ch9 - Chapter 9 Chemical Bonding Pt. 1 Goals • Understand...

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Unformatted text preview: Chapter 9 Chemical Bonding Pt. 1 Goals • Understand the difference between ionic and covalent bonds. • Lattice energy (ionic compounds) • Draw Lewis electron dot structures for small molecules and ions (the octet rule). • Understand the properties of covalent bonds and their influence on molecular structure. • Understand electronegativity and its use in predicting the polarity of bonds and molecules. • Predicting bond length and bond energies. Valence Electrons Lewis Dot Symbols for Atoms forces that hold atoms together in compounds are called chemical bonds. * The electrons involved in bonding are usually those in the outermost (valence) shell. The (inner) core electrons are not involved in chemical electrons behavior. Lewis dot formulas (Lewis dot symbols; Lewis structures) are a convenient method for tracking valence electrons. * Attractive Valence electrons: For the main groups (representative) elements they are the outermost s and p electrons. The # of valence electrons is equal to the group number. * For the transition elements they are the ns and (n−1)d electrons. (n− Types of Bonds When a chemical reaction occurs, the valence electrons of the atoms are reorganized so that net attractive forces (chemical bonds) occur between atoms. Chemical bonds are classified into two types: Ionic bonding results from electrostatic attractions between ions (ions are formed by the transfer of one or more electrons from one atom to another). Covalent bonding results from sharing one or more electron pairs between two atoms. . H . Li .. Be .. He .. .. .. .. .. .. . . . . B . . C . . N . . O . . .. . . Ne . F . . .. Formation of Ionic Compounds • We can also use Lewis dot formulas to represent the neutral atoms and the ions they form. .. .. . + . . Li . + . .. . F F Li . .. . [ ] LiF 3 Ca . . .. + 2 .N. . 3 Ca 2+ 3. .. . 2 [ . N. ] .. Coulombic Ion Attraction and Lattice Energy radius effect QQ E∝ c a d E ∝ d Qc × Qa E = -C ×────── d d Na+ F– negative due Qa E = ionic bond strength Q cQ a d d C: a constant Na+ Cl– NaF has the most negative E and NaI the least d Qc = charge on cation (is positive) Na+ Qa = charge on anion (is negative) I– Br– Na+ d = distance between centers of ions (sum of radii) E = −C × Cmpd Q cQ a charge effect d ions Product of charges NaCl Na+ Cl– −1 CaCl2 Ca2+ Cl– −2 CaS Ca2+ S–2 −4 Al2S3 Al3+ S2– −6 Lattice Energy (∆Hlattice ) kJ/mol (∆ the energy of formation of 1 mol of solid crystalline ionic compound when ions in the gas phase combine. It is always negative. negative. Among several compounds, the one with the most negative ∆Hlattice is said to have the highest lattice energy (absolute value)… and highest Tf value)… Mg2+(g) + 2F–(g) → MgF2(s) ∆Hlattice = −2910 kJ/mol NaF, ∆Hlattice = −911 kJ/mol NaF, KF, ∆Hlattice = −815 kJ/mol Al2S3 has the most negative E and NaCI the least WHY? charge of Mg2+ > Na+ = K+ d, radius of Mg2+ < Na+ < K+ Lattice Energy Lattice Energy Small ions with high charges have more negative lattice energies. Large ions with small charges have less negative lattice energies. Arrange the following compounds in order of decreasing lattice energy (least -ve to most -ve) CaO, MgO, SrO, and BaO CaO, MgO, SrO, The anion (O2−) is common to the four oxides. In the equation QQ Use this information, plus information from Chapter 8, to arrange these compounds in order of decreasing values of lattice energy: KCl, Al2O3, CaO KCl, Ionic radii: 1.33 1.81 ∆Hlattice K+Cl− |∆H | lattice 1.06 1.40 > least negative Ca2+O2− 0.57 > 1.40 Å 2Al3+3O2− most negative (absolute values) Al2O3 > CaO > KCl values) E = −C × c a d Qc and Qa = (+2 and −2). The difference is in d. The four cations are in group 2, so d varies Ba2+>Sr2+>Ca2+>Mg2+. E is negative and inversely proportional to d: least negative most negative Ba2+O2− > Sr2+O2− > Ca2+O2− > Mg2+O2− Ionic Bond Formation • Born-Haber Cycle Born-the formation of an ionic compound from its elements in a series of steps, and considers the energetics of each. Example: formation of NaCl Na(s) + ½ Cl2(g) → NaCl(s) ∆H° = –410.9 kJ (s) (s) The Born-Haber Cycle is an energy level Borndiagram that applies Hess’s law: Hess’ ∆Hrxn = ∑ ∆H of all steps 725.4 E (kJ) 376.4 229.4 107.7 0 Na+(g) + Cl(g) + e– (g) (g) EACl IENa Na+(g) + Cl–(g) (g) Na(g) + Cl(g) (g) (g) Na(g) + ½ Cl2(g) (g) ∆Hatom,Cl Na(s) + ½ Cl2(g) (s) ∆Hlattice ∆Hsub, Na sub, ∆H°f, NaCl –410.9 NaCl(s) (s) The Born-Haber Cycle of NaCl Born- Covalent Bonding ∆H°f = ∆Hsublimation, Na + ∆Hatom, Cl + IE1, Na + EACl + sublimation, atom, ∆Hlattice, NaCl (the two in blue are negative) lattice, Covalent bonds are formed when atoms share electrons. If the atoms share 2 electrons ( a pair) a single covalent bond is formed. If the atoms share 4 electrons (two pairs) a double covalent bond is formed. If the atoms share 6 (three pairs) electrons a triple covalent bond is formed. The attraction between the nuclei and the electrons is electrostatic in nature. Directional. The atoms have a lower potential energy when bound (see upcoming diagram). ∆H°f = 107.7kJ + 121.7kJ + 496.0kJ + (–349.0kJ) (– – 787.3 kJ ∆H°f = – 410.9 kJ If it is not known, ∆Hlattice can be calculated ∆Hlattice = ∆H°f − ∆Hsublimation,, Na − ∆Hatom,, Cl − IE1, Na sublimation atom − EACl Formation of Covalent Bonds Plot shows potential energy of H2 molecule as a function of the distance between the two H atoms. atoms Writing Lewis Formulas: The Octet Rule The octet rule states that representative elements usually attain stable, noble gas electron configurations in most of their compounds/ions. compounds/ions. molecule We must distinguish between bonding (shared) electrons and nonbonding shared) (unshared or lone pairs of) electrons. of) The Octet Rule & Lewis structures For all ions we must adjust the number of electrons available based on charges. charges. i.e., add one e− for each negative charge; subtract one e− for each positive charge. The central atom in a molecule or polyatomic ion is determined by: -the atom that requires the largest number of electrons to complete its octet. It goes in the center. If two atoms are in the same, the less electronegative element goes in the center. Drawing the Lewis Structures (STRUCTURAL; ch 3) Example: SO3. 1) Find the total number of valence electrons 1S = 1(6) = 6 e− 3O = 3(6) = 18 e− available = 24 e− O | O ─S─ O 2) Draw a single bond from the central atom to each of the atoms surrounding it. 3 bonds × 2 e− = 6 e− 24 − 6 = 18 e− 3) Subtract electrons used in bond formation from the total number of electrons. 4) The remaining electrons (in pairs) are placed .. on 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, put them on the central atom. •• Drawing the Lewis Structures Lewis Structures 5) If the central atom does not have an octet, then one or two single bonds are converted to double bonds, or, a double bond is converted to a triple bond. Make double bonds before making a triple bond. Write Lewis dot and dash formulas for the sulfite ion, SO32−; (3 O’s bonded to S) O’ Valence e- = 6 (S) + 3 × 6 (O) = 24 e Charge = (2-) = 2 (2 Total = 26 electrons in bonds = 6 This polyatomic ion has ? BP and ? LP. .. :O: .. | .. :O ─S─ O: •• •• → .. :O: .. .. •• | O=S ─ O: •• •• ·· ·· ·O · S · · ·· ·· · O · ·· Exceptions to the octet rule: (H, Be, B, Al, and other elements in rows 3, 4, 5, and 6). Lewis Structures (practice) CO2, NO2+, NO3−, HNO3, SO42−, H2SO4, PO43−, H3PO4 Cl2O7, CH3CH2OH (ethanol), CH3COOH (acetic acid) PCl3, PCl5, SF6, SOCl2, IF5, IF4+ ·· · O · 2· · ·· · · or Lewis Structures: BBr3 T # electrons: B = 3; Br = 3 × 7 = 24e3 bonds = 6e- ; 18 e- left ( 3 pairs per B atom) Br B Br Br incomplete octet ·· ·O · ·· ·· S ·O· · ·· · ·· 2O· · ·· BBr4T # electrons: B = 3; Br = 4 × 7; +1e= 32e- 4 bonds = 8e- ; 24 e- left (3 pairs per B atom) Br Br B Br Br Lewis Structures: AsF5 (5 F around As) Lewis Structures: C2H6 or CH3CH3 T # electrons: As = 5; F = 5 × 7 = 40e- 3 H’s bond to each C; C bonds to C T # electrons: C = 2 × 4; H = 1 × 6 = 14e- 5 bonds = 10e- ; 30 e- left ( 3 pairs per F atom) F expanded octet H C H As F F Lewis Structures: CH3CH2OH Lewis Structures: CH3COOH 3 H’s bond to C; C bonds to C; O to C… 3 H’s bond to C; C bonds to C; O to C… T # electrons: C = 2 × 4; H = 1 × 6; O = 6 = 20e- T # electrons: C = 2 × 4; H = 1 × 4; 2 × O = 12 = 24e- 8 bonds = 16e- ; 4 left over (2 pairs) 7 bonds = 14e- H O H C C O H H C has octet H has duet H H C H H H H C C O H O C O H H H Resonance Lewis Structures Isoelectronic Species: Molecules and/or ions having the same number of valence electrons and the same Lewis structures. [:N≡N:] :N≡ C H H F F [:N≡O:]+ :N≡ H H 7 bonds = 14e- ; [:C≡O:] :C≡ [:C≡N:]− :C≡ They all have 3 BP and 2 LP (2 atoms) There are three possible structures for SO3. The double bond can be placed in one of three places. ·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 Structures Resonance is a method used to represent molecules. • There are no single or multiple bonds in SO3 (nor CO2) • All bonds in SO3 are equivalent. The best Lewis formulas of SO3 (CO2) that can be drawn are: (average of 3 (average structures) structures) Resonance Hybrids CO2 2 × O = 12 e1 × C = 4 eTotal = 16 e- O C O O C O O Resonance C O S O O O C O O Molecules with odd number of electrons NO has 11 valence electrons Formal Charges on Atoms: N O an accounting tool for electron ownership = (# valence e– in free atom) – (# e– in lone pairs on atom) – ½(# bonded e– on atom) NO2 has 17 valence electrons O N O O N Charge Distribution in Covalent Bonds and Molecules (Formal Charge) O These are known as Free Radicals: they have an Radicals: unpaired e−. Very reactive: e.g. dimerization of NO2 to N2O4. Formal Charge (FC) and Best structure Best Lewis structure has • zero FC on all atoms • lowest FC possible • negative FC on most electronegative atoms and positive FC on least electronegative atoms The most electronegative elements are at the top right of periodic table (except for the noble gases.) Formal charge , FC = group number of atom − LPE - # Bonds Formal Charge (FC) NH4+ H | H─N─H | H O C O N= 5 − 0 - 4 = +1 H = 1 − 0 -1 = 0 C=4−0-4=0 O = 6 − 4 -2 = 0 For example, consider thiocyanate ion SCN– 0 +1 –2 –2 +2 –1 –1 0 0 •• •• •• •• •• •• N C S C S N S N C •• •• •• •• •• •• best structure because lowest FC and –ve FC on most electronegative atom Electronegativity Electronegativity is a measure of the relative tendency of an atom to attract electrons to itself when bonded to another element. Electronegativity is measured on the Pauling scale. scale. F is the most electronegative element. Cs and Fr are the least electronegative elements. For the representative elements, electronegativities usually increase from left to right across periods and decrease from top to bottom within groups. Sum of formal charges = −1 + 0 + 0 = −1 Electronegativity Example: Arrange these elements based on increasing electronegativity values. Se, Ge, Br, As (period # 4) Ge, Arrange these elements based on increasing electronegativity values. Be, Mg, Ca, Ba (group IIA) Electronegativity, polarity, and Ionic bonds Electronegativity, A non-polar bond is formed when the nondifference in electronegativity is between 0.0 – 0.4 A polar bond is formed when the difference in electronegativity is 0.5 – 1.9 (partial charge) Ionic bonds form when the difference of electronegativity is greater than 1.9 (net charge); e- transferred no charge Cl−Cl (0.0) Cl− non-polar non- Bond Polarity and ∆EN Nonpolar covalent polar covalent ionic bond Cl is more electronegative δ+ δ− H−Cl (0.9) Na+ and Cl− (2.1) polar ionic Dipole Moments Molecules whose centers of positive and negative charge do not coincide, have an asymmetric charge distribution, and are polar. These molecules have a net dipole moment (µ). The dipole moment is: 0.0 d, distance separating charges of equal magnitude and opposite sign 0.4 2.0 Electronegativity difference, ∆EN equal sharing unequal sharing µ=δ×d magnitude of the charge e- transferred e41 indicated by crossed arrow pointing from positive end to negative end of dipole Dipole Moments Polarity of Molecules Some nonpolar molecules have polar bonds. Molecules in which dipole moments of the bonds do not cancel, are polar molecules There are two conditions that must be true for a molecule to be polar. 1. There must be at least one polar bond or, one lone pair of electrons present. 2. The polar bonds, if there are more than one, and lone pairs, must be arranged so that their dipole moments do not cancel each other. δ– δ+ δ– µtotal = 0 nonpolar δ+ Cl δ– H δ+ Molecules that do not contain polar bonds, or in which all dipole moments cancel, are nonnonpolar molecules µtotal = 0 (CO2) 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, ..) O O (ClF Bonding Order ClF, CO2 & H2O O C µtotal ≠ 0 H δ+ µtotal ≠ 0 polar # of shared pairs linking atoms Bond order = ────────────────────── # of links in the molecule or ion (no L.P.) (no H – Cl 1 O–O=O 1 2 O=C=O 2 2 H – C≡ N 1 3 δ– F µtotal ≠ 0, polar Bond Strength and Bond Length bond length (shortest) triple < double < single bond strength single < double < triple (strongest) Bond Length (Å) Strength (kJ/mol) (Å C–C 1.54 348 C=C 1.34 614 C≡C 1.20 839 Bond Energy (Enthalpy) -the energy required to break 1 mole of the bond in the gas phase H2(g) → 2H(g) 436 kJ/mol (single bond) O2(g) → 2O(g) 498 kJ/mol (double bond) N2(g) → 2N(g) 946 kJ/mol (triple bond) Average bond energies can be used to estimate the enthalpy change, ∆H of a reaction. Use: ∆Hrxn = Σ (∆Hs bonds broken) + Σ (∆Hs bonds formed) requires energy (+) releases energy (-) Bond Energy (Enthalpy) Bond Energy (Enthalpy) Estimate the enthalpy change of the reaction between iodomethane and water Estimate the enthalpy change of the reaction between iodomethane and water CH3CH2I(g) + H2O(g) → CH3CH2OH(g) + HI(g) CH3CH2I(g) + H2O(g) → CH3CH2OH(g) + HI(g) Plan: which bonds are formed; which bonds are broken add enthalpy changes: bonds formed, positive (+); bonds broken, negative (-) Break (reactants): C-I bond (CH3CH2I) = 238 kJ one O-H bond (H2O) = 463 kJ Form (products): C-O bond (CH3CH2OH) = 360 kJ H-I bond (HI) = 299 kJ ∆Hrxn = +42 kJ Overall, the reaction is endo or exothermic? ...
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