2 REVIEW OF IONIC BONDING Occurs when a metal transfers an electron to a nonmetal forming a charged particle.Intermolecular forces are ElectrostaticNaming: Group IA,IIA,IIIA metals to nonmetals - Name the metal then the nonmetal with ‘ide’ ending.Transition metal to nonmetal - Name the transition metal, then its charge in Roman numerals in parentheses, followed by the nonmetal with ‘ide’ ending.
3 Metal to polyatomic ion – name the metal, then the. polyatomic ion Metal to polyatomic ion – name the metal, then the polyatomic ion. If a transition metal, name the metal, its charge in Roman numerals, then the polyatomic ion.Why does this bonding occur?To achieve lower energy, that is, if the energy of the compound is lower than that of the separate atoms.Bond energy can be calculated using Coulomb’s Law:E = 2.31 x 10-19J nm Q1Q2rIf ‘E’ is negative, then the bonded pair has lower energy than the separate ions or atoms.Calculating for NaCl: E = 2.31 x 10-19K nm (+1)(-1) = x 10-19J0.276nm
4 Covalent BondsCovalent bonds are formed when a nonmetal shares one or more pairs of electrons with another nonmetal. These are neutral molecules.
5 ELECTRONEGATIVITY:ELECTRONEGATIVITY IS THE ABILITY OF AN ATOM IN A MOLECULE TO ATTRACT ELECTRONS TO ITSELF.THERE ARE TABLES OF ELECTRONEGATIVITY VALUES FOR EACH ELEMENT.WHEN THE ELECTRONEGATIVITY DIFFERENCE BETWEEN TWO BONDED ATOMS IS 0.4 OR BELOW, THE MOLECULE IS CONSIDERED TO BE NON POLAR, COVALENT.
6 Polarity in bonds...Because different atoms have different electronegativities, one atom will pull more strongly on the electron pair in a bond.
7 Polarity...This uneven attraction for the electrons will cause the electron cloud, (the area where the electron is residing), to be more dense nearer to the atom with the higher electronegativity.
8 Polarity...The atom with the stronger attraction for the electrons gets a partial negative charge (because the negative electrons are closer to it). The other atom gets a partial positive charge.
9 The F has an electronegativity of 4.0. The H is 2.1. d-d+
11 Polar bonds forming non polar molecules... In a completely symmetrical molecule, the differences in electronegativity cancel each other out.
12 Example...3.0 :Cl:I ..3.0 :Cl : C : Cl: 3.0ICl = C = 2.5
13 Forming of dipole attractions between polar molecules.... Because a polar molecule has a partial positive charge on one atom and a partial negative on another atom, there is a positive / negative attraction to other polar molecules.
14 The TWO POLES ( + and -) are referred to as a dipole The TWO POLES ( + and -) are referred to as a dipole (aka dipole moment)These dipole -dipole attractions occur between polar molecules. The positive end of one attracts to the negative end of another.
17 Hydrogen Bonding...Another important intermolecular force of attraction is Hydrogen bonding.This is a strong force of attraction between atoms of Hydrogen with either N, O, or F.
18 It is a stronger version of a dipole - dipole attraction. Nitrogen, Oxygen, and Fluorine are the most electronegative atoms and so the dipole is more pronounced in a molecule that contains H-N, H-O, or H-F
19 Because of these strong intermolecular attractions... Molecules which contain these bonds will be held together more tightly and thus be more difficult to separate.Molecules with H-bonds will have higher melting and boiling points, and viscosities than other molecules.
20 There are single, double, and triple covalent bonds. A single bond occurs when one pair of electrons is shared between two atoms.A double bond occurs when four electrons (two pairs) are shared between two atoms.continued...
21 continued...A triple bond occurs where six electrons, (3 pairs) are shared between two atoms.Some elements do not form multiple bonds; (The Halogens, and some of the larger nonmetals).
22 Bonding order of some elements one single bond – the halogens, -F, -Cl, -Br, -Itwo bonds – oxygen, -O- , or = Othree bonds – nitrogen, - N - , = N - , º Nfour bonds – carbon, - C - , - C = , = C = , - C º
23 The ‘how to’ of Lewis Dot Structures of Covalent Compounds 1. Count up the valence electrons on all of the atoms in the molecule. Total them up.2. If there is a ‘+‘ charge, subtract the magnitude of the charge from the total.
24 Lewis Dot Structures continued... 3. If it has a ‘-’ charge, add the magnitude of the charge to the total.4. Choose the element that there is less of to be the center of the molecule and place the others symmetrically around it.
25 Lewis Dot Structures continued... 5. Place dots (electrons) around each atom, two on each side, so that each atom has an octet. The two dots between two different atoms count toward the octet for each of those atoms.6. Be sure to use no more or less electrons than the total from step 1.
26 Lewis Dot Structures continued... 7. If you run out of electrons before every atom has an octet, then you must move a pair from the outside of the atom adjacent to the lacking atom, to be shared between the two atoms to form a double (or triple) bond to satisfy the octet rule.
27 Example...CCl4 C = 4 valence electrons4Cl = 7 x 4 valence electronsTotal is 32 electrons to fill in. There is no charge.continued...
28 C is the center of the molecule... continued...C is the center of the molecule...: Cl ::Cl : C : Cl::Cl:Each atom has an octet; 32 total e-
29 SO4-2 S = 6 valence electrons Example...SO S = 6 valence electrons4, O = 6 x 4 valence electronsThe total is 30 electrons plus two for the negative charge = 32 total e-
30 S will be the center... :O: nonbonding .. l .. electrons :O-- S –O: I
31 CO2 C = 4 valence electrons 2, O = 6 x 2=12 valence total = 16 e- Dot Structures of covalently bonded compounds with multiple bonds...Ex.CO2 C = 4 valence electrons2, O = 6 x 2=12 valencetotal = 16 e-:O- C-O:THE ATOMS DO NOTHAVE COMPLETE OCTETS !!!
32 If two pairs of non bonding electrons are moved from the outside of the carbon to a position between C and O, octets will be satisfied.
33 Both of these are correct Both of these are correct. The top one may be more stable than the bottom one.:O=C=O:or:O :::C :O:
34 Dot structures for ionic compounds show the non metal with a noble gas configuration (complete octet) and the metal with an empty outer shell.Na+ :Cl:
35 ResonanceThe last covalent example showed a structure that exhibited resonance. It appears that CO2 could exist it two different forms just by moving electrons around.
36 In a resonane structure the electrons are delocalized. This means that the electrons are shared by all of the atoms, not only the two that a simple diagram illustrates.
37 Other resonance structures... CO NO3-1Draw the dot structures of these ions and determine why they are resonance structures.Note: a single line can be used to represent an electron pair between two atoms, a double line - two pairs, triple - three pairs.
38 ..24 e-2-OO C O::......::::....The double bond could appear between any one of the C-O bonds.
39 l bonding O pairs yourself NO31- will have the same structure as CO32- NO electrons1-O = N - O Put in the non-l bondingO pairs yourself
40 O- N -O O- N =O O=N -O ll l l O O O Write all possible structures when drawing a resonance diagram.1-O- N -OllOO- N =OlOO=N -OlOPut brackets around the structures and place the charge outside the upper right corner.
41 Bond Lengths... Single bonds are longer than double bonds. Double bonds are longer than triple bonds.Triple bonds are the shortest and the strongest of the three.Double bonds are stronger than single bonds.
42 Bond Length is the distance between two bonded nuclei.
43 Bond Lengths In Resonance Structures... Experimentally, it was found that the bonds in a resonance structure are in between the lengths of single and double bonds.
44 Formal Charges... Why does NO31- have a -1 charge? Why does PO43- have a -3 charge?Why does NH3 have a 0 charge?
45 These charges can be determined by defining the formal charges on each individual atom in the particle.
46 how??? Determine the valence on each atom in the compound. Divide the shared electrons so that half of them count for each atom sharing them.Nonbonding electrons count only on the atom that they belong to.
47 That will be the charge on that individual atom. continued...Determine the difference between the valence number of electrons and the electrons that were counted around the atom.That will be the charge on that individual atom.
49 Therefore each has a charge of -1. Counting electrons on each atom and then finding the difference from the valence...The left and bottom Oxygens have 7 electrons which are one more than their valences of 6.Therefore each has a charge of -1.The right Oxygen has 6 electrons which equals its valence, so it has a zero charge.
50 Ex. continued...The Nitrogen has 4 electrons which is one less than its valence therefore it has a +1 charge.
51 That will be the charge on the entire molecule or ion. Add up the charges for all of the individual atoms.That will be the charge on the entire molecule or ion.
52 This is the charge on the ion. Adding up these individual charges gives ; = -1This is the charge on the ion.
54 Solution for CO2..a::::::O C O......bO C O::::::
55 O = -1, C = 0, O = +1 Solutions... (a) 7 e- on O, 4 e- on C, 5 e- on O = 0
56 Solutions (b)6 electrons on each Oxygen gives each of them a zero charge.4 electrons on Carbon gives it a zero charge.Therefore the total charge = 0.
57 Formal charges can be used to determine the most stable resonance structure.
58 CO2 has 3 resonance structures, two of them are the same but with the triple bond on opposite sides. Which one is the most stable of the three ?
59 (a)The triple - single bonded structure has charges of -1, 0, +1(b)The double - double bonded structure has charges of 0, 0, 0
60 (b) will be the most stable because... The structure with the least number of charges or the lowest charges will be the most stable.
61 Lattice energy is the change in energy that takes place when separated gaseous ions are packed together to form an ionic solid.The signs on the energy values will be negative if the process is exothermic and positive if the process is endothermic.Lattice energy can be calculated by using an application of Hess’ Law known as the Born-Fajans-Haber cycle.
62 There are many steps in the formation of an ionic solid from its elements in their standard states: A solid metal of zero oxidation state must go through sublimation and ionization to become a gaseous ion.(enthalpy of sublimation and enthalpy of ionization)A diatomic molecule must dissociate into one atom (dissociation energy)A non metal must gain an electron to become an anion. (electron affinity)The lattice energy (the formation of 1 mole of ionic solid from ions in their gaseous states)Then enthalpy of formation (the enthalpy of the formation of 1 mole of the solid from it elements in their standard states)
63 DHo = DHsub + DHI.E. +1/2 DHB.diss + DHE.A + DHL.E EACH OF THESE VALUES CAN BE OBTAINED FROM TABLES AND THE UNKNOWN VALUE IN THE EQUATION MAY BE SOLVED FOR.example of lattice energy calculations on doc cam.
64 DHrxn CAN ALSO BE APPROXIMATED BY USE OF AVERAGE BOND ENTHALPIES. KEEP IN MIND;BONDS BREAKING HAVE (+) VALUES (IT TAKES ENERGY TO BREAK A BOND)BONDS FORMING IS EXOTHERMIC (-)USING HESS’ LAW, THE SUM OF ALL BONDS BREAKING AND ALL BONDS FORMING = APPROXIMATELY THE ENTHALPY OF REACTION.example on doc cam
65 Lattice energy cannot be directly measured but can be calculated as we have just done. The enthalpy change observed when a solution is prepared is due to lattice energy (breaking of the crystal lattice of the ionic compound) and hydration energy (the energy released when the ions are hydrated; bonds are formed between the ions and solvent)
66 EXCEPTIONS TO THE OCTET RULE AND GEOMETRIES OF MOLECULES: Boron will be deficient in electrons and is thus very reactive toward electron rich species. see diagramFree radicals form in some molecules:NOElements in the third energy level have ‘d’ orbitals. These elements can have expanded octets, utilizing the ‘d’ orbitals.examples follow.
67 Bonding In Hydrocarbons... Hydrocarbons are compounds composed of only Carbon and Hydrogen.The carbons bond to form a ‘Carbon chain’.The hydrogens bond around the carbon chain, to each carbon.
68 Hydrogen can only have 2 electrons bonded to it. Why? Example of a hydrocarbon...H H H H Hl l l l lH - C - C - C - C- C - HH H H H HA 5 carbon chain with single bonds --- Pentane.
69 Carbon always has 4 bonds to it. A carbon may be bonded with 4 single bonds...or 2 double bonds...or 1 single and 1 triple bond...or 2 singles and 1 double
71 Alkanes are hydrocarbons with only single bonded carbons. Their names end in ‘ane’ like the group they belong to, the Alkanes.Some simple alkanes are methane, ethane, propane, butane, pentane, hexane, heptane, octane, and decane.
72 Ex. Methane and propane...H H HH C C C HHH C H........::::::........
73 Alkenes are hydrocarbons with at least 1 double bond. Their names end in ‘ene’ like the group they belong to, the alkenes.Some examples of simple alkenes are ethene, propene, butene, pentene, hexene, heptene, octene, nonene, decene.
74 Alkenes Ethene and Propene H - C = C - HH HllHH - C - C = C - HH H Hllll
75 Alkynes are hydrocarbons with at least 1 triple bond. Their names end with ‘yne’ like the group they belong to, the Alkynes.Some simple alkynes are ethyne, propyne, pentyne, etc.
76 Alkynes, Ethyne, Propyne H - C C - HHH - C - C C - Hll
77 Determining oxidation numbers in compounds. There are a few rules to follow so that different oxidation numbers in a compound can be determined.
78 Things to remember in order to determine the charge on a particle in a compound... Oxygen always has a - 2 charge in compounds, except in compounds called peroxides, where it is -1. (H2O2 is a peroxide).Hydrogen always has a +1charge in compounds except in compounds called hydrides where it is -1. (NaH is a Hydride).Group IA and IIA metals will always have a +1 and +2 charge respectively in compounds.F is always -1
79 ...other charges can be calculated. Knowing the charges on O, H, F, and the group IA and IIA metals......other charges can be calculated.Ex. KMnO4K = +1, Mn = ?, O = -2
80 There is a simple formula to use on neutral molecules. The sum of each atoms (charge x its subscript) must equal zero.K=+1, Mn=?, O= -2KMnO4+1(1) +?(1)+ -2(4) =0The charge on Mn = +7
81 Ex. H3PO4H=+1, P=?, O= -2+1(3) + ?(1) + -2(4)= 0The charge on P = +5