AP Chemistry
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Ch.8 Outline: Bonding : General Concepts
A. Bonding : General Concepts
1. Chemical bond -a force that holds atoms together to make them function as a unit.
a. bond energy - the energy required to break the bond
b. Why do atoms bond?- to achieve the lowest energy state possible
c. ionic bonding - involves transfer between a metal and a nonmetal - results in an ionic compound
- energy of interaction between ions can be calculated from Coulomb's law :
E = -2.31 x 10-19J x nm(Q1Q2/r)
- where : Q1 and Q2 are the numerical ion charges (note - E value increases with greater charge
difference - generally most important factor)
- r is the distance between ion centers in nm (note smaller r value gives greater E value)
- negative sign indicates an attractive force (lower energy state- more stable)
- a positive charge can be calculated by the repulsive forces of like charged ions
d. covalent bonding - involves sharing of electrons
- ex. H2-forces : - attractive forces between nuclei and electrons
- repulsive forces of nuclei
- repulsive forces of electrons
-atoms will position themselves where the total energy of all three forces is the minimum
(most stable) value which is the bond length (know and understand Fig. 8.1 p. 346)
δ+ δ- polar covalent bond - between atoms of unequal attraction for electrons e.g. H-F
2. Electronegativity- the ability of an atom in a molecule to attract shared electrons to itself
a. calculation of electronegativity difference : ∆= (H-X)act - (H-X)exp
- where : ∆ is the electronegativity difference
(H-X)act is the actual measured value
(H-X)exp is the expected value (= (H-H bond energy + X-X bond energy)/2)
- the greater the electronegativity differences, the greater the ∆ value and the greater the ionic
character of the bond and vice-versa (0 = covalent bond)
- periodic trend in electronegativity - increases across a period (F is highest)
- Group trend -decreases down a group-Fr, Cs are lowest
H-F
Sample problem : Arrange the following bonds in order of increasing polarity : C-F, N-F, O-F, F-F
Answer since the electronegativity order is C > N > O > F, the order is F-F, O-F, N-F, C-F
3. Dipoles - dipolar (have a dipole moment) molecules - have a positive center and a negative center
- molecular shape is an important determiner of polarity -linear, planar and tetrahedral molecules
have shapes that cancel out bond polarities
4. Ions : Electron configuration and sizes
a. nonmetallic atoms and representative group metals bond to get noble gas electron configurations
Sample problems :
1. Write the electron configuration of the following ions : Ba2+, Te2-, Al3+, N3Ba2+ : [Kr] 5s24d105p6
Te2- : [Kr] 5s24d105p6
Al3+ : 1s22s22p6
N3- : 1s22s22p6
2. Which of the following ions in each set have noble gas electron configurations?
a. Cr3+, Cr6+
b. Mn3+, Mn4+ , Sn4+, Sn2+, Y3+
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b. ionic compounds are considered to be solids (dissociate in aqueous solution)
c. Ion size plays an important role in determining the structure and stability of ionic solids, properties
of ions in aqueous solution and the biologic effects of ions
- positive ions are smaller than the parent atom
- negative ions are larger than the parent atom
- ion size increases down a Group
- compare O2- , F-, Na+, Mg2+ and Al3+ - all have [Ne] configuration (10e-, isoelectronic), but have
an increasing number of protons and therefore decrease in size
Sample problem : Give three ions that are isoelectronic with Kr and arrange them in order of
increasing size.
Answer : Three from : Y3+, Sr2+, Rb+, Br , Se2-, As3Order is because of the Z (nuclear charge) value, Size depends on two factors : Number of
electrons and nuclear charge. Since these species are all isoelectronic nuclear charge is the
determining factor and so they are arranged in order of decreasing nuclear charge.
5.Formation of binary Ionic compounds
a. lattice energy- the change in energy that takes place when separated gaseous ions are packed
together to form an ionic solid : M+(g) + X (g) --> MX(s)
- energy change is negative if exothermic, positive if endothermic
Sample problems :
1. Predict which of the following pairs of ionic substances will have the most exothermic (negative)
lattice energy. (See 1c. above)
a. LiCl, NaCl
b. MgCl2, MgO
c. KF, CsCl
Answers : a. LiCl - Li+ is smaller than Na+
b. MgO - O2- has a greater charge than Clc. KF - ions are smaller
2. Calculate the ∆Hºf for lithium fluoride :
- example : formation of one mole of LiF
Step 1: sublimation of Li
Li(s) --> Li(g)
Step 2: Ionization of Li atoms Li(g) --> Li+(g) + eStep 3: Dissociation of F2
½F2(g) --> F(g)
Step 4: Ionization of F atom
F(g) + e- --> F-(g)
Step 5: Formation of LiF
Li+(g) + F-(g) --> LiF(s)
Overall
Li(s) + ½F2(g)---> LiF(s)
∆H = 161 kJ/mol
∆H = 520 kJ/mol
∆H = 154 kJ/mol
∆H = -328 kJ/mol
∆H = -1047 kJ/mol (lattice en.)
∆Hºf = -617 kJ/mol
The above is a favorable change with the large, negative lattice energy as the dominate force.
3. Given that O(g) + e- Æ O-(g) has a ∆H value of -141 kJ/mol and O-(g) + e- Æ O2-(g) has a ∆H of
+878 kJ/mol, suggest a reason why sodium oxide is Na+2O2- rather than Na+O- (the ionization
energy of one mole of sodium atoms = 495 kJ/mol ).
Answer : Na2O must have a higher more favorable lattice energy than NaO.
4. Give a reason for the following lattice energies : LiF = 1030 kJ/mol, LiCl = 834 kJ/mol.
AP Chemistry
Answer F- is smaller than Cl-
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5. Arrange the following compounds in order of increasing lattice energy based on size of ions and
charge differences : CuCl, CuCl2, CuO, Cu2O
Answer : CuCl - smallest charge difference(Q1 = +1, Q2 = -1), largest ions
CuCl2 - medium charge difference, Cl- is larger than O2Cu2O - medium charge difference, O2- smaller than ClCuO - largest charge difference(Q1 = +2, Q2 = -2), smallest ions
b. Calculation of lattice energies : Lattice energy = k(Q1Q2/r)
- where Q1 and Q2 are the charges on the ions, k is a proportionality constant and r is the shortest
distance between the anion and cation
6. Partial ionic character of ionic bonds
a. compounds do not have either covalent or ionic bonds - it is a continuum between the two
- no bonds are completely ionic
- bonds with more than 50 % ionic character are said to be ionic
- % ionic character = (measured dipole moment (X-Y)/calculated dipole moment (X+Y-)) x 100%
Sample problem : Arrange the following compounds in order of increasing ionic character. HCl, NaCl,
BCl3, H2
Answer : H2, BCl3, HCl, NaCl
b. definition of ionic compound : any substance which conducts electricity when dissolved or melted
7. The covalent chemical bond : a model
a. Bonds indicate molecular stability - form in order for system to obtain lowest energy state
b. Models - theoretical explanations of observed phenomena
bonds - some data support notion of discrete bonds - shared pairs of e- held in place
- some data support delocalized electrons - free to move about molecule
8. Covalent Bond Energies and Chemical Reactions
a. The strength and type of bond depends on the bond environment (e.g. a C-H bond in CH4 is
different than a C-H bond in CHCl3)
b. Bonds can be single (one shared e- pair), double (two shared e- pairs), triple (3 shared e- pairs)
- as number of shared pairs increase, bond length decreases
c. Breaking bonds is an endothermic process (∞H is positive), forming bonds is exothermic (∞H is
negative - more stable product)
- ∞H can be defined as the sum of the energy needed to break old bonds (+ value) plus the sum of
the energies needed to form the new bonds (neg. value).
- ∞H = ΣD(bonds broken) - ΣD(bonds formed)
- where D = bond energy per mole of bonds
Sample problem : Use bond energy values to estimate the ∆H value of the following reaction. (N≡N =
941 kJ/mol, F-F = 154 kJ/mol, N-F = 272 kJ/mol
N≡N + 3F2 Æ 2 NF3
∆H = ΣD(bonds broken) - ΣD(bonds formed)
= (941 kJ + 3 x 154 kJ) - ( 6 x 272 kJ)
= -229 kJ/mol
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Sample problem : Use the standard enthalpies of formation for C(g) (=716.7 kJ/mol))and H(g) (= 218.0
kJ/mol) and CH4(g) (= -74.6) to estimate the strength of the C-H bond. Compare this to the value given
in Table 8.4 (C-H = 413 kJ/mol).
Answer : CH4(g) Æ C(g) + 4 H(g)
Sample problem (cont'd)
∆Hº = 4 DCH = Σ Hproducts - Σ Hreactants = (716.7 + (4 x 218.0)) - (-74.6) = 1663.3 kJ
DCH = 1663.3 kJ / 4 = 415.83 kJ
This answer agrees very well with the value of 413 kJ/mol given in Table 8.4
9. The Localized Electron (LE) Bonding Model - assumes molecules are composed of atoms that are
bound together by shared pairs of electrons in orbitals of the bound atoms
a. Three parts to localized electron (LE) model :
- Use of Lewis structures to show valence electron arrangement in the molecule
- Prediction of geometry using VSEPR (Valence Shell Electron Pair Repulsion) theory
- Description of atomic orbitals used by atoms to hold shared pairs or lone pairs
10. Lewis Structures - most important factor is to obtain a noble gas electron configuration
a. duet rule - 2 e- [He] - elements - H, Li, Be, B
b. octet rule - 8 e- remaining representative elements
- bonding pairs shown as lines
11. Exceptions to the octet rule
- BF3 - boron has only 6 e- : boron trifluoride reacts with electron rich
substances (substances with unshared of e- pairs )
Be can also have fewer than 8 e- SF6 - sulfur has 12 e- : uses 3 p and 3d orbitals
- third row and heavier elements can exceed the octet rule
- if octet rule must be exceeded, the extra electrons need to be added
to the central atom
Sample problem : Draw the Lewis structure for PCl3
Step 1 : Count number of valence electrons (P = 5, 3Cl = 7 x 3 ; total = 26)
Step 2 : Attach all atoms to the central atom with a single bond (Usually first atom in molecule is the
central atom. H is a common exception to this rule)
2.
3.
Step 3 : Complete octet rule (or duet rule depending on species) Extra electrons will go on the central
atom.
12. Resonance -occurs when more than one valid Lewis structure can be drawn for a particular molecule
a. Nitrate ion :
Lewis structure indicates three possible structures each with two types of bonds : single (longer and
double (shorter). Experimental evidence indicates only one bond length and strength - between a
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single and a double bond. The nitrate ion then is an average of all three structures -indicates
delocalization of electrons.
b. Formal charge - the difference between the number of valence electrons on free atom and number
of valence electrons assigned to the atom in a molecule (valence e- shared - valence assigned)
- lone pairs belong entirely to the atom in question
- shared pairs are divided equally
- atoms try to achieve formal charges as close to zero as possible
- negative formal charges go to the most electronegative element involved in the bond
- the sum of all the formal charges on an ion or molecule equal the overall charge on the species
- formal charges are only estimates - not actual charges - actual charges are determined
experimentally
Sample problems :
1. Calculate the formal charges on the P atom and the chlorine atom in the PCl3 as illustrated above :
P : valence electrons in free atom = 5
valence electrons in molecule = 2 free + ½(6 shared) = 5
Formal charge = 5-5 = 0
Cl : valence electrons in free atom = 7
valence electrons in atom in molecule = 6 + ½(2) = 7
Formal charge on 3 oxygen atoms = 7-7 = 0
2. Calculate the formal charge on the sulfur and oxygen atoms in the sulfate ion.
The sulfate ion has two possible Lewis structures (S can exceed octet rule) :
In a. the formal charge of S = 6 (in free atom) - 4 ( shared in ion) = +2
The formal charge of O = 6 (in free atom) - 7 (in ion) = -1
In b. the formal charge = 6 (in free atom) - 6 (shared in ion) = 0
The formal charge of O = 6 (in free atom) - 7 (in ion) = -1
The Lewis structure in b. is the preferable form due to the lower formal charges
13. VSEPR Model - Valence Shell Electron Pair Repulsion model- molecules arrange themselves to
achieve the minimum electron repulsion possible
- unshared pairs need more space than shared pairs - more repulsion in unshared pairs
- multiple bonds (double or triple) count as a single bond (one shared pair)
- unshared pairs with angles less than 120 degrees will distort other bonds
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Total
Electron
Bonding
Nonbonding
Domains
Electron- Domains
Domains (
(shared
Domains (shared
unshared
or
Geometry pairs of
pairs of e-)
unshared
e-)
pairs of
e-)
2
Linear
2
Molecular
Geometry
0
Example
CO2
Linear
3
3
0
Trigonal
Planar
BF3
Trigonal Planar
2
1
SO2
Bent
4
0
CH4
Tetrahedral
4
Tetrahedral
3
1
NH3
Trigonal Pyramidal
2
2
H2O
Bent
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5
0
PCl5
Trigonal Bipyramidal
4
5
1
SF4
Seesaw
Trigonal
Bipyramidal
3
2
ClF3
T-shaped
2
3
XeF2
Linear
6
0
SF6
Octahedral
6
Octahedral
5
1
BrF5
Square Pyramidal
4
2
XeF4
Square Planar
AP Chemistry
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Chapter 9 Outline-Sections 9.1, 9.2
1. Hybridization and the Localized Electron Model
a. hybridization - mixing of native ("normal") orbitals to form specialized orbitals for bonding
- sp3 hybridization - four effective bonding pairs around the central atom-tetrahedral geometry
- example is methane
- one s and three p orbitals combine to make four equal sp3 orbitals
- needed whenever a set of four equivalent tetrahedral atomic orbitals are required
- like all hybridization the molecule arranges itself to
achieve lowest energy state
- sp2 hybridization - for three effective bonding pairs around
the central atom (120 degree angle)-trigonal planar
geometry
- example : ethylene (C2H4)
- two p (px and py) orbitals hybridize with s orbital, pz
orbital is not hybridized
- involves a sigma (δ) bond - in a straight line between
From Zumdahl's Chemistry,
nuclei of bonding atoms and a pi (π) bond - above and
4th ed. p. 412
below the line joining the atoms - electrons are from parallel p orbitals that are
perpendicular to the sigma bonds
- double bonds always involve one sigma and one pi bond
- used whenever three effective pairs (double bond = one effective pair) are needed around an
atom
- sp hybridization - two effective pairs around the central atom- 180 degree angle - linear geometry
- examples include carbon dioxide and nitrogen (N2)
- one s orbital and one p orbital hybridize leaving two unhybridized p orbitals
- involves one sigma bond and two pi bonds
- dsp3 hybridization- five effective pairs around the central atom resulting in a trigonal bipyramidal
arrangement
- examples include PCl5 and I3- involves one s orbital, three p orbitals and one d orbital
- d2sp3 hybridization - six effective pairs surround central atom resulting in an octahedral geometry
- example : SF6 and XeF4
- involves one s orbital, three p orbitals and 2 d orbitals
b. To describe a molecule follow these steps :
- draw the Lewis structure
- use the VSEPR theory to arrange electron pairs
- determine the hybrid orbitals needed to accommodate the number of electron pairs
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From Zumdahl's
Chemistry, 4th ed.
p. 418
c. Magnetism
- in the presence of a magnetic field, two types of magnetism can be induced - paramagnetism and
diamagnetism
- paramagnetism is associated with unpaired electrons and shows an increase in attraction to a
magnetic field
-diamagnetism is associated with paired electrons and shows a repulsion to a magnetic field
- paramagnetism and diamagnetism can be experimentally shown by determining the mass of a
substance in and out of a magnetic field