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Lecture 21: Ionic to Covalent

Lecture 21: Ionic to Covalent. Reading: Zumdahl 13.4-13.6 Outline Binary Ionic Compounds Partial Ionic Compounds Covalent Compounds. Electronegativity. Electronegativity: The ability of an atom in a molecule to attract shared electrons to itself. Electronegativity.

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Lecture 21: Ionic to Covalent

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  1. Lecture 21: Ionic to Covalent • Reading: Zumdahl 13.4-13.6 • Outline • Binary Ionic Compounds • Partial Ionic Compounds • Covalent Compounds

  2. Electronegativity • Electronegativity: The ability of an atom in a molecule to attract shared electrons to itself.

  3. Electronegativity • Electronegativity can be defined in many ways. Pauling model is the most widely used. • Idea: compare the bond energy of an “HX” molecule to that of the average of an HH bond and an XX bond: Expected energy = [(H-H energy)(X-X energy)]1/2 D = (H-X)experimental - (H-X)expected D > 0: ionic character D = 0: covalent

  4. Electronegativity (cont.) • Pauling used this approach to develop a scale, where F = 4.0 (flourine has largest electronegativity). F = 4 O = 3.4 Cl = 3.2 C = 2.6 H = 2.2 Na = 0.9

  5. Electronegativity (cont.) • The key idea is this: the greater the electronegativity difference between two atoms, the more ionic the bond. • Example: Which of the following compounds is expected to demonstrate intermediate bonding behavior (i.e., polar covalent). Cl-Cl O-H Na-Cl 0 2.3 1.2 Delect

  6. Dipole Moments • The above discussion involved bonds in which electrons were shared, but shared unequally in polar, covalent bonds • In the HF example, when placed in an electric field the HF atoms will align. • This observation demonstrates that the centers of negative and positive charge do not coincide.

  7. Dipole Moments (cont.) • When the centers of negative and positive charge are separated, we say that the molecule has a dipole moment.

  8. Dipole Moments (cont.) • The dipole moment (m) is defined as: m = QR Charge magnitude Separation distance R + center

  9. Dipole Moments (cont.) • The units of dipole moment are generally the Debye (D): 1 D = 3.336 x 10-30 C.m • Example, the dipole moment of HF is 1.83 D. What would it be if HF formed an ionic bond (bond length = 92 pm)? M = (1.6 x 10-19 C)(9.2 x 10-11 m) = 1.5 x 10-29 C.m x (1D/3.336 x 10-30 C.m) = 4.4 D

  10. Dipole Moments (cont.) • Molecular geometry is a critical factor in determining if a molecule has a dipole moment:

  11. Dipole Moments (cont.) • Molecular geometry is a critical factor in determining if a molecule has a dipole moment: No net dipole moment. Dipoles add vectorially!

  12. 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 nobel-gas configuration • Example: NaCl versus Na+Cl- Na: [Ne]3s1 Cl: [Ne]3s23p5 Na+: [Ne] Cl-: [Ne]3s23p6 = [Ar]

  13. Properties of Ions (cont.) • 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.

  14. Properties of Ions (cont.) • One can use this tendency to satisfy the “octet rule” to predict the stoichiometry of ionic compounds. • Example: Ca and O Ca: [Ar]4s2 O: [He]2s22p4 2 e- Ca2+: [Ar] O2-: [He]2s22p6 = [Ne] Formula: CaO

  15. Properties of Ions (cont.) • Ions on figure correspond to nobel-gas electron configurations. • To form ionic binary compounds, one simply combines in proportions such that total charge is zero. • This approach is not to be applied to transition metals.

  16. Properties of Ions (cont.) • Note that size decreases for isoelectronic species. • Mainly a consequence of increased charge of nucleus.

  17. Partial Ionic Compounds • From last lecture, if two atoms forming a bond have differing electronegativities, they will form a bond having 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.

  18. Dipole Moments • The dipole moment (m) is defined as: m = QR Charge magnitude Separation distance R + center

  19. Dipole Moments (cont.) • Example, the dipole moment of HF is 1.83 D. What would it be if HF formed an ionic bond (bond length = 92 pm)? M = (1.6 x 10-19 C)(9.2 x 10-11 m) = 1.5 x 10-29 C.m x (1D/3.336 x 10-30 C.m) = 4.4 D

  20. Partial Ionic Compounds (cont.) • We can define the ionic character of bonds as follows: (dipole moment X-Y)experimental % Ionic Character = x 100% (dipole moment X+Y-)calculated

  21. Partial Ionic Compounds (cont.) Covalent Polar Covalent Increased Ionic Character Ionic

  22. Covalent Compounds • In covalent bonding, electrons are “shared” between bonding partners. • In ionic bonding, Coulombic interactions resulted in the bonding elements being more stable than the separated atoms. • What about covalent bonds…what is the “driving force”?

  23. Covalent Compounds (cont.) • Back to H2.

  24. Covalent Compounds (cont.) • 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.

  25. Covalent Compounds (cont.) • We can quantify the degree of stabilization by seeing how much energy it takes to separate a covalent compound into its atomic constituents.

  26. Covalent Compounds (cont.) • Since we broke 4 C-H bonds with 1652 kJ in, the bond energy for a C-H bond is: • We can continue this process for a variety of compounds to develop a table of bond strengths.

  27. Covalent Compounds (cont.) • Example: It takes 1578 kJ/mol to decompose CH3Cl into its atomic constituents. What is the energy of the C-Cl bond? CH3Cl: 3 C-H bonds and 1 C-C bond. 3 (C-H bond energy) + C-Cl bond energy = 1578 kJ/mol 413 kJ/mol 1239 kJ/mol + C-Cl bond energy = 1578 kJ/mol C-Cl bond energy = 339 kJ/mol

  28. Covalent Compounds (cont.) • We can use these bond energies to determine DHrxn: DH = 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).

  29. Covalent Compounds (cont.) • Example: Calculate DH for the following reaction using the bond enthalpy method. CH4(g) + 2O2 (g) CO2 (g) + 2H2O (g) Go to Table 13.6: 4 x 4 x O-H 467 C=O 745 C-H 413 O=O 495 2 x 2 x

  30. Covalent Compounds (cont.) CH4(g) + 2O2 (g) CO2 (g) + 2H2O (g) = 4D(C-H) + 2D(O=O)- 4D(O-H) - 2D(C=O) = 4(413) + 2(495) - 4(467) - 2(745) = -716 kJ/mol • Exothermic, as expected.

  31. Covalent Compounds (cont.) CH4(g) + 2O2 (g) CO2 (g) + 2H2O (g) • As a check: 0 = DH°f(CO2(g)) + 2DH°f(H2O(g)) - DH°f(CH4(g)) - 2 DH°f(O2(g)) = -393.5 kJ/mol + 2(-242 kJ/mol) - - (-75 kJ/mol) = -802.5 kJ/mol

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