Lecture 23: Ionic to Covalent Bonds
• Reading: Zumdahl 13.4,13.6-8
• Outline
– Binary Ionic Compounds
– Partial Ionic Compounds
– Covalent Compounds
• Problems (Chapter 13 Zumdahl 5th Ed.)
– 17-19 (Review Ch 12), 20-22, 24, 31, 32, 33
(minimalist approach), 34, 40, 41d, 42
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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+ClNa: [Ne]3s1
Cl: [Ne]3s23p5
Na+: [Ne]
Cl-: [Ne]3s23p6 = [Ar]
2
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
Forming Molecules with Ionic Character
• One can use this tendency to satisfy the noble gas
configuration ns2np6 (“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]4s2
O: [He]2s22p4
2 e-
Ca2+: [Ar]
O2-: [He]2s22p6 = [Ne]
Formula: CaO
Do we ever actually find an O 2 − ion as a separate entity?
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Subtle Example, Z12.100
• Usually 2 K + O2 → 2 K 2O based on E.C..
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−O2 −K
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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.
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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:
−19 Q+ Q−
V = ( 2.3 ⋅10 )
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)
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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.
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Dipole Moments (Repeat: Lecture 22)
• The dipole moment (μ) is defined as:
+Q
R
|
-Q
μ = QR
Charge magnitude
Separation distance
R
+ center
μ = 2 Rδ
n
R = dOH cos ⎛⎜ HOH ⎞⎟
2⎠
⎝
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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)?
μcalc = (1.6 ⋅ 10-19 C)(9.2 ⋅ 10-11 m)
= 1.5 ⋅ 10
-29
C⋅m
1.5 ⋅ 10-29 C ⋅ m
=
= 4.4 Debye
-30
3.336 ⋅ 10 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.
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Partial Ionic Character of Compounds
• We can define the ionic character of bonds as follows:
% Ionic Character =
Experimental Dipole Moment (XY)
x100%
+ −
Calculated Dipole Moment for (X Y )
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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.
μexp ≈ EN (Y ) − EN ( X )
•The percent ionic character as a function of the difference in
Electronegativities is approximately (c.f. figure):
EN ( X ) − EN (Y ) )
(
% ionic character ≈
x100%
2
3 + ( EN ( X ) − EN (Y ) )
2
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Orbital for the bonding electrons in Ionic Molecules
Molecular Orbital for the Shared Electrons
Polar Covalent (HF)
Ionic (KF)
Increased Ionic Character
Covalent (H2, F2)
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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”?
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Generalize the H2 Covalent bond (Prev. Lec.)
The energy in a covalent bond:
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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 467
1
2
H2 ( g ) → H ( g )
ΔH = ΔH 0f ( H ) = +217 kJ
mol
H2 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 ⋅ 217 = 434 kJ
2
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)
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so the reaction might go.
Energy for Reactions is in the Bond
The bond of H2 is stable; but the bond of H-O (in H2O) 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;
2 H → H 2 ΔH << 0
H-H (or H2) is stable relative to H
ΔH << 0
O-H (in H2O) is stable relative to H2. H 2 + 12 O2 → H 2O
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).
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18
Graph of the energy for reaction
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) H 2 + 12 O2 → H 2O
(2) H 2 → 2 H
(3)
1
2
O2 → O
ΔH1 = ΔH of ( H 2O )
ΔH 2 = BE ( HH )
ΔH 3 = 12 ⋅ BE ( OO )
(4) 2 H + O → H 2O
ΔH 4 = −2 ⋅ BE ( OH )
ΔH1 = ΔH 2 + ΔH 3 + ΔH 4 = BE ( HH ) + 12 ⋅ BE ( OO ) − 2 ⋅ BE ( OH )
= 434 + 12 ⋅ 495 − 2 ⋅ 467 = −252.5 kJ
mol
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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:
H 2O ( g ) T H + ( g ) + OH − ( g )
• Break an OH bond and then transfer an electron from H to
OH.
ΔH = BE ( O − H ) + IP ( H ) + EA ( OH )
= 467 + 1300 − 180 = 1580 kJ
mol
• Result: A huge positive number; it will not auto-ionize.
• In water ΔH~100 kJ/mol
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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)
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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)
CH4(g)
q
C ( gr ) , 2 H 2 ( g )
This is much larger than the enthalpy of formation,
which is 75kJ/mol; the energy is stored in the bonds.
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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
Covalent Bonds: Build up Bond energies
• Example: It takes 1578 kJ/mol to decompose CH3Cl into
its atomic constituents. What is the energy of the C-Cl
bond?
CH3Cl: Contains 3 C-H bonds and 1 C-Cl bond.
CH 3Cl → C + 3H + Cl
ΔH = 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.
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The Energy is in the Bonds
• We can use these bond energies (D) to determine ΔHrxn:
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).
ΔH rxn = ∑ Dbonds broken − ∑ Dbonds formed
Breaking Bonds Costs Energy
Forming Bonds Releases Energy
Endothermic (+)
Exothermic (-)
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Covalent Bonds and Bond Energies
• Example: Calculate ΔH for the following reaction using
the bond enthalpy method. These are enthalpies.
CH4(g) + 2O2 (g)
CO2 (g) + 2H2O (g)
Go to Table 13.6:
4x
2x
C-H 413
O=O 495
4 x O-H
2 x C=O
467
745 (or 799)
How do you know which type of bond to choose?
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Bond Energy: General Formulation
CH4(g) + 2O2 (g)
CO2 (g) + 2H2O (g)
ΔH rxn = ∑ Dbonds broken − ∑ Dbonds formed
= 4D(C-H) + 2D(O=O) - 4D(O-H) - 2D(C=O)
= 4(413) + 2(495) - 4(467) - 2(745+54)
= -(716 +108)kJ/mol
• Exothermic, as expected.
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Approximate Enthalpy from B.E. c.f. Z13.34
How is this different from using Hess’s law and Heats of Formation?
CH4(g) + 2O2 (g)
CO2 (g) + 2H2O (g)
• As a check:
o
ΔH rxn
= ∑ ΔH of ( prod.) − ∑ ΔH of (react.)
0
= ΔH°f(CO2(g)) + 2ΔH°f(H2O(g))
- ΔH°f(CH4(g)) - 2 ΔH°f(O2(g))
= -393.5 kJ/mol + 2(-242 kJ/mol) - (-75 kJ/mol)
= -802.5 kJ/mol or ~3% error in this method
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Z13.33 Reaction Enthalpy
• Use the bond energies to estimate ΔH for the (isomerization)
reaction:
CH 3 N ≡ C ( g ) S CH 3C ≡ N ( g )
• 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).
ΔH = BE (C − N ) − BE (C − C ) = 305 − 347 = −42 kJ
mol
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The NI3 molecule
• Will NI3 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 NI3 and NH3 with NF3.
• Use the B.E.s to predict whether NI3 is stable:
N 2 + 3I 2 R 2 NI 3
ΔH = 2ΔH 0f
ΔH = BE ( N ≡ N ) + 3 ⋅ BE ( I − I ) − 2 ⋅ 3 ⋅ BE ( N − I )
Species
N≡N
I −I
N −I
BE
941
151
170
ΔH 0f = +190 kJ
mol
It is NOT stable!
http://www.chm.bris.ac.uk/motm/ni3/ni3j.htm
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