Electrons in Molecules

1. 

Ions in ionic solids
An electron can be transferred from one neutral (uncharged) atom to another. When this happens, the atom from which the electron is transferred becomes a positive ion or cation and the atom to which the electron is transferred becomes a negative ion or anion.
The transfer of an electron to or from an atom involves energy. The energy required to remove one electron from an atom is characteristic of the atom and is its ionisation energy. The energy released when an electron is acquired by an atom is its electron affinity. The electron transfer can take place with a release of energy only when the electron affinity of the receiving species is greater than the ionisation energy of the donating species, which is rarely the case, or, most often, when the resulting ions associate themselves into a new configuration of lower energy. Most electron transfer reactions occur when the ions, once formed, associate themselves into structured lattices called ionic crystals.
Energy Considerations in Ionic Structures
The energy required to produce any cation from an atom, the ionisation energy of the atom, is always larger, and usually much larger, than the energy which is released when an electron is added to any atom, the electron affinity. This is true even if an electron is being removed from an atom, which loses it relatively easily, like sodium, and is being added to an atom, which readily acquires it, such as chlorine. The ionisation energy of sodium is +496 kJ/mole, while the electron affinity of chlorine is only -349 kJ/mole. The reaction Na(g) + Cl(g)   Na+(g) + Cl-(g) would require +147 kJ/mole to proceed, while the reaction Na(s) + 1/2 Cl2(g)  Na+(g) + Cl-(g)  requires +377 kJ/mole. Electron affinity alone cannot provide sufficient energy to form ions or ionic structures; the energy must come from the assembly of isolated ions into stable multi-ion structures.
Ions with the same type of charge repel each other, but ions of opposite charge attract each other. The simplest possible ionic structure which might be stable is the gas-phase ion pair, which consists of one cation and one anion held together by electrostatic attraction. It is relatively simple to calculate how much energy would be gained by this association using the Coulomb law of electrostatic attraction. The energy of the attraction is given by
E = (2.31 x 10-16 J-pm) Z+Z-/d
where Z is the charge on the cation and anion, and d is the distance between the ions, in pm.  The energy of the two associated ions will be less than the energy of the two isolated ions by this amount if the ions are of opposite charge. For the sodium ion, the ionic radius is 97 pm, and for the chloride ion it is 181 pm, so the distance of separation of the centres of the two ions is 278 pm. The energy for one ion pair, multiplied by the Avogadro number NA, gives the molar energy of [Na+Cl-](g) relative to the molar energy of the isolated ions as:
E = -8.31 x 10-19 J/molecule x 6.022...x 1023 molecules/mole
This is -500 kJ/mole, so the standard molar enthalpy of formation of the ion pair estimated using the Coulomb law is -123 kJ/mole (-500 kJ/mole + 377 kJ/mole). Even for a single sodium ion and a chloride ion in the gas phase, it is the lower energy available through the association of ions of opposite charge that drives the formation of ionic compounds.
The ionic radii used in the calculation above were the radii of sodium and chloride ions found in ionic crystals. They are, however, very similar to the radii of these ions under other conditions. The actual distance between the ions in Na+Cl-(g) has been measured and found to be 236.1 pm.
Association of ions of opposite charge is not normally into ion pairs. It is far more common to find ions in the form of the solid ionic crystals, which are large ordered three-dimensional arrays of ions.
The diagram below is the Born-Haber cycle for the formation of an ionic compound from the reaction of an alkali metal (Li, Na, K, Rb, Cs) with a gaseous halogen (F2, Cl2). The Born-Haber thermochemical cycle is named after the two German physical chemists, Max Born and Fritz Haber, who first used it in 1919.

The energies of the cycle above are used to get energy tables for all the alkali metal halides.
The enthalpy change in the formation of an ionic lattice from the gaseous isolated sodium and chloride ions is -788 kJ/mole. That enthalpy change, which corresponds to the reaction Na+(g) + Cl-(g)   NaCl(s), is called the lattice energy of the ionic crystal. Although the lattice energy is not directly measurable, there are various ways to estimate it from theoretical considerations and some experimental values. For all known ionic crystals, the lattice energy has a large negative value. It is ultimately the lattice energy of an ionic crystal which is responsible for the formation and stability of ionic crystal structures.
For sodium chloride, the Born-Haber cycle is:

A cycle of this type is an example of Hess's Law.  It can be used to calculate any of the six enthalpies, given the other five.
Lattice Energy: The Born-Haber cycle
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Table of contents
Ionic solids tend to be very stable compounds. The enthalpies of formation of the ionic molecules cannot alone account for this stability. These compounds have an additional stability due to the lattice energy of the solid structure. However, lattice energy cannot be directly measured. The Born-Haber cycle allows us to understand and determine the lattice energies of ionic solids.
Introduction
This module will introduce the idea of lattice energy, as well as one process that allows us to calculate it: the Born-Haber Cycle. In order to use the Born-Haber Cycle, there are several concepts that we must understand first.
Lattice Energy
Lattice Energy is a type of potential energy that may be defined in two ways. In one definition, the lattice energy is the energy required to break apart an ionic solid and convert its component atoms into gaseous ions. This definition causes the value for the lattice energy to always be positive, since this will always be an endothermic reaction. The other definition says that lattice energy is the reverse process, meaning it is the energy released when gaseous ions bind to form an ionic solid. As implied in the definition, this process will always be exothermic, and thus the value for lattice energy will be negative. Its values are usually expressed with the units kJ/mol.
Lattice Energy is used to explain the stability of ionic solids. Some might expect such an ordered structure to be less stable because the entropy of the system would be low. However, the crystalline structure allows each ion to interact with multiple oppositely charge ions, which causes a highly favorable change in the enthalpy of the system. A lot of energy is released as the oppositely charged ions interact. It is this that causes ionic solids to have such high melting and boiling points. Some require such high temperatures that they decompose before they can reach a melting and/or boiling point.
Born-Haber Cycle
There are several important concept to understand before the Born-Haber Cycle can be applied to determine the lattice energy of an ionic solid; ionization energy, electron affinity, dissociation energy, sublimation energy, heat of formation, and Hess's Law.
• Ionization Energy is the energy required to remove an electron from a neutral atom or an ion. This process always requires an input of energy, and thus will always have a positive value. In general, ionization energy increases across the periodic table from left to right, and decreases from top to bottom. There are some excepts, usually due to the stability of half-filled and completely filled orbitals.
• Electron Affinity is the energy released when an electron is added to a neutral atom or an ion. Usually, energy released would have a negative value, but due to the definition of electron affinity, it is written as a positive value in most tables. Therefore, when used in calculating the lattice energy, we must remember to subtract the electron affinity, not add it. In general, electron affinity increases from left to right across the periodic table and decreases from top to bottom.
• Dissociation energy is the energy required to break apart a compound. The dissociation of a compound is always an endothermic process, meaning it will always require an input of energy. Therefore, the change in energy is always positive. The magnitude of the dissociation energy depends on the electronegativity of the atoms involved.
• Sublimation energy is the energy required to cause a change of phase from solid to gas, bypassing the liquid phase. This is an input of energy, and thus has a positive value. It may also be referred to as the energy of atomization.
• The heat of formation is the change in energy when forming a compound from its elements. This may be positive or negative, depending on the atoms involved and how they interact.
• Hess's Law states that the overall change in energy of a process can be determined by breaking the process down into steps, then adding the changes in energy of each step. The Born-Haber Cycle is essentially Hess's Law applied to an ionic solid.
Using the Born-Haber Cycle
The values used in the Born-Haber Cycle are all predetermined changes in enthalpy for the processes described in the section above. Hess' Law allows us to add or subtract these values, which allows us to determine the lattice energy.

Step 1
Determine the energy of the metal and nonmetal in their elemental forms. (Elements in their natural state have an energy level of zero.) Subtract from this the heat of formation of the ionic solid that would be formed from combining these elements in the appropriate ration. This is the energy of the ionic solid, and will be used at the end of the process to determine the lattice energy.
Step 2
The Born-Haber Cycle requires that the elements involved in the reaction are in their gaseous forms. Add the changes in enthalpy to turn one of the elements into its gaseous state, and then do the same for the other element.
Step 3
Metals exist in nature as single atoms and thus no dissociation energy needs to be added for this element. However, many nonmetals will exist as polyatomic species. For example, Cl exists as Cl2 in its elemental state. The energy required to change Cl2 into 2Cl atoms must be added to the value obtained in Step 2.
Step 4
Both the metal and nonmetal now need to be changed into their ionic forms, as they would exist in the ionic solid. To do this, the ionization energy of the metal will be added to the value from Step 3. Next, the electron affinity of the nonmetal will be subtracted from the previous value. It is subtracted because it is a release of energy associated with the addition of an electron.
*This is a common error due to confusion caused by the definition of electron affinity, so be careful when doing this calculation.
Step 5
Now the metal and nonmetal will be combined to form the ionic solid. This will cause a release of energy, which is called the lattice energy. The value for the lattice energy is the difference between the value from Step 1 and the value from Step 4.
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The diagram below is another representation of the Born-Haber Cycle.

Equation
The Born-Haber Cycle can be reduced to a single equation:
Heat of formation= Heat of atomization+ Dissociation energy+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy
*Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive.
Rearrangement to solve for lattice energy gives the equation:
Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities)
Problems
1. Define lattice energy, ionization energy, and electron affinity.
2. What is Hess' Law?
3. Find the lattice energy of KF(s).
Note: Values can be found in standard tables.
4. Find the lattice energy of MgCl2(s).
5. Which one of the following has the greatest lattice energy?
a. A) MgO
b. B) NaC
c. C) LiCl
d. D) MgCl2
6. Which one of the following has the greatest Lattice Energy?
a. NaCl
b. CaCl2
c. AlCl3
d. KCl

Solutions
1. Lattice energy: The difference in energy between the expected experimental value for the energy of the ionic solid and the actual value observed. More specifically, this is the energy gap between the energy of the separate gaseous ions and the energy of the ionic solid.
Ionization energy: The energy change associated with the removal of an electron from a neutral atom or ion.
Electron affinity: The release of energy associated with the addition of an electron to a neutral atom or ion.
2. Hess' Law states that the overall energy of a reaction may be determined by breaking down the process into several steps, then adding together the changes in energy of each step.
3. Lattice Energy= [-436.68-89-(0.5*158)-418.8-(-328)] kJ/mol= -695.48 kJ/mol
4. Lattice Energy= [-641.8-146-243-(737.7+1450.6)-(2*-349)] kJ/mol= -2521.1 kJ/mol
5. MgO. It has ions with the largest charge.
6. AlCl3. According to the periodic trends, as the radius of the ion increases, lattice energy decreases.


Ionization Energy and Electron Affinity
The First Ionization Energy
Patterns In First Ionization Energies

Exceptions to the General Pattern of First Ionization Energies
2nd, 3rd, 4th, and Higher Ionization Energies

Electron Affinity
Consequences of the Relative Size of Ionization Energies and Electron Affinities

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The First Ionization Energy
The energy needed to remove one or more electrons from a neutral atom to form a positively charged ion is a physical property that influences the chemical behavior of the atom. By definition, the first ionization energy of an element is the energy needed to remove the outermost, or highest energy, electron from a neutral atom in the gas phase.
The process by which the first ionization energy of hydrogen is measured would be represented by the following equation.
H(g)   H+(g) + e- Ho = -1312.0 kJ/mol

Practice Problem 3:
Use the Bohr model to calculate the wavelength and energy of the photon that would have to be absorbed to ionize a neutral hydrogen atom in the gas phase.
Click here to check your answer to Practice Problem 3
Click here to see a solution to Practice Problem 3

The magnitude of the first ionization energy of hydrogen can be brought into perspective by comparing it with the energy given off in a chemical reaction. When we burn natural gas, about 800 kJ of energy is released per mole of methane consumed.
CH4(g) + 2 O2(g)   CO2(g) + 2 H2O(g) Ho = -802.4 kJ/mol
The thermite reaction, which is used to weld iron rails, gives off about 850 kJ of energy per mole of iron oxide consumed.
Fe2O3(s) + 2 Al(s)   Al2O3(s) + 2 Fe(l) Ho = -851.5 kJ/mol
The first ionization energy of hydrogen is half again as large as the energy given off in either of these reactions.

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Patterns in the First Ionization Energies
The first ionization energy for helium is slightly less than twice the ionization energy for hydrogen because each electron in helium feels the attractive force of two protons, instead of one.
He(g)   He+(g) + e- Ho = 2372.3 kJ/mol
It takes far less energy, however, to remove an electron from a lithium atom, which has three protons in its nucleus.
Li(g)   Li+(g) + e- Ho = 572.3 kJ/mol
This can be explained by noting that the outermost, or highest energy, electron on a lithium atom is in the 2s orbital. Because the electron in a 2s orbital is already at a higher energy than the electrons in a 1s orbital, it takes less energy to remove this electron from the atom.
The first ionization energies for the main group elements are given in the two figures below.
 
Two trends are apparent from these data.
• In general, the first ionization energy increases as we go from left to right across a row of the periodic table.
• The first ionization energy decreases as we go down a column of the periodic table.
The first trend isn't surprising. We might expect the first ionization energy to become larger as we go across a row of the periodic table because the force of attraction between the nucleus and an electron becomes larger as the number of protons in the nucleus of the atom becomes larger.
The second trend results from the fact that the principal quantum number of the orbital holding the outermost electron becomes larger as we go down a column of the periodic table. Although the number of protons in the nucleus also becomes larger, the electrons in smaller shells and subshells tend to screen the outermost electron from some of the force of attraction of the nucleus. Furthermore, the electron being removed when the first ionization energy is measured spends less of its time near the nucleus of the atom, and it therefore takes less energy to remove this electron from the atom.

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Exceptions to the General Pattern of First Ionization Energies
The figure below shows the first ionization energies for elements in the second row of the periodic table. Although there is a general trend toward an increase in the first ionization energy as we go from left to right across this row, there are two minor inversions in this pattern. The first ionization energy of boron is smaller than beryllium, and the first ionization energy of oxygen is smaller than nitrogen.

These observations can be explained by looking at the electron configurations of these elements. The electron removed when a beryllium atom is ionized comes from the 2s orbital, but a 2p electron is removed when boron is ionized.
Be: [He] 2s2
B: [He] 2s2 2p1
The electrons removed when nitrogen and oxygen are ionized also come from 2p orbitals.
N: [He] 2s2 2p3
O: [He] 2s2 2p4
But there is an important difference in the way electrons are distributed in these atoms. Hund's rules predict that the three electrons in the 2p orbitals of a nitrogen atom all have the same spin, but electrons are paired in one of the 2p orbitals on an oxygen atom.

Hund's rules can be understood by assuming that electrons try to stay as far apart as possible to minimize the force of repulsion between these particles. The three electrons in the 2p orbitals on nitrogen therefore enter different orbitals with their spins aligned in the same direction. In oxygen, two electrons must occupy one of the 2p orbitals. The force of repulsion between these electrons is minimized to some extent by pairing the electrons. There is still some residual repulsion between these electrons, however, which makes it slightly easier to remove an electron from a neutral oxygen atom than we would expect from the number of protons in the nucleus of the atom.
Practice Problem 4:
Predict which element in each of the following pairs has the larger first ionization energy.
(a) Na or Mg
(b) Mg or Al
(b) Mg or Al
(c) F or Cl
Click here to check your answer to Practice Problem 4


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Second, Third, Fourth, and Higher Ionization Energies
By now you know that sodium forms Na+ ions, magnesium forms Mg2+ ions, and aluminum forms Al3+ ions. But have you ever wondered why sodium doesn't form Na2+ ions, or even Na3+ ions? The answer can be obtained from data for the second, third, and higher ionization energies of the element.
The first ionization energy of sodium, for example, is the energy it takes to remove one electron from a neutral atom.
Na(g) + energy   Na+(g) + e-
The second ionization energy is the energy it takes to remove another electron to form an Na2+ ion in the gas phase.
Na+(g) + energy   Na2+(g) + e-
The third ionization energy can be represented by the following equation.
Na2+(g) + energy   Na3+(g) + e-
The energy required to form a Na3+ ion in the gas phase is the sum of the first, second, and third ionization energies of the element.
First, Second, Third, and Fourth Ionization Energies
of Sodium, Magnesium, and Aluminum (kJ/mol)

It doesn't take much energy to remove one electron from a sodium atom to form an Na+ ion with a filled-shell electron configuration. Once this is done, however, it takes almost 10 times as much energy to break into this filled-shell configuration to remove a second electron. Because it takes more energy to remove the second electron than is given off in any chemical reaction, sodium can react with other elements to form compounds that contain Na+ ions but not Na2+ or Na3+ ions.
A similar pattern is observed when the ionization energies of magnesium are analyzed. The first ionization energy of magnesium is larger than sodium because magnesium has one more proton in its nucleus to hold on to the electrons in the 3s orbital.
Mg: [Ne] 3s2
The second ionization energy of Mg is larger than the first because it always takes more energy to remove an electron from a positively charged ion than from a neutral atom. The third ionization energy of magnesium is enormous, however, because the Mg2+ ion has a filled-shell electron configuration.
The same pattern can be seen in the ionization energies of aluminum. The first ionization energy of aluminum is smaller than magnesium. The second ionization energy of aluminum is larger than the first, and the third ionization energy is even larger. Although it takes a considerable amount of energy to remove three electrons from an aluminum atom to form an Al3+ ion, the energy needed to break into the filled-shell configuration of the Al3+ ion is astronomical. Thus, it would be a mistake to look for an Al4+ ion as the product of a chemical reaction.
Practice Problem 5:
Predict the group in the periodic table in which an element with the following ionization energies would most likely be found.
1st IE = 786 kJ/mol
2nd IE = 1577
3rd IE = 3232
4th IE = 4355
5th IE = 16,091
6th IE = 19,784
Click here to check your answer to Practice Problem 5


Practice Problem 6:
Use the trends in the ionization energies of the elements to explain the following observations.
(a) Elements on the left side of the periodic table are more likely than those on the right to form positive ions.
(b) The maximum positive charge on an ion is equal to the group number of the element
Click here to check your answer to Practice Problem 6


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Electron Affinity
Ionization energies measure the tendency of a neutral atom to resist the loss of electrons. It takes a considerable amount of energy, for example, to remove an electron from a neutral fluorine atom to form a positively charged ion.
F(g)   F+(g) + e- Ho = 1681.0 kJ/mol
The electron affinity of an element is the energy given off when a neutral atom in the gas phase gains an extra electron to form a negatively charged ion. A fluorine atom in the gas phase, for example, gives off energy when it gains an electron to form a fluoride ion.
F(g) + e-   F-(g) Ho = -328.0 kJ/mol
Electron affinities are more difficult to measure than ionization energies and are usually known to fewer significant figures. The electron affinities of the main group elements are shown in the figure below.

Several patterns can be found in these data.
• Electron affinities generally become smaller as we go down a column of the periodic table for two reasons. First, the electron being added to the atom is placed in larger orbitals, where it spends less time near the nucleus of the atom. Second, the number of electrons on an atom increases as we go down a column, so the force of repulsion between the electron being added and the electrons already present on a neutral atom becomes larger.
• Electron affinity data are complicated by the fact that the repulsion between the electron being added to the atom and the electrons already present on the atom depends on the volume of the atom. Among the nonmetals in Groups VIA and VIIA, this force of repulsion is largest for the very smallest atoms in these columns: oxygen and fluorine. As a result, these elements have a smaller electron affinity than the elements below them in these columns as shown in the figure below. From that point on, however, the electron affinities decrease as we continue down these columns.

At first glance, there appears to be no pattern in electron affinity across a row of the periodic table, as shown in the figure below.

When these data are listed along with the electron configurations of these elements, however, they make sense. These data can be explained by noting that electron affinities are much smaller than ionization energies. As a result, elements such as helium, beryllium, nitrogen, and neon, which have unusually stable electron configurations, have such small affinities for extra electrons that no energy is given off when a neutral atom of these elements picks up an electron. These configurations are so stable that it actually takes energy to force one of these elements to pick up an extra electron to form a negative ion.
Electron Affinities and Electron Configurations for the First 10 Elements in the Periodic Table
Element Electron Affinity (kJ/mol) Electron Configuration
H 72.8 1s1
He <0 1s2
Li 59.8 [He] 2s1
Be <0 [He] 2s2
B 27 [He] 2s2 2p1
C 122.3 [He] 2s2 2p2
N <0 [He] 2s2 2p3
O 141.1 [He] 2s2 2p4
F 328.0 [He] 2s2 2p5
Ne <0 [He] 2s2 2p6


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Consequences of the Relative Size of Ionisation Energies and Electron Affinities
Students often believe that sodium reacts with chlorine to form Na+ and Cl- ions because chlorine atoms "like" electrons more than sodium atoms do. There is no doubt that sodium reacts vigorously with chlorine to form NaCl.
2 Na(s) + Cl2(g)   2 NaCl(s)
Furthermore, the ease with which solutions of NaCl in water conduct electricity is evidence for the fact that the product of this reaction is a salt, which contains Na+ and Cl- ions.
         NaCl(s) H2O                 Na+(aq) + Cl-(aq)
The only question is whether it is legitimate to assume that this reaction occurs because chlorine atoms "like" electrons more than sodium atoms.
The first ionization energy for sodium is one and one-half times larger than the electron affinity for chlorine.
Na: 1st IE = 495.8 kJ/mol
Cl: EA = 328.8 kJ/mol
Thus, it takes more energy to remove an electron from a neutral sodium atom than is given off when the electron is picked up by a neutral chlorine atom. We will obviously have to find another explanation for why sodium reacts with chlorine to form NaCl. Before we can do this, however, we need to know more about the chemistry of ionic compounds.