Today we looked at the three general types of strong forces that hold together individual atoms or ions. We collectively refer to these strong forces as chemical bonds.
1. Metallic bonds
a. building blocksmetal cations b. nature of the bondmetal cations are held together by the sea of delocalized valence electrons c. examples Zn (s), Au(s), alloys such as bronze or brass
a. building blocksmetal cations
b. nature of the bondmetal cations are held together by the sea of delocalized valence electrons
c. examples Zn (s), Au(s), alloys such as bronze or brass
2. Covalent bonds
a. building blocksnon-metal atoms b. nature of the bondThe desire for the atoms to aquire 8 valence electrons results in the sharing of electrons between two neighboring atoms. The shared electrons are attracted to both nuclei and act as a bridge to hold them together. c. examples O2(g), CH4(g), C(graphite)
a. building blocksnon-metal atoms
b. nature of the bondThe desire for the atoms to aquire 8 valence electrons results in the sharing of electrons between two neighboring atoms. The shared electrons are attracted to both nuclei and act as a bridge to hold them together.
c. examples O2(g), CH4(g), C(graphite)
3. Ionic bonds
a. building blocksmetal cations + non-metal anions b. nature of the bondHere, the desire to obtain 8 valence electrons results in an electron transfer between the metal (low IE) and the non-metal (high EA). The resulting cations and anions are held together by electrostatic interaction. c. examples NaCl(s), Mg(OH)2(s), KNO3(s)
a. building blocksmetal cations + non-metal anions
b. nature of the bondHere, the desire to obtain 8 valence electrons results in an electron transfer between the metal (low IE) and the non-metal (high EA). The resulting cations and anions are held together by electrostatic interaction.
c. examples NaCl(s), Mg(OH)2(s), KNO3(s)
In all three types of bonds, it is the valence electrons that are chemically significant (they influence the chemistry of that element). With that in mind we defined valence electrons as the electrons outside of the previous noble gas, not including any electrons in full d-blocks.
Try this: How many valence electrons do each of the following elements have? 1. Nickel, Ni answer: the e- configuration for Ni is: [Ar](4s^2)(3d^8) the d-block is not full, so we count those electron => 10 valence e- 2. Tin, Sn answer: the e- configuration for Sn is: [Kr](5s^2)(4d^10)(5p^2) the d-block is now full, so we do not count those electron => 4 valence e- 3. Fluorine, F answer: the e- configuration for F is: [He](2s^2)(2p^5) => 7 valence e-
If we ignore the d-block elements, then the maximum number of valence electrons an element can have is 8 (ns^2 np^6). This configuration is extremely stable (isoelectronic with the noble gases). This desire for the s and p-block elements to have 8 valence electrons dictates its bonding (both the type and number of bonds) and is the basis for the octet rule. We can conveniently represent information concerning how many valence electrons an element has using Lewis dot symbols:
We begin with the symbol for the element and then place the given number of valence electrons around it with a maximum of 2 above, 2 below, 2 on the right and 2 on the left.
Note: it is more important that you have the right number of valence electrons then putting them in some special order.
Today we focused on ionic bonds. We learned in class that the for ionic compounds can not be explained solely in terms of the electron transfer that occurs. We must also consider the stabalization due to the electrostatic attraction between the ions. This stabalization can be understood if we look at the reverse process: how much energy does it take to separate the ions in one mole of an ionic compound. This reverse process is called the lattice energy. The larger the lattice energy, the harder the ions are to separate and therefore, the more stable the ionic compound is.
The following lattice energies were taken from Table 8.2 in your text:
compound reaction lattice energy NaF NaF (s)Na+(g) + F-(g) 910 kJ/mol NaCl NaClNa+(g) + Cl-(g) 788 kJ/mol MgCl2 MgCl2Mg^+2(g) + 2Cl-(g) 2326 kJ/mol
The lattice energy values are related to the potential energy (E in the following equation) of the interacting ions which is described by the equation:
It is seen that the E holding the ions together is directly related to the charges and inversely related to the internuclear distance. The data in the table above makes sense. NaF is expected to have a greater lattice energy than NaCl, since F- is smaller than Cl-, so the ions can get closer together (smaller d). The smaller d means the E we calculate is greater. This is the energy that must be overcome when we measure the lattice energy. Similarly, MgCl2 has the greatest lattice energy of the three because the Mg^2+ has a greater charge (greater Q).
Try this: 1. For each pair of compounds choose the one with the GREATER lattice energy (without looking at a table of values). a) LiI vs NaI answer: LiI is greater since Li+ is smaller than Na+ b) NaBr vs CsI answer: NaBr is greater since both Na+ and Br- are smaller than Cs+ and I- c) MgO vs MgCl2 answer: MgO is greater since both the O(-2) and Mg(+2) have a larger charge (Q). In MgCl2, the Cl are -1.
Try this: 1. For each pair of compounds choose the one with the GREATER lattice energy (without looking at a table of values).
a) LiI vs NaI answer: LiI is greater since Li+ is smaller than Na+ b) NaBr vs CsI answer: NaBr is greater since both Na+ and Br- are smaller than Cs+ and I- c) MgO vs MgCl2 answer: MgO is greater since both the O(-2) and Mg(+2) have a larger charge (Q). In MgCl2, the Cl are -1.
While we are talking about the sizes of ions, let's review our four rules:
1. Anions are always larger than their parents. Why? We have added an electron without increasing Z(eff). 2. Cations are always smaller than their parents. Why? We have removed an electron without decreasing Z(eff). We have also removed some electron-electron repulsion. 3. In an isoelectronic series, the size of the ion decreases as the nuclear charge increases. Why? Here, we have the same number of electrons, but the higher nuclear charge pulls more tightly on the electrons. 4. For ions with the same charge, size increases as we go down a group. Why? As with atomic radii, as we go down a given group, we are in a higher n-value shell. Try this: 1. For each pair of species choose the one with the GREATER radius (without looking at a table of values). a) I vs I- answer: I- is greater since anions are larger than their parent b) Mg^2+ vs Mg answer: Mg is greater since cations are smaller than their parent c) Cl- vs Br- answer: Br- is greater since they are in the same group d) Rb+ vs Se^2- answer: Se2- is greater since they are isoelectronic (36 electrons) the one with greater nuclear charge is smaller.
1. Anions are always larger than their parents. Why? We have added an electron without increasing Z(eff). 2. Cations are always smaller than their parents. Why? We have removed an electron without decreasing Z(eff). We have also removed some electron-electron repulsion. 3. In an isoelectronic series, the size of the ion decreases as the nuclear charge increases. Why? Here, we have the same number of electrons, but the higher nuclear charge pulls more tightly on the electrons. 4. For ions with the same charge, size increases as we go down a group. Why? As with atomic radii, as we go down a given group, we are in a higher n-value shell.
1. Anions are always larger than their parents. Why? We have added an electron without increasing Z(eff).
2. Cations are always smaller than their parents. Why? We have removed an electron without decreasing Z(eff). We have also removed some electron-electron repulsion.
3. In an isoelectronic series, the size of the ion decreases as the nuclear charge increases. Why? Here, we have the same number of electrons, but the higher nuclear charge pulls more tightly on the electrons.
4. For ions with the same charge, size increases as we go down a group. Why? As with atomic radii, as we go down a given group, we are in a higher n-value shell.
Try this: 1. For each pair of species choose the one with the GREATER radius (without looking at a table of values).
a) I vs I- answer: I- is greater since anions are larger than their parent b) Mg^2+ vs Mg answer: Mg is greater since cations are smaller than their parent c) Cl- vs Br- answer: Br- is greater since they are in the same group d) Rb+ vs Se^2- answer: Se2- is greater since they are isoelectronic (36 electrons) the one with greater nuclear charge is smaller.
Today we focused on covalent bonds. Below is a figure showing the the potential energy diagram for a H-H covalent bond.
At large internuclear distances (right side of graph) there is no overlap of electron clouds. This state is designated with PE = 0.
As the nuclei approach each other (moving to the left on the graph), the atoms become stabalized due to sharing of their electron with the other atom. We see the PE decrease in this region.
The minimum on the curve corresponds to the optimized balancing of attraction (nuclear-electron) and repulsion (nuclear-nuclear and electron-electron). The energy at the minimum is the bond strength which represents the amount that the molecule is stabalized over the free atoms. If you want to break the bond, you need to supply that much energy. The internuclear distance at this minimum is the bond length for that molecule.
If the nuclei are brought any closer together, the nuclear-nuclear repulsion dominates and the PE for the molecule shoots up.
In the case of the hydrogen, H-H (or any homonuclear diatomic molecule), the electrons are equally shared by each of the atoms in the bond. This is called a non-polar covalent bond. But this is not always the case. The majority of covalent bonds are made up of atoms that have a different amount of pull (measured by electronegativity, EN) on the shared electrons. These are called polar covalent bonds.
Below is a drawing representing the general trend (there are exceptions) in EN based on the periodic table:
The greater the value for the electronegativity, the greater the pull that atom has on the shared electrons. If the pull is strong enough relative to the pull exhibited by the other atom in the bond, then the electron can actually be removed from its parent atom (ionic bond). Of course, the key word here is "relative" and so we compare the relative electronegativity (EN) of the bonded atoms (using the table in the book). The type of bond depends on the value of EN we calculate. Way back in chapter 2, we learned to identify ionic bonds by finding a metal and a non-metal atom in the comopund. Now we know why that is true. The metals with their low IE tend to have small values for EN while the non-metals with their large negative EA have large value for EN.
The diagram below show a rough relationship (the cut-offs are not rigid) between the calculated EN and the type of bond. For the first time we begin to look at bonding as a continuum (depending on the atoms involved) with non-polar bonds on one end and ionic bonds on the other extremem.
Try this: For each bond, calculate EN and name the type of bond using the diagram above. 1. N - O answer: delta EN = 3.5- 3.0 = 0.5 boarderline, non-polar covalent and polar covalent 2. C - C answer: delta EN = 2.5 - 2.5 = 0 non-polar covalent bond 3. Ca O answer: delta EN = 3.5 - 1.0 = 2.5 ionic bond
Where atoms are in a molecule and how those atoms are connected in 3-D space impacts the chemistry of that molecule. Life could not exist on earth if water was not a polar molecule. The key to understanding why water is polar lies in understanding the shape of the water molecule. Lewis Dot Structures are an important first step in determining the 3-D shape of a molecule.
Step 1: Draw the Skeleton Be sure to follow these simple rules:
1. If possible, make the skeleton symmetric 2. H and halogens are usually not the central atom 3. Least EN element is usually in the center 4. O does not bond to O (except in O2 and O3)
Step 2: Calculate # shared electrons using S = N - A
S = # shared electrons N = # valence electrons needed = 8 (except H=2, Be=4, B=6) A = # valence electrons available
One last brief topic formal charges (FC):
1. What are formal charges? The FC is the charge that an atom in a molecule would have if all atoms had the same electronegativity. 2. What formal charges aren't. They aren't real charges on the atoms, they are just a bookkeeping method to track valence electrons. 3. How do we find them? We must individually find the FC on each atom in a molecule or ion using, FC = (# valence electrons) - (# non-bonding electrons) - (1/2 # of bonding electrons) 4. What can we do with them once we have them? We can use them to decide which of several possible Lexis Dot Structure is the more reasonable. The most reasonable structure will have: a) atoms with the smallest formal charges and b) any negative charges on the most electronegative element. Try this: Determine the formal charge for each element in CH2O. Hint: Make sure you draw the Lewis dot structure first. If you are having trouble it is the same as problem 8.48e in your text book. (or 8.42d in the 7th edition) answer: C = (4) - (0) - 1/2(8) = 0 H = (1) - (0) - 1/2(2) = 0 O = (6) - (4) - 1/2(4) = 0
1. What are formal charges? The FC is the charge that an atom in a molecule would have if all atoms had the same electronegativity. 2. What formal charges aren't. They aren't real charges on the atoms, they are just a bookkeeping method to track valence electrons. 3. How do we find them? We must individually find the FC on each atom in a molecule or ion using,
4. What can we do with them once we have them? We can use them to decide which of several possible Lexis Dot Structure is the more reasonable. The most reasonable structure will have: a) atoms with the smallest formal charges and b) any negative charges on the most electronegative element.
Try this: Determine the formal charge for each element in CH2O. Hint: Make sure you draw the Lewis dot structure first. If you are having trouble it is the same as problem 8.48e in your text book. (or 8.42d in the 7th edition) answer: C = (4) - (0) - 1/2(8) = 0 H = (1) - (0) - 1/2(2) = 0 O = (6) - (4) - 1/2(4) = 0
Exceptions to the octet rule: It is important to keep in mind that the octet rule is not always valid. It should be considered a useful guideline that helps us get an appropriate structure about 80% of the time.
1. Radicals: molecules or ions that have an odd number of valence electrons available. The easy way to tell you are dealing with a radical is if the value you calculate for "A" (available valence electrons) is odd then you must have a radical. Radicals tend to be highly reactive and are the culprit in everything from food spoillage, the aging process and the depletion of the ozone layer. You can image how important they are for scientists to study. 2. Electron deficient: these are molecules or ions that contain H, Be or B which, as we saw on Monday, do not require a full octet. Lewis Dot Structures for electron deficient species are created the same way as the examples from Monday. The only change is that when you fill in the table for N, remember the H = 2, Be = 4, B = 6. 3. Expanded octet: these molecules and ions have central atoms that can accomodate more than 8 electron due to their relatively large size and having empty d-orbitals in the same shell as their valence electrons. To draw Lewis Dot Structures for these compounds, focus on the number of valence electrons available (A). Begin by placing these in bonding positions using as many as are needed based on the number of atoms surrounding the central atom (rather than basing it on "S" like we usually due). Continue with the remaining available electrons as normal.
1. Radicals: molecules or ions that have an odd number of valence electrons available. The easy way to tell you are dealing with a radical is if the value you calculate for "A" (available valence electrons) is odd then you must have a radical. Radicals tend to be highly reactive and are the culprit in everything from food spoillage, the aging process and the depletion of the ozone layer. You can image how important they are for scientists to study.
2. Electron deficient: these are molecules or ions that contain H, Be or B which, as we saw on Monday, do not require a full octet. Lewis Dot Structures for electron deficient species are created the same way as the examples from Monday. The only change is that when you fill in the table for N, remember the H = 2, Be = 4, B = 6.
3. Expanded octet: these molecules and ions have central atoms that can accomodate more than 8 electron due to their relatively large size and having empty d-orbitals in the same shell as their valence electrons. To draw Lewis Dot Structures for these compounds, focus on the number of valence electrons available (A). Begin by placing these in bonding positions using as many as are needed based on the number of atoms surrounding the central atom (rather than basing it on "S" like we usually due). Continue with the remaining available electrons as normal.
Calculating Hrxn using bond energies: Once we are able to draw Lewis Dot Structures, we are able to add a fourth method to our calculations of Hrxn (with calorimetry, Hess's law and ), This newest method allows us to relate the enthalpy change for a reaction to the bonds that are broken and formed during the process. After drawing the Lewis Dot Structures for each reactant and product molecule, we add up the bond enthalpies (Table 8.3 in the 7th edition and Table 8.4 in the 8th edition) for all the bonds broken and subtract from it the sum of all the bond enthalpies (from the same Table) of all the bonds formed.
For example: Determine Hrxn for the reaction, N2(g) + 3F2(g) 2NF3(g) Step 1: Draw the Lewis Dot Structures for each reactant and each product.
Step 2: Make a table of bond enthalpies for each bond broken and each bond formed.
Total
Don't forget that each NF3 has 3 N-F bonds. Also be careful, double and triple bonds have their own values and are not simple 2x or 3x the corresponding single bond.