Review Chapter 9: The Basics of Chemical Bonding Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop.

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Presentation transcript:

Review Chapter 9: The Basics of Chemical Bonding Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop

Chapter 9 Concepts 2  Know the difference between an ionic and covalent bond  Electronegativity:  Identify polar bonds and estimate relative dipole moments  Ionic Bond:  Energetics of ionic bonding: lattice energy  Predict possible ionic compounds  For transitions metals predict ions that will have half, full, or empty d-orbitals  Covalent Bond:  Draw lewis dot structures  Predict most reasonable lewis dot structure  Formal charges  Electronegativity  Identify possible reasonable resonance structures  Draw hybrid resonance structures  Understand electron delocalization & stability from it.

Covalent vs Ionic Bonds Ionic Bonds result from electrostatic attraction between a cation and anion: metal-nonmetal (with the exception of NH 4 + and H 3 O + cations). Covalent bonds result from the sharing of electrons between two atoms: nonmetal-nonmetal. Li F Ionic Bonds Covalent Bonds

Ionic Bonds 4 Coulomb’s law determines the potential energy of two ions (q 1 and q 2 ) separated by a distance (r), where k is a proportionality constant. Lattice Energy For a stable ionic compound to form the potential energy must be lowered.  Lattice Energies are usually very large negative numbers.  Lattice Energies always exothermic Transition Metals Hard to predict ions electron configurations, but transition metal ions with exactly filled or half-filled d subshells are extra stable and therefore tend to form.  This is transition metals form multiple oxidation states

Electronegativity & Polar Bonds  +  – H——Cl  + on H =  – on Cl = –0.17  EN = |EN 1 – EN 2 | μ = q × r Atoms participating in covalent bonds will not share electron density equally if they have a difference in electronegativities of between 0.5 and 1.7.

Drawing Lewis Dot Structures Step [1] Determine the valence electrons for each element and draw the lewis symbol. Step [2] Count the valence electrons. The sum gives the total number of e − that must be used in the Lewis structure. For each atom the number of bonds = 8 – valence electrons. Step [3] Arrange the atoms next to each other that you think are bonded together. Place H and halogens on the periphery, since they can only form one bond. Step [4] Arrange the electrons around the atoms. Place one bond (two e − ) between every two atoms. Use all remaining electrons to fill octets with lone pairs, beginning with atoms on the periphery.

Drawing Lewis Dot Structures 1.Decide how atoms are bonded – Skeletal structure = arrangement of atoms. – Central atom Usually given first Usually least electronegative 2.Count all valence electrons (all atoms) 3.Place two electrons between each pair of atoms – Draw in single bonds 4.Complete octets of terminal atoms (atoms attached to central atom) by adding electrons in pairs 5.Place any remaining electrons on central atom in pairs 6.If central atom does not have octet – Form double bonds – If necessary, form triple bonds

The Octet Rule Exceptions –Holds rigorously for second row elements like C, N, O, and F –B and Be sometimes have less than octet BeCl 2, BCl 3 –2 nd row can never have more than eight electrons –3 rd row and below, atoms often exceed octet When atoms form covalent bonds, they tend to share sufficient electrons so as to achieve outer shell having eight electrons

Formal Charges & Reasonable Structures FC = # valence e – – [# bonds to atom + # unshared e – ] Most Stable Lewis Structure 1.Lowest possible formal charges are best 2.All FC   1  3.Any negative FC on most electronegative element

Resonance Stabilization = Resonance Structures Resonance Hybrid Drawing Good Resonance Structures: 1.All must be valid Lewis structures 2.Only electrons are shifted - Usually double or triple bond and lone pair - Nuclei can't be moved - Bond angles must remain the same Resonance Structures are equivalent lewis dot structures: - Atom connectivity does not change (ie, nuclei stay the same) - Only electrons change, ie, # of bonds between atoms can rearrange Hybrid Resonance Structures show how electrons are delocalized over the molecule, the more atoms that are participating in the delocalization the more stable the molecule is. 4.Number of unpaired electrons, if any, must remain the same 5.Major contributors are the ones with lowest potential energy (see above) 6.Resonance stabilization is most important when delocalizing charge onto two or more atoms

Coordinate Covalent Bonds

Problem Set A 1.Identify the covalent and ionic bonds: a.CaF 2 b.CCl 4 c.NaOH d.NH 4 NO 3 2.Which ionic solid is likely to have the smallest exothermic lattice energy? LiCl; CsCl; NaCl, KCl 3.What ion will form for each element? Draw the elements lewis symbol and write the ion’s electron configuration. a.I b.Rb c.P 4.Predict the ionic compound that will form between the following: a. Aluminum (Al) and Chlorine (Cl) b.Strontium (Sr) and Bromine (Br)

Problem Set B/C 5.Draw lewis dot structures for the following molecules: a.C 6 H 6 b.C 5 H 10 c.C 2 HCl 3 O 2 d.C 2 H 7 N e.CH 2 O 3 f.NH 4 CN 6.Which has the least polar bond? HCl, HF, HI, HBr. 7.What are 2 possible structures for XeO4? Please indicate the formal charge on Xe in both structures and decide which is the most reasonable. 8.Draw 3 resonance structures of NCO -. Evaluate the formal charges on each and decide which is the best structure. 9.Draw the best lewis structure for the following: a.HClO 4 b.XeF 4 c.I 3 – d.BrF 5

Solutions 1.Identify the covalent and ionic bonds: a.CaF 2 Ionic b.CCl 4 Covalent c.NaOH Ionic, OH - a polyatomic ion with covalent bonds d.NH 4 NO 3 Ionic, NH 4 + and NO 3 - are polyatomic ions with covalent bonds 2.CsCl, because Cs has the largest radius, therefore r large and E less negative. 3.What ion will form for each element? Write the ion’s electron configuration and the lewis symbol. a.[Kr] 5s 2 4d 10 5p 6 b.[Kr] or [Ar] 4s 2 3d 10 4p 6 c.[Ne] 3s 2 3p 6 4.Predict the ionic compound that will form: a.AlCl 3 b.SrBr 2 I Rb        

Solutions