Chem 150 - Chapter 1

Aufbau Principle

Electrons must fill lower energy orbitals before filling higher levels.

Degenerate

Exactly energetically equivalent.

Penetration

How well and efficiently an electron in an a given orbital can approach the nucleus.

Polarizability

Ability of the electron cloud to be distorted by a point charge. If the cloud is more distorted, then it is more polarizable.

Explain why atomic radius decreases as we move from left to right across a row in the periodic table.

As Z-eff increases, the orbitals contract due to the higher effective nuclear charge, which results in increased coulombic attraction between the nucleus and the electrons.Thus the electrons are drawn closer to the nucleus and a decrease in atomic radius.

The electronic configuration of an element and its ions has a profound effect on reactivity and stability. Understanding how to write out the configuration will be very helpful in predicting reactivity and tying together the concepts above. For practice, write out the electronic configuration if the following elements and ions: As, Cr, Cr+1, Cu-1 , and S-2.

As: [Ar]4s23d104p3 Cr: [Ar]4s13d5 Cr+: [Ar]3d5 Cu-1: [Ar]4s23d10 S-2: [Ar]

Diffuse orbital

Distributed over a greater area, used when comparing two things-in this class when discussing different kinds of orbitals.

Compact Orbital

Distributed over a smaller area, used when comparing two things-in this class when discussing different kinds of orbitals.

Electron Affinity

Electron affinity is the energy change that occurs when an electron is added to an isolated atom in the gaseous state. Or as your book describes the energetic change associated with removing an electron from a negatively charged ion.

Does the electron seem to be held more or less tightly to the nucleus as Z increases?

Electron is held more tightly to the nucleus as Z increases.

Metallic Radius

Half the experimentally determined distance between the centers of nearest-neighbor atoms in the solid for a metallic element

Covalent Radius

Half the internuclear distance between neighboring atoms of the same element in a molecule

Consider ions of elements Is it easier or harder to remove an electron from an anionic element? Why?

It is easier to remove an electron from an anionic element than its neutral counterpart because the Zeff of the anion is lower (because more shielding) than the neutral element, meaning the outermost electrons of the anion are held less tightly by the nucleus (less coulombic attraction), and thus easier to remove.

Consider ions of elements Is it easier or harder to remove an electron from a cationic element? Why?

It is harder to remove an electron from a cationic element than its neutral counterpart because the cation has a higher Zeff (less shielding) than the neutral element. The outermost electrons of the cation are held more tightly by the nucleus than the outermost electrons of the neutral species (more coulombic attraction), making the outermost electrons of the cation harder to remove.

In detail, explain the concepts of penetration and shielding and relate these concepts to each of the three kinds of orbitals (s, p, d). Which penetrates the best/worst and why? Which shields the best/worst and why?

Penetration is how well electrons in particular orbitals penetrate the nucleus (interact with it). The s penetrates the best, followed by p, and then d. The reason why p does not penetrate as well because p orbitals contain one angular node, where there is zero probability of finding an electron, more disperse away from the nucleus, feels Z effective less. d orbitals are larger than s and p, they contain two nodal planes, so the electron density is more dispersed, spends more time farther from the nuclear charge. Shielding is due to electron/electron repulsions, which reduces the actual nuclear charge felt specific electrons (the effective nuclear charge). The order of the ability of the orbitals to shield is: s > p > d. s orbitals shield the best because of its shape, compact nature, and the lack of angular nodes. p and d shield worse because of the presence of angular nodes and larger size, so there electron density is more disperse (d orbital being the most disperse), resulting in shielding poorer than s.

Electronegativity

Power of an atom of the element to attract electrons when it is part of a compound

Consider 3d, 4d, and 5d elements. As we move down a given column in the d-block the atomic radius increases significantly on going from 3d to 4d elements. However, going from 4d to 5d the increase in atomic radius is negligible. Clearly explain why this occurs.

Recall that Zeff = Z - shielding .This occurs because the inclusion of the protons and electrons of the lanthanide series. These contribute electrons from the f orbitals, which are the least effective at shielding when compared to s, p, and d. They essentially don't shield anything, so Z increases very rapidly. 14 protons are gained in the nucleus through the lanthanide series, but the f electrons don't shield. Thus Z-eff dramatically increases for all electrons, and the nucleus pulls electrons closer, resulting in reduced radius. This is called the lanthanide contraction and is responsible for the negligible increase seen in atomic radius on going from 4d to 5d elements.

Consider ions of elements. Explain why the radius of a given element is larger when it is negatively charged and smaller when it is positively charged?

Recall that Zeff = Zactual - Electron Shielding. In comparison to a neutral atom, one with a negative charge would have one more electron, resulting in an increase in coulombic repulsion. Since there is more electron-electron repulsion, Z-eff decreases, thus resulting in a larger radius. For a positive ion, the loss of an electron results in less coulombic repulsion, meaning there is less electron shielding occurring. There is no change in the value of Z-actual since the number of protons doesn't change, thereby increasing the effective nuclear charge and decreasing the radius.

Shielding

Reduction of true nuclear charge because of coulombic repulsions from other electrons

Consider the observed trend. Does the 1s-orbital seem to be contracting or expanding with increased Z?

The 1s orbital is contracting as the value of Z increases, therefore the radius is decreasing as Z increases.

Consider an electron in a 3d orbital of Pd. In this case the effective nuclear charge this electron experiences is not equal to Z-total. Explain why this is, in detail, using the concepts of penetration and shielding.

The Pd 3d electrons are shielded by the 1s, 2s, 2p, 3s and 3p electrons, resulting in a reduced effective nuclear charge. Zeff = Z - shielding.

Consider electrons in p and d orbitals, which have a zero probability of being found at the nucleus. Discuss how strongly electrons in these orbitals are bound to the nucleus, relative to electrons in an s orbital of the same shell.

The electrons in the p orbitals are not as tightly bound to the nucleus as the s orbital of the same shell due to less efficient penetration and shielding by the inner core electrons as well as the s electrons of the same shell. d electrons in the same shell are the even less tightly bound, since they penetrate the worst, and are shielded by the core electrons as well as the s and p electrons of the same shell.

Do the energies of the 5th shell diverge or converge more than those of the 1st shell?

The energies of the 5th shell converge more than those of the 1st shell

Consider ions of elements Explain why the second ionization energy of an element is always larger than the first ionization energy.

The first ionization energy can be summarized with the equation: M + energy -> M^(+)+ e^(-) . The second ionization energy can be summarized with the equation: M^(+) + energy -> M^(2+) + e^(-). The second ionization energy requires more energy because it is much harder to pull an electron away from a positively charged ion (more coulombic attraction) than from a neutral atom. Since the cation has a higher Zeffthan its neutral counterpart, the outermost electron of the cation is held more tightly by the nucleus than the outermost electron of the neutral species, making the outermost electron of the cation harder to remove.

Z-effective

The net positive charge felt by an electron in a multi-electron atom; the shielding effect of electrons prevents higher orbital electrons from experiencing the full nuclear charge due to the repelling effect of lower energy level electrons Z-eff = Z-actual - electron shielding

Explain why atomic radius usually increases significantly as we move down a column in the periodic table.

The radius is determined by the farthest electrons from the nucleus. If we go down a group and look at the electronic structure of atoms, the outermost electrons will be occupying a higher energy level further from the nucleus. Consider the valence electron in Li (2s) and Na (3s). 3s is bigger and farther away from the nucleus than 2s so the radius is bigger for Na.

Explain why the relative energies of orbitals of different elements decrease as we move from left to right on the periodic table. What happens to the relative energies of these orbitals as we go down a column?

The relative energies of orbitals of different elements decrease from left to right due to the increase in Z-eff. Increasing the fraction of the nuclear charge that an electron experiences results in increased coulombic attraction. So relative to a free electron, an electron occupying the orbital would have a lower potential energy since the attractive force has increased.The relative energies of these orbitals also decrease dramatically down a colomn because Z is increasing by more than 1 and thus the coulombic attraction increases dramatically down a group. The graph in 5b illustrates this point.

Consider ions of elements In question 5d you explained how the relative energies of orbitals changed as we move to the right across a row and down a column of the periodic table. Considering this trend, describe how it relates to ionization energies of the elements.

The trend of the first ionization energy of elements generally increases from left to right in the periodic table can be explained by Zeff. Since Zeff is increasing as we going from left to right, the electron is held more tightly as we move from left to right in the periodic table, resulting in a higher first ionization energy needed to remove the electron. However, there are exeptions such as going from Be to B (in B highest energy electron occupies a p-orbital and at this point the nuclear charge isn't sufficient enough to lead to an increase in ionization energy.) Also on going from N to O. O has a paired electron in its pshell which shields it and makes it easier to remove. After O the nuclear charge is sufficient to lead to an increase in ionization energy even though electrons are paired. The trend of the first ionization energy of elements generally decrease as we go down a group can be explained through orbital energy levels and orbital size. If we are going down a group, we are increasing the number of electrons and orbitals. The electrons in the other orbitals shield the outermost electron from the nucleus, making the outermost electron held less loosely and easier to remove. Another reason is when we go down a group, the n value (orbital energy level) increases (Ex: 2p -> 3p), making the orbital of the outermost electron larger in size. The larger size orbital allows for the electron to be easier to remove as the lectron is held more loosely to the nucleus. Coulombic attraction decreases exponentially with distance

Pauli Exclusion Principle

There is maximum of two allowed electrons per orbital. If two electrons are present in an atomic or molecular orbital, the spin must be opposite. Remember we can have 0, 1, or 2 electrons in an orbital. But never more than 2.

Hund's Rule

Unoccupied orbitals will be filled before occupied orbitals. For example, one electron of the same spin will be assigned to each p orbital before a second electron can occupy the same orbital. This is explained by the amount of energy required to place two electrons in the same orbital (pairing energy). Electrons will occupy degenerate orbitals before sticking together in the same orbital to avoid this energetic penalty(although it is very small) of pairing the electrons. In addition, maximizing parallel spins leads to more exchange pairs, which is a secondary stabilizing effect.

Ionic Radius

the distance between the centers of neighboring cations and anions in an ionic compound