Transcript Bonding - Berkeley City College

Chemical Bonds
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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 Bonds
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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 and
decreases down a group;
2. electronic shell gets smaller left-to-right across
period and gets bigger down a group;
3. Atomic size decreases left to right across period
and increases top to bottom down a group:
Review of Atomic Properties
• Atomic Size and Ionization Energy:
1. L-to-R: atomic size decreases; ionization energy
increases;
2. Top-to-bottom: atomic size increases; ionization
energy decreases;
3. Ionization energy increases across period (L-toR), but decreases down a group;
Review of Atomic Properties
• Electron affinity increases left to right and decreases
top to bottom:
smaller atoms have stronger attraction of added electron
than larger atoms
Nonmetals have higher tendency to gain electrons than
metals and become anion
Review of Atomic Properties
• Atomic Size and Electron Affinity:
1. L-to-R: atomic size decreases, electron affinity
increases;
2. Top-to-bottom: atomic size increases, electron
affinity decreases;
3. Electron affinity increases across period (L-to-R),
but decreases down a group;
Ionic bonds
• Attractions between cations and anions;
• Bonds formed between metals and nonmetals
Formation of Cations
• Ions formed when metals react with nonmetals metal atoms lose valence electrons to nonmetals;
• Atoms of representative metals lose valence
electrons to acquire the noble gas electron
configuration;
• Cations of representative group have noble gas
electron configurations;
Formation of Cations
• From the alkali metals (1A):
M  M+ + e• From the alkaline Earth metals (2A):
M  M2 + + 2e• From Group 3A metals:
M  M + 3e- ;
3+
Formation of Ions
• The nonmetal atoms gain electrons to the noble gas
electron configuration;
• Anions have noble gas electron configuration;
Formation of Anions
• From the halogen family (VIIA):
X + e-  X • From the oxygen family (VIA):
X + 2e-  X2• From N and P (in Group VA):
X + 3e-  X3-
Common Ions of the Representative Elements
• Ions isoelectronic to He (1s2): Li+ & H-
• Ions isoelectronic to Ne (1s2 2s2 2p6):
Na+, Mg2+, Al3+, F-, O2-, and N3• Ions isoelectronic to Ar (1s2 2s2 2p6 3s2 3p6):
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:
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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
• Transition metal atoms lose variable number of electrons;
• Cations have variable charges;
• Cations do not acquire noble gas electron configurations
Electron Configurations of
Transition Metal Cations
• 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 Strength Ionic Bond
• Charge density = charge/size of ion
Greater charge but 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
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Li+(g) + F- (g)  LiF(s);
UL = -1047 kJ/mol
Lattice energy
Lattice energy  k(q1q2/r2)
q1 and q2 = charge magnitude on ions;
r = distance between nuclei, and
k = proportionality constant.
Lattice energy increases with charge
magnitude but decreases with ionic size
Lattice Energies of Some Ionic Compounds
• Lattice Energy, UL(kJ/mol)
The energy required to separate a mole of ionic solids into the
gaseous/vapor ions;
MX(s)  M+(g) + X-(g)
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Mn+/XnLi+
Na+
K+
Mg2+
Ca2+
F1047
923
821
2957
2628
Cl853
787
715
2526
2247
Br807
747
682
2440
2089
I757
704
649
2327
2059
O22942
2608
2311
3919
3570
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The Born-Haber Cycle for NaCl
• Na+(g) + Cl(g) _______________
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-349 kJ
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+496 kJ _______ Na (g) + Cl (g)
• Na(g) + Cl(g)___________
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+121 kJ
• Na(g) + ½Cl2(g)________
? kJ
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+108 kJ
• Na(s) + ½Cl2(g)________
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-411 kJ
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NaCl(s)_________________
Chemical Processes in the Formation of NaCl
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Na(s)  Na(g);
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½Cl2(g)  Cl(g);
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Na(g)  Na+(g) + e-;
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Cl(g) + e-  Cl-(g);
• Na+(g) + Cl-(g)  NaCl(s);
• Na(s) + ½Cl2(g)  NaCl(s);
DHs = +108 kJ
½BE = +121 kJ
IE = +496 kJ
EA = -349 kJ
UL = ? kJ
DHf = -411 kJ
UL = DHf – (DHs + ½BE + IE + EA)
(DHs = Enthalpy of sublimation; IE = Ionization energy;
BE = Bond energy; EA = Electtron affinity; UL = Lattice
energy; DHf = Enthalpy of formation)
The Born-Haber Cycle for LiF
• Li+(g) + F(g) _______________
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-328 kJ
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+520 kJ _______Li (g) + F (g)
• Li(g) + F(g)___________
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+77 kJ
• Li(g) + ½F2(g)________
? kJ
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+161 kJ
• Li(s) + ½F2(g)________
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-617 kJ
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LiF(s)_________________
Chemical Processes in the Formation of LiF
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Li(s)  Li(g);
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½F2(g)  F(g);
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Li(g)  Li+(g) + e-;
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F(g) + e-  F-(g);
• Li+(g) + F-(g)  LiF(s);
• Li(s) + ½F2(g)  LiF(s);
DHs = +161 kJ
½BE = +77 kJ
IE = +520 kJ
EA = -328 kJ
UL = ?
DHf = -617 kJ
UL = DHf – (DHs + ½BE + IE + EA)
(DHs = Enthalpy of sublimation; IE = Ionization energy;
BE = Bond energy; EA = Electtron affinity; UL = Lattice
energy; DHf = Enthalpy of formation)
The Born-Haber Cycle for MgO
• Mg2+(g) + O2-(g) _____________
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+737 kJ
• Mg2+(g) + O(g)________
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+2180 kJ
• Mg(g) + O(g)_________
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+247 kJ
• Mg(g) + ½O2(g)________
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? kJ
+150 kJ
• Mg(s) + ½O2(g)________
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-602 kJ
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MgO(s)_________________
Chemical Processes in the Formation of MgO
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Mg(s)  Mg(g);
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½O2(g)  O(g);
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Mg(g)  Mg2+(g) + 2e-;
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O(g) + 2e-  O2-(g);
• Mg2+(g) + O2-(g)  MgO(s);
• Mg(s) + ½O2(g)  MgO(s);
DHs = +150 kJ
½BE = +247 kJ
IE = +2180 kJ
EA = +737 kJ
UL = ? kJ
DHf = -602 kJ
UL = DHf – (DHs + ½BE + IE + EA)
(DHs = Enthalpy of sublimation; IE = Ionization energy;
BE = Bond energy; EA = Electron affinity; UL = Lattice
energy; DHf = Enthalpy of formation)
Covalent Bonds
• Bonds between two nonmetals or between a
semimetal and a nonmetal atoms
• Bonds formed by sharing electron pairs;
• One, two or three pairs of electrons shared
between two atoms;
• A pair of atoms may form single, double, or triple
covalent bonds;
Potential energy of H-atoms during the
formation of H2 molecule
Polarity of Covalent Bonds
1. Covalent bonds - 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 DEN;
DEN = difference in electronegativity of bonded
atoms
Electronegativity
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Electronegativity = relative ability of bonded
atom to pull shared electrons.
Electronegativity Trend:
increases left-to-right across periods;
decreases down the group.
Electronegativity
Most electronegative element is at top right corner of
Periodic Table
Fluorine is most electronegative with EN = 4.0
Least electronegative element is at bottom left corner
of 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
Compound
Electronegativity
Difference
Type of Bond
F2
HF
LiF
4.0 - 4.0 = 0
4.0 - 2.1 = 1.9
4.0 - 1.0 = 3.0
Nonpolar
covalent
Polar covalent
Ionic (noncovalent)
•In F2 electrons are shared equally and bond is nonpolar
•In HF the fluorine is more electronegativity than hydrogen electrons are drawn closer to fluorine.
• H―F bond is very polar
Electronegativity and bond polarity
The H-F bond can thus be represented as:
•The 'd+' and 'd-' 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 Electronegativity
General rule of thumb for bonds:
• DEN = 0-0.5, bond is non-polar covalent;
• DEN > 0.5, but < 2.0, bond is polar covalent
• DEN > 2.0, bond is considered ionic.
Potential Energy Diagram for Covalent
Bond Formation
Bond Length
Bond length - distance between the nuclei of
bonded atoms.
The larger the atoms that are bonded, the greater
the bond length.
Bond length:
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single bonds > double bonds > triple bonds
Bond Energy
Bond energy - the energy required to break the
bonds between two atoms.
The shorter the bond, the greater the 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
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H─H
C─C
N─N
O─O
F─F
Cl─Cl
Br─Br
I─I
C─S
C─C
C─N
C─O
O─O
O=O
C=O
C=N
74
154
145
148
142
199
228
267
182
154
147
143
148
121
123
138
432
347
160
146
154
243
193
149
259
347
305
358
146
495
745
615
H─C
H─N
H─O
H─F
H─Cl
H─Br
H─I
C─F
C─Cl
C─Br
C─I
C─C
C=C
C≡C
N=N
N≡N
109
101
96
92
127
141
161
135
177
194
214
154
134
120
?
110
413
391
467
565
427
363
295
485
339
276
240
347
614
839
418
945
Bond Breaking and Bond Formation in the
Reaction to form H2O
Using Bond Energy to Calculate Enthalpy
• Chemical reactions in the gaseous state only involve:
the breaking of covalent bonds in reactants and the
formation of covalent bonds in products.
• Bond breaking requires energy
• Bond formation releases energy
DHreaction = S(Energy of bond breaking) + S(Energy of
bond formation)
Calculating Enthalpy Reaction
Using Bond Energy
Example: use bond energy to calculate DH
for the following reaction in gaseous
state:
CH3OH + 2 O2  CO2 + 2H2O;
DHreaction = S{BE(in reactants)} - S{BE(in products)}
Using bond energy to calculate enthalpy
S{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
S{BE(in products)
= 2 x BE(C═O)* + 4 x BE(O─H)
= (2 x 799) + (4 x 495) = 3578 kJ
DHreaction = S{BE(in reactants)} - S{BE(in products)}
DHreaction = 3054  3578 = -524 kJ
Lewis Structures for Molecules or Polyatomic
ions
Step-1:
• Calculate 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 (the least
electronegative atom)
(Hydrogen and Fluorine cannot become central atoms)
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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.
• Place remaining pairs of electrons (if present)
on central atom as lone pairs.
Octet State of Central Atom
Step-4:
• If central atom has not acquired octet state but
no more electrons available, move lone-pair
electrons from terminal atoms, one pair at a
time, to form double or triple bonds to
complete octet of the central atom.
Lewis Symbols and Formation of Covalent Molecules
Lewis Structures of CH4, NH3 and H2O
Lewis Structures of CO2, HCN, and C2H2
Resonance Lewis Dot Structures for CO32-
Exception to Octet Rule
1. If central atom is from group 2A or 3A, octet state
is not acquired - the central atom has incomplete
octet.
2. Central atoms from periods 3, 4, 5, …may have
more than 8 valence electrons (expanded octet)
3. Molecules with odd number of electrons will
contain unpaired electrons.
Covalent Molecules with Central Atoms
have 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 =
(Atom’s group #) – (# 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
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If two or more Lewis dot structures that satisfy the
octet rule can be drawn, the most stable one will be
the structure in which:
1. The formal charges are as small as possible.
2. Any negative charges are located on the more
electronegative atoms.
Which Lewis structures of CO2 & N2O are correct?
The Shape of Water Molecules
Molecular Shapes of BeI2, HCl, IF2-, ClF3, and NO3-
Lewis Structures, Molecular Shapes & Polarity
The Shapes of Methane and Ammonia Molecules
The VSEPR Shapes
Linear and Trigonal Planar Electron-Pair Geometry
The Tetrahedral Electron-Pair Geometry
Trigonal Bipyramidal Electron-Pair Geometry
The Octahedral Electron-Pairs Geometry
Lewis Structures of HF, H2O, NH3, & CH4
Lewis Symbols for O, F, and Na
Lewis Model for the Formation of Covalent
Bonds and Covalent Molecules
Covalent Bonds and Lewis Structures Some Molecules
Resonance Lewis Structures of PO43-
Assigning Appropriate Formal Charges
Structures and Shapes of Formaldehyde and Ethylene