Lecture 21: Ionic to Covalent Reading: Zumdahl 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 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  = (H-X) experimental - (H-X) expected  = 0: covalent  > 0: ionic character

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

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-ClO-HNa-Cl  elect

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.

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

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

Dipole Moments (cont.) The units of dipole moment are generally the Debye (D):  D = x 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 C)(9.2 x m) = 1.5 x C.m x (1D/3.336 x C.m) = 4.4 D

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

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!

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]3s 1 Cl: [Ne]3s 2 3p 5 Na + : [Ne]Cl - : [Ne]3s 2 3p 6 = [Ar]

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.

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]4s 2 O: [He]2s 2 2p 4 2 e - Ca 2+ : [Ar] O 2- : [He]2s 2 2p 6 = [Ne] Formula: CaO

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.

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

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.

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

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 C)(9.2 x m) = 1.5 x C.m x (1D/3.336 x C.m) = 4.4 D

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

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

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”?

Covalent Compounds (cont.) Back to H 2.

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.

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.

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.

Covalent Compounds (cont.) 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: 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

Covalent Compounds (cont.) We can use these bond energies to determine  H rxn :  H = 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).

Covalent Compounds (cont.) Example: Calculate  H for the following reaction using the bond enthalpy method. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) Go to Table 13.6: C-H413 O=O495 O-H467 C=O745 4 x 2 x 4 x 2 x

Covalent Compounds (cont.) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (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.

Covalent Compounds (cont.) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) As a check: =  H° f (CO 2 (g)) + 2  H° f (H 2 O(g)) -  H° f (CH 4 (g)) - 2  H° f (O 2 (g)) 0 = kJ/mol + 2(-242 kJ/mol) - - (-75 kJ/mol) = kJ/mol