Lecture 23: Ionic to Covalent Bonds

Transcription

1 Lecture 23: Ionic to Covalent Bonds Reading: Zumdahl 13.4, Outline Binary Ionic Compounds Partial Ionic Compounds Covalent Compounds Problems (Chapter 13 Zumdahl 5 th Ed.) (Review Ch 12), 20-22, 24, 31, 32, 33 (minimalist approach), 34, 40, 41d, 42 1

2 Properties of Ions When will a stable bond be formed? When one exams a series of stable compounds, it becomes evident that in the majority of compounds, bonding is achieved such that atoms can achieve a noble-gas configuration Example: NaCl versus Na + Cl - Na: [Ne]3s 1 Cl: [Ne]3s 2 3p 5 Na + : [Ne] Cl - : [Ne]3s 2 3p 6 = [Ar] 2

3 Ionic Compounds In this example involving NaCl, we have a metal (Na) bonding to a non-metal (Cl). Metal/non-metal binding generally results in ionic bonding. 3

4 Forming Molecules with Ionic Character One can use this tendency to satisfy the noble gas configuration ns 2 np 6 ( octet rule ) to predict the stoichiometry of ionic (or nearly ionic) compounds. Example: Ca and O: Use Noble Gas E.C. as a guide to moving electrons. Move 2 electrons from Ca to O Ca: [Ar]4s 2 O: [He]2s 2 2p 4 2 e - Ca 2+ : [Ar] O 2- : [He]2s 2 2p 6 = [Ne] Formula: CaO 2 Do we ever actually find an O ion as a separate entity? 4

5 Subtle Example, Z Usually 2K + O based on E.C.. 2 2K2O Each K contributes one electron, to make the K+ ion, and each O takes two electrons. However, sometimes one can form KO. Is this K(II)O (like CaO) or K peroxide. Why? K O K 2 5

6 Size of Ions: Isoelectronic Series Ions shown for nobel-gas electron configurations. To form ionic binary compounds, combine in proportions so that total charge is zero. Ion size decreases as charge increases when the number of electrons does not change (an isoelectronic series), due to increased nuclear attraction. Finding noble-gas E.C. does not extend to transition metals, which use inner shell electrons. 6

7 Z13.22 Lattice Energy: Which ionic substance has the more exothermic lattice energy? LiF or CsF? How did we calculate lattice energy? Used Coulombs law: ( ) QQ + V = r Apply this rule qualitatively to these two cases. F - anion is same size for both (and larger than either Li + cation, but slightly smaller than Cs + cation); lattice type is the same Q± = ± 1 Q+ Q = 1 so the Cs-F distance must be larger than the Cs-Li distance, therefore the lattice energy for the LiF must be more exothermic (a larger negative number) ± 7

8 Partial Ionic Compounds From last lecture, if two atoms forming a bond have differing electronegativities, they will form a bond having (partial) ionic character. But where is the dividing line between ionic bonding and polar covalent bonding? In the end, total ionic bonding is probably never achieved, and all ionic bonds can be considered polar covalent, with varying degrees of ionic character. 8

9 Dipole Moments (Repeat: Lecture 22) The dipole moment (μ) is defined as: μ = QR R +Q -Q Charge magnitude Separation distance R + center μ = 2Rδ R= doh cos HOH 2 9

10 Dipole Moment and Ionic Character Example, the dipole moment of HF is 1.83 D. This is for HF as a diatomic molecule in the gas phase (not as part of a solid). What would it be if HF formed an ionic bond (bond length = 92 pm)? μ calc = ( C)( m) -29 = C m C m -30 = = 4.4Debye C m/d So the actual dipole moment is about half as large as the calculated one, which assumes that the electron completely moved from the H to be centered on the F. 10

11 Partial Ionic Character of Compounds We can define the ionic character of bonds as follows: Experimental Dipole Moment (XY) % Ionic Character = x100% + Calculated Dipole Moment for (X Y ) 11

12 Partial Ionic Character (for simple binary molecules) Assume the experimental distance between X and Y to compute the calculated dipole, so the ratio is the fraction of the charge that really did move from X to Y. The experimental dipole moment is very close to the difference in (Pauling) Electronegativities. μ EN( Y ) EN( X ) exp The percent ionic character as a function of the difference in Electronegativities is approximately (c.f. figure): ( EN( X ) EN( Y )) % ionic character x100% EN( X ) EN( Y ) ( ) 2 12

13 Orbital for the bonding electrons in Ionic Molecules Molecular Orbital for the Shared Electrons Covalent (H 2, F 2 ) Polar Covalent (HF) Ionic (KF) Increased Ionic Character 13

14 Covalent Compounds In covalent bonding, electrons are shared between bonding partners. In ionic bonding, Coulombic interactions resulted in the bonding elements (the ions in close proximity) being more stable than the separated neutral atoms. What about covalent bonds what is the driving force? 14

15 Generalize the H 2 Covalent bond (Prev. Lec.) The energy in a covalent bond: 15

16 Covalent Bonds and Bond Energies Example: Calculate ΔH for the following reaction using the bond energy (enthalpy) method. These are enthalpies. The energy of a bond: If you break a bond it will cost you energy; the molecules will absorb the energy and have a less stable bond (endothermic). Conversely, going to a more stable bonding situation produces a stronger bond (exothermic). Bond Energies (Table 13.6) H-H 434 O=O 495 O-H ( ) ( ) ( ) H g H g Δ H =Δ H H = f 217 kj mol H 2 is more stable than 2 H atoms separated. It has a bond, formed by the two electrons. The energy of that bond: BE ( H 2 ) = = 434 kj mol The B.E. (because it is a positive number) corresponds to the endothermic, or bond breaking process. The larger the B.E. then, the more stable (or lower energy) the bond. BE(O-H) > BE(H-H) 16 so the reaction might go.

17 Energy for Reactions is in the Bond The bond of H 2 is stable; but the bond of H-O (in H 2 O) is more stable, so the reaction is exothermic (and spontaneous). There is a hierarchy of stability: H is stable relative to proton and electron separated; H-H (or H 2 ) is stable relative to H 2H H2 Δ H << 0 1 O-H (in H 2 O) is stable relative to H 2. H2 + 2 O2 H2O Δ H << 0 A similar hierarchy applies to O. The more stable the bond, the lower in energy is the chemical system. Chemical systems relax to lower energy states by giving off energy (either as heat or light). 17

18 18 Graph of the energy for reaction

19 Calculating Reaction Energies from Bond Energies Using Hess s Law, we add up reactions to make the net reaction. Use independent atoms as the intermediates. Apply to the energy of making water (all species are in the gas phase) The net reaction (1) is the sum of the three reactions below it (2,3, and 4). Reactions 2 and 3 break bonds; 4 forms bonds. (1) (2) H 2H Δ H = BE HH (3) H + O HO Δ H =ΔH HO 2 2 O O Δ H = BE OO (4) 2H + O H O Δ H = 2 BE OH o f ( ) ( ) ( ) ( ) 1 ( ) ( ) 2 ( ) Δ H =Δ H +Δ H +Δ H = BE HH + BE OO BE OH 1 = = kj mol 19

20 Will water auto-ionize (in gas phase)? Z13.41d From the bond energies just developed for water formation, what is the ΔH for the reaction: 2 + ( ) ( ) + ( ) H O g H g OH g Break an OH bond and then transfer an electron from H to OH. Δ H = BE O H + IP H + EA OH ( ) ( ) ( ) = = 1580 kj mol Result: A huge positive number; it will not auto-ionize. In water ΔH~100 kj/mol 20

21 Covalent Molecules The same concept can be envisioned for other covalent compounds: Think of the covalent bond as the electron density existing between the C and H atoms. C-H is nearly even sharing (a nonpolar bond) 21

22 Covalent Bonds We can quantify the degree of stabilization by seeing how much energy it takes to separate a covalent compound into its atoms (i.e. atomic constituents). C(g) + 4H(g) q CH 4 (g) ( ),2 ( ) C gr H g 2 This is much larger than the enthalpy of formation, which is 75kJ/mol; the energy is stored in the bonds. 22

23 Covalent Bonds Since we broke 4 C-H bonds with 1652 kj in, the bond energy for a C-H bond is: 1652 kj mol 4 = 413 kj mol This represents an average value; this process is not exact. We can continue this process for a variety of compounds to develop a table of bond strengths. The energy is in the bond. The unequal sharing of a pair of electrons is the reason for molecular stability. When bonds are broken it costs energy (endothermic) When bonds are formed energy is released (exothermic) 23

24 Covalent Bonds: Build up Bond energies Example: It takes 1578 kj/mol to decompose CH 3 Cl into its atomic constituents. What is the energy of the C-Cl bond? CH 3 Cl: Contains 3 C-H bonds and 1 C-Cl bond. ( ) ( ) CH3Cl C + 3H + Cl Δ H = 3BE CH + BE CCl 1578 kj/mol 413 kj/mol BE(CCl) = C-Cl bond energy = 339 kj/mol All bond energies (D) are listed as positive numbers. 24

25 The Energy is in the Bonds We can use these bond energies (D) to determine ΔH rxn : ΔH = sum of energy required to break bonds (positive.heat into system) plus the sum of energy released when the new bonds are formed (negative.heat out from system). Δ Hrxn = Dbonds broken Dbonds formed Breaking Bonds Costs Energy Endothermic (+) Forming Bonds Releases Energy Exothermic (-) 25

26 Covalent Bonds and Bond Energies Example: Calculate ΔH for the following reaction using the bond enthalpy method. These are enthalpies. CH 4 (g) + 2O 2 (g) Go to Table 13.6: CO 2 (g) + 2H 2 O (g) 4 x 2 x C-H 413 O=O x 2 x O-H 467 C=O 745 (or 799) How do you know which type of bond to choose? 26

27 Bond Energy: General Formulation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) ΔH rxn = D bonds broken D bonds formed = 4D (C-H) + 2D (O=O) -4D (O-H) -2D (C=O) = 4(413) + 2(495) - 4(467) - 2(745+54) = -( )kJ/mol Exothermic, as expected. 27

28 Approximate Enthalpy from B.E. c.f. Z13.34 How is this different from using Hess s law and Heats of Formation? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) As a check: o ΔH rxn = o ΔH f (prod.) o ΔH f (react.) = ΔH f (CO 2 (g)) + 2ΔH f (H 2 O(g)) 0 - ΔH f (CH 4 (g)) - 2 ΔH f (O 2 (g)) = kj/mol + 2(-242 kj/mol) - - (-75 kj/mol) = kj/mol or ~3% error in this method 28

29 Z13.33 Reaction Enthalpy Use the bond energies to estimate ΔH for the (isomerization) reaction: CH N C g CH C N g ( ) ( ) 3 3 You could list all the bonds that are formed and all that are broken. But lets try to list the ones that are different. (There is no need to add the 3 CH bonds that contribute equally to the product and the reactant, same for CN triple bond.) So you make a C-C (single bond) in the product, and break a C-N (single bond) of the reactant. {Making bonds stabilizes, lowers energy so the C-C bond is used as the negative of the BE(C-C). Δ H= BEC ( N) BEC ( C) = = 42 kj mol 29

30 The NI 3 molecule Will NI 3 have a dipole moment? What direction will the dipole moment point? Compare with the EN of N and I: Which is the more electronegative? EN(N)=3; EN(I)=2.7; EN(H)=2.2 Compare the expected dipole moment of NI 3 and NH 3 with NF 3. Use the B.E.s to predict whether NI 3 is stable: N + 3I 2NI Δ H = 2ΔH f Δ H = BE N N + 3 BE I I 2 3 BE N I Species N N I I N I ( ) ( ) ( ) BE Δ H f =+ 190 kj mol It is NOT stable! 30