1: Below you can see a computer generated image of a 2s orbital from Rutgers University website.

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1 Quantum Mechanics of an H atom: 1: Below you can see a computer generated image of a 2s orbital from Rutgers University website. 2: Since the rest of the problems include alteration to the plot, it is attached at the end of this section. 3: We remember that the expectation value of any function f is given by the following equation. Note that the integral is taken over the entire domain of the function in question. From this formula we can easily calculate the expectation value of r from the probability function P. 4: In addition, we can also calculate the most probable location easily enough by taking the derivative of the probability function. This equation has four roots shown below. ( ) ( ) These roots will correspond to the maxima of the function; therefore, one of them will correspond to the function s maxima. Examination gives the probability value of the electron at each of the 4 points.

2 (( ) ) ( ) ( ( )) (( ) ) ( ) ( ( )) Therefore, the most probable location of the electron occurs at the point. ( ) 5: We know that at the limit of the classical region in the 2s (or any other) orbital the potential energy of the electron s location is equal to the total energy of the orbital. We remember that the energy of an orbital level N in the hydrogen atom is given by the following equation. We compare this value to the potential energy term in the Hamiltonian operator. Lastly, we set them equal to each other and solve for r. This will give us the distance from the nucleus at which the potential energy is equal to the total energy of the orbital. We substitute in the bohr radius below, which yields the radial solution. Therefore, the forbidden region occurs for. 6: Since the electron can exist at any point from r=0 to, and the probability is normalized over this region, we can determine the probability that the electron is in the forbidden region by integrating over it (r > 8*a 0 ). 7: For an electron to transition between orbitals there must be a change of ±1 in l. For any s orbital, l=0. Additionally, in a neutrally charged hydrogen atom, all orbitals are degenerate with the s orbital of equal level. This means that effectively only the s orbital at each level can be reached (1s, 2s, 3s, etc.) Since this will never correspond to a change in the value of l, the electron cannot transition out of the 2s orbital.

3 Probability vs. Radius in the 2s Orbital for the Hydrogen Atom Probability Most Probable r Expected r Forbidden region 0.14 Probability 1/a Distance From Nucleus r/a 0

4 Many-electron atoms: 1) The ground state electron of configuration of an oxygen atom is shown below. Orbital # electrons 1s 2 2s 2 2p 4 2) A possible configuration of quantum numbers for the oxygen atom ground state is shown below. n l m l m s core or valence /2 core /2 core /2 valence /2 valence /2 valence /2 valence /2 valence /2 valence We know that the spin of an atom is calculated by summing up the values of ms over the entire atom (since the core states will always be full, the sum of ms over them will always equal 0, so effectively we only need to sum over the valence electrons). We find that the ground state oxygen atom has a spin of 1. We can then determine the multiplicity by multiplying the spin by 2, and adding 1 to it. Therefore the ground state spin multiplicity of an oxygen atom is 3 (this will always be equal to the number of unpaired electrons +1). 3) It will require much more energy to remove the 1s electron from the sulfur atom than from an oxygen atom. This is due to the great increase in the coulomb potential by increasing the number of protons in the nucleus. Additionally, since we are explicitly considering the 1s orbital, which is not shielded, the electrons will feel the full effect of the extra charge in the nucleus, while the effect will be decreased in further out orbitals due to shielding (i.e. the difference in the energy levels of the 2p orbitals will not be as great as the difference in the 1s orbitals). 4) Sulfur has a lower first ionization energy than oxygen. Ionization is easiest from the highest energy orbitals, and the n = 3 p orbitals of sulfur are higher in energy than the n = 2 p orbitals of oxygen. 5) The first ionization energy of fluorine is greater than oxygen. Fluorine has one more proton and one more 2p electron than oxygen. Electrons in the same subshell are ineffective at shielding each other, so the 2p electrons in fluorine feel a greater effective nuclear charge than do those in oxygen. 6) Lastly, we compare the chances of a 2s vs. a 2p orbital being within a bohr radius of the nucleus.

5 Therefore, an electron in a 2s orbital is much more likely to be within 1 bohr radius of the nucleus, than an electron in a 2p orbital.