4/15/13 Chemical Bonding The Octet Rule ! There are basically two types of ! Atoms want to have a filled valence chemical bonds: shell—for the main group elements, this 1. Covalent bonds—electrons are means having filled s and p orbitals. shared by more than one nucleus Hence the name “octet rule” because 2. Ionic bonds—electrostatic attraction when the valence shell is filled, they between ions creates chemical bond have a total of eight valence electrons. Hydrogen bonding and van der Waals’ ! We use “dot structures” to represent bonding are subsets of electrostatic atoms and their electrons. bonding Lewis Dot Structures Lewis Dot Structures ! Dots around the elemental symbol ! None of the atoms in the previous represent the valence electrons. example contained full valence shells. Examples ! When creating bonds, atoms may share electrons in order to complete their Hydrogen (1s1) H· valence shells. · Carbon (1s2 2s2 2p2) · C · · ·· Chlorine ([Ne] 3s2 3p5) : Cl · ·· Lewis Dot Structures Lewis Dot Structures Examples Examples H2 molecule: H2 molecule: H:H -’ H needs two e s to fill its valence shell. The result is a “covalent bond” in which Each hydrogen atom shares its electron two electrons are shared between nuclei with the other in order to fill their and create a chemical bond in the valence shells. process. H· + ·H ⇒ H:H When two e-’s are shared, it makes a “single” bond. 1 4/15/13 Lewis Dot Structures Lewis Dot Structures Examples Examples F2 molecule: CH4 molecule: F has seven valence e-’s, but wants eight C has four valence e-’s; H has one e-’s to fill its valence shell. valence e-. H ·· ·· ·· ·· ·· ·· · H :F· + ·F: ⇒ :F:F: ⇔ :F – F: · · ·· ·· ·· ·· ·· ·· ·· ·C· + 4 ·H ⇒ H· ·C· ·H ⇒ H : C : H · · ·· H A line is often used to represent shared · H ı H H – C – H electrons. ı H Lewis Dot Structures Lewis Dot Structures Examples Examples C2H4 molecule: C2H4 molecule: H Is this complete— H · H H ı · · ·· · do all atoms have ·· H· ·C· ·C· ·H ⇒ H : C : C : H filled valence H : C ·· C : H ⇒ H – C – C – H · · · ·· ·· ı H H shells? H ⇓ H H ·· Sharing of four electrons between two H : C ·· C : H ·· nuclei results in a “double” bond. H Lewis Dot Structures Lewis Dot Structures Examples Examples NH3 molecule: CO molecule: . .. .. .. .. C. O.. ⇒ C. O.. .N. 3 H . Not all electrons in a C has 5 e-’s; O has 7 e-’s Lewis dot structure .. .. .. .. need to be part of a . .... - H N. H ⇒ H-N-H chemical bond— ⇒ C .. O ⇒ :C .. O : ⇒ :C-O:- . | .. H H some electrons may C has 6 e-’s; O has 8 e-’s CO has a covalent triple be in the form of bond—6 e-’s are shared to “lone pairs”. form the bond 2 4/15/13 Formal Charge of Atoms Formal Charge of Atoms The formal charge of an atom in a molecule is Rules (con’t.): the charge the atom would have if all electrons 3. The sum of formal charges for each atom in the were shared equally. molecule must equal the actual charge of the Rules: molecule: 0 for a neutral molecule; ionic charge 1. All lone pair electrons are assigned to the atom for an ion. to which they are associated. 2. Half of the bonding electrons are assigned to each atom comprising that bond. Example: carbon monoxide The sum of these electrons is subtracted from the C O :C O: Valence electrons 4 6 number of valence electrons to determine the ≡ Lone pair electrons 2 2 formal charge ½ shared electrons 3 3 Formal charge -1 +1 Formal Charge of Atoms Formal Charge of Atoms Example: carbon dioxide Example: Nitrous oxide, N2O .. .. .. .. .. O=N=N :O-N≡N: O=C=O.. .. .. .. .. 0 +1 -1 -1 +1 0 structure #1 structure #2 C O O Valence electrons 4 6 6 Lone pair electrons 0 4 4 The formal charge can help predict which structure is ½ shared electrons 4 2 2 preferred when multiple Lewis structure can be made: Formal charge 0 0 0 • Smaller formal charges are preferred • Negative formal charge should reside on more electronegative atoms • Like charge should not be on adjacent atoms Formal Charge of Atoms Formal Charge of Atoms Example: Sulfuric Acid, H SO Example: Sulfuric Acid, H SO .. 2 4 2 4 :O : :O : .. 3 .. .. 3 .. H-O..1 -S-O..2-H H-O..1 -S-O..2-H :O.. 4: :O4: This is the atom S O1 O2 O3 O4 H atom S O1 O2 O3 O4 H correct Lewis valence e-’s 6 6 6 6 6 1 valence e-’s 6 6 6 6 6 1 ½ shared e-’s 4 2 2 1 1 1 ½ shared e-’s 6 2 2 2 2 1 structure for - - lone pair e ’s 0 4 4 6 6 0 lone pair e ’s 0 4 4 4 4 0 H2SO4 formal charge 2 0 0 -1 -1 0 formal charge 0 0 0 0 0 0 3 4/15/13 Resonance Structures Electronegativity and Bonding Suppose you can draw two different Lewis ! When nuclei share electrons to form a structures that are equally valid (same formal covalent chemical bond, the electrons are not charges). Which structure is correct? necessarily shared equally—a shared electron Example: NO2 may spend more time closer to one of the .. .. N structure 1 N structure 2 nuclei. ! The electronegativity of the nuclei determines :O : :O : .. :O : :O : 1 .. 2 N 1 .. 2 how the electron is shared. each N-O bond is ! Electronegativity is a measure of how strongly :O1: :O2: really a 1.5 bond, not a “bound” electron participating in a chemical a single or double resonance structure bond is attracted to a nucleus. bond Electronegativity and Bonding Electronegativity and Bonding ! Electronegativity is related to electron affinity and ionization energy. ! Electronegativity (denoted by the greek symbol χ) is highest for elements in the upper right hand side of the Periodic Table and increases from left to right and from bottom to top. Polar Covalent Bonds Polar Covalent Bonds ! When two elements with different ! The molecule has a polar bond meaning the electronegativities bond, the resulting electrical charge is not equally distributed covalent bond will be polar, i.e., the shared between the nuclei involved in the chemical electrons will spend more time closer to the bond. The molecule also has a dipole nucleus with the higher χ, so that end of the moment—an uneven distribution of electrical bond will be slightly more negative, and the charge. other end will be slightly more positive. δ+ δ- δ+ δ- H - F H - F χ=2.1 χ=4.0 χ=2.1 χ=4.0 4 4/15/13 Polar Covalent Bonds Polar Covalent Bonds Other Examples Examples Water: H2O is a bent molecule with two pairs of Carbon monoxide: CO unshared e-’s in p orbitals. Is CO a polar molecule? Is water a polar molecule? δ- δ- .. χ(O) = 3.5 δ+ δ- χ(O) = 3.5 ..O :C ≡ O: (H) = 2.1 (C) = 2.5 H H χ χ δ+ δ+ Polar Covalent Bonds Bond Lengths ! The bonding between atoms can have a Examples significant effect on the bond distance Carbon dioxide: CO2 is a linear triatomic between atoms. molecule. ! Multiple bonds between two atoms have Is CO2 a polar molecule? shorter bond lengths compared to single bonds involving the same elements: δ- δ+ .. .. O = C = O χ(O) = 3.5 .. .. (C) = 2.5 δ+ δ- χ Bond Lengths Bond Energies Examples ! The bond energy is the amount of average bond lengths energy it takes to pull two atoms apart single bond C-C 154 pm C-O 143 pm and break the chemical bond. double bond C=C 133 pm C=O 120 pm ! For some diatomic gas phase species, triple bond C≡C 120 pm C≡O 113 pm we know the bond energies exactly through measurement. 5 4/15/13 Bond Energies Bond Energies ! For polyatomic molecules, we can calculate Examples (see Table 8-2 for a more complete list) the average bond energy for a given type of Average bond energy (kJ mol-1) bond by measuring bond energies in a wide single bond double bond triple bond variety of molecules containing that specific C-H 416 type of bond and averaging the results. C-C 356 598 813 ! Just as in bond lengths, the type of bonding C-O 336 750 1073 has a dramatic effect on the bond energy in C-N 285 616 866 molecules. N-N 160 418 946 ! Multiple bonds are much harder to break than N-O 201 605 are single bonds. C-S 272 575 Bond Energies Bond Energies Example: Calculate the reaction enthalpy for the Example: Calculate the reaction enthalpy for the combustion of methane combustion of methane Step 1—Write balanced chemical equation: Step 3—Determine energy released in forming new bonds: CH4 + 2 O2 → CO2 + 2 H2O 2 C=O bonds: 2 x 750 kJ/mol = 1500 kJ 4 O-H bonds: 4 x 467 kJ/mol = 1868 kJ Step 2—Determine energy needed to break bonds: total energy to form bonds = 3368 kJ 4 C-H bonds: 4 x 416 kJ/mol = 1664 kJ Step 4—Determine enthalpy of reaction: 2 O=O bonds: 2 x 498 kJ/mol = 996 kJ ΔErxn = ΣBEreact - ΣBEprod total energy to break bonds = 2660 kJ = 2660 kJ – 3368 kJ = -708 kJ ΔHrxn = -802 kJ (actual value) Reaction Energies Reaction Energies Example: combustion of acetylene Example: combustion of acetylene 2 C2H2 + 5 O2 → 4 CO2 + 2 H2O 2 C2H2 + 5 O2 → 4 CO2 + 2 H2O • Which bonds are broken? • Which bonds are formed? -1 -1 4 x C-H 4 (415 kJ mol ) = 1660 kJ 8 x C=O 8 (800 kJ mol ) = 6400 kJ -1 -1 2 x C≡C 2 (835 kJ mol ) = 1670 kJ 4 x O-H 4 (460 kJ mol ) = 1840 kJ -1 5 x O=O 5 (495 kJ mol ) = 2475 kJ total energy released = 8240 kJ total energy needed = 5805 kJ ΔErxn = ΣBEreact - ΣBEprod = 5805 kJ – 8240 kJ mol = -2435 kJ (-2512 kJ actual) 6 4/15/13 Reaction Energies Reaction Energies C H + O Exothermic reaction 2 2 2 Example: reacants 6 H2O + 2 N2 → 4 NH3 + 2 O2 • Which bonds are broken? 12 x O-H 12 (460 kJ mol-1) = 5520 kJ 2435 kJ 2 x N≡N 2 (945 kJ mol-1) = 1890 kJ Energy total energy needed = 7410 kJ CO2 + H2O products Reaction Energies Reaction Energies Example: Endothermic reaction NH + O 6 H2O + 2 N2 → 4 NH3 + 3 O2 3 2 • Which bonds are formed? products 12 x N-H 12 (390 kJ mol-1) = 4680 kJ 1245 kJ -1 3 x O=O 3 (495 kJ mol ) = 1485 kJ Energy total energy released = 6165 kJ H2O + N2 reacants ΔErxn = ΣBEreact - ΣBEprod = 7410 kJ – 6165 kJ = 1245 kJ (1267 kJ actual) Molecular Orbitals Molecular Orbitals When atomic orbitals (AO’s) overlap to create a Because electrons can be described as waves, covalent bond, the result is the formation of when the AO’s overlap, the waves may either molecular orbitals (MO’s).