Lecture 22: Ionic to Covalent Reading: Zumdahl 13.4-13.5 Outline –Binary Ionic Compounds –Partial Ionic Compounds –Covalent Compounds.

Lecture 22: Ionic to Covalent

• Reading: Zumdahl 13.4-13.5

• Outline– Binary Ionic Compounds– Partial Ionic Compounds– Covalent Compounds

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]

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]4s2 O: [He]2s22p4

2 e-

Ca2+: [Ar] O2-: [He]2s22p6 = [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.

Magnesium (Mg) forms compounds of the form MgX2. What group is X from?

A. Group 1

B. Group 2

C. Group 6

D. Group 7

Mg

In forming MgX2. What bonding partner will result in the Mg and X ions being isoelectronic?

A. Period 2

B. Period 3

C. Period 4

D. Period 5

Mg

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 (Lecture 21)• 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)?

= (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

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

Where should NaI be on the following graph?

AB

CD

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 H2.

Covalent Compounds (cont.)

• The same concept can be envisioned for other covalent compounds:

Think of the covalent bond as theelectron density existingbetween 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.

q

CH4(g)

C(g) + 4H(g)

Covalent Compounds (cont.)• Since we broke 4 C-H bonds with 1652 kJ in, the bond

energy for a C-H bond is:

€

1652kJ mol4

= 413kJ mol

• 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 CH3Cl into

its atomic constituents. What is the energy of the C-Cl bond?

CH3Cl: 3 C-H bonds and 1 C-Cl 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 Hrxn:

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).

€

Hrxn = Dbonds broken∑ − Dbonds formed∑

Covalent Compounds (cont.)• Example: Calculate H for the following reaction using

the bond enthalpy method.

CH4(g) + 2O2 (g) CO2 (g) + 2H2O (g)

Go to Table 13.6:

C-H 413O=O 495

O-H 467C=O 745

4 x2 x

4 x2 x

Covalent Compounds (cont.)CH4(g) + 2O2 (g) CO2 (g) + 2H2O (g)

€

Hrxn = Dbonds broken∑ − Dbonds formed∑

= 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.)CH4(g) + 2O2 (g) CO2 (g) + 2H2O (g)

• As a check:

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Hrxno = ΔH f

o∑ (prod .) − ΔH fo∑ (react .)

= H°f(CO2(g)) + 2H°f(H2O(g)) - H°f(CH4(g)) - 2 H°f(O2(g))

0

= -393.5 kJ/mol + 2(-242 kJ/mol) - - (-75 kJ/mol)

= -802.5 kJ/mol