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Unformatted text preview: Chapter 9 Chapter Chemical Bonding Chapter Goals Chapter • Understand the difference between ionic and Understand covalent bonds. covalent • Draw Lewis electron dot structures for small Draw molecules and ions. molecules • Use the valence shell electron-pair repulsion Use theory (VSEPR) to predict the shapes of simple molecules and ions and to understand the structure of more complex molecules. structure • Use electronegativity to predict the charge Use distribution in molecules and ions and to define the polarity of bonds. the • Predict the polarity of molecules • Understand the properties of covalent bonds and Understand their influence on molecular structure. their Introduction: Valence Electrons * Attractive Attractive forces that hold atoms together in compounds are called chemical bonds. compounds * The electrons involved in bonding are usually those in the outermost (valence) shell. The (inner) those shell. core electrons are not involved in chemical core lectrons behavior. behavior. Valence electrons: For the main groups (representative) elements they For are the outer most s and p electrons. The # of and valence electrons is equal to the group number. valence * For the transition elements they are the ns and For ns (n−1)d electrons. (n Lewis Dot Symbols for Atoms Lewis * Lewis dot formulas or Lewis dot symbols are a Lewis convenient bookkeeping method for tracking valence electrons. valence Valence electrons are those electrons that are Valence transferred or involved in chemical bonding. They are chemically important. They Symbol considered to have 4 sides. Two dots per side maximum. . H . Li .. Be .. He .. .. .. .. .. .. . . C . . N . . O . . F . . Ne . B . . . . .. . .. . Lewis Dot Symbols for Atoms Lewis Elements that are in the same periodic group Elements have the same Lewis dot structures. have . . Li & Na .. .. . N. & .P . . . .. .. .. .. F . .. & . Cl .. Formation of Bonds Formation When a chemical reaction occurs, the valence When electrons of the atoms are reorganized so that net attractive forces −chemical bonds− occur between atoms. between Chemical bonds are classified into two types: Ionic bonding results from electrostatic attractions between ions, which are formed by the transfer of one or more electrons from one transfer atom to another. atom Covalent bonding results from sharing one or sharing more electron pairs between two atoms. more Comparison of Ionic and Covalent Compounds Compounds Melting point comparison Ionic compounds are usually solids with high melting points high Typically > 400oC Covalent compounds are gases, liquids, Covalent or solids with low melting points or Typically < 300oC Solubility in polar solvents (such as water) Ionic compounds are generally soluble Covalent compounds are generally Covalent insoluble insoluble Comparison of Ionic and Covalent Compounds Compounds Solubility in nonpolar solvents Ionic compounds are generally insoluble Covalent compounds are generally soluble Conductivity in molten solids and liquids Ionic compounds generally conduct electricity They contain mobile ions Covalent compounds generally do not Covalent conduct electricity conduct Comparison of Ionic and Covalent Compounds Compounds Conductivity in aqueous solutions Ionic compounds generally conduct electricity They contain mobile ions Covalent compounds are poor conductors of Covalent electricity electricity Formation of Compounds Ionic compounds are formed between elements Ionic with large differences in electronegativity large Often a metal and a nonmetal Covalent compounds are formed between Covalent elements with similar electronegativities with Usually two or more nonmetals Ionic Bonding Ionic Formation of Ionic Compounds Formation An ion is an atom or a group of atoms An ion possessing a net electrical charge. possessing Ions come in two basic types: positive (+) ions or cations positive cations These atoms have lost 1 or more These electrons. electrons. negative (−) ions or anions ions anions These atoms have gained 1 or more These electrons. electrons. That applies to binary compounds That binary Formation of Ionic Compounds Formation Ionic bonds are formed by the attraction of cations for anions usually to form solids. cations Commonly, metals react with nonmetals to Commonly, form ionic compounds. form The formation of NaCl is one example of an The ionic compound formation. ionic Formation of Ionic Compounds Formation Reaction of Group IA Metals with Group VIIA Nonmetals VIIA IA metal VIIA nonmetal 2 Li (s) + F2(g) silver yellow solid gas Formation of Ionic Compounds Formation Reaction of Group IA Metals with Group VIIA Nonmetals VIIA IA metal VIIA nometal 2 Li (s) + F2(g) → 2 LiF(s) silver solid yellow gas white solid o with an 842 C melting point Formation of Ionic Compounds Formation Reaction of Group IA Metals with Group VIIA Nonmetals VIIA The underlying reason for the formation of LiF The lies in the electron configurations of Li and F. lies 1s 2s 2p 1s 2s 2p Li ↑↓ ↑ lose lose ↑↓ s one electron F ↑↓ ↑↓ ↑↓ ↑↓ ↑ gains one ↑↓ ↑↓ ↑↓ ↑↓ electron electron These atoms form ions with these configurations. + 2 Formation of Ionic Compounds Formation Reaction of Group IA Metals with Group VIIA Reaction Nonmetals *We can also use Lewis dot formulas to *We represent the neutral atoms and the ions they form. they *electrons are transferred until the metal loses all its valence *electrons electrons and the nonmetal has an octet (Octet Rule) electrons (Octet Li . + .. .. .F .. + Li .. .F . . .. . Formation of Ionic Compounds Formation For the reaction of IA metals with VIA nonmetals, a good example is the reaction of lithium with oxygen. reaction The reaction equation is: + 2 2 Li (s) + 1 / 2O 2(g) → Li O 2( s) Formation of Ionic Compounds Formation Draw the electronic configurations for Li, O, and their appropriate ions. O, 2s 2p 2s 2p 2s 2p 2s Li [He] ↑ → Li + O [He] ↑↓ ↑↓ ↑ ↑ → O2- ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ octet rule octet Draw the Lewis dot formula representation Formation of Ionic Compounds Formation Draw the electronic representation of the Ca and N Draw reaction. reaction. Ca [Ar] N [He] 3 Ca . . 4s 4p 4p 2+ ↑↓ → Ca2+ ↑↓ 2s 2p 2s 2p ↑↓ ↑ ↑ ↑ → N3↑↓ + .. 2 .N. . 4s 4s 4p 2s 2p 2s ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 2+ 3 Ca .. 3. N. 2 [. .. . ] Other IIA and VA elements behave similarly. Symbolically, this reaction can be represented as: Symbolically, 2+ 3 M(s) + 2 X(g) → M32+ X23M can be the IIA elements Mg to Ba. X can be the VA elements N to As. Formation of Ionic Compounds Formation Simple Binary Ionic Compounds Table Reacting Groups Compound General Formula Compound Example IA + VIIA IIA + VIIA MX MX MX2 MX NaF NaF BaCl2 BaCl IIIA + VIIA MX3 MX AlF3 AlF IA + VIA IA M2X Na2O Na IIA + VIA IIIA + VIA MX MX M2X3 BaO BaO Al2S3 Al Ion Attraction and Lattice Energy Ion Ionic Bond • electrostatic attraction between oppositely electrostatic charged ions charged • non-directional • strength directly proportional to charges strength of ions and inversely proportional to distance between ion centers distance Coulombic Ion Attraction and Lattice Energy Coulombic Q cQ a E∝ d Qc × Qa E = C × ────── is negative because of charges ────── is d 1 E = ionic bond strength C (a constant) = —— ionic —— 4πε 0 Qc = charge on cation (is positive) Qa = charge on anion (is negative) d = distance between centers of ions Q cQ a E∝ d Product of charges NaCl Na+ Cl– −1 CaCl2 Ca2+ Cl– −2 CaS Ca2+ S–2 −4 Al2S3 Al3+ S2– −6 Qc x Qa is the heaviest factor. Product is much more negative for Al2S3. The least negative product is NaCl’s Q cQ a E∝ d shortest d Na+ F– 0.095 nm E ∝ − 4.33 4.33 d Na+ Cl– 0.136 nm longest d NaF has the most negative E and NaI the least d Na+ 0.095 nm E ∝ − 3.22 3.22 I– 0.216 nm Na+ Br– Lattice Energy (∆ Hlattice ) Lattice the energy of formation of 1 mol of solid the crystalline ionic compound when ions in the gas phase combine. It is always negative. It Among several compounds, the one with the most negative ∆ Hlattice is said to have the highest lattice energy (absolute value), and highest Tfusion lattice Mg2+(g) + 2F–(g) → MgF2(s) ∆ Hlattice = −2910 kJ/mol Mg for NaF, ∆ Hlattice = −911 kJ/mol for for KF, ∆ Hlattice = −815 kJ/mol for WHY? charge of Mg2+ (2) > Na+= K+, # of F− (1) d, radius of Mg2+ < Na+ < K+ (see chapt. 8) Mg Lattice Energy Lattice Small ions with high ionic charges have more Small negative lattice energies (higher lattice energies). negative ). Large ions with small ionic charges have less negative lattice energies. negative Use this information, plus the periodicity rules from Chapter 8, to arrange these compounds in order of Chapter decreasing values of lattice energy KCl, Al2O3, CaO Ionic radii: Ionic ∆ Hlattice 1.33 1.81 1.06 1.40 0.57 1.40 Å K+Cl− > Ca2+O2− > 2Al3+3O2− least negative least most negative Lattice Energy Lattice Arrange the following compounds in order of Arrange decreasing value of lattice energy (least negative to value most) CaO, MgO, SrO, and BaO The anion (O2−) is common to the four is oxides. In the equation Q c Qa E = C × ───── ───── d Qc and Qa are the same (+2 and −2). Then, the difference is made by d. The four cations are in the The same group, so d will be for Ba2+>Sr2+>Ca2+>Mg2+. E is negative and inversely proportional to d: the least negative the most negative the Ba2+O2− > Sr2+O2− > Ca2+O2− > Mg2+O2− (highest E) Ba (highest Ionic Bond Formation Ionic • Born-Haber Cycle Breaks formation of an ionic compound Breaks from its elements into a series of theoretical steps and considers the energetics of each Example: consider formation of NaCl Example: Na(s) + 1/2Cl2(g) → NaCl(s) ∆ H° = –410.9 kJ The Born-Haber Cycle is a level diagram that The Born-Haber applies Hess’s law (chapter 6): ∆ H of reaction = ∑ ∆ H of all theoretical steps of of 725.4 E (kJ) 376.4 229.4 107.7 0 Na+(g) + Cl(g) + e– EACl IENa Na(g) + Cl(g) Na(g) + 1/2Cl2(g) ∆ Hatom,Cl Na(s) + 1/2Cl2(g) ∆ Hsub, Na ∆ H°f, NaCl –410.9 Na+(g) + Cl–(g) NaCl(s) ∆ Hlattice The Born-Haber Cycle of NaCl The Born-Haber The Born-Haber Cycle of NaCl The Born-Haber ∆ H°f = ∆ Hsublimation, Na + ∆ Hatom, Cl + IE1, Na + EACl + EA ∆ Hlattice, NaCl (the two in blue are negative) ∆ H°f = 107.7kJ + 121.7kJ + 496.0kJ + (–349.0kJ) + (–787.3 kJ) (all values are given here). (–787.3 ∆ H°f = –410.9 kJ If unknown, ∆ Hlattice can be calculated If ∆ Hlattice = ∆ H°f − ∆ Hsublimation,Na − ∆ Hatom,Cl − IE1, Na − EACl EA Covalent Bonding Covalent Covalent bonds are formed when atoms Covalent share electrons. share If the atoms share 2 electrons (a pair) a single If covalent bond is formed. covalent If the atoms share 4 electrons (two pairs) a double If double covalent bond is formed. If the atoms share 6 (three pairs) electrons a triple If covalent bond is formed. covalent The attraction between the nuclei and the The electrons is electrostatic in nature. Directional. Directional. The atoms have a lower potential energy when The bound. bound. Formation of Covalent Bonds Formation This figure shows the potential energy of an This H2 molecule as a function of the distance between the two H atoms. between Unstable, Unstable, because of high energy energy Bond Bond energy energy Sharing two electrons Bond is formed, E < 0 Free atoms, Free Energy = 0 Energy Formation of Covalent Bonds Covalent Representation Representation of the formation of an H2 molecule from H atoms from Formation of Covalent Bonds Formation We can use Lewis dot formulas to show We covalent bond formation. H2 molecule formation representation. H. + H. . H . H or H2 duet rule duet rule .. .. . . . H .. HCl molecule formation . Cl . or HCl H . + . Cl . .. Formation of Covalent Bonds Formation Homonuclear diatomic molecules . H. H or · N · · · N· · ··· · HH or .. .. . .F .F. . . .. .. or .. . .F .. .. . F. .. ·N N· · · Heteronuclear diatomic molecules: Heteronuclear hydrogen halides ·· .F· H. · ·· . ·· · H . Br· ·· or ·· H F· · ·· or ·· H Br· · ·· . ·· · H . Cl· ·· or ·· H Cl· · ·· + H• H• H • H• y y x • • •• • •F•F• •• •• • • Writing Lewis Formulas: Writing The Octet Rule The octet rule states that representative The octet elements usually attain stable noble gas elements electron configurations in most of their compounds. compounds. Lewis dot formulas are based on the octet Lewis rule. rule. We need to distinguish between bonding (or We bonding shared) electrons and nonbonding (or shared electrons nonbonding unshared or lone pairs of) electrons. lone The Octet Rule The N - A = S rule NAS rule Simple mathematical relationship to help us Simple write Lewis dot formulas. write N = number of electrons needed to achieve a noble gas needed configuration. configuration. N usually has a value of 8 for representative elements. usually N has a value of 2 for H atoms. has A = number of electrons available in valence shells of the available atoms. atoms. A is equal to the periodic group number for each is element. A is equal to 8 for the noble gases. is S = number of electrons shared in bonds (in bonding pairs.) shared bonding A-S = number of electrons in unshared, lone, pairs. lone The Octet Rule The For ions we must adjust the number of electrons For available, A. available, Add one e− to A for each negative charge. Subtract one e− from A for each positive charge. The central atom in a molecule or polyatomic ion is determined by: is The atom that requires the largest number of e-s to complete its octet. It goes in the center. to H is never central atom, it shares two e-s only. is For two atoms in the same periodic group, the less electronegative element goes in the center (the one towards the bottom of the group.) Lewis Structures of Covalent Compounds and Polyatomic Ions Polyatomic Drawing Lewis structures by this method, use the following as a guide: a) Draw skeletal Lewis structure. b) Draw the Lewis electron dot structure for each atom. (Use the method in b) which the electrons are spread to all four sides of an imaginary square before being paired.) For the sake of keeping the drawing as neat as possible, direct single electrons on adjacent atoms towards each other. possible, c) Draw a line from a single unpaired electron on the central atom to a single c) unpaired electron on the surrounding atom. This constitutes the formation of a covalent bond.) Continue doing this until each atom has an octet (exceptions are H, Be, B, Al, elements on rows 3, 4, 5, and 6.) No electrons should be left unpaired (only in rare cases will a species contain an unpaired electron.) For those atoms that can have more than an octet, if all of its single electrons are used in a covalent bond, and there are surrounding atoms with electrons still to be paired, then lone electron pairs are used in bonding. The electron pair(s) being shared must be placed between the two atoms forming the bond. must d) For polyatomic ions, -1 charge, add 1 electron to the most electronegative atom -2 charge, add 2 electrons (one to each of the most electronegative atoms) -3charge, add 3electrons (one to each of the most electronegative atoms) +1 charge, remove 1 electron from the least electronegative atom. Lewis Structures of Covalent Compounds and Polyatomic Ions Polyatomic e) Coordinate Covalent or Dative Bond Formation (a bond formed when both of the electrons in a bond are supplied by the same atom): If all the single electrons of one atom (atom 1) are used in binding and the adjacent atom (atom 2) has single electrons which need to be shared, then the electrons on atom 2 (the one still having single electrons) are paired and atom 1 donates a pair of electrons to atom 2 thus forming a coordinate covalent bond. A coordinate covalent bond is represented by →. The arrowhead is pointed to The the atom to which the electron pair has been donated. the Exceptions to Octet: a) Some atoms have less than an octet (Be, B, and Al are metals, but Some they can form covalent compounds.) | | H─ ─Be─ ─B─ ─Al─ H─ b) Central atoms of elements of rows 3, 4, 5, and 6 of the periodic table, b) elements can have more than an octet due to the availability of low laying d-orbitals. These elements can have up to 18 electrons surrounding d-orbitals. them. A Quick-N-Dirty Method for Drawing the Lewis Structure of Covalent Compounds and Ions Covalent The method will be illustrated by drawing the Lewis structure of SO 3. 1) Find the total number of valence electrons 1) by adding the number of valence electrons 1S = 1(6) = 6 by from each atom in the formula. 3O = 3(6) = 18 (For polyatomic ions add (anions) or subtract 24 (cations) the charge to or from the total number of valence electrons.) O 2) Draw a single bond from the central atom | 2) to each of the atoms surrounding it. O ─S─ O to 3) Subtract the electrons used so far in the 3 bonds x 2 e − = 6 e− structure from the total number of electrons. 24 − 6 = 18 e− structure Each bond contains two electrons. Each 4) The remaining electrons are spread, as .. 4) pairs, to the surrounding atoms first. Each :O: surrounding atom (except H) will receive .. | .. enough electron pairs to have an octet. If :O ─S─ O: •• • any electrons are left over, give them to the central atom. ••• any A Quick-N-Dirty Method for Drawing the Lewis Structure of Covalent Compounds and Ions Covalent 5) If the central atom does not have an octet, then 5) one or two single bonds are converted to double bonds or one double bond is converted to triple bond. If possible, make two double bonds before making one triple bond. making .. .. :O: :O: .. | .. .. :O ─S─ O: •• •• → .. :O: | .. :O=S ─ O: =S O: •• •• Remember exceptions to the octet rule (H, Be, B, Al, and elements from rows 3, 4, 5, and 6). Al, Lewis Structures of Covalent Compounds and Polyatomic Ions Polyatomic Example: Write Lewis dot and dash formulas for hydrogen cyanide, HCN. formulas A = 1 (H) + 4 (C) + 5 (N) = 10 H −C−N S (so far) = 2 bonds x 2 = 4 electrons to be spread = 10 − 4 = 6 electrons .. .. We put the 6e- over N, H−C−N: ¨ But, to fulfill the octet of C and N, or or H−C≡ N: N: H:C:::N: with four BP and one LP over N BP LP Lewis Structures of Covalent Compounds and Polyatomic Ions Polyatomic Example: Write Lewis dot and dash Example: formulas for the sulfite ion, SO32−. formulas A = 6(S) + 3 x 6 (O) + 2 (- charge) = 26 (S (so far) = 3 bonds x 2 = 6 e-s to be spread = 26 − 6 = 20 to 20 with an overall 2− charge 2− O−S−O | O .. .. .. .. :O−S−O: −O: ¨ |¨ :O: ¨ Lewis Structures of Covalent Compounds and Polyatomic Ions Polyatomic Other examples: CO2, NO2+, NO3−, HNO3, SO42−, H2SO4, PO43−, H3PO4 PO Cl2O7, CH3CH2OH (ethanol), CH3COOH (acetic Cl COOH acid) acid) PCl3, PCl5, SF6, SOCl2, IF5, IF4+ Lewis Structures of Covalent Compounds and Polyatomic Ions Polyatomic Other examples: BBr3, BBr4− less than octet less ·· · Br 6 e- on B · ·· AsF5 ·· . As . . ·· ·F· ·· · ·· · ·· · · F · As · F · · ·· · · ·· ·· ·· · F· · F · · ·· · ·· ·· ·· .B ·· · B · Br · · · · ·· ·· · Br · · · ·· or ·· · Br . · ·· or ·· · Br · ·· B ·· Br · · ·· · Br · · · ·· ·· · F .more than octet · more ·· ·· · F · 10 valence e- on As ·· · · ·· · ·F · ·· As F · ·· ·F · · F · · ·· · ·· ·· Lewis Structures of Covalent Compounds and Polyatomic Ions Polyatomic Isoelectronic Species: Molecules and/or ions having the same Molecules number of valence electrons and the same Lewis structures. Lewis [:N≡ O:]+ [:N≡ N:] N:] [:C≡ O:] O:] They all have 3 BP and 2 LP BP LP [:C≡ N:]− Resonance Resonance Example: Write Lewis dot and dash Example: formulas formulas for sulfur trioxide, SO3. N = 8 (S) + 3 x 8 (O) (S) A = 6 (S) + 3 x 6 (O) S A-S · ·· · · · · · O · S· · O · ·· ·· ·· ·O· · · ·· or = 32 32 = 24 24 =8 ·· = 1·6 16 ·O ·· S ·O · · ·· · 4BP, 8 LP 4BP, · O· ·· Resonance Resonance There are three possible structures for SO3. The double bond can be placed in one of three places. three ·O · ·· S ·O · · ·· · ·· · O· ·· ·· ·O · ·· S ·O· ·· ·· O· · ·· ·· ·O · ·· S ·O · · ·· · O· · ·· When two or more Lewis formulas are necessary to show the bonding in a molecule, we must use equivalent resonance structures to show the molecule’s structure. Double-headed arrows are used to indicate resonance formulas. Resonance Resonance Resonance is a flawed method of Resonance representing molecules. representing • There are no single or double bonds in There SO3. SO In fact, all of the bonds in SO3 are equivalent. equivalent. The best Lewis formula of SO3 that can be • drawn is: drawn O S O O CO2 CO •• ••• • O C O• ••• •• C 4 e– 2 O 12 e– 16 e– 2 bonds – 4 e– 12 e– – 12 e– 12 0 e– Three Resonance structures for CO2 •• ••• • O C O• • ••• • O C O• • •• •• • O C O• • •• • Resonance Resonance Other examples: O3, C6H6 (benzene), NO3− (nitrate ion) Molecules with odd number of electrons Molecules NO has 11 valence electrons NO .. .. ⋅ N= O ⋅⋅ NO2 has 17 valence electrons NO ⋅⋅ ⋅ ⋅⋅ :O ─N= O ⋅⋅ ⋅⋅ ↔ ⋅⋅ ⋅ ⋅⋅ O = N─ O : ⋅⋅ ⋅⋅ They are members of a family of substances They called Free Radicals: they have an unpaired e−. Free Very reactive: e.g. dimerization of NO2 to N2O4. 2 NO2(g) NO (g) → N2O4(g) Central Atoms with Single-Bond Pairs and Lone Pairs Lone Valence Shell Electron-Pair Repulsion theory Valence (VSEPR Theory): (VSEPR Regions of high electron density around the central atom are arranged as far apart as possible to atom minimize repulsions. minimize Lone pairs of electrons require more volume than shared pairs. Hence, _Lone pair to lone pair is the strongest repulsion. _Lone pair to bonding pair is intermediate repulsion. _Bonding pair to bonding pair is the weakest repulsion. Mnemonic for repulsion strengths lp/lp > lp/bp > bp/bp VSEPR Theory VSEPR There are five basic molecular shapes based There on the number of regions of high electron density (r.h.e.d) around the central atom. density Several modifications of these five basic Several shapes will also be examined. shapes VSEPR Theory VSEPR 1. Two regions of high electron density (BP + 1. Two BP LP) around the central atom. LP 2. Three regions of high electron density (BP + BP LP) around the central atom. LP VSEPR Theory VSEPR 3. Four regions of high electron density 3. (BP + LP) around the central atom. around VSEPR Theory VSEPR 4. Five regions of high electron density 4. (BP + LP) around the central atom. around VSEPR Theory VSEPR 5. Six regions of high electron density 5. (BP + LP) around the central atom. around VSEPR Theory VSEPR Frequently, we will describe two geometries for Frequently, each molecule or ion. or Electronic geometry is determined by the locations Electronic of regions of high electron density around the central atom(s). central Molecular geometry determined by the arrangement of atoms around the central atom(s). atom(s). Electron pairs are not used in the molecular Electron geometry determination, just the positions of the atoms in the molecule are used. the VSEPR Theory VSEPR An example of a molecule that has the same An electronic and molecular geometries is methane - CH4. methane Electronic and molecular geometries are Electronic tetrahedral. tetrahedral. H HC H H VSEPR Theory VSEPR An example of a molecule that has different An electronic and molecular geometries is water- H2O. waterElectronic geometry is tetrahedral. Molecular geometry is bent or angular. H HC H H Predicting Molecular Geometries and Bond Angles Predicting Write the Lewis structure. Count the number of bonding pairs and lone pairs Count number around the central atom (regions of high electron around Density = groups of electrons). groups Treat a double or triple bond as one region h.e.d. Treat region Use Table provided to predict geometry. When lone pairs are present, bond angles are When slightly less than those given (except, linear = 180° and squared planar = 90°.) When only two atoms are present, it is inappropriate to speak of bond angles. bond Predicting Molecular Geometries and Bond Angles Predicting 2 BP CO2, CS2, BeCl2 linear 3 BP BF3, CO32− BF trigonal planar 2 BP 1 LP SO2, NOCl SO bent 4 BP CH4, NH4+, BF4− tetrahedral 3 BP 1 LP BP NH3, H3O+ trigonal pyramidal 2 BP 2 LP BP H2O, NH2− V-shaped or bent 5 BP PCl5, AsF5 PCl AsF trigonal bipyramid Charge Distribution in Covalent Bonds and Molecules Molecules The way the electrons are distributed in the The molecule is called its charge distribution. charge Formal Charges on Atoms: an accounting tool for electron ownership = (# valence e– in free atom) – (# e– in lone pairs on atom) – ½(# bonded e– on atom) electrons electrons Formal charge = FC = group number of atom − [LPE + ½(BE)] [LPE FC = group number − LPElectr − BP FC Formal Charge (FC) and Best structure Formal Best structure has • zero FC on all atoms • lowest FC possible • negative FC on most electronegative negative atoms and positive FC on least electronegative atoms electronegative The most electronegative elements are at the The most top right of periodic table (except for the top noble gases.) noble For example, consider thiocyanate ion SCN– SCN •• N •• •• CS •• •• C •• •• SN •• •• S •• •• NC •• For example, consider thiocyanate ion, SCN – For What is the best structure? •• N •• •• CS •• •• C •• FCN = 5 – 4 – 2 = –1 •• SN •• •• S •• •• NC •• For example, consider thiocyanate ion For SCN– •• N •• •• CS •• •• C •• FCN = 5 – 4 – 2 = –1 FCC = 4 – 0 – 4 = 0 FCS = 6 – 4 – 2 = 0 •• SN •• •• S •• •• NC •• For example, consider thiocyanate ion For SCN– –1 0 0 •• •• •• •• •• •• SNC NCS CSN •• •• •• •• •• •• FCN = 5 – 4 – 2 = –1 FCC = 4 – 4 – 2 = –2 FCS = 6 – 0 – 4 = +2 For example, consider thiocyanate ion For SCN– –1 0 0 –2 +2 –1 •• •• •• •• •• •• NCS CSN SNC •• •• •• •• •• •• FCN = 5 – 0 – 4 = +1 FCC = 4 – 4 – 2 = –2 FCS = 6 – 4 – 2 = 0 For example, consider thiocyanate ion For SCN– –1 0 0 –2 +2 –1 0 +1 –2 •• •• •• •• •• •• NCS CSN SNC •• •• •• •• •• •• For example, consider thiocyanate ion For SCN– –1 0 0 –2 +2 –1 0 +1 –2 •• •• •• •• •• •• NCS CSN SNC •• •• •• •• •• •• best structure because lowest FC and negative FC on most electronegative atom Sum of formal charges = −1 + 0 + 0 = −1, Sum −1, that is the charge of the ion. Formal Charge (FC) Formal NH4+ NH CO2 CO H | H─N─H | H .. .. O= C= O .. .. C = 4 − 8/2 = 0 N= 5 − 8/2 = +1 8/2 H = 1 − 2/2 = 0 2/2 O | .. .. SO42− O─S─O: SO . | ... O S = 6 − 8/2 = 6 −4 =+ 2 O = 6 − (6 + 2/2) = −1 O = 6 − (4 + 4/2) = 0 -1 +1 0 0 +1 -1 .. .. .. .. O3 (ozone) :O─O= O O= O─O: Bond Polarity and Electronegativity Polar and Nonpolar Covalent Bonds Polar Covalent bonds in which the electrons are Covalent shared equally are designated as nonpolar shared nonpolar covalent bonds. covalent Nonpolar or pure covalent bonds have a Nonpolar pure symmetrical charge distribution. To be pure covalent, the two atoms involved in the bond must be the same element to in share equally. . H. H or HH · N · · · N· · ··· · or ·N N· · · Pure Covalent Bonds + H• H• H • H• y y x • • •• • •F•F• •• •• • • Bond Polarity and Electronegativity Polar and Nonpolar Covalent Bonds Polar Covalent bonds in which the electrons are Covalent not shared equally are designated as polar not polar covalent bonds. covalent Polar covalent bonds have an asymmetrical Polar charge distribution. charge To be a polar covalent bond the two atoms To involved in the bond must have different electronegativities. electronegativities. y y Br F •• •• • Br• F • ••• Polar Covalent Bonds Electronegativity Electronegativity Electronegativity is a measure of the relative tendency Electronegativity of an atom to attract electrons to itself when chemically combined with another element. combined Electronegativity is measured on the Pauling scale. Pauling Fluorine is the most electronegative element. Fluorine Cesium and francium are the least electronegative Cesium elements. elements. For the representative elements, electronegativities For usually increase from left to right across periods increase and decrease from top to bottom within groups. decrease Electronegativity Electronegativity Electronegativity Electronegativity Example: Arrange these elements based on Example: their electronegativity. their Se, Ge, Br, As (period # 4) Ge < As < Se < Br Arrange these elements based on their Arrange electronegativity. electronegativity. Be, Mg, Ca, Ba (group IIA) Ba < Ca < Mg < Be Electronegativity, polarity, and Ionic bonds A non-polar bond is formed when the non-polar difference of electronegativity is between 0.0 – 0.4 0.0 A polar bond is formed when the difference of polar electronegativity is 0.5 – 1.8 (partial charge) 0.5 Ionic bonds form when the difference of Ionic electronegativity is greater than 1.8 (net charge) greater no charge Se−H (0.3) Se non-polar non-polar Cl is more electronegative Cl δ+ δ− C−Cl (0.5) K+ and Br− (1.9) polar ionic polar covalent ionic CsF (3.3) pure covalent Bonding Continuum Bonding pure ionic 1.8 nonpolar covalent 0.4 BrF (1.3) ∆ E’neg E’neg 0 H2 (0) (0) Molecular Polarity Molecular The higher the diff. of electronegativity the more The polar the bond is polar Electroneg ativities H I 2.1 2.5 0 .4 Difference = 0.4 slightly polar bond H Electroneg ativities F 2.1 4.0 1 .9 Difference = 1.9 very polar bond Polar and Nonpolar Covalent Bonds Polar Polar molecules can be attracted by magnetic Polar and electric fields. and Dipole Moments Dipole Molecules whose centers of positive and Molecules negative charge do not coincide, have an asymmetric charge distribution, and are polar. These molecules have a dipole moment. dipole The dipole moment has the symbol µ ; µ is the The product of the distance, d, separating charges of equal magnitude and opposite sign, and the magnitude of the charge, δ . It is indicated by crossed arrow pointing from It positive end to negative end of dipole positive µ =δ × d Dipole Moments Dipole Molecules that have a small separation of Molecules charge have a small µ.. µ Molecules that have a large separation of Molecules charge have a large µ.. µ For example, HF and HI (longer d): δ H - Fδ 1.91 Debye units + - δ H -Iδ 0.38 Debye units + - Ratio Ratio 5.0 5.0 Dipole Moments Dipole There are some nonpolar molecules that have There nonpolar polar bonds. There are two conditions that must be true There for for a molecule to be polar. 1. There must be at least one polar bond 1. present or one lone pair of electrons. present 2. The polar bonds, if there are more than 2. one, and lone pairs must be arranged so that their dipole moments do not cancel one not Polarity of Molecules Polarity Molecules in which dipole moments of the Molecules bonds do not cancel are polar molecules bonds µ total ≠ 0 Molecules that do not contain polar bonds or Molecules in which all dipole moments cancel are nonin polar molecules µ total = 0 CO2 CO δ–δ+δ– O C O nonpolar µ total = 0 vs H2O δ– O H δ+ H δ+ polar µ total ≠ 0 Dipole Moments of Polyatomic molecules Dipole Some models H H HC H H HC H H BeCl2 BF 3 1 lone pair H HC H H .. H HC H H H H HC H H CH4 CH NH 3 N H H pyramidal H 2O Dipole Moments of Polyatomic molecules Dipole Some models BeCl2 BF3 BF H2O Dipole Moments of Polyatomic molecules Dipole Some examples to work… CCl4 (µ = 0) CHCl3 (µ = 1.04 D) CHCl No net dipole Net dipole Net CH2Cl2 (µ = 1.60 D) CH3Cl (µ = 1.92 D) CH NH3 (µ = 1.47 D) NH CH4 … CCl4 CH Polar or Not? Only CH4 and CCl4 are NOT polar. These are the only two molecules that are “symmetrical.” only Bonding Order Bonding Order of a bond is the number of bonding electron pairs shared by two atoms in a molecule. There are bond orders of 1 (single), 2 (double), 3 (triple), and fractional (1.5, 2.5, ..) Average # of shared pairs linking atoms verage Bond order = ──────────────────────── ond rder ──────────────────────── # of links in the molecule or ion (no L.P.) (no H – Cl Cl 1 O–O=O 12 O=C=O 22 H – C≡ N 13 For the molecule (around the central atom): A.B.O. 1/1 = 1 3/2 = 1.5 4/2 = 2 4/2 = 2 Bond Strength and Bond Length Bond bond length (shortest) triple < double < single (shortest) bond strength bond single < double < triple (strongest) Bond Length (Å) Strength (kJ/mol) Bond C–C 1.54 348 348 C=C 1.34 614 (not 2x) C≡ C 1.20 1.20 839 (not 3x) Bond Energies Bond • chemical reactions involve breaking bonds in chemical reactant molecules and making new bond to create the products create • the ∆ H°reaction can be calculated by comparing the the cost of breaking old bonds to the profit from making new bonds making • the amount of energy it takes to break one mole the of a bond in a compound is called the bond energy energy – in the gas state – homolytically – each atom gets ½ homolytically bonding electrons bonding Trends in Bond Energies Trends • the more electrons two atoms share, the the stronger the covalent bond stronger – C≡C (837 kJ) > C=C (611 kJ) > C−C (347 kJ) – C≡N (891 kJ) > C=N (615 kJ) > C−N (305 kJ) • the shorter the covalent bond, the stronger the the bond bond – Br−F (237 kJ) > Br−Cl (218 kJ) > Br−Br (193 kJ) – bonds get weaker down the column of bonds periodic table. periodic Using Bond Energies to Estimate ∆ H°rxn • the actual bond energy depends on the the surrounding atoms and other factors surrounding • we often use average bond energies to average estimate the ∆ Hrxn – works best when all reactants and works products in gas state products • bond breaking is endothermic, bond ∆ H(breaking) = + • bond making is exothermic, ∆ H(making) = − bond H(making) ∆ Hrrxn = ∑ (∆ H(bonds broken)) + ∑ (∆ H(bonds H(bonds H(bonds xn formed)) formed)) Estimate the Enthalpy of the Following Reaction Estimate H H + O O H O O H Estimate the Enthalpy of the Following Reaction Estimate H H + O O H O O H H2(g) + O2(g) → H2O2(g) reaction involves breaking 1mol H-H and 1 mol reaction O=O and making 2 mol H-O and 1 mol O-O O=O bonds broken (energy cost) (+436 kJ) + (+498 kJ) = +934 kJ bonds made (energy release) bonds 2(−464 kJ) + (−142 kJ) = -1070 ∆ Hrxn = (+934 kJ) + (-1070. kJ) = -136 kJ (Appendix ∆ H°f = -136.3 kJ/mol) (Appendix Bond Lengths Bond • the distance between the the nuclei of bonded atoms is called the bond length called bond • because the actual bond because length depends on the other atoms around the bond we often use the bond average bond length average – averaged for similar averaged bonds from many compounds compounds Trends in Bond Lengths Trends • the more electrons two atoms share, the the shorter the covalent bond shorter – C≡C (120 pm) < C=C (134 pm) < C−C (154 pm) – C≡N (116 pm) < C=N (128 pm) < C−N (147 pm) • decreases from left to right across period – C−C (154 pm) > C−N (147 pm) > C−O (143 pm) • increases down the column – F−F (144 pm) < Cl−Cl (198 pm) < Br−Br (228 Br pm) • in general, as bonds get longer, they also get in weaker weaker Bond Lengths Bond ...
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This note was uploaded on 11/07/2011 for the course CHM 2045 taught by Professor Geiger during the Fall '08 term at University of Central Florida.

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