is a wave function? What is a node?

Answer 1

it is a function that maps out what a wave looks like in three dimensions, it may also be a function of time.

a node is the point described at the zero point of the wave function

Question 2

What is the derivation of the wave function and how does it give us The Schrödinger Equation in one dimension?

Answer 2

notes

Question 3

what is the stationary state?

Answer 3

the stationary state is when a wave function is independent of the time, which occurs when a system is confined in a space by potential V(x) that does not vary with time and the solutions to Schrodinger’s equation can be found only for discrete energy values

Question 4

how do we measure the value of the wave function?

Answer 4

Measuring the wave function directly is impossible, as seen similarly with wave-like concepts such as electromagnetism.

To solve this issue, it was suggested that the square of the modulus of the wave function for a particle is equal to the probability density of the particle, giving the following equation:

|𝚿^2(x, y, z)|dV - (the probability of a particle being found in a small volume).

Question 5

What is the probability density function? What are its conditions?

Answer 5

the probability density function is when the wave function probability ( |𝚿^2(x, y, z)|dV) is continuous across all values giving as a probability function at which its integral tells us the probability that a particle is found in a interval dx.

The conditions that are required for the probability density to take place is:

1. the probability density must be normalized, meaning that all the possible outcomes are included giving a value of 1 for the area under the graph
2. P(x) must be continuous at every point of x, with both 𝚿(x) and 𝚿’(x) are continuous
3. if a particle is at a bound state (𝚿(x) =0) the corresponding x value must be very large (approaching infinity)

note: The requirement that 𝚿(x) be continuous results in boundary conditions.

Question 6

How do you solve Schrodinger’s equation using the one-dimensional particle in a box model?

Answer 6

notesss

Question 7

what is the zero-point energy principle?

Answer 7

it is a principle that states that a particle in the box cannot have zero energy and the probability of the particle at the nodes isn’t zero as there would then be a contradiction that there isn’t a particle in the box to begin with

Question 8

What is the Correspondence principle?

Answer 8

it is a principle that states the results of quantum mechanics reduce to those of classical mechanics for large values of quantum number

Question 9

What is the energy equation for a particle in a three-dimensional box? (also explain how we got there)

Answer 9

when working with three dimensions there are 3 directions of motion to consider the x, y, and z.

Same with two dimensional models the potential energy is zero at the interior or the cube and the potential energy is infinite within the box which then allows for each direction of motion of the particle to be independent, the wave function is the result of the product of the wave functions of each independent direction and the energy is the sum of total energies of motion of all the particles giving us an overall equation of:

E = ((h^2)/8mL^2) * (nx^2 + ny^2 + nz^2)
where nz, ny, and nx are independent quantum numbers.

we can obtain the two-dimensional model by setting nz to zero

Question 10

what is a degenerate

Answer 10

an energy level diagram where energy levels correspond to more than one quantum state.

Question 11

what is the wave function for three and two-dimensional boxes?

Answer 11

page 215 and 218 on the pdf

Question 12

what are the spherical coordinates? Why do we use them?

Answer 12

the spherical coordinates are (r, θ, Φ)

r- is the distance between the nucleus and the electron.

θ - the angle related to latitude between z and r.

Φ- the angle related to the longitude (it’s between the x-axis and the projection onto the x-y plane of the arrow from the origin to P).

The reason we use them is because the solution to the Schrodinger equation is easiest when the coordinates represent the natural symmetry of a potential energy function whereas the spherical coordinates tend to do that better.

Question 13

what are the quantum numbers introduced to help us solve the Schrodinger equation?

Answer 13

1. The angler momentum quantum number (l)- which is due to the quantization of the square of the angular momentum
2. magnetic quantum number (m)- which is due to teh quantization of the projection along the z-axis and it determines the energy shift of an electron when it is placed in an external magnetic field.

Both exist due to Schrodinger equation and both are needed solve the Schorfinger’s equation

Question 14

What are the allowed values for L^2 and Lz?

Answer 14

page 233

Question 15

What is a quantum state? What is a degenerate in this context? How do we label such degenerates and what are the allowed values?

Answer 15

notesssss

Question 16

for each quantum number(n,l,m), the solution of the Schrodinger equation provides the wave function in what form? Explain what the equation means.

Answer 16

𝚿(n,l,m)(r, θ, Φ) = Rnl(r) Yml(θ, Φ)

where Rnl(r) describes the radial part and Yml(θ, Φ) describes the angular part, and this is a result of the symmetric potential energy function that then allows us to examine each part separately.

Yml(θ, Φ) is called spherical harmonics.

Question 17

How do we measure the wave function in 3 dimensions( in the spherical and cartesian equation)

Answer 17

the wavefunction itself is not measured similarly to one diminution we find the possibility of finding the particle in said coordinates

the equations of both the cartesian and spherical are found in page 234

Question 18

what are the different angular and radial parts of a wave function for one-electron atoms?

Answer 18

table 5.2 in page 235 which you don’t need to know the equations just understand what it means

Question 19

How can you calculate where the electron density is greatest in a molecule

Answer 19

first, solve the integral for the probability wave distribution which since we are solving for a very small value is then reduced to the following equation:

(𝚿(xo,yo,zo))^2 x volume of the place.

after you find teh probability you can then multiply by the charge of e.

Question 20

What are the 3 features of orbital shapes and sizes

Answer 20

1. for a given value of l, as n increases the distance of the electron from the nucleus increases, and the orbital size increases.
2. for an orbital with quantum numbers n,l, it has l angular node and n - l - 1 radial nodes with overall n-1 nodes. The angular nodes appear at the angular plane in the plot of p orbitals. The wave function values change signs as you cross the nodes in the angular or radial part. The energy for a one-electron atom only depends on the nodes that are on n and energy increases as teh nodes increase.
3. when r approach zero, the wave function disappears for all orbitals except s, thus only the s orbit can penetrate the nucleus

Question 21

what is the probability distribution function of the radial part at different orbits?

Answer 21

in slides 108-111

Question 22

why are new methods being used instead of the Schrodinger equation?

Answer 22

because when we start working with multiple electrons in an orbit the solution to the Schrodinger equation becomes much more complicated, to the point where an event using a numerical method won’t give a good approximation to the energy of an orbital

Question 23

What are the three simplified assumptions made by the SCF orbital approximation method?

Answer 23

For any atom, Hartree’s method begins with the exact Schrödinger equation in which each electron is attracted to the nucleus and repelled by all the other electrons, simplified by these 3 points:

1. an electron moves in an effective field created by the nucleus and all other electrons, and this effective field for an electron i depends on its position ri
2. the effective field of an electron i is obtained by averaging the coulomb potential interactions with all the other electrons over their position so the ri is the only coordinate in the description
3. the effective field is spherically symmetric, therefore no angular dependence

Question 24

what is the orbital approximation of atoms

Answer 24

it is a wave function given by the product of these one-electron orbitals in an atom defined as Φari = ( Φbr1)(Φbr2)…..

Question 25

why is the probability density of the same orbital that Hydrogen has is more than other elements with the same orbital?

Answer 25

because other elements have a large nuclear charge so the pull in electrons is great so the probability of finding it at the exact orbital is less

Question 26

what is proportional to the radial charge density distribution function?

Answer 26

The radial density function upon integration gives us the probability of finding an electron near the distance r regardless of its orbital, then the function that is made from the sum of all the radial probability density of the occupied orbitals is then proportional to radial charge density distribution.

Question 27

why is the radial charge density distribution function important?

Answer 27

because the number of peaks represents the number of occupied shells an element/atom contains.

Question 28

what is the difference between the energy level obtained by the Hartree method for many electrons that resemble the Hydrogen atom and the energy level of the hydrogen atom?

Answer 28

1. orbitals with different n and l no longer have the sam energy
2. energies of the orbital have shifted due to a stronger nuclear change

Question 29

what is the approximation of Coulomb’s potential for an electron moving in shell n

Answer 29

Vn = - Zeff (n) *e^2/r

Where Zeff(b) is the effective nuclear charge

Question 30

What is the effective nuclear charge? How do we calculate the orbital energy from it/

Answer 30

it is the net reduced nuclear charge experienced by an electron due to the screening from other electrons

where Zeff(n) = Z - S
S=screaning constant

e(n) = (Zeff(n))^2/n^2