lect 01, 19pp atoms,config,IP,EA,Xpsbeauchamp/pdf/314_lect_01.pdfThe size of an atom is the size of...

1 Lecture 1 Chemistry is all about atoms. For chemists, atoms are composed of three fundamental particles: protons, neutrons and electrons. Atoms mostly appear to be gregarious creations. They love being around other atoms, and do so in a variety of ways that we will study, usually in molecules, ions or radicals. Since our charge is to learn organic chemistry, we can dispense with most of the atoms of the periodic table and focus on the chemistry of mainly one element, carbon. Of course any element that associates with carbon will be important to us as well. These additional elements include hydrogen, nitrogen, oxygen, sulfur and the halogens. We will use other elements in organic chemistry, but mainly as tools to produce some desired change in a structure composed of the atoms mentioned above. Atoms are incredibly small, so small in fact that we cannot actually see them with our eyes or even in a microscope (although this has changed in recent years with electron microscopy). We use a variety of instrumental techniques to study atoms and molecules. We also use our imaginations to create theoretical and mathematical models of their invisible world. In this book, we will use some of these tools in a qualitative way to increase our understanding of how and why organic chemistry happens the way it does. An added benefit of learning the fundamentals of organic chemistry is that you will also be learning the fundamentals of biochemistry, which explain how life works. As incredibly small as atoms are, the number of them in something as common as a gallon of gasoline, is incredibly large. How many molecules are in a gallon of gasoline? Since gasoline is a complicated mixture, let’s say we have a gallon of octane, a term associated with gasoline. Octane is a simple organic molecule composed of eight carbon atoms and 18 hydrogen atoms. It belongs to a family called the alkanes, a subgroup of a larger family called hydrocarbons (molecules composed of only carbon and hydrogen). We need to do a few freshman chemistry conversions to get from a gallon of octane to the number of molecules of octane in a gallon of octane. These are shown below.

1.0 gallon of octane litersgallon

mlliter

grams ml

molesgrams

molecules mole

1000 1.0

3.81.0

0.70 1.0

1.0114

6.02x1023

1.0

= 7,000,000,000,000,000,000,000,000 molecules in a gallon of octane (gasoline)

= 7.0 x 1024

molecules

What does a number like this mean? The truth is that it’s beyond our comprehension. It’s just REALLY BIG! Can you even say the number?

= 7,000,000,000,000,000,000,000,000 molecules

millionsbillions

trillionsquadrillions

quintillionssextillions

septillions

thousandshundreds

There's about seven septillion octane molecules in a gallon of octane.

So what we are studying is:

Something incredibly small An incredibly large number of them =x something we can see and try to understand

Get out your imagination. You are going to need it.

2 Lecture 1 The structure of atoms…provides a basis for the structure of molecules

We will begin our discussion of organic chemistry with atoms. You probably have a pretty good idea about the structure of an atom from a prior chemistry course, but just to make sure let’s review what are the essential features of atoms? They consist of incredibly small and dense nuclei surrounded by relatively huge and light electron clouds. Essentially, all of the mass of an atom is in its positively charged nucleus and all of the volume is its negatively charged electron cloud. A simplistic picture of an atom is shown below.

What is the relative volume of an electron cloud (Ve) compared to volume of a nucleus (Vn)?

p = protons = This number is constant for a particular type of atom and defines an element. If there are six protons, the element has to be carbon.

n = neutrons = This number can vary in an element; it defines an isotope. Some isotopes are stable and some are radioactive. Carbon-12 has six protons and six neutrons and is stable. Carbon-13 has six protons and seven neutrons and is stable. Carbon-14 has six protons and eight neutrons and is unstable (and radioactive) with a half life of almost 6000 years.

e = electrons = The number of electrons can vary. It can increase or decrease, depending on an atom's position in the periodic table.

Electron clouds determine the overall volume of atoms.

Protons and neutrons determine the mass of an atom.

Ve 4 π Vn 3 = (1.33)(3.14)(100,000)3 = 4 x 1015 = =

rern

3

If electrons = protons? (same number of electrons and protons) neutral atom

If electrons < protons? (deficiency of electrons) posistively charged cation

If electrons > protons? (excess of electrons) negatively charged anion

associated term

What is the relative mass of the electrons compared to the mass of nuclear particles?

4,000,000,000,000,000 1

The size of an atom is the size of its electron cloud.

The mass of an atom is mostly the mass of its incredibly small nucleus.

mass protons (or neutrons) 1840 mass electrons 1

evalp,n

ecore

d = 1d = 100,000

Valence electrons (eval) are the outermost layer of electrons. They determine the bonding patterns of an atom and the usual goal is to attain a Noble gas configuration. This is accomplished by losing electrons (becoming cations) or gaining electrons (becoming anions) or sharing electrons in covalent bonds (in neutral molecules or in complex ions). Core electrons (ecore) are electrons in completely filled inner-shells. They are held too tightly for bonding (sharing with another atom) and they are not usually considered in the bonding picture. They are important because they cancel a portion of the nuclear charge so that the valence electrons only see an effective nuclear charge, Zeffective.

3 Lecture 1 The effective nuclear charge, Zeffective, is the net positive charge felt by the valence electrons (bonding and lone pair electrons). It can be estimated by subtracting the number of core electrons from the total nuclear charge.

=approximate positive charge felt by the valence electronsZeffective = (Ztotal) - (# core electrons)

Zeffective is an important parameter in determining how tightly an atom pulls electrons to itself and in determining the polarity of its bonds. Polarity is one of our more important concepts.

All of the above information about different atoms can symbolically be represented with an atomic symbol and numbers written about it that can show the number of protons, neutrons, electrons an total number of atoms present. Not all of these numbers are listed every time an atom is written, but it is good to know their meaning when they are listed.

Some examples

Atomp + n

p

charge

# of atoms

35Br379

80Hg2+2200

35Br79

35Br28112C6

13C6 8O-216

8O217

7N-314

7N215

How to symbolically represent atoms.

1H12H1

3H1

Problem 1 – Fill in the appropriate values in the following table. Stable isotopes only make up about 10% of all known isotopes. However, because they are stable, these are the isotopes we mostly study and know best.

core valence abundance oratom/Ion protons neutrons electrons electrons Zeff half life

1H 99.99% 2H 0.01% 3H 12.3 years 6Li 7.4%" 7Li 92.6% 11C 20 min 12C 98.9% 13C 1.1% 14C 5730 years 14N 99.63% 15N 0.37 16O 99.76% 17O 0.04% 18O 0.20% 19F 100.0%24Mg+2 78.7%25Mg+2 10.1%26Mg+2 11.2%35Cl 75.5%37Cl 24.5%79Br 50.5%81Br 49.5%

-3

-2

4 Lecture 1 Atomic Orbitals

Atomic orbitals represent regions in space, about an atom, where there is a high probability of finding up to two electrons. The shapes of atomic orbitals are predicted by the mathematics of quantum mechanics. However, we don’t need to understand the mathematics to understand the pictorial shapes predicted for these orbitals. The electrons in these orbitals can be core electrons or valence electrons. The valence electrons are most relevant for us, because they will largely determine the shape of an atom and its chemistry. This is where the bonds, lone pairs and free radical sites will be indicated. The shapes of atoms are important because they contribute to the shape, polarity and chemistry of molecules. The core electrons are important in the sense that they shield the valence electrons from part of the positively charged nucleus and affect the electron attracting power of an atom (later we will refer to this as electronegativity).

s and p atomic orbitals (…most important to organic students)

+z +z

1s atomic orbitalhas a spherical shape, no nodes. This is the orbital used by hydrogenin bonding.

2s atomic orbitalspherical shape,one node

a single 2p atomic orbital, artificially separated from the other two 2p orbitals, dumbell shape at 90o angle to the other two p orbitals,there is a single node at the nucleus

all 2p atomic orbitals together, 2px, 2py, 2pz,when completely filledthe 2p orbitals have spherical symmetry

node = a region in space where the probability of finding electron density goes to zero.

Shaded and white regions of orbitals (above) represent regions with high probability of finding electron density when occupied by one or two electrons. The different colors represent the opposite phase of the wave nature of an electron in such an orbital. Similar phases add electron density together in a constructive manner (bonding), and opposite phases add electron density together in a destructive manner (antibonding). In our course we are mainly concerned with 1s orbitals of hydrogen and 2s and 2p orbitals of carbon, nitrogen, oxygen and fluorine. Also in our course the 3s and 3p valence orbitals of sulfur [n=3] and the larger halogens [n = 3 (Br), 4 (Cl) and 5 (I)] are viewed in a similar manner to the 2s and 2p atomic orbitals of the second row elements. It’s important to us to know where the electrons are because that’s where bonds and lone pairs of electrons will be. Bonds hold atoms together in molecules and there’s a lot of chemistry that occurs because of lone pairs of electrons.

d orbitals look more complicated (…only occasionally invoked in organic chemistry)

lobes in the yz plane

lobes in the xz plane

lobes in the xy plane

lobes in the xy planealong axes

lobes along the z axis

x

y

zdyz dxz dxy dx2-y2 dz2

x x x x

y y y y

z z z z

We rarely have occasion to discuss d orbitals, but we will use them briefly in discussions of sulfur, phosphorous and a few transition metals. Even in those discussions we will mainly look at them as bigger versions of p-like orbitals.

5 Lecture 1 Atomic Configuration

The atomic orbital model is mainly used to describe isolated atoms. It must be modified for atoms bonded to other atoms in molecules, where the model changes slightly by mixing orbitals together in various possible ways (we will mainly use hybridization). In atoms there can be one to many layers of atomic orbitals containing electrons (n = 1, 2, 3, 4, etc.). The Aufbau Principle is the set of rules describing the minimum energy configuration of the electrons around an atom. As mentioned earlier, core electrons are found in completely filled inner electron shells (we will repeat ourselves over and over until it sticks). The core electrons are considered to shield the outermost valence electrons from the positive charge of the nucleus by an amount equal to their total negative charge. The amount of residual positive charge (total nuclear charge minus core electrons) is the effective nuclear charge, Zeffective. Atoms in a column of the periodic table have a similar Zeff, which is mainly what gives them their similar properties and makes them a family.

Valence electrons, in the outermost occupied electron shell, are the ones we are mainly interested in, because they largely determine the observed bonding patterns that we observe. The following figure is a schematic of the first three energy levels. We mainly work with the first energy level (hydrogen) and the second energy level (carbon, nitrogen, oxygen and the halogens). The farther an electron is from the nucleus, the greater its potential energy. Electrons fill into the atomic orbitals in a manner that minimizes their energy about an atom. {Think of stretching a rubber band around your wrist and letting it snap back. Where does the rubber band have more energy: close or far?}

1. Aufbau Principle - Orbitals are filled with electrons in order of increasing energy, from the innermost orbitals (lower energy) towards the outermost orbitals (higher energy).

2. Pauli Exclusion Principle - Only two electrons may occupy any orbital and those electrons must have opposite spins (applicable to s, p, d and f orbitals).

3. Hund's Rule - Electrons entering a subshell containing more than one orbital will spread themselves out over all of the available orbitals with their spins in the same direction, until the subshell is over half filled.

n = 1

n = 2

n = 3

+Z = total nuclear charge = number protons+Z

s

p

d

s

sp

Zeff = residual positive charge felt by the outer most valence electrons = (Ztotal - core electrons)

Problem 2 – Fill in the core electrons, valence electrons, the total nuclear charge and effective nuclear charge. We will use all of the atoms shown below at some point in our discussions, though we will study the atoms in molecules. In molecules we will have to modify the “atomic” pictures below.

s

p

dp

s

s

n=1 n=2 n=3Zeffective =

atomicboron

An atom of boron would be pretty unusual, considering that pure boron sublimes in our world at 2550oC.

s

p

dp

s

s


atomiccarbon

An atom of carbon would be pretty unusual, considering that pure carbon is graphite or diamond (bp ≈ 2550oC).

6 Lecture 1

s

p

dp

s

s


atomicnitrogen

An atom of nitrogen would be pretty unusual, considering that pure nitrogen in our world is an inert diatomic gas.

s

p

dp

s

s


atomicoxygen

An atom of oxygen is also pretty unusual, considering that pure in our world is mainly diatomic oxygen or ozone. Oxygen loves to bond with less electronegative elements, like carbon and hydrogen, which makes it very reactive with organic compounds.

s

p

dp

s

s


atomicfluorine

An isolated atom of fluorine is also pretty unusual. Even more than oxygen, fluorine loves to bond with less electronegative elements and forms some of the strongest bonds with other atoms of any atom in the periodic table. It hates to bond with itself. Pure fluorine is dangerously reactive with almost everything.

s

p

dp

s

s


atomicneon

An isolated atom of neon is about the only way we ever see neon. It prefers its Noble gas configuration to almost any other possibility.

The goal of every atom is to attain a Noble Gas Configuration. We rarely find isolated atoms in nature. Noble gases are examples where this does occur, but they are often referred to as “inert gases” because they almost never react with anything. Occasionally, pure atomic materials are encountered, such as graphite, diamond, gold or silver arranged in lattice structures of uncountable numbers of atoms. Nitrogen and oxygen are also present as pure elemental diatomic molecular gases.

The most common way for atoms to arrange themselves in nature is to imitate the Noble gas configuration of electrons. They can do this by bonding with other atoms, forming either ionic, covalent or metallic bonds (we will not consider metallic bonds). Chemistry is pretty much about how atoms shift from one arrangement to another in this constant quest to look like a Noble gas. The octet rule describes how the second row elements of organic chemistry fill their valence shell. Often non-transition elements of the third through higher rows also obey the octet rule, but there are other ways to bond when there are available d orbitals. We will pretty much stick to the octet rule in this book, with a few exceptions.

7 Lecture 1 Hydrogen is also an exception to the octet rule, following a doublet rule to attain the helium atomic configuration. That gives hydrogen a full n = 1 shell, 1s2. Except for hydrogen, all of the elements have two possible Noble gases they could emulate, one lower and one higher. Atoms can attain Noble gas configurations by losing electrons (becoming cations), gaining electrons (becoming anions) or sharing electrons (forming covalent bonds). They tend to do this in the lowest energy way possible. Atoms on the left edge can more easily lose one or two electrons and are usually found in nature as +1 or +2 cations. Atoms on the right edge can gain electron density either by capturing additional electrons, forming anions, or sharing electron density, forming covalent bonds. Atoms in the middle, mostly tend to form covalent bond, often in a variety of ways. Sitting right in the middle of the second row, carbon is the master of ways of forming covalent bonds and altering its valency. It can form bonds as single, double, triple or two doubles with bonding partners of hydrogen, other carbons, nitrogens, oxygens and halogen atoms. It can be a neutral atom, a cation, an anion or a free radical and it can do all of this in an infinite number of ways. Carbon’s covalent bonds are usually strong enough for its compounds to be stable long enough in our world to be useful in nature and to be studied by organic chemists (biochemists study them too). Yet, they are reactive enough to be broken with an input of energy, and reconfigure the molecular architecture, activities essential for life to exist. Carbon is a really special element!

The figure below shows how hydrogen and the second row elements can attain a Noble gas configuration. Only valence electrons are shown in the diagram below. All of the second row elements have two core electrons shielding the nucleus (1s2), forming an effective nuclear charge, Zeffective, two less than the total nuclear charge, Ztotal. (For second row elements Zeffective = Ztotal – 2.)

H Hegain 1 e-lose 1 e- (almost never)

no Noble gas Zeff = +1

Li NeHegain 7 e-s (never)lose 1 e-

Zeff = +1 Be NeHe

gain 6 e-s (never)lose 2 e-s Zeff = +2 B NeHe

gain 5 e-slose 3 e-s Zeff = +3

C NeHegain 4 e-slose 4 e-s Zeff = +4

N NeHe

gain 3 e-slose 5 e-s Zeff = +5

NeOHegain 2 e-slose 6 e-s (never) Zeff = +6

NeHe

lose 7 e-s (never) gain 1 e-F

Zeff = +7

8 Lecture 1

Part of the rest of the periodic table is shown below. Many of these elements are used in organic chemistry. We will use a number of them in our discussions (those in bold), though often only briefly.

H HeLi Be B C N O F NeNa Mg Al Si P S Cl ArK Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br KrRb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I XeCs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

Usually = +1 +2 ? ? ? -2 -1 0Transition state elements tend to have several oxidation states, mostly positive.

Elements on the left side tend to lose electrons forming cations, while elements on the right side tend to gain electrons forming anions or to share electrons forming covalent bonds. Carbon, in the middle, usually shares electrons in covalent bonds.

1A 2A 3B 4B 5B 6B 7B 8B 1B 2B 3A 4A 5A 6A 7A 8A

A Noble gas configuration is the goal of most atoms.

1. Ionization potential as a measure of an atom’s electron attracting power.

Ionization potential is the amount of energy needed to remove an electron from an atom, molecule or ion and provides a measure of an atom’s electron attracting power. An atom’s electron attracting power is a key idea in establishing the concept of polarity. The curved arrows shown in the equation below may be new to you. In organic chemistry, we use curved arrows to show how key electrons change in a chemical reaction. As electrons leave one location and move to a new location, curved arrows keep track this movement. When you learn how to use curved arrows, you will understand, pretty well, how organic chemistry and biochemistry work. We will include them at every opportunity to maximize your exposure to these new tools. The half-headed arrow in the following equation indicates single electron movement. Later we will show two electron movement with full-headed arrows.

starting energy

final energy

H +314 (He) +568Li +124 Be +215 B +192 C +261 N +335 O +315 F +402 (Ne) +499Na +118 Mg +177 Al +138 Si +189 P +242 S +239 Cl +300 (Ar) +363K +99 Ca +141 Ga +138 Ge +182 As +226 Se +225 Br +273 (Kr) +323

Group 1A Group 2A Group 3A Group 4A Group5A Group 6A Group 7A Group 8A

Energy to ionize an electron from neutral atoms = IP1 (units are kcal/mole). Compare rows and compare columns.

Zeff = +1 Zeff = +2 Zeff = +3 Zeff = +4 Zeff = +5 Zeff = +6 Zeff = +7 Zeff = +8

∆ Energy = Potential Energy

greater PE (less stable)

lower PE (more stable)

Ionization always has an energy cost to strip an electron from an atom.

ionization potential(kcal/mole)

Atom

electron is lost

Atom + electron

Atom

Atom e-

9 Lecture 1

The general trend, as one progresses across a row, is that the ionization potential gets larger and the hold on electrons is stronger. Why? The answer is found in the size of Zeffective. As we move from Li (IP1 = +124) to carbon (IP1 = +261) to fluorine (IP1 = +402), the valence shell stays the same (n = 2), but the effective nuclear charge holding those electrons keeps increasing, from +1 to +4 to +7. Because each extra electron goes into the same shell (n = 2), there is essentially no shielding of the nucleus by any of the additional electrons (the electrons are doing their best to avoid one another). It’s a lot harder to pull an electron away from a +7 charge than a +1 charge. It seems reasonable that fluorine has a stronger attraction for electrons than the other second row elements, even when those electrons are shared in a chemical bond.

The core 1s2 electrons form a -2 shield of the +3 nucleus forming a Zeff of +1 pulling on the single 2s1 valence electron.

+3 +9

The core 1s2 electrons form a -2 shield of the +9 nucleus forming a Zeff of +7 pulling on the seven 2s22p5 valence electrons.

lithium fluorinevalence electron sees a Zeff = 3 - 2 = +1

valence electrons see a Zeff = 9 - 2 = +7

-2 -2

There are a few surprises as one moves across a row (Be B and N O) due to completely filled or half filled subshells (Be = 2s2, N = 2s2,2p3). A good freshman chemistry textbook will explain these exceptions. We will ignore them. Also, it’s easier for lithium to lose a single electron, attaining the helium configuration than to gain seven additional electrons having the neon configuration. On the other hand it’s easier for fluorine to gain a single extra electron, attaining the neon configuration, rather than losing seven valence electrons having the helium configuration.

All of the second row elements have a full n = 1 shell, so there is effectively a minus two shield for the total nuclear charge, Ztotal. The electrons in the second shell see an effective nuclear charge, Zeffective, of (Ztotal – 2).

carbonZtotal = +6Zeffective = (6 - 2) = +44 valence electronsneeds 4 more for Noblegas configuration

+6 +7

nitrogenZtotal = +7Zeffective = (7 - 2) = +55 valence electronsneeds 3 more for Noblegas configuration

boronZtotal = +5Zeffective = (5 - 2) = +33 valence electronsneeds 5 more for Noblegas configuration

+5 -2 -2 -2

+8 +9 +10

oxygenZtotal = +8Zeffective = (8 - 2) = +66 valence electronsneeds 2 more for Noblegas configuration

fluorineZtotal = +9Zeffective = (9 - 2) = +77 valence electronsneeds 1 more for Noblegas configuration

neonZtotal = +10Zeffective = (10 - 2) = +88 valence electronsneeds 0 more, it is a Noble gas

-2 -2 -2

10 Lecture 1

Where the data are available, it appears to get easier to ionize an atom when it is bonded to other atoms. A methyl carbon, CH3, with three hydrogen atoms attached to it has a much lower ionization potential (+229) than a bare carbon atom (+260). Changing one of the methyl hydrogen atoms for another CH3 lowers the ionization potential even more (+201). The extra electrons in the additional bonds evidently help compensate for the loss of an electron in ionization. The more extra electrons there are from additional bonded groups, the easier it is to ionize an electron from that atom. Later in the course we will recognize this property as an electron donating inductive effect.

C

C

H

H

H

C

H

H

H

C

H

H

OHNH

H

electron ionized

C

electron ionized

electron ionized

electron ionized

+260

+229 +261 +296

electron ionized+335


N ON O

NH

HOHC

H

H

H


C

H

H

H

C

H

HOC

H

H

H


OC

H

H

H

electron ionized+402F F

carbon nitrogen oxygen fluorine

Problem 3 – Provide an explanation for the different energy values of the ionization potential of the oxygen series above.

How does the ionization potential of the halogen atoms vary down a column? The entire halogen family has a constant Zeffective of +7.

F

Cl

Br

I

F

Br

I

Cl

e-+

e-+

e-+

e-+

electronionized +402




Largerionization potential

The general trend as one progresses down a column is that the ionization potential gets smaller and the hold on an electron is weaker. Why? As we move from fluorine (+402) to chlorine (+298) to bromine (+272) to iodine (+241), the valence shell changes from n=2 to n=3 to n=4 to n=5. In each case an electron is lost from an atom with Zeff = +7, but the shell it is removed from is farther and farther from the nucleus. In chlorine the work of pulling the electron from n=2 to n=3 has already been done and in bromine the work of pulling the electron from n=3 to n=4 has already been done and in iodine the work of pulling the electron from n=4 to n=5 has already been done. Iodine does not have as strong an attraction for its outer most shell (n=5) as does bromine (n=4). In a similar vein bromine has a weaker attraction for its outermost electron than chlorine (n=3) and chlorine has a weaker attraction than fluorine (n=2). The

11 Lecture 1 trend in any column is that atoms of similar Zeff have a weaker pull for the outer most electrons as one moves down the column, because the valence electrons are farther from the same effective nuclear charge.

The result of these two trends is that atoms on the right and towards the top of the periodic table have the strongest hold on their valence electrons and point to the corner of the periodic table where we find fluorine, the most electronegative atom in the periodic table.

Problem 4 – What is the total nuclear charge and effective nuclear charge for each of the atoms below? How does this affect the electron attracting ability of an atom? H atom Li atom Na atom K atom B atom C atom N atom O atom F atom Ne atom

Ztotal =

Zeff =

Problem 5 - Which atom in each pair below has the larger ionization potential and why? Are any of the comparisons ambiguous? Why? Check your answers with the data in the I.P. table above.

a. S vs Se b. Si vs Al c. N vs F d. S vs Br e. Cl vs O

2. Electron Affinity

The potential energy change when a free electron enters an atom’s, molecule’s or ion’s valence shell is usually lowered (potential energy is released = more negative potential energy = more stable result). Similar trends are observed to those observed in ionization potential, but the energy changes are much smaller in magnitude. Also, when a free electron is captured and a subshell becomes exactly half full (carbon) or completely full (lithium, fluorine), the magnitude of the energy released is somewhat greater than expected. Often, when an electron is added to a second row element, the energy released is slightly smaller in magnitude than expected (e.g. fluorine vs. chlorine), most likely due to greater electron/electron repulsion in the smaller volume of the n=2 shell.

∆ EnergyPotential Energy

greater PE (less stable)

lower PE (more stable)

Energy is usuallylowered when anelectron is capturedby an atom.

starting energy

final energy

electron affinity(kcal/mole)

Atom

electron is gained

Atom

e- Atom

Atom

H -17.4 (He) -Li -14.3 Be - B -6.4 C -29.2 N +4.1 O -33.7 F -78.4 (Ne) -Na -12.6 Mg - Al -10.5 Si -32.0 P -17.1 S -47.8 Cl -83.4 (Ar) -K -11.6 Ca - Ga -6.9 Ge -27.8 As -18.4 Se -46.6 Br -77.6 (Kr) -


Energy released when an electron combines with an atom (units are kcal/mole). Compare rows and compare columns.


Group 2A elementshave a full s subshell

Group 5A elements have a half full p subshell

12 Lecture 1 Problem 6 - Which atom in each pair below has the larger electron affinity and why?

a. S vs Se b. Si vs Al c. N vs F d. Cl vs Br

Both of these properties (IP and EA) suggest that atoms in the upper right corner of the periodic table have the largest attraction for electrons. The rationale in each case is similar: greater effective nuclear charge (in a row) and closer penetration to that charge (in a column).

3. Electronegativity - Why is it important for us?

Electronegativity defines the relative attraction an atom has for electrons in chemical bonds with other atoms. A larger numerical value indicates a larger attraction for electrons in bonds. Many scales have been created to specify electronegativity, but we will consider it to be based on a simplified Mulliken Model. Mulliken’s model uses ionization potential (IP) and electron affinity (EA) as the measure an atom’s attraction for electrons, which is why we covered them, above.

χ (IP + EA) 2Electronegativity = ≈

Mulliken Model of Electronegativity ≈ Average of ionization potential (IP) and electron affinity (EA).

attracting power of atoms for electrons in chemical bonds=

Electronegativity will determine nonpolar, polar and ionic characteristics of bonds, and when shapes are included it determines the same attributes in molecules. Compare rows and compare columns in the following table. The largest numbers are found in the upper right corner and the smallest numbers are found in the lower left corner. Group 8A elements don’t have a value listed because they don’t make bonds.

H 2.2 (He) -Li 1.0 Be 1.5 B 1.8 C 2.5 N 3.0 O 3.5 F 4.0 (Ne) -Na 0.9 Mg 1.2 Al 1.5 Si 1.9 P 2. 2 S 2.6 Cl 3.2 (Ar) -K 0.8 Ca 1.0 Ga 1.6 Ge 1.9 As 2.0 Se 2.4 Br 3.0 (Kr) - I 2.7



Table of electronegativities.

When two atoms share electrons in a covalent bond, there are two possibilities. Either the electrons are shared evenly (a pure covalent bond) or they are not shared evenly (a polar covalent bond). Elemental hydrogen, oxygen and nitrogen are examples of atoms sharing electrons evenly because the two atoms competing for the bonded electrons are the same. Each line drawn between two atoms symbolizes a two electron bond. These examples also illustrate that pure covalent bonding can occur with single, double and triple bonds.

H H OO NNhydrogen molecules with a

single bond, each H has a duet of electrons (like helium)

oxygen molecules with a double bond, each O has an octet

of electrons (like neon)

nitrogen molecules with a triple bond, each N has an octet

of electrons (like neon)

If the two bonded atoms are not the same, then there will be different attractions for the electrons. One atom will have a greater pull for the electrons and will claim a greater portion of the shared electron density. This will make that atom polarized partially negative, while the atom on the other side of the

13 Lecture 1 bond will be polarized partially positive by a similar amount. Because there are two opposite charges separated along a bond, the term “dipole moment” (µ) is used to indicate the magnitude of charge separation. Bond dipoles depend, not just on the amount of charge separated, but also the distance by which the charges are separated, as indicated by their bond lengths. Electronegativity is a concept that explains the direction and degree of polarity. (See J. Chem. Ed., vol. 82, p. 325, 2005 for an interesting discussion of electronegativity used to explain covalent, ionic and metallic bonds).

The symbols + and - represent qualitative charge separation forming a bond dipole. Alternatively, an arrow can be drawn pointing towards the negative end of the dipole and a positive charge written at the positive end of the dipole.

A B A B

d d

µδ+ δ-

or...........

Two qualitative pictures of a bond dipole are represented below. B is assumed to be more electronegative than A.

µ =amount of charge separated

x distance between charges in cm

= units of Debye (D = 10-18 esu-cm)

µ = (e)(d) = dipole moment

e = electrostatic charge (sometimes written as q)

d = distance between the opposite charges

The abosolute value of a unit charge on an electron or proton is 4.8x10-10 esu

This is often given in angstrums, but converted to cm for use in calculations (1A = 10-8 cm, A = angstrum)

The quantitative measure of dipole moment requires that one know the magnitude of the charges separated (the absolute value of a full positive or negative charge corresponds to 4.80 x 10-10 esu) and the distance that they are separated (usually in cm). The units of dipole moments are usually listed in Debye = D (1D = 10-18 esu-cm). A typical example might be 0.1 unit of charge separated by 1.5 A (A = angstrum = older unit on the order of a bond length, 1.5 x10-8 cm in this example) as shown in the calculation below.

µ = (q)(d) = (magnitude of charge)(distance apart) = dipole moment

µ = (0.1) x (4.80 x 10-10 esu) x (1.5 x 10-8 cm) = 0.72 x 10-18 esu-cm = 0.72 D

Quantitative calculation of dipole moment from partial charge information. The partial charge is assumed to be 0.1.

partial charge

full unit charge

distance between charges

14 Lecture 1 Because there are two factors that make up bond dipole moments (charge and distance), bonds that appear to be more polar based on electronegativity differences might have similar dipole moments if they are shorter than less polar bonds. The methyl halides provide an example of this aspect.

H3C X H3C X

d d

µδ+ δ-

or...........

µ = (partial charge)(4.8x10-10 esu)(distance in cm)

CH3 = methyl

X = halogen atom (F, Cl, Br, I)

F is more electronegative, but smaller

I is less electronegative, but larger

Depending on the magnitude of the difference in electronegativity between the bonded atoms, we can classify the polarity or charge separation, qualitatively, as nonpolar bonds (slight difference in electronegativity), as polar bonds (moderate difference in electronegativity) or as ionic bonds (large difference in electronegativity). One arbitrary calculation used for making these distinctions is shown below. We will use this calculation to classify bond polarity, even though many exceptions are known.

χ∆ = χa χb ≤

≥

<

_

χ∆ = χa χb_

χ∆ = χa χb_

0.4

1.4-2.0

1.4-2.0

classify bond as nonpolar covalent

classify bond as polar covalent

classify bond as ionic

atom a

atom b

bond

χa χb

χ∆ = χa χb_

χa symbolizes the electronegativity of atom a.

Problem 7 – The following examples illustrate many variations of polar covalent bonds. These can occur as single, double or triple bonds between different atoms having moderate to significantly different electronegativities. Use the “special” polarity arrow to indicate if and in what direction bond dipoles exist. The atom with the greater electron attracting power (greater electronegativity) will have a partially negative charge leaving the less electronegative atom with a partial positive charge. This idea of polarity will be an important factor when we begin to study how chemical reactions occur. Similar charges will repel and opposite charges will attract. Attraction will bring key atoms close to one another and allow for electron transfer producing chemical change. Shape is a crucial factor in determining the dipole moment of an entire molecule, but we will defer that aspect until our discussion of physical properties.

NH

H

HOH

FHCH

H

H

H

CH NC

H

H

O

CH

H

H

OCO O

H

H

a b c d

hgfe

15 Lecture 1

Types of Bonding

Ionic bonding

The elements at the edges of the periodic table tend to lose or gain valence electrons, depending on which side they are. A complete transfer of electrons forms ions (cations lose electrons and anions gain electrons). Neutral salts require charge balance: (total negative charge) = (total positive charge), so that the net charge is zero.

If one were to mix diatomic chlorine gas, Cl2, with metallic sodium, Na, quite likely an explosion would occur in a violent transfer of electrons. The metallic sodium atoms would give up their electrons to the chlorine diatomic molecules, breaking the covalent bond between the chlorine atoms. In so doing, each element would attain the desired Noble gas configuration as oppositely charged ions and a large amount of energy would be released (the lattice energy), as the reaction is exothermic. The metallic lattice structure of sodium atoms would disintegrate and the gaseous chlorine would disappear. The ions formed would surround themselves with opposite charge and avoid similar charge, forming an ionic lattice of table salt. There is a dramatic change in physical and chemical properties when this reaction occurs.

Na

Na

ClClNa

NaCl

table salt lattic structure,both Na and Cl ionsattaine a Noble gas configuration

"explosion?"with transfer of electrons Cl

χ∆ = χa χb_ = 3.2 - 0.9 = 2.3

Na (metal)mp = 98oCbp = 883oC

NaCl (salt)mp = 801oCbp = 1413oC

Cl2 (gas)mp = -101oCbp = -35oC

Our arbitrary rules classify this difference in electronegativity as ionic.

(curved arrows show electron movement)

Ionic substances, in general, tend to have “omni directional bonding” (omni = “all”), which tends to produce strongly bonded lattice structures that are difficult to break down, leading to high melting and boiling points. There are many types of lattice structures, depending on the size of the charges and the size of the ions. We will simplistically represent all ionic substances with the figure below.

++

++

++

+

Each ion is surrounded on many sides by oppositely charged ions. To introduce the disorder of a liquid (melt) or a gas (boil) requires a very large input of energy (mp indicates the amount of energy required to breakdown the ordered lattice structure and boiling point indicates the amount of energy required to completely remove an ion pair from the influence of the lattice structure). Ionic bonds (ionic attractions on all sides) can only be broken at a great expense in energy.

Lattice structure - depends onthe size and charge of the ions.

+

+

+ +

+

+

+

+

+

+

+

+

+

+

16 Lecture 1 Examples Melting point (oC) Boiling point (oC) ∆χ NaCl 801 1465 2.3

Na2O 1275 sub 2.5 NaOH 318 1390 2.5? sub = sublimation Na3N 300 dec 2.1 dec = decomposed AlN >2200 1.4* FeCl2 674 sub 1.4* Notice the very high melting FeCl3 306 315 dec 1.2* points and boiling points of MgCl2 714 1412 1.9* ionic substances. MgO 2800 3600 2.1 Mg3(PO4)2 1184 2.1? NH4Br 452 ? NH4NO3 170 210 ? * = exception to our electronegativity rules about bond polarity

Problem 8 - Predict the formula for the combination of the following pairs of ions. What kinds of melting points would be expected for these salts?

F NO2 NO3 O -2 HCO3 PO4 -3

K

Ba+2

Zn+2

Al+3

Covalent Bonding (Molecules)

A single neutral hydrogen atom would have a single valence electron. This is not a common occurrence in our world because such a hydrogen atom would be too reactive. However, if one were able to generate a source of hydrogen atoms, they would quickly join together in simple diatomic molecules having a single covalent bond with the two hydrogen atoms sharing the two electrons and attaining the helium Noble gas configuration. Such a reaction would be very exothermic.

χ∆ = χa χb_ = 2.2 - 2.2 = 0

H H HH

The line symbolizes a two-electron, pure-covalent bond based on the calculation below.

If two hydrogen atoms should find one another, they wouldfrom a diatomic molecule with a tremendous release of energy,about 104 kcal/mole.

The atoms of organic chemistry, H, C, N, O, S and halogens, tend to attain a Noble gas configuration by sharing electrons in covalent bonds of molecules. Simple formulas often have only one choice for joining the atoms in a molecule (CH4, NH3, H2O, HF). As the number of atoms increase,

17 Lecture 1 however, there are many more possibilities, especially for carbon structures (sometimes incredible numbers of possibilities!). These possibilities may require single, double and/or triple bonds (σ and π bonds will be discussed soon), or rings of atoms. Carbon, nitrogen, oxygen and fluorine can be bonded in all combinations, according to their valencies. We study most of the bond types below.

C

H

H

H

H NH

H

H OH H FH

χ∆ = χa χb_

= 2.5 - 2.2 = 0.3

pure covalent bond based on the calculation below

χ∆

χ∆ = χa χb_

= 3.0 - 2.2 = 0.8

polar covalent bond based on the calculation below

χ∆

χ∆ = χa χb_

= 3.5 - 2.2 = 1.3


χ∆

χ∆ = χa χb_

= 4.0 - 2.2 = 1.8


χ∆

C

H

H

H

H C

H

H

H

C

H

H

H

C

H

H

H

N H

H

C

H

H

H

O

H

C

H

H

H

F

N

H

H N H

H

OH N H

H

OH O H NF

F

F O

F

F

single bonds

C C

H

H

H

H

C N

H

H

H

C O

H

H

N O

H3C

N N

H

H

O O

double bonds

two double bonds

C C

H

H

OC C

H

H

C

H

H

O C OC C

H

H

N

H

C C HH C NH N N

triple bonds

C C HC

H

H

H

C NC

H

H

H

18 Lecture 1 rings

H2C

H2CCH2

NH

H2C

H2C

H2C

CH2

CH2

CH2

CH2

H2C

H2C

H2CO

C C

C

CC

C

HH

H

H

H

H

Problem 9 – Supply dipole arrows to any polar bonds above (according to our arbitrary rules). Make sure they point in the right direction.

Carbon tends to share electron density with 2-4 atoms to form its octet. Bond energies to carbon also tend to be strong, which leads to an infinite variety of possible stable chains and stable rings with itself and other atoms. Hydrogen and the halogen atoms only form one bond with carbon, when they are present, and are found covering the surface of the chains and rings (they are like the skin of a molecule). Oxygen atoms and nitrogen atoms, on the other hand, form two and three bonds, respectively. They can be found in the interior of chains and rings or on the surface of chains and rings. The physical properties (mp, bp, etc.) of molecular substances are very different from ionic substances. The strong attractions in covalent bonds do not change when a substance melts or boils since none of those bonds break. In contrast to ionic salts, there are much weaker forces that attract one molecule to another in covalent substances, and these will be discussed later.

Examples Mol. Wt. Melting point (oC) Boiling point (oC) NaCl (ionic) 58.4 +801 +1465 (ionic salt for comparison)H-H 2.0 -259 -253 CH4 16.0 -182 -164 NH3 17.0 -78 -33 H2O 18.0 0 100 HF 20.0 -83 +20 CH3CH3 30.0 -183 -89 CH3NH2 31.0 -94 -6 CH3OH 32.0 -98 +65 CH3F 34.0 -142 -78 Absolute zero = 0 K = -273oC

molecular substances

∆T = 80oC

∆T = 131oC∆T = 133oC

Most of the time when we see numbers such as the temperatures in the above table, our eyes glaze over and we move on to the next blurry group of words. But this time, let’s do a thought experiment to give those temperatures more meaning. Let’s fill two imaginary pots with water. One we merely set in front of us and one we put over our imaginary stove burner and bring to a boil. The pot in front of us, at room temperature, is about 25oC, while the pot boiling on the stove is about 100oC, a mere 75oC higher. Now for the thought experiment: stick your imaginary hand into each of the pots of water. What? You say I’m crazy? You know that 75oC is a huge difference in temperature. The imaginary cold water does nothing to your imaginary hand, while the imaginary boiling water cooks it. If you want to try this as a real experiment, substitute a hot dog for your hand. Now, think about what the much larger differences in temperature in the table above are telling us while you are eating your real hot dog.

19 Lecture 1 Ions with covalent bonds

Sometimes there is a mix of covalent and ionic bonds. Anions and/or cations can be connected with covalent bonds and still have a net overall negative or positive charge, and both charges are even possible in the same molecule (e.g. amino acids). The following ions provide additional examples of ions with covalent bonds from your prior chemistry course.

NO

O

NO

O

O

S

O

O

O S

O

O

OO C

O

O O

Cl

O

O

OO Cl OCl

O

OO Cl OO C

O

O OH

CN NH

H

OH O

H

H H N

H

H H

H

P

O

O

OO

nitrite nitrate sulfite sulfate carbonate

perchlorate chlorate chlorite hypochlorite bicarbonate phosphate

cyanide amide anion hydroxide hydronium ion ammonium ion

20 Lecture 1 Problem 1 – Fill in the appropriate values in the following table. Stable isotopes only make up about 10% of all known isotopes. However, because they are stable, these are the isotopes we mostly study and know best.

atom/ion protons neutrons electrons electrons Zeff or halflifeabundancecore valence

1H2H3H+

6Li7Li+11C12C13C14C14N15N-3

16O17O18O-2

19F24Mg+2

25Mg+2

26Mg+2

35Cl-37Cl-79Br-

81Br-

99.99%.0.01%12.3 years7.4%92.6%20 min98.9%1.1%5730 years99.63%0.37%99.76%0.04%0.20%100.0%78.7%10.1%11.2%75.5%24.5%50.5%49.5%.

11133666677888912121217173535.

1233456787889101012131418204446.

00022222222222210101010101818.

1111044445866870008888.

1111144445566672227777.

Problem 2 – Fill in the core electrons, valence electrons, the total nuclear charge and effective nuclear charge. We will use all of the atoms shown below at some point in our discussions, though we will study the atoms in molecules. In molecules we will have to modify the “atomic” pictures below.

s

p

dp

s

s


atomicboron

core electrons = 1s2

valence electrons = 2s2 2p1

s

p

dp

s

s


atomiccarbon

+5 +6

+3 +4core electrons = 1s2


21 Lecture 1

s

p

dp

s

s




s

p

dp

s

s


+7 +8



atomicnitrogen

atomicoxygen

s

p

dp

s

s




s

p

dp

s

s


+9 +10



atomicfluorine

atomic neon

Problem 3 – Provide an explanation for the different energy values of the ionization potential of the oxygen series above.

O O OH C

H

H

H

O+315 +296 +285

OH OC

H

H

H

+ e- + e-+ e-

It's easier for oxygen to give up an electron when there are more electrons in its valency. It can pull electron density toward itself from the greater number of nearby electrons in bonds.

Problem 4 – What is the total nuclear charge and effective nuclear charge for each of the atoms below? How does this affect the electron attracting ability of an atom?

H atom Li atom Na atom K atom B atom C atom N atom O atom F atom Ne atom

Ztotal = +1 +3 +10 +19 +5 +6 +7 +8 +9 +10

Zeff = +1 +1 +1 +1 +3 +4 +5 +6 +7 +8

The higher the effective nuclear charge, the stronger the attraction for electrons.

22 Lecture 1 Problem 5 - Which atom in each pair below has the larger ionization potential and why? Are any of the comparisons ambiguous? Why? Check your answers with the data in the I.P. table.

Se

S

SiAl

N F

Br

Cl

O

S

Elements in the same column (S,Se) have the same Zeff, but the one higher up (S) has its valence electrons closer to that Zeff so they are harder to ionize an electron from sulfur.

Both Al and Si are in the same row and use the same n=3 valence shell. Si has a Zeff = +4 (like carbon) and holds onto its electrons tighter than Al with a Zeff = +3.

Both N and F are in the same row and use the same n=2 valence shell. F has a Zeff = +7 and holds onto its electrons tighter than N with a Zeff = +5.

Br is farther to the right and has a larger Zeff = +7 than S (Zeff = +6). However, S is higher in its column so its valence electrons are closer to its nucleus. This one is ambiguous. The data shows IPS = 239 and IPBr = 272 so Br attracts its valence electron more strongly.

Cl is farther to the right and has a larger Zeff = +7 than O (Zeff = +6). However, O is higher in its column so its valence electrons are closer to its nucleus. This one is ambiguous. The data shows IPO = 314 and IPCl = 299 so O attracts its valence electron more strongly.

Problem 6 - Which atom in each pair below has the larger electron affinity and why?

Se

S

SiAl

N F

Br

S

Elements in the same column (S,Se) have the same Zeff, but the one higher up (S) has its valence electrons closer to that Zeff. An electron is attracted more strongly to sulfur (-48) than to selenium (-47).

Both Al and Si are in the same row and use the same n=3 valence shell. Si has a Zeff = +4 (EA = -32) and attracts electrons more strongly than Al with a Zeff = +3 (EA = -10).

Both N and F are in the same row and use the same n=2 valence shell. F has a Zeff = +7 (EA = 0) and attracts electrons more strongly than N with a Zeff = +5 (EA = -78).

Br is farther to the right and has a larger Zeff = +7 than S (Zeff = +6). However, S is higher in its column so its valence electrons are closer to its nucleus. This one is ambiguous. The data shows EAS = -48 andEABr = -77 so Br attracts its valence electron more strongly, probably because it acquires a Noble gas configuration.

, probably because it acquires a Noble gas configuration

23 Lecture 1 Problem 7 –Use the “special” polarity arrow to indicate if and in what direction bond dipoles exist.

NH

H

HOH

FHCH

H

H

H

CH NC

H

H

O

CH

H

H

OCO O

H

H

a b c d

hgfe

no polar bonds

bond dipoles cancel Problem 8 - Predict the formula for the combination of the following pairs of ions. What kinds of melting points would be expected for these salts?

KF KNO2 KNO3 K2O KHCO3 K3PO4

BaF2 Ba(NO2)2 Ba(NO3)2 BaO Ba(HCO3)2 Ba3(PO4)2

ZnF2 Zn(NO2)2 Zn(NO3)2 ZnO Zn(HCO3)2 Zn3(PO4)2

AlF3 Al(NO2)3 Al(NO3)3 Al2O3 Al(HCO3)3 AlPO4 Problem 9 – Supply dipole arrows to any polar bonds (according to our arbitrary rules). Make sure they point in the right direction.

C

H

H

H

N H

H

C

H

H

H

O

H

C

H

H

H

F

N

H

H N H

H

OH N H

H

OH O H NF

F

F O

F

F

C N

H

H

H

C O

H

H

N O

H3C

N N

H

H

C C

H

H

OC C

H

H

N

H

H2C

H2CCH2

NH

H2C H2C

H2CO

C NH C NC

H

H

H

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