This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 9
Chemical Bonding Pt. 1 Goals
• Understand the difference between ionic and
• Lattice energy (ionic compounds)
• Draw Lewis electron dot structures for small
molecules and ions (the octet rule).
• Understand the properties of covalent bonds
and their influence on molecular structure.
• Understand electronegativity and its use in
predicting the polarity of bonds and molecules.
• Predicting bond length and bond energies. Valence Electrons Lewis Dot Symbols for Atoms forces that hold atoms together in
compounds are called chemical bonds.
* The electrons involved in bonding are usually
those in the outermost (valence) shell. The (inner)
core electrons are not involved in chemical
behavior. Lewis dot formulas (Lewis dot symbols;
Lewis structures) are a convenient method
for tracking valence electrons. * Attractive Valence electrons:
For the main groups (representative) elements they
are the outermost s and p electrons. The # of
valence electrons is equal to the group number.
* For the transition elements they are the ns and
(n− Types of Bonds
When a chemical reaction occurs, the valence
electrons of the atoms are reorganized so that
net attractive forces (chemical bonds) occur
Chemical bonds are classified into two types:
Ionic bonding results from electrostatic
attractions between ions (ions are formed by
the transfer of one or more electrons from one
atom to another).
Covalent bonding results from sharing one or
more electron pairs between two atoms. .
B . . C . . N . . O . . .. . . Ne .
.. Formation of Ionic Compounds
• We can also use Lewis dot formulas to
represent the neutral atoms and the ions
+ . .
Li . + . .. .
Li . .. . [ ]
LiF 3 Ca .
. .. + 2 .N.
. 3 Ca 2+ 3. .. .
2 [ . N. ]
.. Coulombic Ion Attraction and Lattice Energy radius effect QQ
E∝ c a
d E ∝ d Qc × Qa
E = -C ×──────
d d Na+ F– negative due Qa E = ionic bond strength Q cQ a
d d C: a constant Na+ Cl– NaF has the most negative E and NaI the least d Qc = charge on cation (is positive) Na+ Qa = charge on anion (is negative) I– Br– Na+ d = distance between centers of ions (sum of radii) E = −C ×
Cmpd Q cQ a charge effect
d ions Product of charges NaCl Na+ Cl– −1 CaCl2 Ca2+ Cl– −2 CaS Ca2+ S–2 −4 Al2S3 Al3+ S2– −6 Lattice Energy (∆Hlattice ) kJ/mol
(∆ the energy of formation of 1 mol of solid
crystalline ionic compound when ions in the
gas phase combine. It is always negative.
Among several compounds, the one with the
most negative ∆Hlattice is said to have the highest
lattice energy (absolute value)… and highest Tf
Mg2+(g) + 2F–(g) → MgF2(s) ∆Hlattice = −2910 kJ/mol
NaF, ∆Hlattice = −911 kJ/mol
KF, ∆Hlattice = −815 kJ/mol Al2S3 has the most negative E and NaCI the
least WHY? charge of Mg2+ > Na+ = K+
d, radius of Mg2+ < Na+ < K+ Lattice Energy Lattice Energy Small ions with high charges have more negative
Large ions with small charges have less negative
lattice energies. Arrange the following compounds in order of
decreasing lattice energy (least -ve to most -ve)
CaO, MgO, SrO, and BaO
CaO, MgO, SrO,
The anion (O2−) is common to the four
oxides. In the equation
QQ Use this information, plus information from
Chapter 8, to arrange these compounds in order of
decreasing values of lattice energy: KCl, Al2O3, CaO
KCl, Ionic radii: 1.33 1.81 ∆Hlattice K+Cl− |∆H | lattice 1.06 1.40 > least negative Ca2+O2− 0.57 > 1.40 Å 2Al3+3O2− most negative (absolute values) Al2O3 > CaO > KCl
values) E = −C × c a d Qc and Qa = (+2 and −2). The difference is in d. The
four cations are in group 2, so d varies
E is negative and inversely proportional to d:
Ba2+O2− > Sr2+O2− > Ca2+O2− > Mg2+O2− Ionic Bond Formation
• Born-Haber Cycle
Born-the formation of an ionic compound from its
elements in a series of steps, and considers
the energetics of each.
Example: formation of NaCl
Na(s) + ½ Cl2(g) → NaCl(s) ∆H° = –410.9 kJ
The Born-Haber Cycle is an energy level
Borndiagram that applies Hess’s law:
∆Hrxn = ∑ ∆H of all steps 725.4 E
0 Na+(g) + Cl(g) + e–
IENa Na+(g) + Cl–(g)
(g) Na(g) + Cl(g)
Na(g) + ½ Cl2(g)
(g) ∆Hatom,Cl Na(s) + ½ Cl2(g)
(s) ∆Hlattice ∆Hsub, Na
sub, ∆H°f, NaCl
(s) The Born-Haber Cycle of NaCl
Born- Covalent Bonding ∆H°f = ∆Hsublimation, Na + ∆Hatom, Cl + IE1, Na + EACl +
∆Hlattice, NaCl (the two in blue are negative)
lattice, Covalent bonds are formed when atoms
If the atoms share 2 electrons ( a pair) a single
covalent bond is formed.
If the atoms share 4 electrons (two pairs) a double
covalent bond is formed.
If the atoms share 6 (three pairs) electrons a triple
covalent bond is formed.
The attraction between the nuclei and the
electrons is electrostatic in nature. Directional.
The atoms have a lower potential energy when
bound (see upcoming diagram). ∆H°f = 107.7kJ + 121.7kJ + 496.0kJ + (–349.0kJ)
– 787.3 kJ
∆H°f = – 410.9 kJ
If it is not known, ∆Hlattice can be calculated ∆Hlattice = ∆H°f − ∆Hsublimation,, Na − ∆Hatom,, Cl − IE1, Na
− EACl Formation of Covalent Bonds
Plot shows potential energy of H2 molecule as a
function of the distance between the two H atoms. atoms Writing Lewis Formulas:
The Octet Rule
The octet rule states that representative
elements usually attain stable, noble gas
electron configurations in most of their
compounds/ions. molecule We must distinguish between bonding
(shared) electrons and nonbonding
(unshared or lone pairs of) electrons.
of) The Octet Rule & Lewis structures
For all ions we must adjust the number of
electrons available based on charges.
i.e., add one e− for each negative charge;
subtract one e− for each positive charge. The central atom in a molecule or polyatomic
ion is determined by:
-the atom that requires the largest number of
electrons to complete its octet. It goes in the
If two atoms are in the same, the less
electronegative element goes in the center. Drawing the Lewis Structures (STRUCTURAL; ch 3)
Example: SO3. 1) Find the total number of valence electrons
1S = 1(6) = 6 e−
3O = 3(6) = 18 e−
available = 24 e−
O ─S─ O 2) Draw a single bond from the central atom
to each of the atoms surrounding it. 3 bonds × 2 e− = 6 e−
24 − 6 = 18 e− 3) Subtract electrons used in bond formation
from the total number of electrons. 4) The remaining electrons (in pairs) are placed
on the surrounding atoms first. Each
surrounding atom (except H) will receive
.. | ..
enough electron pairs to have an octet. If
:O ─S─ O:
any electrons are left over, put them on the central atom. •• Drawing the Lewis Structures Lewis Structures 5) If the central atom does not have an octet, then
one or two single bonds are converted to double
bonds, or, a double bond is converted to a triple
bond. Make double bonds before making a triple
bond. Write Lewis dot and dash formulas for the sulfite
ion, SO32−; (3 O’s bonded to S)
Valence e- = 6 (S) + 3 × 6 (O)
electrons in bonds
This polyatomic ion has ? BP and ? LP. ..
.. | ..
:O ─S─ O:
•• •• → ..
O=S ─ O:
•• •• ·· ··
·O · S
·· Exceptions to the octet rule: (H, Be, B, Al, and
other elements in rows 3, 4, 5, and 6). Lewis Structures (practice)
CO2, NO2+, NO3−, HNO3, SO42−, H2SO4,
Cl2O7, CH3CH2OH (ethanol), CH3COOH (acetic
PCl3, PCl5, SF6, SOCl2, IF5, IF4+ ··
· O · 2·
· or Lewis Structures:
T # electrons:
B = 3; Br = 3 × 7
= 24e3 bonds = 6e- ;
18 e- left ( 3 pairs
per B atom)
· ·· · ·· 2O·
·· BBr4T # electrons:
B = 3; Br = 4 × 7; +1e= 32e- 4 bonds = 8e- ;
24 e- left (3 pairs
per B atom)
Br Br Lewis Structures: AsF5 (5 F around As) Lewis Structures: C2H6 or CH3CH3 T # electrons:
As = 5; F = 5 × 7
= 40e- 3 H’s bond to each C; C bonds to C
T # electrons:
C = 2 × 4; H = 1 × 6
= 14e- 5 bonds = 10e- ;
30 e- left ( 3 pairs
per F atom) F
octet H C H As F F Lewis Structures: CH3CH2OH Lewis Structures: CH3COOH 3 H’s bond to C; C bonds to C; O to C… 3 H’s bond to C; C bonds to C; O to C… T # electrons:
C = 2 × 4; H = 1 × 6; O = 6
= 20e- T # electrons:
C = 2 × 4; H = 1 × 4; 2 × O = 12
= 24e- 8 bonds = 16e- ;
4 left over (2 pairs) 7 bonds = 14e- H O H C C O H H C has octet
H has duet H
H H H H C C O H O
C O H H H Resonance Lewis Structures
Molecules and/or ions having the same
number of valence electrons and the same
:N≡ C H H F F [:N≡O:]+
:N≡ H H
7 bonds = 14e- ; [:C≡O:]
:C≡ They all have 3 BP and 2 LP (2 atoms) There are three possible structures for SO3.
The double bond can be placed in one of three
· ·· · ··
· · ··
·· S O·
· ·· · When two or more Lewis formulas are necessary
to show the bonding in a molecule, we must use
equivalent resonance structures to show the
Double-headed arrows are used to indicate
resonance formulas. Resonance Structures Resonance is a method used to represent
• There are no single or multiple bonds in
SO3 (nor CO2)
• All bonds in SO3 are equivalent.
The best Lewis formulas of SO3 (CO2) that
can be drawn are: (average of 3
Resonance Hybrids CO2
2 × O = 12 e1 × C = 4 eTotal = 16 e- O C O O C O O Resonance C O S O O O C O O Molecules with odd number of electrons
NO has 11 valence electrons Formal Charges on Atoms: N O an accounting tool for electron ownership
= (# valence e– in free atom) – (# e– in lone
pairs on atom) – ½(# bonded e– on atom) NO2 has 17 valence electrons O N O O N Charge Distribution in Covalent Bonds and
Molecules (Formal Charge) O These are known as Free Radicals: they have an
Very reactive: e.g. dimerization of NO2 to N2O4. Formal Charge (FC) and Best structure
Best Lewis structure has
• zero FC on all atoms
• lowest FC possible
• negative FC on most electronegative
atoms and positive FC on least
The most electronegative elements are at the
top right of periodic table (except for the
noble gases.) Formal charge , FC =
group number of atom − LPE - # Bonds Formal Charge (FC)
H O C O N= 5 − 0 - 4 = +1
H = 1 − 0 -1 = 0 C=4−0-4=0
O = 6 − 4 -2 = 0 For example, consider thiocyanate ion
0 +1 –2
–2 +2 –1
–1 0 0
N C S
C S N
S N C
•• best structure because lowest FC and
–ve FC on most electronegative atom Electronegativity
Electronegativity is a measure of the relative tendency
of an atom to attract electrons to itself when bonded
to another element.
Electronegativity is measured on the Pauling scale.
F is the most electronegative element.
Cs and Fr are the least electronegative elements.
For the representative elements, electronegativities
usually increase from left to right across periods
and decrease from top to bottom within groups. Sum of formal charges = −1 + 0 + 0 = −1 Electronegativity
Example: Arrange these elements based on
increasing electronegativity values.
Se, Ge, Br, As (period # 4)
Ge, Arrange these elements based on increasing
Be, Mg, Ca, Ba (group IIA) Electronegativity, polarity, and Ionic bonds
A non-polar bond is formed when the
nondifference in electronegativity is between 0.0 – 0.4
A polar bond is formed when the difference in
electronegativity is 0.5 – 1.9 (partial charge)
Ionic bonds form when the difference of
electronegativity is greater than 1.9 (net charge);
e- transferred no charge
non- Bond Polarity and ∆EN
covalent polar covalent ionic bond Cl is more electronegative
Na+ and Cl− (2.1)
ionic Dipole Moments
Molecules whose centers of positive and
negative charge do not coincide, have an
asymmetric charge distribution, and are polar.
These molecules have a net dipole moment (µ).
The dipole moment is: 0.0 d, distance separating charges of equal
magnitude and opposite sign 0.4
Electronegativity difference, ∆EN equal sharing unequal sharing µ=δ×d magnitude of the charge e- transferred
e41 indicated by crossed arrow pointing from
positive end to negative end of dipole Dipole Moments Polarity of Molecules Some nonpolar molecules have polar bonds. Molecules in which dipole moments of the
bonds do not cancel, are polar molecules There are two conditions that must be true for
a molecule to be polar.
1. There must be at least one polar bond
or, one lone pair of electrons present.
2. The polar bonds, if there are more than
one, and lone pairs, must be arranged so
that their dipole moments do not cancel
each other. δ– δ+ δ– µtotal = 0
δ+ Cl δ– H
δ+ Molecules that do not contain polar bonds, or
in which all dipole moments cancel, are nonnonpolar molecules µtotal = 0 (CO2) Order of a bond is the number of bonding electron
pairs shared by two atoms in a molecule. There are
bond orders of 1 (single), 2 (double), 3 (triple), and
fractional (1.5, 2.5, ..) O O (ClF Bonding Order ClF, CO2 & H2O O C µtotal ≠ 0 H
δ+ µtotal ≠ 0
polar # of shared pairs linking atoms
Bond order = ──────────────────────
# of links in the molecule or ion (no L.P.)
H – Cl
1 2 O=C=O
2 2 H – C≡ N
1 3 δ– F µtotal ≠ 0, polar Bond Strength and Bond Length
(shortest) triple < double < single
single < double < triple (strongest)
C–C 1.54 348 C=C 1.34 614 C≡C 1.20 839 Bond Energy (Enthalpy)
-the energy required to break 1 mole of the bond in the
H2(g) → 2H(g) 436 kJ/mol (single bond) O2(g) → 2O(g) 498 kJ/mol (double bond) N2(g) → 2N(g) 946 kJ/mol (triple bond) Average bond energies can be used to estimate the
enthalpy change, ∆H of a reaction. Use:
∆Hrxn = Σ (∆Hs bonds broken) + Σ (∆Hs bonds formed)
requires energy (+) releases energy (-) Bond Energy (Enthalpy) Bond Energy (Enthalpy) Estimate the enthalpy change of the reaction between
iodomethane and water Estimate the enthalpy change of the reaction between
iodomethane and water CH3CH2I(g) + H2O(g) → CH3CH2OH(g) + HI(g) CH3CH2I(g) + H2O(g) → CH3CH2OH(g) + HI(g) Plan:
which bonds are formed; which bonds are broken
add enthalpy changes: bonds formed, positive (+);
bonds broken, negative (-)
Break (reactants): C-I bond (CH3CH2I) = 238 kJ
one O-H bond (H2O)
= 463 kJ Form (products): C-O bond (CH3CH2OH) = 360 kJ
H-I bond (HI)
= 299 kJ ∆Hrxn = +42 kJ Overall, the reaction is endo or exothermic? ...
View Full Document