Chapter 2 Chemical Bonding The interaction between atoms that leads to a rearrangement of the electrons to a more stable state is what we define as chemical bonding. All atoms, except those of the noble gases*, readily engage in chemical bonding either with atoms of their own kind (in elements) or with atoms of a different kind (in compounds). *) This statement is not absolutely true. Some of the heavier noble gases such as xenon and krypton can be persuaded to form compounds with very strong oxidizers. Compounds of helium and neon are presently not known. Three types (extremes) of chemical bonds exist: –

Ionic bond - electrostatic forces hold together oppositely charged ions.

–

Covalent bond - two atoms sharing electron pair(s). Electrons are mutually attracted by the nuclei of adjacent atoms.

–

Metallic bond -is a complicated type of bond. At the simples level metal atoms are envisaged of being metal cations in a “sea” of electrons. The electrons are shared between all the ions simultaneously. A better description of metals is obtained from the band theory.

Ionic bond Ionic bonds form between oppositely charged ions, generally, but not always, between metals and nonmetals. Metals versus Nonmetals Metals are conductors of heat and electricity, have a lustre (shiny) appearance, are ductile (can be drawn in to wirers) and are malleable (can be hammered out into thin foils). With the exception of hydrogen, which is a non-metal, metals are located to the left in the periodic table. Non-metals are generally non conductors, do not have a metallic lustre and are brittle. They are located to the right in the periodic table. The elements near the “step ladder” have both, metallic and non-metallic properties and are called metalloids or semi metals.

1

2

3

4

5

6

7

1 1 H 3 Li 11 Na 19 K 37 Rb 55 Cs 87 Fr

2

4 Be 12 Mg 20 Ca 38 Sr 56 Ba 88 Ra

* **

57 La 89 Ac

3

4

5

6

21 22 23 24 Sc Ti V Cr 39 40 41 42 Y Zr Nb Mo 57* 72 73 74 La Hf Ta W 89** 104 105 106 Ac Rf Db Sg 58 Ce 90 Th

59 Pr 91 Pa

60 Nd 92 U

61 Pm 93 Np

7

8

9

10

11

12

13

14

15

16

17

6 C 14 Si 32 Ge 50 Sn 82 Pb

7 N 15 P 33 As 51 Sb 83 Bi

8 O 16 S 34 Se 52 Te 84 Po

9 F 17 Cl 35 Br 53 I 85 At

25 26 27 Mn Fe Co 43 44 45 Tc Ru Rh 75 76 77 Re Os Ir 107 108 109 Bh Hs Mt

28 Ni 46 Pd 78 Pt

29 Cu 47 Ag 79 Au

30 Zn 48 Cd 80 Hg

5 B 13 Al 31 Ga 49 In 81 Tl

62 Sm 94 Pu

65 Tb 97 Bk

66 Dy 98 Cf

67 Ho 99 Es

68 69 70 71 Er Tm Yb Lu 100 101 102 103 Fm Md No Lr

63 Eu 95 Am

64 Gd 96 Cm

To understand chemical bonding trends in three atomic properties must be considered: electron affinity, ionization energy and atomic radius

Atomic radius Atomic radius is difficult to define – there is a small probability of finding an electron at an infinite distance from the nucleus. Often atomic radius is defined in terms of internuclear distance in a compound. Methods of representing Size covalent radii: half of the internuclear distance in a homonuclear X-X single bond. ionic radii:

The ionic radius, rion, is the size of an ion (due to changes in effective nuclear charge; for cations, rion < rcov; whilst for anions, rion > rcov). These values are estimated from electron density maps determined from crystallographic data.

van der Waals Radii:

half of the distance of closest approach of two non-bonded atoms. The forces holding the atoms together are weak London Dispersion forces (more to come…).

18 2 He 10 Ne 18 Ar 36 Kr 54 Xe 86 Rn

Source: petrucci Periodic Trends in Atomic Radii Trends in atomic radii can be rationalized if we consider the shielding or screening effect of the core electrons.

Effective nuclear charge: Zeff = Z – S Z = atomic number, S = number of inner electrons that screen an outer electron The more electronic shells in an atom, the larger the atom. Atomic radius increases from top to bottom in a group. The atomic radius decreases from left to right through a period of elements.

Class exercise: Arrange the following atoms in order of increasing atomic radius: Mg, N, Rb, Be

Source Petrucci Ions Ions are charged atoms or groups of atoms •

Ionic Radius: Cations are smaller than the atoms from which they are formed. Isoelectronic cations have the same number of electrons in identical configurations. Na+ and Mg2+ both have closed shells (both have 10 electrons but the magnesium nucleus has more protons to attract the 10 electrons then the sodium nucleus). For isoelectronic cations, the more positive the ionic charge, the smaller the ionic radius.

•

Anions are larger than the atoms from which they are formed. For isoelectronic anions, the more negative the charge, the larger the ionic radius.

Source Petrucci

Class exercise: Arrange the following ions in order of increasing ionic radius: Al3+, Cl-, Na+, F-

Electron affinity and ionization energy: As we have seen an atom can gain one electron or more. This fundamental atomic property is known as electron affinity and is usually exothermic. H + e- → H- + energy An atom can also lose one electron (or more). This fundamental atomic property is called ionization or ionization energy. Atoms are stable species. It requires energy to release an electron from an atom so ionization is an endothermic process and ionization energies are positive H + energy → H+ + eIt is important to understand that electron gain and ionization are truly atomic properties. They are properties of isolated gas phase atom (not bulk phase atoms as in a chunk of copper). Metals tend to lose electrons and form cations →

Mg+(g) + e-

I1 = 738 kJ/mol

Mg+(g) →

Mg2+(g)+ e-

I2 = 1451 kJ/mol

Mg(g)

I1 is called the first ionization energy, I2 the 2nd ionization energy. Trends of ionization energy within a group in the periodic table: As you go down a group (chemical family) the ionization energy decreases. It is not difficult to understand the rationale behind that. With every new row (period) in the periodic table a new electronic shell is being filled with electrons (atoms get bigger!) The further away the electron is from the nucleus, the smaller is its attractive interaction with the nucleus and the easier it is to remove it. Chemically speaking, the easier it is to remove the electron the greater the reactivity of the element toward an electron accepting element (an oxidizer). The table below lists the first ionization energies of the alkaline metals (group 1 in the periodic table) Element Lithium Sodium Potassium Rubidium Cesium

Radius in pm 152 186 227 248 265

I1 in kJ/mol 520.2 495.8 418.8 403.0 375.7

Trends of ionization energy within a period: As you go from the left to the right ionization energy increases. Electrons enter the same shell within a perioid. This leads to an increase in repulsion between the electrons in the outermost electronic shell. At the same time as the number of electrons increases so does the number of protons in the nucleus and the greater the positive charge of the nucleus the

greater the attraction towards the electrons. It is this increased attraction that dominates over electron repulsion. Thus ionization energies generally increase from the left to the right within a period. We will take a closer look at the exceptions later. This nucleus having a “tighter grip” on the outermost electrons also means that atomic radius decreases from the left to the right in a period. Strictly speaking this is only true for gas phase atoms. In the bulk phase elements show different bonding arrangements that might stabilize the atoms and make them less prone to react e.g. the reactivity of the 3rd row elements toward O2 decreases from Na to Si but hugely increases at P! Why? The table below lists ionization energies of the 3rd row elements in KJ/mol

I1 I2 I3 I4 I5 I6 I7

Na 496 4562

Mg 738 1451 7733

Al 578 1817 2745 11580

Si 787 1577 3232 4356 16090

P 1012 1903 2912 4957 6274 21270

S 1000 2251 3361 4564 7013 8496 27110

Cl 1251 2297 3822 5158 6542 9362 11020

Ar 1521 2666 3931 5771 7238 8781 12000

Consider the reactivity of the following elements Na and K as well as Na and Mg!

There are two more points worth noting in this table. First there is a huge increase in ionization energy after the last valence electron has vacated the valence shell and a core electron from the inner electronic shell is removed e.g. it takes 496 kJ /mol to remove one electron from sodium but it takes 4562 kJ/mol to remove a second electron which is why no compounds are known that contain Na2+ ions but essentially all compounds of sodium contain Na+ ions. Second, by the time you get to the middle of the 3rd period (e.g. Si) ionization energies become quite large and it is less feasible for atoms to lose electrons. Nonmetals which are located to the right in the periodic table do not lose but tend to gain electrons and form anions. We called this atomic property electron affinity and the following trend is observed within the periodic table: Electron affinity generally increases from the left to the right within a period and decreases from the top the bottom in a group. Chemically speaking this means the element most eager to gain an electron is the most reactive non metal element which is fluorine. Fluorine reacts with most elements in the periodic table. Class exercise: Arrange the following atoms in order of increasing ionization energy: Si, F, Na, O

Arrange the following atoms in order of increasing electron affinity: Cl, As, P, Ba

Octet Rule In taking a closer look at the number of electrons that metals lose and nonmetals gain it becomes apparent that many try to obtain the same electron count as their nearest noble gas, especially the elements close to the right or the left. So in compounds group 1 elements such as sodium or potassium always lose just one electron, never two or more. In doing so they obtain the same number of electrons as the nearest noble gas. Similarly in compounds with metals group 17 elements such as fluorine or bromine always gain just one electron never two or more to achieve a noble gas electronic configuration (more to come).

1 2

3

4

5

6

1 2 +/1 H 1+ 2+ Li Be 1+ 2+ Na Mg 1+ K 1+ Rb 1+ Cs

2+ Ca 2+ Sr

3

4

5

6

7

8

9

10

11

12

13

14

3+ Al 3+ Sc

4+ Ti

2,3,4, 5+

3, 6+

2,4, 7+

2,3, 6+

V

Cr

Mn

Fe

-1 to +7

2+ Co

2+ Ni

3,4,6, 8+

1,2+ Cu 1+ Ag

2+ Zn 2+ Cd

2+ 3,4+ 2,3,4+ 1,3+ 1,2+ Ba Re Os Ir Pt Au Hg Common oxidation states/oxidation numbers of selected elements

15

16

17

3N 3P

2O 2S

1F 1Cl

2Se 2Te

1Br 1I

Ga 2,4+

Sn 1,3+

2,4+

3,5+

Tl

Pb

Bi

Since metals tend to lose electrons and non-metals tend to gain electrons combining a metal and a nometal leads to a chemical reaction in which electrons are completely transferred from the metal to the non-metal resulting in metal cations and non-metal anions that feel a strong electrostatic attraction. This might be represented by the equation below, which uses Lewis dot formulas or Lewis structures including the number of valence electrons (electrons in the outermost electronic shell) as dots around the element symbol.

Li

+

F

Li

+

F

Compounds between metals and non-metals are called salts (e.g. sodium chloride, magnesium bromide) or ionic compounds (e.g. magnesium oxide, copper sulfide). Many important materials are ionic including ordinary table salt, ceramics, many ores and gypsum. There is a fundamental difference between an atom and an ion. Sodium ions cannot be isolated and placed in a jar on a shelf like sodium atoms which come in form of bars or chunks of metal under

mineral oil in a tin can. They are associated with a counter ion (chloride, fluoride, sulfate) to preserve overall electrical neutrality. A sodium ion is a stable species abundant in large quantity in the human body. Sodium metal is a reactive element that vigorously reacts with water to produce hydrogen and sodium hydroxide, a strong base. Class exercise Write balanced chemical equations for the reaction of:

sodium with bromine

magnesium with chlorine

aluminum with sulfur

strontium with fluorine

hydrogen with calcium

nitrogen with lithium

Remember that electrical neutrality must be preserved in the final product and that the charges of the ions cancel properly.

Energetic of ionic bonding As we have seen it requires 496 kJ/mol to remove electrons from sodium (first ionization energy of sodium). We get 349 kJ/mol back by giving the electrons to chlorine (electron gain is exothermic). But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic! What is as of yet unaccounted for is the electrostatic attraction between the newly formed ions. Negative anions are attracted to positive cations. The result is an ionic bond and a three-dimensional crystal lattice of anions and cations. The binding energy might be calculated from the following experimental data (Born Haber Cycle:

Na+(g) + e- + Cl(g)

Na+(g) + Cl-(g) Na(g) + Cl(g)

+496 kJ

Na(g) + 0.5Cl2(g)

+122 kJ

Na(s) + 0.5Cl2(g)

+107 kJ

Start

-787 kJ

-411 kJ

NaCl(s) End

-349 kJ

Step 1 (Start) Sublime one mole of sodium atoms Na(s) → Na(g)

+107 kJ/mol

Step 2 Dissociate half a mole of chlorine molecules 0.5 Cl2(g) → Cl(g)

+122 kJ/mol

Step 3 Ionize one mole of sodium atoms Na(g) → Na+(g)

+496 kJ/mol

Step 4 Convert one mole of chlorine atoms to one mole of chlorine ions Cl(g) + e- → Cl-(g)

-349 kJ/mol

Step 5 Combine one mole of chloride ions with one mol of sodium ions to form one mole of sodium chloride Na+(g) + Cl-(g) → NaCl(s)

(lattice energy unknown)

Step 6 (End) Measure the heat of formation for the overall reaction Na(s) + 0.5 Cl2(g) → NaCl(s)

-411 kJ/mol

The energy change between reactants and products is independent of the path that the reaction takes (more to come in thermochemistry). The heat of the reaction that we obtain by directly combining the elements should thus be equal to the energy change if we take the reactants clockwise through the Born-Haber cycle because we end up with the same product in the same state. -411 kJ/mol = 107 kJ/mol + 122 kJ/mol + 496 kJ/mol +(-349 kJ/mol) +x kJ/mol x = -787 kJ/mol As you can see it is the hugely negative lattice energy that is largely responsible for the exothermic reaction between the elements sodium and chlorine that releases 411 kJ/mol. Relationship between Bonding and Properties Now why is an understanding of chemical bonding important? There is a relationship between structure and properties of compounds (materials!) If you know and understand the structure of a material, the bonding at the atomic level, you can understand and often predict its physical properties. This is why chemists have been so successful in engineering materials with properties tailored to specific applications. Ionic compounds have very largely negative lattice energies. This tells us that they are very stable with respect to the elements from which they have formed and explains why ionic compounds are usually hard materials (ceramics, “rock salt”). Another property of ionic compounds is their very high melting

and boiling points which is related directly to the high lattice energy. Although anions and cations can and do form ion pairs in solution and in the gas phase in the solid state they form crystalline solids which can be described by a three dimensional crystal lattice. The lattice energy is defined as the energy given off when oppositely charged isolated gaseous ions come together to form one mole of ionic solid. The greater the lattice energy the harder it is to vaporize an ionic compound. Alumina and magnesia have such high lattice energies that they make excellent materials for furnace bricks or crucibles. Zirconium oxide on top of having high melting and boiling points has exceptional fracture toughness along with very good chemical resistance. Another ceramic, zirconium nitride is a cement like refractory material. A coating of zirconium nitride is commercially applied on parts that are subject to high wear and tear, corrosive environments or to improve refraction. Ta4HfC5 is presently holding the record with the highest melting point of any compound at 4215 °C.

Class exercise Given below are calculated lattice energies as well as measured melting points, boiling points and aqueous solubility of selected ionic compounds from Handbook of Chemistry and Physical 90th Edition 2009. Rationalize the trends in lattice energies, melting and boiling points.

Compound LiF NaCl KBr RbI

Compound Na2O MgO Al2O3 ZrN

Lattice energy (kJ/mol) 1,049 787 691 632

Melting point (°C) 848 801 734 656

Boiling Point (°C) 1673 1465 1435 1300

Solubility (g/100mL H2O) 0.13 36 67 165

Lattice energy (kJ/mol) 2481 3795 15916 7633

Melting point (°C) 1134 2825 2054 2952

Boiling Point (°C) decomposition 3600 2977 _____

Solubility (g/100mL H2O) decomposition insoluble insoluble insoluble

Picture source: Wikipedia The sodium chloride structure in which each sodium ion is surrounded by six chloride ions and each chloride ion is surrounded by six sodium ions in octahedral geometry is adopted by many other compounds which have a stoichiometry of 1:1 and where anions and cations don’t differ to greatly in ionic radius.

Covalent bond For some elements it is not feasible to gain the number of electrons required to reach an octet. e.g. carbon would have to gain 4 electrons to have the same number of electrons as neon. Carbon (and many nonmetals) can share valence electrons so that they are surrounded by 8 electrons (hydrogen by 2 electrons). This sharing of electrons is called covalent bonding.

It is not immediately obvious why the sharing of electrons leads to a more stable state. Electrons after all are negatively charged particles and as such repel another. Let’s consider the simplest case, a dihydrogen molecule. Hydrogen possesses only one electron. To obtain the same number of electrons as its nearest noble gas (helium) it would have to gain one electron. In reactions with active metals this is exactly what hydrogen does! It acquires the electron of a sodium or potassium atom and forms ionic metal hydrides which contain the hydride ion H-. Dihydrogen though is not a combination of two hydride ions which would give the H22- ion but a neutral molecule. In H2 each of the two electrons is shared equally between the two atoms. There are several electrostatic interactions in this bond: Attractions between electrons and nuclei Repulsions between electrons Repulsions between nuclei

Atractive Interactions

Repulsive Interactions

If the electrons are shared, which means they physically spend a lot of time in between the two nuclei, the attractive interactions outnumber the repulsive interaction. Furthermore, the electrons effectively shield the nuclei from their positive charge. The net effect is that the electrons pull the two nuclei closer together resulting in a covalent bond which in the case of dihydrogen is 436 kJ/mol more stable than the isolated hydrogen atoms.

Hypothetical “snapshot” of a hydrogen molecule. The same photographic plate is exposed over and over again to show where it is most likely to encounter the electrons in the molecule.

Electron Dot Formulas (Lewis structures) We have written simple Lewis structures of elements and ions to describe electron transfer in ionic compounds by placing the number of valence electrons (electrons in the outermost electronic shell) as dots around the element symbol. It is essential that we become familiar in drawing Lewis structures of molecules and polyatomic ions to describe covalent bonding between atoms. First let’s make sure we can easily identify the number of valence electrons for an element. The number of valence electrons for an element follows directly from its position in the periodic table of the elements. The elements are organized into eighteen groups. Each group shares similar chemical behaviour. The 18 groups are organized into four blocks. H and He are unique and are not generally placed in a “block”

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

1

2

3

4

5

6

sblock

pblock dblock

7

fblock

Valence Electrons are referred to as the number of s and p electrons in the outermost electronic shell. d and f electrons are not counted as valence electrons. For s-block elements: The number of valence electrons is the group number. For p-block elements: The number of valence electrons is the group number minus 10.

Class exercise: Determine the number of valence electrons for each of the following elements. Potassium Selenium Iodine Barium

Class exercise: Write Lewis structures Electron dot formulas) for the following elements: –

Calcium

–

Boron

–

Bromine

–

Sulfur

Lewis Structures of Molecules Lewis structures are representations of molecules showing all valence electrons - bonding and nonbonding. Each bonding electron pair is represented as a single line between the bound atoms. Each nonbonding electrone pair (also known as lone pair) is represented as a pair of dots. Drawing Lewis structures of simple molecules illustrated on the examples CH4, NH3, H2O, HF and O2 Draw each atom along with its valence electrons. H

H H

C

H

H

N

H

H

O

H

H

F

O

O

H

Make a bond between the central atom and each terminal atom by pairing up electrons. H

H H

C H

H

H

N

H

O H

H

H

F

O

O

Represent each shared electron pair with a single line between the bonded atoms. Make multiple bonds if necessarily to complete octets for all atoms (maximum of triple bond and no more than four bonds per atom). Hydrogen only requires two electrons. H H

C

H H

H

N

H

H

O

H

H

F

O

O

H

Fluorine only needs to share one of its seven valence electrons to be surrounded by an octet, oxygen needs to share 2 of its 6 valence electrons and nitrogen must share three of its 5 valence electrons to obtain an octet. Carbon must share all four valence electrons to gain an octet (carbons four bond requirement). In the case of CH4, NH3 and H2O you might have wondered which atom to choose as the central atom. Note that hydrogen would not make a very good central atom since it would have to form more than one bond and thus be surrounded by more than 2 electrons. Hydrogens are never central atoms. How about CO2, SO2 or N2O though? The following guidelines and some practice will greatly aid you in drawing Lewis structures of hundreds of species. Being able to draw them quickly will greatly aid you in succeeding in the chapters to come. 1. Usually, but not always, the least electronegative element is the central atom. Electronegativity correlates very well with electron affinity. It decreases from the top to the bottom in a group and it increases within a period. For example in all oxoanions such as [CO3]2-, [ClO4]-, [PO4]3-or [SO4]2- the non oxygen atom is the central atom. 2. Generally if there are four atoms or more, small compact structures are preferred over linear arrangements. For example the molecule SO3 does not consist of a chain of four atoms but carries a central sulfur atom (less electronegative then oxygen) with three oxygen atoms surrounding it. Class exercise: Draw the Lewis structures for the following molecules: CH2O (formaldehyde), CO2, SO2 and HCN (hydrogen cyanide). Do the examples above illustrate every possible way how carbon can meet its four bond requirement?

Lewis structures of polyatomic ions The concept of Lewis structures also applies to polyatomic ions. We just have to take into account the extra electron(s) for anions, or the missing electron for cations. The following exercises will work you through two example: Practice example: Draw the Lewis structure for the ammonium ion. Start by drawing the central nitrogen atom with all its valence electrons surrounded by four hydrogen atoms with one valence electron each.

Remove an electron from the central atom. Indicate the absence of that electron by placing a circled positive charge sign on the nitrogen atom.

Make four bonds between the central nitrogen atom and the terminal hydrogen atoms.

Practice example: Draw the Lewis structure for the nitrite ion. Start by drawing the central nitrogen atom with all its valence electrons surrounded by two oxygen atoms with six valence electrons each.

Add one electron to a terminal atom. Indicate the presence of the extra electron with a circled negative charge sign on that atom.

Make two bonds between the central and the terminal atom.

Form an additional bond to satisfy the octet for all atoms.

Resonance The nitrite example raises an important point. How did you decide which of the two oxygen atoms in the nitrite ion was to receive the extra electron. The one on the right or the one on the left? And what would have happened had you chosen the other oxygen instead. We could have come up with two different Lewis structures as shown below:

O

N

O

O

N

O

But how are they different. Can we distinguish the right from the left oxygen atom? Of course we cannot. They are equivalent which is why we call the two Lewis structures above equivalent Lewis structures. They both represent extreme forms of what the ion could look like. When one actually determines the bond distance between the nitrogen and the two oxygen in the nitrite ion the observation is made that they are both identical within experimental error and that they are somewhere in between the expected bond length for nitrogen to oxygen single and double bonds. The observed structure is said to be a resonance hybrid with the two resonance structures contributing equally (50 % each). The double headed arrow is used for resonance structures. Sometimes chemists like to draw the resonance hybrids by using dashed lines for partial bonds that result from delocalized electrons. Note that the charge of – 0.5 would reside on average on each oxygen in the resonance hybrid depicted in the centre. In Chem 150 we will only draw full bonds (e.g. structure on the right and left with a resonance arrow in between them!)

O

N

O

O

N

O

O

N

O

Formal charges: In the example above one of the two oxygen atoms in the equivalent Lewis structures carries a charge of -1. This is called a formal charge. Of course there must be a charge on some atom since we have an ion but as we shall see formal charges can also arise in neutral molecules. Consider the molecule NOF3. The least electronegative element is nitrogen which is surrounded by three fluorine atoms and one oxygen atom.

O F O

N

O

F

F

c) F

N

F

F

N

F O

F

F

a)

b)

F

N

F

F

d) One might be persuaded to pair the two electrons between nitrogen and oxygen in b) and draw a double bond as in c) but that would violate the octet rule for nitrogen now being surrounded by 10 electrons instead of 8. To get to the correct Lewis structure d) the electron from nitrogen is transferred to oxygen. This will lead to formal charges since nitrogen now has ownership of 4 rather than 5 valence electrons and oxygen now owns 7, one more than its normal 6 valence electrons. Nitrogen is said to have a formal charge of +1 and oxygen has a formal charge of -1. Omitting formal charges invalidates a Lewis structure. Do not confuse formal charges with oxidation numbers (oxidation states) commonly used to balance redox equations.

NOF3 Formal Charge Oxidation Number

N +1 +5

O -1 -2

F O -1

The formal charge (FC) might be calculated using the equation: FC = number of valence electrons – 0.5 x number of bonding electrons – number of lone pair electrons. e.g. the formal charge of nitrogen in NOF3(structure d) is: FCN = 5 - (0.5 x 8) – (0.5 x 0) = +1

In neutral molecules the formal charges of all atoms must add up to zero. In polyatomic ions the formal charges of all atoms must add up to the charge of the ion.

Class exercise: Draw the Lewis structure of the azide ion [N3]Let’s use a general algorithm applicable to any polyatomic ion or molecule. 1. Draw each atom along with the number of valence electrons

2. Make one bond between the central atom and each terminal atom by pairing electrons

3. If overall charge (n-) add n electrons to a terminal atom(s), if overall charge n+ remove n electrons from the central atom.

4. Pair electrons to make multiple bonds (maximum of four bonds per atom) and complete octet for all atoms.

5. If necessary complete octets by forming additional bonds or transferring electrons from one atom to another atom. This will lead to formal charges on some atoms.

Non equivalent Lewis structures. In the nitrite example, which we used to introduce the concept of resonance, we saw that two equivalent Lewis structures contributed equally to the resonance hybrid. Consider now the molecule N2O, dinitrogen monoxide, also known as nitrous oxide or laughing gas because of its use in general anesthesia. Draw all valid Lewis structures for the molecule

Which one is the best Lewis structure? To answer this, consider the following rules: The best Lewis structures has the lowest possible formal charges (e.g. zero formal charges!) If formal charges are necessarily the best Lewis structure will be the one that has the negative formal charge on the more electronegative atom and the positive charge on the less electronegative atom. Formal charges of the same sign on adjacent atoms are very unlikely.

On the basis of these rules which of the three resonance structures above is the worst and which one is the best? When we have non-equivalent Lewis structures like in the nitrous oxide case above the different resonance structures will not contribute equally to the resonance hybrid but the best Lewis structure will contribute most strongly. Would you expect the nitrogen-nitrogen bond in N2O resemble that of a) b) c) d) e)

a single bond a double bond a triple bond in between a single and double bond in between a double and a triple bond

Justify your answer!

Exceptions to the octet rule Although the octet rule is very useful there are quite a few exceptions we need to be aware of. 1. Odd-electron species (e.g. NO) No matter how you try to share the 11 valence electrons in NO you cannot obey the octet for both atoms simultaneously. Two even numbers cannot add up to give an odd number. 2. Incomplete octets (e.g. BF3) Rare but important case. Even if it shares all of its three valence electrons it can only form 3 bonds! 3. Expanded valence shells e.g. PCl5, SF6, SO2, [SO4]2- and [ClO4]- (Most important!) The very existence of compounds such as PCl5 with one central and five terminal atoms shows that certain elements can exceed the octet rule. The bottom line is: Elements of the 2nd period can never exceed the octet. Elements of the 3rd or higher periods can exceed the octet rule.

Draw as many valid Lewis structures for the sulfate ion as you can think of. Does any of them obey the octet rule and if yes at what cost? Identify the best Lewis structure/structures. Identify equivalent Lewis structures.

In Chem 150 we will always minimize formal charges if we can to get to the best Lewis structure. Remember that this simplified two dimensional drawing of the sulfate ion is a poor representation of its three dimensional shape which is tetrahedral. (Chapter 3)

Energetics of covalent bonding: The energetics of covalent compounds are generally more complex than those of ionic solids. Nevertheless we can make some qualitative predictions about material properties based on bond energies and structure. For example a double bond between two atoms is stronger (has a greater bond energy) than a single bond and a triple bond is stronger yet. We can measure bond energies by measuring the energy it takes to break a bond. The following table gives a selection of bond energies obtained experimentally from a variety of sources including calorimeter and spectroscopy and more recently theoretically using sophisticated computational methods. The bond energies of diatomic molecules such as H-H, NΞN or Cl-Cl are actually bond dissociation energies and are known much more precisely.

bond

bond energy

bond

bond energy

bond

Bond energy (in kJ/mol)

H-H

436

C-C

347

N-N

163

H-C

414

C=C

611

N=N

418

H-N

389

CΞC

837

NΞN

946

H-O

464

C-N

305

N-O

222

H-S

368

C=N

615

N=O

590

H-F

565

CΞN

891

O-O

142

H-Cl

431

C-O

360

O=O

498

HBr

364

C=O

736

F-F

159

H-I

297

C-Cl

339

Cl-Cl

243

Br-Br

193

I-I

151__________________________________________

We can estimate the energy absorbed or given off by a reaction (mostly heat) using these bond energy values. Consider the formation of water from the elements: 2H2 + O2

2H2O

For the reaction to proceed two H-H bonds as well as one O=O bond need to be broken and four O-H bonds form.

Energy required for bond breaking:

Energy gained from bond formation:

2 mol of H-H bonds

+ 872 kJ

1 mol of O=O bonds

+ 498 kJ

4 mol of O-H bonds

- 1856 kJ

Sum:

- 486 kJ

486 kJ are released in this reaction which forms two mol of water indicating the exothermic character of the combustion of hydrogen gas. The energy (or heat) of formation for one mol of gaseous water would thus be - 243 kJ/mol indicating that one mole of water is 243 kJ more stable than the elements it is composed of.

The experimentally determined value is 242 kJ/mol measured for gas phase formation (steam). More to come in thermochemistry…