Chemical Bonding and Atomic Properties
Chemical Bonding
and Molecular Geometry
•
Ionic Bond
•
Formation of Ions
•
Electron Configurations of Ions
•
Ionic Size and Charge density,
•
Relative Strength of Ionic Bonds
•
Lattice Energy
•
Steps in the Formation of an Ionic Compound
•
The Born-Haber Cycle
Chemical Bonding
and Molecular Geometry
•
Covalent Bonds
•
Electronegativity
•
Polarity of Covalent Bonds
•
Lewis Structures and the Octet Rule
•
Exceptions to the Octet Rule
•
Resonance Lewis Structures
•
Bond Energies
•
Calculating Enthalpy using Bond Energy
•
Molecular Shape - The VSEPR Model
Review of Atomic Properties
•
Effective nuclear charge & Atomic Size:
1.
Effective nuclear charge increases left-to-right
across periods, and decreases top-to-bottom down
a group;
2.
Atomic size decreases left-to-right across periods,
and increases top-to-bottom down a group:
3.
Ionization energy incresses left-to-right across a
period, and decreases top-to-bottom down a
group.
Review on Atomic Properties
•
Atomic Size, Ionization Energy, and Electron
Affinity:
1.
L-to-R: atomic size decreases; ionization energy
increases; electron affinity increases;
2.
Top-to-bottom: atomic size increases; ionization
energy decreases; electron affinity decreases;
Ionic Compound
•
The atoms in sodium chloride (common table salt) are arranged to
(a)
maximize opposite charges interacting. The smaller spheres
represent sodium ions, the larger ones represent chloride ions. In
the expanded view
(b)
, the geometry can be seen more clearly.
Note that each ion is “bonded” to all of the surrounding ions—six in
this case.
Ionic bonds
•
Attractions between cations and anions
;
•
Bonds formed between metals and
nonmetals
Formation of Cations
•
Ions are formed when metals react with nonmetals
– the metal atom loses one or more of its valence
electrons to the nonmetal atom;
•
Atoms of representative metals lose valence
electrons to acquire the noble gas electron
configurations;
•
Cations of representative metals have noble gas
electron configurations;
Formation of Cations
•
All Alkali metals (1A):
M
M
+
+ e
-
•
All Alkaline Earth metals (2A)
:
M
M
2
+
+ 2e
-
•
Certain Group 3A metals:
M
M
3
+
+ 3e
-
;
Formation of Anions
•
Nonmetal atoms gain electrons to acquire the noble
gas electron configurations;
•
Simple anions have noble gas electron
configurations;
Formation of Anions
•
The halogen family (VIIA):
X + e
-
X
-
•
The oxygen family (VIA):
X + 2e
-
X
2-
•
N and P (in Group VA):
X + 3e
-
X
3-
Common Ions of the Representative Elements
•
Ions that are
isoelectronic
to He
(1
s
2
): Li
+
& H
-
•
Ions
isoelectronic
to Ne
(1s
2
2s
2
2p
6
):
Na
+
, Mg
2+
, Al
3+
, F
-
, O
2
-
, and N
3
-
•
Ions
isoelectronic to Ar
(1s
2
2s
2
2p
6
3s
2
3p
6
):
K
+
, Ca
2+
, Sc
3+
, Cl
-
, S
2
-
, and P
3
-
Common Ions of the Representative Elements
•
Ions
isoelectronic to Kr
(1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
10
4p
6
):
Rb
+
, Sr
2+
, Y
3+
, Br
-
, and Se
2
-
;
•
Ions
isoelectronic to Xe
(1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
10
4p
6
5s
2
4d
10
5p
6
)
Cs
+
, Ba
2+
, La
3+
, I
-
, and Te
2
-
;
Ionic Radii
Relative size of isoelectronic ions
:
•
Al
3+
< Mg
2+
< Na
+
< Ne < F
-
< O
2-
< N
3-
;
•
Sc
3+
< Ca
2+
< K
+
< Ar < Cl
-
< S
2-
< P
3-
;
Trend of ionic radii within a group:
•
Li
+
< Na
+
< K
+
< Rb
+
< Cs
+
;
•
F
-
< Cl
-
< Br
-
< I
-
;
Cations From Transition Metals
•
Atoms of the transition metals lose variable
number of electrons;
•
Cations have variable charges;
•
Cations derived from transition metal group
do not
acquire the noble gas electron
configuration;
Electron Configurations of
Selected Cations from Transition Metals
•
Examples:
Cr: [Ar]
4s
1
3d
5
Cr
Cr
2+
+ 2e
-
;
Cr
2+
: [Ar] 3d
4
Cr
Cr
3+
+ 3e
-
;
Cr
3+
: [Ar] 3d
3
Fe: [Ar]
4s
2
3d
6
Fe
Fe
2+
+ 2e
-
;
Fe
2+
: [Ar] 3d
6
Fe
Fe
3+
+ 3e
-
;
Fe
3+
: [Ar] 3d
5
Charge Density and
the Strength of Ionic Bond
Lattice Energy (
U
L
)
•
Lattice energy
- energy released when gaseous ions
combine to form solid ionic compound:
M
+
(g)
+ X
-
(g)
MX
(s)
;
U
L
=
Lattice energy
•
Examples:
Na
+
(g)
+ Cl
-
(g)
NaCl
(s)
;
U
L
= -787 kJ/mol
•
Li
+
(g)
+ F
-
(g)
LiF
(s)
;
U
L
= -1047 kJ/mol
Lattice energy
Lattice energy
k
(
q
1
q
2
/r
2
)
q
1
and
q
2
= charge magnitude on ions;
r = distance between nuclei, and
k
= proportionality constant.
Lattice energy
increases as charge magnitude
increases and ionic size decreases.
Lattice Energies of Some Ionic Compounds
•
Lattice Energy,
U
L
(kJ/mol)
The energy required to separate a mole of ionic
solids into gaseous ions
;
MX
(s)
M
+
(g)
+ X
-
(g)
•
M
n+
/X
n-
F
-
Cl
-
Br
-
I
-
O
2
-
•
Li
+
1047
853
807
757
2942
•
Na
+
923
787
747
704
2608
•
K
+
821
715
682
649
2311
•
Mg
2+
2957
2526
2440
2327
3919
•
Ca
2+
2628
2247
2089
2059
3570
The Born-Haber Cycle for the Formation of NaCl
•
Na
+
(g)
+ Cl
(g)
_______________
•
-349 kJ
•
+496
kJ
_______ Na
+
(g)
+ Cl
-
(g)
•
Na
(g)
+ Cl
(g)
___________
•
+121 kJ
•
Na
(g)
+ ½Cl
2
(g)
________
?
kJ
•
+108 kJ
•
Na
(s)
+ ½Cl
2
(g)
________
•
-411 kJ
•
NaCl
(s)
_________________
Chemical Steps in the Formation of NaCl
•
Na
(s)
Na
(g)
;
H
s
= +108 kJ
•
½Cl
2
(g)
Cl
(g)
;
½BE = +121 kJ
•
Na
(g)
Na
+
(g)
+ e
-
;
IE = +496 kJ
•
Cl
(g)
+ e
-
Cl
-
(g)
;
EA = -349 kJ
•
Na
+
(
g)
+ Cl
-
(g)
NaCl
(s)
;
U
L
= ? kJ
•
Na
(s)
+ ½Cl
2
(g)
NaCl
(s)
;
H
f
= -411 kJ
H
f
=
H
s
+ ½BE + IE + EA +
U
L
H
s
=
Enthalpy of sublimation
; IE =
Ionization energy
;
BE =
Bond energy
; EA =
Electtron affinity
;
U
L
=
Lattice
energy
;
H
f
=
Enthalpy of formation
)
The Born-Haber Cycle for LiF
•
Li
+
(g)
+ F
(g)
_______________
•
-328 kJ
•
+520
kJ
_______Li
+
(g)
+ F
-
(g)
•
Li
(g)
+ F
(g)
___________
•
+77 kJ
•
Li
(g)
+ ½F
2
(g)
________
? kJ
•
+161 kJ
•
Li
(s)
+ ½F
2
(g)
________
•
-617
kJ
•
LiF
(s)
_________________
Chemical Steps in the Formation of LiF
•
Li
(s)
Li
(g)
;
H
s
= +161 kJ
•
½F
2
(g)
F
(g)
;
½BE = +77 kJ
•
Li
(g)
Li
+
(g)
+ e
-
;
IE = +520 kJ
•
F
(g)
+ e
-
F
-
(g)
;
EA = -328 kJ
•
Li
+
(
g)
+ F
-
(g)
LiF
(s)
;
U
L
= ?
•
Li
(s)
+ ½F
2
(g)
LiF
(s)
;
H
f
= -617 kJ
H
f
=
H
s
+ ½BE + IE + EA +
U
L
H
s
=
Enthalpy of sublimation
; IE =
Ionization energy
;
BE =
Bond energy
; EA =
Electtron affinity
;
U
L
=
Lattice
energy
;
H
f
=
Enthalpy of formation
)
The Born-Haber Cycle for MgO
•
Mg
2+
(g)
+ O
2-
(g)
_____________
•
+737 kJ
•
Mg
2+
(g)
+ O
(g)
________
•
+2180 kJ
•
Mg
(g)
+ O
(g)
_________
•
+247 kJ
•
Mg
(g)
+ ½O
2
(g)
________
? kJ
•
+150 kJ
•
Mg
(s)
+ ½O
2
(g)
________
•
-602
kJ
•
MgO
(s)
_________________
Chemical Steps in the Formation of MgO
•
Mg
(s)
Mg
(g)
;
H
s
= +150 kJ
•
½O
2
(g)
O
(g)
;
½BE = +247 kJ
•
Mg
(g)
Mg
2+
(g)
+
2
e
-
;
IE = +2180 kJ
•
O
(g)
+
2
e
-
O
2-
(g)
;
EA = +737 kJ
•
Mg
2+
(g)
+
O
2-
(g)
MgO
(s)
;
U
L
= ? kJ
•
Mg
(s)
+ ½O
2
(g)
MgO
(s)
;
H
f
= -602 kJ
H
f
=
H
s
+ ½BE + IE + EA +
U
L
H
s
=
Enthalpy of sublimation
; IE =
Ionization energy
;
BE =
Bond energy
; EA =
Electron affinity
;
U
L
=
Lattice
energy
;
H
f
=
Enthalpy of formation
)
Covalent Bonds
•
Bonds between two nonmetal atoms, or between
an atom of semimetal and an atom of nonmetal;
•
Bonds are formed by sharing one or more
electron pairs;
•
One, two or three pairs of electrons may be
shared between two atoms to form a single,
double, or triple covalent bonds, respectively.
Potential Energy Change during the
Formation of H
2
Molecule
Potential Energy for the Formation of H
2
The potential energy of two separate hydrogen atoms (right)
decreases as they approach each other, and the single electrons on
each atom are shared to form a covalent bond. The bond length is the
internuclear distance at which the lowest potential energy is achieved.
Polarity of Covalent Bonds
1.
Covalent bonds can be polar or nonpolar
;
2.
Nonpolar covalent bonds
- bonds between identical
atoms or atoms having the same
electronegativity
.
3.
Polar covalent bonds
- bonds between atoms with
different
electronegativity
;
Polar Covalent Bonds
1.
Bonds have
partial ionic
character
2.
Bond
polarity
depends on
EN;
EN =
difference
in electronegativity
of bonded
atoms
Electronegativity
•
Electronegativity
=
relative ability
of bonded
atom to pull shared electrons.
Trends of Electronegativity:
increases left-to-right across periods;
decreases top-to-bottom down the group.
Electronegativity
Most electronegative
element is at top right corner of
the Periodic Table (
noble gas is excluded
):
Fluorine is most electronegative with EN = 4.0
Least electronegative
element is at bottom left corner
of the Periodic Table:
Francium is least electronegative with EN = 0.7
General trends:
•
Electronegativity
increases from left to right
across a period
•
For the representative elements (
s
and
p
block) the
electronegativity
decreases
as one goes down
a group
•
Electronegativity trend for transition metals is less predictable.
Electronegativity and Bond Polarity
•
In F
2
molecule, electrons are shared equally and the bond is
nonpolar;
•
In HF molecule, F-atom is more electronegative than H-
atom; electrons are drawn closer to the fluorine atom.
•
H
―
F bond is very polar;
Electronegativity and bond polarity
The H-F bond can thus be represented as:
•
The '
+
' and '
-
' symbols indicate
partial
positive and negative charges.
•
The arrow indicates the "pull" of electrons off the hydrogen and towards
the more
electronegative
atom.
•
In lithium fluoride the much greater
relative
electronegativity of the
fluorine atom completely strips the electron from the lithium and the result
is an ionic bond (no sharing of the electron)
Predicting Bond Type From
EN
General rule of thumb for bond polarity
•
EN
=
0-0.4
, bond is
non-polar covalent
;
•
EN
> 0.4,
but
< 1.9
, bond is
polar covalent
•
EN
>
1.9
, bond is considered
ionic.
Lewis Symbols and Lewis Structures
•
Lewis symbols illustrating the number of valence electrons for each
element in the third period of the periodic table.
Lewis Structure for Cl
2
•
Individual Cl atom (left) has incomplete octet; in Cl
2
molecule, each
atom acquires the octet state.
•
H
2
molecule does not have lone-pair or nonbonding pair; Cl
2
molecule has one bonding pair represented by a ”dash” and 3
lone-pairs on each atom.
Using “dash” to represent bonding
pair and dot-pairs for lone-pairs
Drawing Lewis Structures for Molecules
or Polyatomic Ions
Step-1:
•
Calculate the total number of valence
electrons;
•
For polyatomic ions, add one additional
electron for each negative charge, or subtract
one for each positive charge on the ion.
Lewis Structures for Molecules and
Polyatomic ions
Step-2:
•
Choose a central atom; normally the least
electronegative atom in the group;
(Hydrogen and Fluorine cannot become central
atoms)
•
Connect other atoms to the central atom with
single bonds (a pair of electrons).
Lewis Structures for Molecules
and Polyatomic ions
Step-3:
•
Complete the octet state of all terminal atoms,
except hydrogen, using available valence
electron pairs;
•
Place remaining pairs of electrons (if present)
on central atom as lone pairs.
Octet State of Central Atom
Step-4:
•
If the central atom has not achieved the octet
state, but all valence electrons have been used,
move one or more lone-pair electrons from
“terminal” atoms, one pair at a time, to form a
double or triple bonds to complete octet of the
central atom.
Lewis Model for the Formation of Covalent
Bonds and Covalent Molecules
Lewis Structures of CH
4
, NH
3
and H
2
O
•
A double-bond can be represented by a double lines or
two pairs of dots.
Lewis Structures of
Formaldehyde and Ethylene
•
A triple bonds can be represented by a triple lines or
three pairs of dots.
Triple Bonds
Lewis Structures of CO
2
, HCN, and C
2
H
2
Resonance Lewis Dot Structures for CO
3
2-
Resonance Lewis Structures of PO
4
3-
Assigning Appropriate Formal Charges
Exception to Octet Rule
1.
If the central atom is from group 2A or 3A, and the
octet state is not achieved, the central atom is said
to have
incomplete octet
.
2.
Central atoms from periods 3, 4, 5, …may have
more than 8 valence electrons, and it is said to have
an
expanded octet
;
3.
Some molecules contain odd numbers of electrons;
one of the atoms in it will contain unpaired
electrons.
Covalent Molecules with the Central
Atoms having an Expanded Octet State.
Evaluate Formal Charge
•
Evaluate
formal charges
(fc) on each atom in the
molecule to determine best correct or best Lewis
structures.
•
Formal charge
is apparent charge on an atom
in a Lewis formula; it is determined as follows:
•
Formal charge
(fc) =
(Group # of atom) – (# of lone-pair electrons on the atom)
– (# of covalent bonds the atom forms)
Assigning Formal Charges
Choosing the correct or best Lewis structures
based on formal charges
•
If two or more Lewis structures that satisfy the octet
rule can be drawn for a given molecule or
polyatomic ions, the preferred structure would be
one in which:
1.
The formal charges used are the smallest;
2.
Negative fc are located on more electronegative
atoms, and positive fc on less electronegative
atoms.
Which Lewis structures of CO
2
& N
2
O are correct?
Bond Length and Bond Energy
Bond Length
.
.
.
•
Bond length
-
distance between nuclei of
bonded atoms
.
•
Bond length
increases with atomic size;
Bond length
:
single bonds > double bonds > triple bonds
Bond Energy
•
Bond energy
-
the energy required to break a
covalent bond
.
•
Shorter covalent bonds, higher
bond energy
.
Bond energy:
Triple bonds > double bonds > single bond
Bond Length and Bond Energies
•
Bond length (pm) and bond energy (kJ/mol)
•
Bond Length Energy Bond Length Energy
_________________________________________________________________________________________________________
•
H─H
74 432
H─C
109
413
•
C─C
154 347
H─N
101
391
•
N─N 145 160
H─O
96
467
•
O─O 148 146
H─F
92
565
•
F─F
142 154
H─Cl
127
427
•
Cl─Cl 199 243
H─Br
141
363
•
Br─Br 228 193
H─I
161
295
•
I─I
267 149
C─F
135
485
•
C─S
182 259
C─Cl
177
339
•
C─C
154 347
C─Br 194 276
•
C─N 147 305
C─I
214 240
•
C─O 143 358
C─C
154
347
•
O─O 148 146
C=C
134
614
•
O=O
121 495
C
≡
C
120
839
•
C=O
123
745
N=N
?
418
•
C=N
138
615
N
≡
N
110
945
Using Bond Energy to Calculate Enthalpy
•
Chemical reactions in the gaseous state involve:
Breaking of covalent bonds in reactants and the formation
of covalent bonds in products.
•
Bond breaking requires energy;
•
Bond formation releases energy;
H
reaction
=
(
Energy of bond breaking
) +
(
Energy
of bond formation
)
H
reaction
=
{BE(in reactants)} -
{BE(in products)}Bond Breaking and Bond Formation in the
Reaction to form H
2
O
Calculating Enthalpy Reaction
Using Bond Energy
Example
: use bond energy to calculate
H
for the
following reaction in gaseous state:
2H
2
(g)
+ O
2
(g)
2H
2
O
(g)
;
H
reaction
=
{BE(in reactants)} -
{BE(in products)}Using bond energy to calculate enthalpy
BE(in reactants) = 2 x BE(H─H) + BE(O═O)
= (2 x 435) + 495 = 1365 kJ
BE(in products) = 4 x BE(O─H)
= (4 x 467) = 1868 kJ
H
reaction
=
{BE(in reactants)} -
{BE(in products)}
H
reaction
= 1365
1868 = -503 kJ
Calculating Enthalpy Reaction
Using Bond Energy
Example
: use bond energy to calculate
H
for the
following reaction in gaseous state:
CH
3
OH + 2O
2
CO
2
+ 2H
2
O
;
H
reaction
=
{BE(in reactants)} -
{BE(in products)}Using bond energy to calculate enthalpy
BE(in reactants) =
3 x BE(C─H) + BE(C─O) + BE(O─H) + 2 x BE(O═O)
= (3 x 413) + 358 + 467 + (2 x 495) = 3054 kJ
BE(in products)
= 2 x BE(C═O)* + 4 x BE(O─H)
= (2 x 799) + (4 x 495) = 3578 kJ
H
reaction
=
{BE(in reactants)} -
{BE(in products)}
H
reaction
= 3054
3578 = -524 kJ
Molecular Shapes of BeI
2
, HCl, IF
2
-
, ClF
3
, and NO
3
-
Lewis Structures, Molecular Shapes & Polarity
The Shape of Water Molecules
Structures and Shapes of Formaldehyde and Ethylene
The Shapes of Methane and Ammonia Molecules
The VSEPR Shapes
Explore the formation of ionic and covalent bonds, electron configurations of ions, and molecular geometry. Learn about ionic compound formation, atomic properties like effective nuclear charge, atomic size, ionization energy, and electron affinity. Discover the essential concepts of cations formation and the interactions in ionic compounds.
Uploaded on Sep 24, 2024 | 1 Views
Download Presentation
Please find below an Image/Link to download the presentation.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. Download presentation by click this link. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.
E N D
Presentation Transcript
Chemical Bonding and Molecular Geometry Ionic Bond Formation of Ions Electron Configurations of Ions Ionic Size and Charge density, Relative Strength of Ionic Bonds Lattice Energy Steps in the Formation of an Ionic Compound The Born-Haber Cycle
Chemical Bonding and Molecular Geometry Covalent Bonds Electronegativity Polarity of Covalent Bonds Lewis Structures and the Octet Rule Exceptions to the Octet Rule Resonance Lewis Structures Bond Energies Calculating Enthalpy using Bond Energy Molecular Shape - The VSEPR Model
Review of Atomic Properties Effective nuclear charge & Atomic Size: 1. Effective nuclear charge increases left-to-right across periods, and decreases top-to-bottom down a group; 2. Atomic size decreases left-to-right across periods, and increases top-to-bottom down a group: 3. Ionization energy incresses left-to-right across a period, and decreases top-to-bottom down a group.
Review on Atomic Properties Atomic Size, Ionization Energy, and Electron Affinity: 1. L-to-R: atomic size decreases; ionization energy increases; electron affinity increases; 2. Top-to-bottom: atomic size increases; ionization energy decreases; electron affinity decreases;
Ionic Compound The atoms in sodium chloride (common table salt) are arranged to (a) maximize opposite charges interacting. The smaller spheres represent sodium ions, the larger ones represent chloride ions. In the expanded view (b), the geometry can be seen more clearly. Note that each ion is bonded to all of the surrounding ions six in this case.
Ionic bonds Attractions between cations and anions; Bonds formed between metals and nonmetals
Formation of Cations Ions are formed when metals react with nonmetals the metal atom loses one or more of its valence electrons to the nonmetal atom; Atoms of representative metals lose valence electrons to acquire the noble gas electron configurations; Cations of representative metals have noble gas electron configurations;
Formation of Cations All Alkali metals (1A): M M++ e- All Alkaline Earth metals (2A): M M2 ++ 2e- Certain Group 3A metals: M M3++ 3e-;
Formation of Anions Nonmetal atoms gain electrons to acquire the noble gas electron configurations; Simple anions have noble gas electron configurations;
Formation of Anions The halogen family (VIIA): X + e- X- The oxygen family (VIA): X + 2e- X2- N and P (in Group VA): X + 3e- X3-
Common Ions of the Representative Elements Ions that are isoelectronic to He (1s2): Li+& H- Ions isoelectronic to Ne (1s2 2s22p6): Na+, Mg2+, Al3+, F-, O2-, and N3- Ions isoelectronic to Ar (1s2 2s22p63s23p6): K+, Ca2+, Sc3+, Cl-, S2-, and P3-
Common Ions of the Representative Elements Ions isoelectronic to Kr (1s22s22p63s23p64s23d104p6): Rb+, Sr2+, Y3+, Br-, and Se2-; Ions isoelectronic to Xe (1s22s22p63s23p64s23d104p65s24d105p6) Cs+, Ba2+, La3+, I-, and Te2-;
Ionic Radii Relative size of isoelectronic ions: Al3+< Mg2+< Na+< Ne < F-< O2-< N3-; Sc3+< Ca2+< K+< Ar < Cl-< S2-< P3-; Trend of ionic radii within a group: Li+< Na+< K+< Rb+< Cs+; F-< Cl-< Br-< I-;
Cations From Transition Metals Atoms of the transition metals lose variable number of electrons; Cations have variable charges; Cations derived from transition metal group do not acquire the noble gas electron configuration;
Electron Configurations of Selected Cations from Transition Metals Examples: Cr: [Ar] 4s13d5 Cr Cr2++ 2e-; Cr2+: [Ar] 3d4 Cr Cr3++ 3e-; Cr3+: [Ar] 3d3 Fe: [Ar] 4s23d6 Fe Fe2++ 2e-; Fe Fe3++ 3e-; Fe2+: [Ar] 3d6 Fe3+: [Ar] 3d5
Charge Density and the Strength of Ionic Bond Charge density = ????????? ?? ? ???? ?? ??? ????? ???? Greater charge magnitude and small ionic radius High charge density stronger ionic bond; Stronger ionic bond High lattice energy; Stronger ionic bond High melting point;
Lattice Energy (UL) Lattice energy - energy released when gaseous ions combine to form solid ionic compound: M+(g) + X-(g) MX(s); UL= Lattice energy Examples: Na+(g) + Cl-(g) NaCl(s); UL= -787 kJ/mol Li+(g) + F-(g) LiF(s); UL= -1047 kJ/mol
Lattice energy Lattice energy k(q1q2/r2) q1and q2= charge magnitude on ions; r = distance between nuclei, and k = proportionality constant. Lattice energy increases as charge magnitude increases and ionic size decreases.
Lattice Energies of Some Ionic Compounds Lattice Energy, UL(kJ/mol) The energy required to separate a mole of ionic solids into gaseous ions; MX(s) M+(g) + X-(g) Mn+/Xn- Li+ Na+ K+ Mg2+ Ca2+ O2- 2942 2608 2311 3919 3570 F- Cl- 853 787 715 2526 2247 Br- 807 747 682 2440 2089 I- 757 704 649 2327 2059 1047 923 821 2957 2628
The Born-Haber Cycle for the Formation of NaCl Na+(g) + Cl(g) _______________ -349 kJ _______ Na+(g) + Cl-(g) +496 kJ Na(g) + Cl(g)___________ Na(g) + Cl2(g)________ ? kJ Na(s) + Cl2(g)________ NaCl(s)_________________ +121 kJ +108 kJ -411 kJ
Chemical Steps in the Formation of NaCl Na(s) Na(g); Cl2(g) Cl(g); Na(g) Na+(g) + e-; Cl(g) + e- Cl-(g); Hs= +108 kJ BE = +121 kJ IE = +496 kJ EA = -349 kJ UL= ? kJ Hf= -411 kJ Na+(g) + Cl-(g) NaCl(s); Na(s) + Cl2(g) NaCl(s); Hf = Hs+ BE + IE + EA + UL ( Hs= Enthalpy of sublimation; IE = Ionization energy; BE = Bond energy; EA = Electtron affinity; UL= Lattice energy; Hf= Enthalpy of formation)
The Born-Haber Cycle for LiF Li+(g) + F(g) _______________ -328 kJ +520 kJ _______Li+(g) + F-(g) Li(g) + F(g)___________ Li(g) + F2(g)________ ? kJ Li(s) + F2(g)________ LiF(s)_________________ +77 kJ +161 kJ -617 kJ
Chemical Steps in the Formation of LiF Li(s) Li(g); F2(g) F(g); Li(g) Li+(g) + e-; F(g) + e- F-(g); Hs= +161 kJ BE = +77 kJ IE = +520 kJ EA = -328 kJ UL= ? Hf= -617 kJ Li+(g) + F-(g) LiF(s); Li(s) + F2(g) LiF(s); Hf = Hs+ BE + IE + EA + UL ( Hs= Enthalpy of sublimation; IE = Ionization energy; BE = Bond energy; EA = Electtron affinity; UL= Lattice energy; Hf= Enthalpy of formation)
The Born-Haber Cycle for MgO Mg2+(g) + O2-(g) _____________ Mg2+(g) + O(g)________ +737 kJ Mg(g) + O(g)_________ Mg(g) + O2(g)________ ? kJ Mg(s) + O2(g)________ MgO(s)_________________ +2180 kJ +247 kJ +150 kJ -602 kJ
Chemical Steps in the Formation of MgO Mg(s) Mg(g); O2(g) O(g); Mg(g) Mg2+(g) + 2e-; O(g) + 2e- O2-(g); Hs= +150 kJ BE = +247 kJ IE = +2180 kJ EA = +737 kJ UL= ? kJ Hf= -602 kJ Mg2+(g) + O2-(g) MgO(s); Mg(s) + O2(g) MgO(s); Hf= Hs+ BE + IE + EA + UL ( Hs= Enthalpy of sublimation; IE = Ionization energy; BE = Bond energy; EA = Electron affinity; UL= Lattice energy; Hf= Enthalpy of formation)
Covalent Bonds Bonds between two nonmetal atoms, or between an atom of semimetal and an atom of nonmetal; Bonds are formed by sharing one or more electron pairs; One, two or three pairs of electrons may be shared between two atoms to form a single, double, or triple covalent bonds, respectively.
Potential Energy Change during the Formation of H2Molecule
Potential Energy for the Formation of H2 The potential energy of two separate hydrogen atoms (right) decreases as they approach each other, and the single electrons on each atom are shared to form a covalent bond. The bond length is the internuclear distance at which the lowest potential energy is achieved.
Polarity of Covalent Bonds 1. Covalent bonds can be polar or nonpolar; Nonpolar covalent bonds - bonds between identical atoms or atoms having the same electronegativity. 2. Polar covalent bonds - bonds between atoms with different electronegativity; 3.
Polar Covalent Bonds 1. Bonds have partial ionic character Bond polarity depends on EN; EN = difference in electronegativity of bonded atoms 2.
Electronegativity Electronegativity = relative ability of bonded atom to pull shared electrons. Trends of Electronegativity: increases left-to-right across periods; decreases top-to-bottom down the group.
Electronegativity Most electronegative element is at top right corner of the Periodic Table (noble gas is excluded): Fluorine is most electronegative with EN = 4.0 Least electronegative element is at bottom left corner of the Periodic Table: Francium is least electronegative with EN = 0.7
General trends: Electronegativity increases from left to right across a period For the representative elements (s and p block) the electronegativity decreasesas one goes down a group Electronegativity trend for transition metals is less predictable.
Electronegativity and Bond Polarity Compound Electronegativity Difference F2 HF LiF 4.0 - 4.0 = 0 4.0 - 2.1 = 1.9 4.0 - 1.0 = 3.0 Nonpolar covalent Ionic (non- covalent) Type of Bond Polar covalent In F2 molecule, electrons are shared equally and the bond is nonpolar; In HF molecule, F-atom is more electronegative than H- atom; electrons are drawn closer to the fluorine atom. H F bond is very polar;
Electronegativity and bond polarity The H-F bond can thus be represented as: The ' +' and ' -' symbols indicate partial positive and negative charges. The arrow indicates the "pull" of electrons off the hydrogen and towards the more electronegative atom. In lithium fluoride the much greater relative electronegativity of the fluorine atom completely strips the electron from the lithium and the result is an ionic bond (no sharing of the electron)
Predicting Bond Type From EN General rule of thumb for bond polarity EN = 0-0.4, bond is non-polar covalent; EN > 0.4, but < 1.9, bond is polar covalent EN > 1.9, bond is considered ionic.
Lewis Symbols and Lewis Structures Lewis symbols illustrating the number of valence electrons for each element in the third period of the periodic table.
Lewis Structure for Cl2 Individual Cl atom (left) has incomplete octet; in Cl2 molecule, each atom acquires the octet state.
Using dash to represent bonding pair and dot-pairs for lone-pairs H2 molecule does not have lone-pair or nonbonding pair; Cl2 molecule has one bonding pair represented by a dash and 3 lone-pairs on each atom.
Drawing Lewis Structures for Molecules or Polyatomic Ions Step-1: Calculate the total number of valence electrons; For polyatomic ions, add one additional electron for each negative charge, or subtract one for each positive charge on the ion.
Lewis Structures for Molecules and Polyatomic ions Step-2: Choose a central atom; normally the least electronegative atom in the group; (Hydrogen and Fluorine cannot become central atoms) Connect other atoms to the central atom with single bonds (a pair of electrons).
Lewis Structures for Molecules and Polyatomic ions Step-3: Complete the octet state of all terminal atoms, except hydrogen, using available valence electron pairs; Place remaining pairs of electrons (if present) on central atom as lone pairs.
Octet State of Central Atom Step-4: If the central atom has not achieved the octet state, but all valence electrons have been used, move one or more lone-pair electrons from terminal atoms, one pair at a time, to form a double or triple bonds to complete octet of the central atom.
Lewis Model for the Formation of Covalent Bonds and Covalent Molecules
Lewis Structures of Formaldehyde and Ethylene A double-bond can be represented by a double lines or two pairs of dots.
Triple Bonds A triple bonds can be represented by a triple lines or three pairs of dots.
