1 CHEMICAL BONDING Cocaine

2 Chemical Bonding Problems and questions — How is a molecule or polyatomic ion held together? Why are atoms distributed at strange angles? Why are molecules not flat? Can we predict the structure? How is structure related to chemical and physical properties?

3 Chemical Bonding: Objectives Objectives are to understand: 1. valence e- distribution in molecules and ions. 2. molecular structures 3. bond properties and their effect on molecular properties.

4 Forms of Chemical Bonds There are 2 extreme forms of connecting or bonding atoms: Ionic —complete transfer of 1 or more electrons from one atom to another Covalent —some valence electrons shared between atoms.Most bonds are somewhere in between.

5 Ionic Bonds Is electron transfer from an element of low IE (metal) to an element of high affinity for electrons (nonmetal) 2 Na(s) + Cl 2 (g) ---> 2 Na Cl - Ionic compounds are primarily between metals (1A and 2A and transition) and nonmetals (O and halogens).

6 Electrostatic Forces COULOMB’S LAW As ion charge increases, the attractive force _______________. As the distance between ions increases, the attractive force ________________. This idea is important and will come up many times in future discussions! increases decreases

7 Importance of Coulomb’s Law NaCl, Na + and Cl -, m.p. 804 o C MgO, Mg 2+ and O 2- m.p o C

8 Lattice Energy Energy released when one mole of solid crystal formed when ions in the gas phase combine Energy related to melting point and solubility Only calculatedOnly calculated not measure directly Net energy is independent of the path so use “Born – Haber cycle” to calculate Energy released (exothermic) 1 Na(g) + ½ Cl 2 (g) ---> 1 Na + Cl -

9 Covalent Bonding The bond arises from the mutual attraction of 2 nuclei for the same electrons. Electron sharing results. Bond is a balance of attractive and repulsive forces.

10 Electron Distribution in Molecules Electron distribution is depicted with Lewis electron dot structures Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. G. N. Lewis

11 Bond and Lone Pairs Valence electrons are distributed as shared or BOND PAIRS and unshared or LONE PAIRS. HCl lone pair (LP) shared or bond pair. This is called a LEWIS ELECTRON DOT structure.

12 Bond Formation A bond can result from a “head-to-head” overlap of atomic orbitals on neighboring atoms. Cl HH + Overlap of H (1s) and Cl (2p) Note that each atom has a single, unpaired electron.

13 Valence Electrons Electrons are divided between core and valence electrons B 1s 2 2s 2 2p 1 Core = [He], valence = 2s 2 2p 1 Br [Ar] 3d 10 4s 2 4p 5 Core = [Ar] 3d 10, valence = 4s 2 4p 5

14 Rules of the Game # of valence electrons = Group number For Groups 3A - 4A # of bond pairs = group number. For Groups 5A -7A, BP’s = 8 - Group #.

15 Rules of the Game Except for H (and sometimes atoms of 3rd and higher periods), Except for H (and sometimes atoms of 3rd and higher periods), BP’s + LP’s = 4 This observation is called the OCTET RULE

16 Building a Dot Structure Decide on the central atom; never H. Central atom is atom of lowest affinity for electrons. Therefore, N is central 2. Count valence electrons H = 1 and N = 5 Total = (3 x 1) + 5 = 8 electrons / 4 pairs Ammonia, NH 3

Form a single bond between the central atom and each surrounding atom H H H N Building a Dot Structure H H H N 4. 4.Remaining electrons form LONE PAIRS to complete octet as needed. 3 BOND PAIRS and 1 LONE PAIR. Note that N has a share in 4 pairs (8 electrons), while H shares 1 pair.

18 10 pairs of electrons are now left. 10 pairs of electrons are now left. Sulfite ion, SO 3 2- Step 1. Central atom = S Step 2. Count valence electrons S = 6 3 x O = 3 x 6 = 18 Negative charge = 2 TOTAL = 26 e- or 13 pairs Step 3. Form bonds

19 Sulfite ion, SO 3 2- Remaining pairs become lone pairs, first on outside atoms and then on central atom. OO O S. Each atom is surrounded by an octet of electrons.

20 Carbon Dioxide, CO 2 1. Central atom = _______ 2. Valence electrons = __ or __ pairs 3. Form bonds.. 4. Place lone pairs on outer atoms. This leaves 6 pairs. C 168

21 Carbon Dioxide, CO Place lone pairs on outer atoms. The second bonding pair forms a pi (π) bond. 5. So that C has an octet, we shall form DOUBLE BONDS between C and O.

22 Double and even triple bonds are commonly observed for C, N, P, O, and S H 2 CO SO 3 C2F4C2F4C2F4C2F4

23 Sulfur Dioxide, SO 2 1. Central atom = S 2. Valence electrons = 18 or 9 pairs bring in left pair OR bring in right pair 3. Form double bond so that S has an octet — but note that there are two ways of doing this. OS O

24 RESONANCE STRUCTURES This leads to the following structures. These equivalent structures are called RESONANCE STRUCTURES.. The true structure is a HYBRID of the two.

25 Urea, (NH 2 ) 2 CO 1. Number of valence electrons = 24 e- 2. Draw sigma bonds.

26 3. Place remaining electron pairs in the molecule. Urea, (NH 2 ) 2 CO

27 4. Complete C atom octet with double bond. Urea, (NH 2 ) 2 CO

28 Violations of the Octet Rule. Usually occurs with B and elements of higher periods. BF 3 SF 4

29 Boron Trifluoride Central atom = _____________ Valence electrons = __________ or electron pairs = __________ Assemble dot structure The B atom has a share in only 6 e (or 3 pairs). Boron in many molecules is electron deficient. The B atom has a share in only 6 e (or 3 pairs). Boron in many molecules is electron deficient. B 3 + 3(7) 12

30 Sulfur Tetrafluoride, SF 4 Central atom = Valence electrons = ___ or ___ pairs. Form sigma bonds and distribute electron pairs. 5 pairs around the S atom. A common occurrence outside the 2nd period. S 1734

31 Formal Charges The predominant resonance structure of a molecule is the one with charges as close to 0 as possible. Formal chargeFormal charge = Group # – 1/2 (# of BE) - (# LP electrons) - (# LP electrons)

32 OOC 4 - (1/2)(8) - 0 = (1/2)(4) - 4 = 0 Carbon Dioxide FC = Group no. – 1/2 (no. of BE) - (no. of LP)

33 Thiocyanate Ion, SCN - SNC SNC SNC Which is the most important resonance form? FC = Group no. – 1/2 (no. of BE) - (no. of LP)

34 Thiocyanate Ion, SCN (1/2)(2) - 6 = (1/2)(6) - 2 = (1/2)(8) - 0 = 0 SNC FC = Group no. – 1/2 (no. of BE) - (no. of LP)

35 Isoelectric, CS 2, CO 2. OO C SS C These equivalent structures are called ISOELECTRIC STRUCTURES. Different atoms with the same structure