1. Basics of Chemical Bonding AP Chemistry Properties of substances are largely dependent on the bonds holding the material together.

2. Basics of Bonding A chemical bond occurs when atoms or ions are strongly attached to each other. Ionic bonds involve the transfer of e – and the subsequent electrostatic attractions. --“metal/nonmetal”(“cation/anion”) Covalent bonds involve the sharing of e – between two atoms. --“nonmetal/nonmetal” metallic bonds: each metal atom is bonded to several neighboring atoms -- bonding e – (i.e., valence e – ) are free to move throughout the material

3. Lewis symbols show ONLY the valence e – (i.e., the ones involved in bonding). octet rule: atoms “want” 8 v.e – -- several exceptions (a topic for later) SCN Gilbert Lewis (1875–1946)

4. Ionic Bonding Na(s) + ½ Cl 2 (g) NaCl(s) “Salts” are brittle solids with high melting points.  H f o = –410.9 kJ/mol lattice energy: the energy required to separate 1 mole of solid ionic compound into gaseous ions --a measure of stability NaCl(s) Na + (g) + Cl – (g)  H latt = +788 kJ/mol enthalpy (i.e., heat) of formation Na Cl +Na + + [ Cl ] In general, ionically bonded substances have... …big, (+) lattice energies. –

5. Because lattice energies are electrostatic in nature, two variables are involved in how big they are: 1. the magnitude of the charges 2. the separation between the ions (“wears the pants”) Put the following in order of increasing lattice energy: LiBr, FeN, CdO. LiBr < CdO < FeNMgCl 2 < MgS < MgO Now these: MgS, MgCl 2, MgO. Charge can easily be twice as large (i.e., 2+ and 2– vs. 1+ and 1–), but the separation never varies by that much.

6. With transition-metal ions, the e – lost first come from the subshell with the largest value of n. e.g., Ni = [Ar] 4s 2 3d 8 Ni 2+ =[Ar] 3d 8 Ni 3+ =[Ar] 3d 7 Recall that polyatomic ions are groups of atoms that stay together and have a net charge. -- their atoms are held to each other by… covalent bonds -- e.g., NO 3 – CH 3 COO –

7. Cations are smaller than the neutral atoms from which they are derived. e.g., Li 1s 2 2s 1 Li + 1s 2 -- orbitals might be completely vacated -- e – /e – repulsion always decreases FeFe 2+ Fe 3+

8. Anions are larger than the neutral atoms from which they are derived. -- ClCl – 17 p +, 17 e – 17 p +, 18 e – An isoelectronic series is a list of species having an identical electron configuration. more e – /e – repulsion e.g., EX.Which has the Rb + Sr 2+ Y 3+ largest radius? N 3–, O 2–, F –, Ne, Na +, Mg 2+, Al 3+ (small radius)(big radius) (smallest Z eff )(largest Z eff )

9. Covalent Bonding -- atoms share e – -- covalent (molecular) compounds tend to be solids with low melting points, or liquids or gases -- three shared pairs = a triple covalent bond of e – (i.e., 6 e – ) -- two shared pairs = a double covalent bond of e – (i.e., 4 e – ) -- one shared pair = a single covalent bond of e – (i.e., 2 e – ) ++ (–) charge density

10. max. EN = 4.0 (F) bond polarity: nonpolar covalent bond: polar covalent bond: electronegativity (EN): the ability of an atom in a molecule to attract e – to itself describes the sharing of e – between atoms e – shared equally e – NOT shared equally -- A bonded atom w /a large EN has a great ability to attract e –. -- A bonded atom w /a small EN does not attract e – very well. min. EN = 0.7 (Cs) -- EN values have been tabulated. (see p. 308)

11. nonpolar covalent The  EN between bonded atoms approximates the type of bond between them.  EN < 0.5 0.5 <  EN < 2.0  EN > 2.0 polar covalent ionic bond As  EN increases, bond polarity... increases.  EN for a C–H bond = 0.4, so all C–H bonds (such as those in candle wax) are nonpolar.

12. “Oh, man… I forgot what the most electronegative elements are.” “Shee-oot… Ow teh ye… for a nickel.” FO’ NCl.” F = 4.0 O = 3.5 N = Cl = 3.0 C = 2.5 H = 2.0

13. Dipole Moments Polar covalent molecules have a partial (–) and a partial (+) charge and are said to have a dipole moment. H–F  +  – partial charge O HH –– ++ ++ big  EN = _____ polarity = _____ dipole moment bigbig

14. Polar molecules tend to align themselves with each other and with ions. H–F NO 3 – NH 4 + ** Nomenclature tip: For binary compounds, the less electronegative element comes first. -- Compounds of metals w /high ox. #’s (e.g., 4+ or higher) tend to be molecular rather than ionic. e.g., TiO 2, ZrCl 4, Mn 2 O 7

15. Lewis Structures (or “electron-dot structures”) 1. Sum the valence e – for all atoms. If the species is an ion, add one e – for every (–); subtract one e – for every (+). 2. Write the element symbols and connect the symbols with single bonds. 3. Complete octets for the atoms on the exterior of the structure, but NOT for H. -- If your LS doesn’t have enough e –, place as many e – as needed on central atom. -- If LS has too many e – OR if central atom doesn’t have an octet, use multiple bonds. 4. Count up the valence e – on your L.S. and compare that to the # from Step 1.

16. 26 e – Draw Lewis structures for the following species. PCl 3 P–ClCl– Cl.. HCN 10 e – H–C–N H–C=N H–C–N.. [ ] PO 4 3– P–OO–.. O O 32 e – 3–

17. H2H2 CH 3 CH 2 OH CO 2 As the # of bonds between two atoms increases, the distance between the atoms... 2 e – H–H 20 e – H H–C–C–O–H H HH 16 e – O=C=O O–C–O.. decreases. This CO bond is longer (and weaker) than this CO bond.

18. formal charge: the charge a bonded atom would have if all the atoms had the same electronegativity; to find it, you must first draw the Lewis structure formal charge # of v.e – in = the isolated – atom # of e – assigned to the atom in the Lewis structure When several Lewis structures are possible, the most stable is the one in which: (1) the atoms have the smallest formal charges, and (2) the (–) charges reside on the most electronegative atoms all unshared e – plus half of the bonding e –

19. [ ] Find the formal charge on each atom in the following species. SO 3 2– [ ] S–OO–.. O 26 e – 2–.. S O v.e – assigned e – 6 5 6 7 +1–1 S: +1 O: –1 NCS – 16 e – N–C–S.. N=C=S.. N–C–S.. – – – 546 745 2–01+ 546 646 1–00 546 547 00 (N is more EN than S.)

20. resonance structures: the two or more Lewis structures that are equally correct for a species e.g., For NO 3 –... [ ] N=OO–.. O – 24 e – [ ] N–OO=.. O – [ ] N–OO–.. O – Resonance structures are a blending of two or more Lewis structures. In drawing resonance structures, e – s move. Atoms don’t move.

21. Aromatic compounds are based on resonance structures of the benzene molecule. In the interest of efficiency, benzene is often represented like this… C C C C CC H H H H H H C C C C CC H H H H H H

22. Exceptions to the Octet Rule There are a few cases (other than for H) in which the octet rule is violated. These are: 1. particles with an odd number of valence e – -- e.g., 2. atoms with less than an octet -- e.g., ClO 2, NO, NO 2 B, Be BF 3 (24 e – ) B–FF=.. F B F F–.. –F.. -- would require resonance -- F is WAY more EN than B and won’t share extra e – -- inconsistent w /chemical behavior

23. 3. atoms with more than an octet -- this occurs when an atom gains an expanded valence shell -- other e.g., PCl 5 (40 e – ) P.. Cl Cl–.. –Cl.. –Cl.. Cl–.. AsF 6 –, ICl 4 –, SF 4 Expanded valence shells occur only for atoms in periods > 3. -- unfilled d orbitals are involved -- large central atom = -- small exterior atoms = fit more easily around central atom more space to accept e – pairs

24. The strengths of a molecule’s covalent bonds are related to the molecule’s stability and the amount of energy required to break the bonds. bond enthalpy: the  H req’d to break 1 mol of a particular bond in a gaseous substance -- -- big  H = -- e.g., always + strong bond (g) I–II–I.. I (g)2.  H = +149 kJ/mol At room temp., I 2 is a silver-purple solid.

25. atomization: the process of breaking a molecule into its individual atoms H H–C–Cl H.. methylchloride (chloromethane) (g) C (g) + 3 H (g) + Cl (g)........ --  H for a given bond (e.g., the C–H bond) varies little between compounds. e.g., C–H bonds in CH 4 vs. those in CH 3 CH 2 CH 3 have about the same  H CH 3 Cl was once used as a refrigerant, but it is toxic and so is now used primarily in the production of silicone polymers. It is also used as an organic solvent.

26. -- Typical values of bond enthalpies for specific bonds have been tabulated. (p. 326) -- To find bond enthalpy for atomization, add up bond enthalpies for each bond broken. Calculate the bond enthalpy for the atomization of dichloromethane. H H–C–Cl Cl (i.e., break 2 C–H bonds and 2 C–Cl bonds)  H atm = 2 (413 kJ / mol ) + 2 (328 kJ / mol ) = 1482 kJ/mol

27. Approximate the reaction enthalpy for the combustion of propane. Broken: You can approximate reaction enthalpy using Hess’s law and tabulated bond enthalpies.  H rxn  (  H broken bonds ) –  (  H formed bonds ) C 3 H 8 + 5 O 2 4 H 2 O + 3 CO 2 2 C–C: 2 (348) 8 C–H: 8 (413) 5 O=O: 5 (495) Formed: 6 C=O: 6 (799) 8 O–H: 8 (463) 6475 kJ 8498 kJ H–C–C–C–H HHH HHH O=O.. O=C=O.. H–O H.. –2023 kJ (Actual = –2220 kJ)

28. bond length: the center-to-center distance between two bonded atoms -- fairly constant for a given bond (e.g., the C–H bond), no matter the compound e.g., C–H bonds in CH 4 are about the same length as those in CH 3 CH 2 CH 3 -- Average bond lengths have been tabulated for many bonds. (p. 329) -- As the number of bonds between two atoms increases, bond length… and bond enthalpy… e.g., C–CC=CC–C 1.54 A 348 kJ / mol 1.34 A 614 kJ / mol 1.20 A 839 kJ / mol