Chemistry 1061

2nd Trend: Ionization Energy these are our 2 trends.. left to right, increasing nuclear charge, smaller atoms going down a group, same nuclear charge, but further away from the nucleus be able to explain that also, be able to rank things by their ionization energy rank things by their 1st ionization energy metals-low ionization energies nonmetals-high ionization energies noble gasses- really stable, they don't want to gain or lose electrons, they should have extremely high ionization energies

(IE) is the minimum energy (in kJ/mol) required to remove an electron from a gaseous atom in its ground state. Ionization energy is the amount of energy in kilojoules needed to strip 1 mole of electrons from 1 mole of gaseous atoms. Gaseous atoms are specified in this definition because an atom in the gas phase is virtually uninfluenced by its neighbors and so there are no inter-molecular forces (forces between molecules) to take into account when measuring ionization energy. From left to right across the periodic table ionization energy will increase As we go down a group ionization energy will decrease As we go from lower left to upper right ionization energy will increase Li=520 kJ/mol Z eff= +1 F=1680 kJ/mol Z eff= +7 Why would an atom resist losing an electron? because it has a larger effective nuclear charge (Z eff) the valence electrons of Fluorine fills 7 times as much charge attracting it to the nucleus that's going to make it harder to take that away but that's only half the story for this trend so what's the other thing that's different between Lithium and Fluorine? Fluorine is also smaller than Lithium, so not only is there a +7 charge, but the valence e⁻ are closer to that charge... the closer you are to the attractive force, the stronger that force is. So since those valence e⁻ are much closer to that +7 charge, they're much more attracted than 7 x the attraction With the higher nuclear charge and smaller nuclear size, make it much more difficult to ionize Fluorine. higher effective nuclear charge, electrons are closer to the nucleus, it's going to be harder to take them away the larger an atom is in a group, the easier it is to ionize, because the e⁻s are further away from the attraction of the positive charge

First Exception for Ionization Energies Why does Boron have a lower ionization than we expect? Why does the energy increase Beryllium to Boron?

Although the general trend in the periodic table is for first ionization energies to increase from left to right, some irregularities do exist. The first exception occurs between Group 2A and 3A elements in the same period. examples include: (Be and B) and (Mg and Al) Group 2A Be ↑↓ ↑↓ 1s 2s Group 3A B ↑↓ ↑↓ ↑ 1s 2s 2p The Group 3A elements have lower first ionization energies than 2A elements because they all have a single e⁻ in the outermost p sub-shell (ns²np¹), which is well shielded by the inner e⁻ and the ns² e⁻. Therefore, less energy is needed to remove a single p e⁻ than to remove an s e⁻ from the same principal energy level. In other words, when we go from Beryllium to Boron, we have a decrease in ionization energy even though Boron is to the right of Beryllium. when we lose that first electron, what do we end up with? We end up with Beryllium... why would Boron want to end up looking like Beryllium? remember what we were talking about with our group 6 and group 11 exceptions last time? we found that having half-filled sub-shells was really stable and having a fully-filled sub-shell was really stable so when boron loses this e⁻, it really doesn't want to lose it, but when it loses it, it ends up in this fairly stable configuration where it has this fully-filled sub-shell in all of these examples, they end up in that state, with fully-filled sub-shells, which is more stable than we expect, and so the ionization energy is lower than we would predict. We have a decrease in energy instead of an increase. a fully-filled sub-shell is just a stable configuration for an atom to be in. Noble gasses have completely filled shells, and that's what makes them so stable.

diamagnetic

Atom or compound that is un-attracted to the magnetic field

paramagnetic

Atom or substance containing unpaired electrons and is consequently attracted by a magnet.

First Trend: Atomic radii

Bonding atomic radius or Covalent radius- measured by bonding two identical atoms together and measuring the distance between the two nuclei, half of that distance is the atomic radius of an atom We define the size of an atom in terms of its atomic radius, which is one-half the distance between the two nuclei in two adjacent metal atoms or in a diatomic molecule. Trend: atomic radius decreases (atoms get smaller) as we go from left to right across the periodic table atomic radius increases (atoms get bigger) as we go down a group toward the bottom of the periodic table from the lower left corner of the periodic table to the upper right corner of the periodic table atomic size decreases (atoms get smaller)

Third Trend for Ionization Energies Electron Affinity (EA)

Electron affinity is a property that greatly influences the chemical behavior of atoms and their ability to accept one or more e⁻. EA is the negative energy change that occurs when an e⁻ is accepted by an atom in the gaseous state to form an anion. e- on the reactant side=fluoride ion e- on the product side= fluorine atom ∆E= - sign means it's releasing energy, becoming more stable ∆E= + sign means it's absorbing energy, to make the atom accept the electron, because you're going from the 1/2 filled sub-shell, and you're forcing another electron into that orbital into the original amount of space, so it resists it and it takes energy to make it happen X(g) + e⁻ → X⁻ (g) EA is negative if the reaction is endothermic. F(g) + e⁻ → F⁻ (g) ∆H= -328 kJ/mol EA is positive if the reaction is exothermic. F⁻(g) → F(g) + e⁻ ∆H=+328 kJ/mol The more positive is the electron affinity of an element, the greater is the affinity of an atom of the element to accept an electron. Another way of viewing electron affinity is to think of it as the energy that must be supplied to remove an electron from the anion. A large positive electron affinity means that the negative ion is very stable. The overall trend is an increase in the tendency to accept electrons (electron affinity values become more positive) from left to right across a period. The electron affinities of metals are generally lower than those of nonmetals. There is a general correlation between electron affinity and effective nuclear charge, which also increases from left to right in a given period. Know the following: Know what the process is: the energy change from when you add an electron Know that most things give away energy when you give them an electron Know that things like Nitrogen and Nobel gasses resist getting an electron, it takes energy to make them take the electron

valence electrons

Electrons on the outermost energy level of an atom (group number)

Ionization energy trends

First ionization energies of elements in a period increase with increasing atomic number. This trend is due to the increase in effective nuclear charge from left to right. A larger effective nuclear charge means a more tightly held valence electron, and a higher first ionization energy

How does a full valence-shell affect an atom?

Full valence-shell electron configurations are associated with an inherent degree of chemical stability. The high ionization energies of the noble gases, stemming from their large effective nuclear charge, comprise one of the reasons for this stability.

How does a half filled valence-shell affect an atom?

Half filled valence-shells are also very stable for an atom.

Ionization energy and Electron affinity help chemists understand the types of reactions that elements undergo and the nature of the elements' compounds. These two measures are related in a simple way...

Ionization energy measures the attraction of an atom for its own electrons Whereas Electron affinity expresses the attraction of an atom for an additional electron from some other source. Together they give us insight into the general attraction of an atom for electrons. With these concepts we can survey the chemical behavior of the elements systematically, paying attention to the relationship between their chemical properties and their electron configurations.

Is the process of Ionization and Endothermic or Exothermic Process?

Ionization is always an Endothermic process. Energy absorbed by atoms (or ions) in the ionization process has a positive value. Ionization energies are all positive quantities.

Irregularities of Electron Affinity

Irregularities: Electron affinity of a Group 2A element is lower than that for the corresponding Group 1A element The electron affinity of a Group 5A element is lower than that for the corresponding Group 4A element. These exceptions are due to the valence electron configurations of the elements involved. Group 2A elements must end up in higher-energy np orbitals, where they are effectively shielded by the ns² electrons, causing them to experience a weaker attraction to the nucleus. In turn they will have a lower electron affinity than the corresponding Group 1A elements. It is harder to add an electron to a Group 5A element (ns²np³) than to the corresponding 4A element (ns²np²). This is because the electron added to the Group 5A element must be placed in a np orbital that already contains an electron and will therefore experience a greater electrostatic repulsion. In spite of the fact that Nobel gases have high effective nuclear charges, they have extremely low electron affinities (zero or negative values). The reason is that an electron added to an atom with an ns² np⁶ configuration has to enter an (n+1)s orbital, where it is well shielded by the core electrons and will only be very weakly attracted by the nucleus. This analysis also explains why species with complete valence shells tend to be chemically stable.

Magnitude of Ionization Energy

Magnitude of ionization energy: a measure of how tightly the electron is held in the atom. The higher the ionization energy, the more difficult it is to remove the electron. Valence electrons are relatively easy to remove from an atom Core electrons are much harder to remove This causes a large jump in ionization energy between the last valence electron and the first core electron. First ionization energy (IE₁): For a many-electron atom, the amount of energy required to remove the first electron from the atom in its ground state energy + X(g)→Z⁺(g) + e⁻ (X) represents an atom of any element (e⁻) represents an electron Second ionization energy (IE₂): energy + X⁺ (g)+e⁻→X²⁺(g)+e⁻ Third ionization energy (IE₃): energy + X²⁺(g)→X³⁺(g) +e⁻ The pattern continues for the removal of subsequent electrons.

Zeff (effective nuclear charge)

The effective nuclear charge (Z eff) is the nuclear charge felt by an electron when both the actual nuclear charge (Z) and the repulsive effect (shielding) of the other electrons are taken into account In general Z eff is given by Z eff = Z- σ (sigma) where (sigma) is called the shielding constant (also called the screening constant) represents the approximate number of core electrons. The shielding constant is greater than zero but smaller than Z. the positive charge/attraction of the nucleus experienced by the valence electrons (increases moving across the periodic table w/ increased # of protons - makes the atom smaller because the force holds it tight)

core electrons

The electrons in the inner shells of an atom; these electrons are not involved in forming bonds.

Ionic radius

The radius of a cation or an anion When a neutral atom is converted to an ion, we expect a change in size. If the atom forms an anion, its size (or radius) increases, because the nuclear charge remains the same but the repulsion resulting from the additional electron(s) enlarges the domain of the electron cloud. Removing one or more electrons from an atom reduces the electron-electron repulsion bur the nuclear charge remains the same, so the electron cloud shrinks, and the cation is smaller than the atom. For isoelectronic ions, the size of the ion is based on the size of the electron cloud, not on the number of protons in the nucleus.

Second Exception for Ionization Energies

The second irregularity occurs between Groups 5A and 6a. (N and O) and (P and S) At Oxygen, the ionization energy is lower than we would expect, it's lower than Nitrogen. Group 5A N ↑↓ ↑↓ ↑↑↑ 1s² 2s² 2p³ Group 6A O ↑↓ ↑↓ ↑↓↑↓ 1s² 2s² 2p⁴ So when we lose our one electron in our Oxygen atom, we end up with this Nitrogen configuration with the half-filled sub-shell. It also ends up with the group 6 exception that having a half-filled sub-shell is also very stable for an atom. In the Group 5A and 6A elements (ns² and np³): The p e⁻ are in three separate orbitals according to Hund's rule. In Group 6A (ns² np⁴), the additional e⁻ must be paired with one of the p electrons. The proximity of two electrons in the same orbital results in greater electrostatic repulsion, which makes it easier to ionize an atom of the Group 6a element, even though the nuclear charge has increased by one unit. thus, the ionization energies for Group 6A elements are lower than those for Group 5A elements in the same period. From Magnesium to Aluminum, we have a decrease in ionization energy. From Phosphorus to Sulfur we have a decrease in ionization energy instead of an increase. core electrons are closer in proximity to the nucleus and therefore have a more positive charge from the nucleus in the case where there is no screening, when you attempt to take away a core electron, it will become a lot harder to remove because it requires more energy so any time you see a jump in ionization energies, it's because we're starting to take away core electrons

Ionization Energy: Describe what happens to the remaining electrons once an electron is removed from an atom.

When an electron is removed from an atom, the repulsion among the remaining electrons decreases. This is because the nuclear charge remains constant, and more energy is needed to remove another electron from the positively charged ions. Thus, ionization energies always increase in the following order: 1E₁<1E₂<1E₃....

be able to explain the reasoning behind these trends be able to rank atoms by their size

from left to right across a period in the periodic table, what properties change? lithium vs fluorine they're in the same period, plus fluorine is a lot smaller than lithium why would that be the case? we should look at their electron configurations what is the effective nuclear charge for Lithium? 1 Li 3 protons - 2 core electrons=1 F 9 protons - 2 core electrons = 7 how would that explain Fluorine being smaller than Lithium? if you go left to right on the periodic table, the effective nuclear charge increases the valence electrons yield an increasing positive charge so they get pulled in closer to the nucleus that is true across any period the number of core electrons stays the same but the number of protons in that nucleus increases and that's just going to pull those electrons in closer and closer

effective nuclear charge

the positive charge that an electron experiences from the nucleus, equal to the nuclear charge but reduced by any shielding or screening from any intervening electron distribution

effective nuclear charge (Zeff)

we want to figure out the charge felt by the valence electron for Sodium so we first write out the electron configuration for sodium Na 1s2 2s2 2p6 3s1 how many core electrons does Sodium have? 10 10 core electrons and 1 valence electrons to find the # of core electrons for a main group element, is to look at the group number (how many valence electrons it has) , and subtract that from the atomic number (which is the total number of electrons) to figure out how much positive charge does this electron feel by calculating the effective nuclear charge or (Z eff) we use Z (which is the atomic number) and subtract off the number of core electrons (which is the atomic number minus the group number) Z eff = Z - σ (sigma) Z eff = Z (atomic number) - σ (atomic number - group number) Z eff = 11 (Na atomic #)- (11-1) (atomic # -group #) Z eff = 11 - 10 Z eff = 1