Bonding Electron Pairs

Chapter 1 – 3

Important people:
Rutherford-discovered the nucleus
Bohr- created the Bohr model and discovered how electron orbit around the nucleus.
Heisenburg- asserted the uncertainty of principle of quantum mechanics.
Pauli- created the Pauli Exclusion Priniciple (no two electrons in an atom can have the same quantum number).
Levosier- discovered the role of oxygen in combustion, law of conservation of mass and creation of the first textbook.
Stoichiometry:
Used to determine the moles of a molecule in a reaction when given the moles of a different molecule.
Write a balanced equation.
Set up a BCA table.
Convert to moles.
Fill in BCA table.
Percent Composition of Compounds:
The part, divided by the whole, multiplied by 100
Determining the empirical formula:
Determine the percentage of each element in your compound
Treat % as grams, and convert grams of each element to moles ofeach element
Find the smallest whole number ratio of atoms
If the ratio is not all whole number, multiply each by an integer so that all elements are in whole number ratio
Determining the molecular formula:
Find the empirical formula mass
Divide the known molecular mass by the empirical formula mass,deriving a whole number, n
Multiply the empirical formula by n to derive the molecular formula
Calculating Percent Yield:
Actual yield - what you got by actually performing the reaction
Theoretical yield - what stoichiometric calculation says the reaction SHOULD have produced
(Actual Yield/Theoretical Yield) x 100% = percent yield
Limiting Reactants:
Convert moles like normal.
Assume one reactant is the limiting reactant.
If there isn’t enough of the of the other reactant, then it isn’t the limiting reactant.

Chapter 5

Boyle’s Law: PV=k

The product of pressure times volume is a constant— P is inversely related to k

Used to find new volume or pressure at a constant temp

P1V1=P2V2

Charles Law: V=bT

Temperature must be measured in Kelvin.

The volume of each gas is directly proportional to temperature.

V1/T1=V2/T2

Avogadro’s Law: V=an

Gas has to be at constant temperature and pressure. Volume is directly proportional to number of moles.

V1n1/V2n2

Ideal Gas Law: PV=nRT

Constants from previous 3 laws (k,b,a) are combined to make a universal constant R.

R=.0821 (L*atm)/(mol*k)

It can be used to solve for pressure, number of moles, volume, or temberature when all other variables are held constant.

At STP (0C and 1atm), the molar volume of an ideal gas is 22.42 L. (Constant)

Dalton’s Law of Partial Pressures: Ptotal=ntotal(RT/V)

Used to find pressure of a gas that is in a mixture of gases in a container. Only total number of moles matters.

Mole Fraction: X1=P1/Ptotal (Pressure is directly related to moles)

Kinetic Molecular Theory:
The particles are so small compared with the distances between them

that the volume of the individual particles can be assumed to be zero

The particles are in constant motion. Collisions of the particles with the

walls of the container cause pressure

Assume that the particles exert no forces on each other.
The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

Relationship between Temperature and Average Kinetic Energy: (KE)avg=(3/2)RT

Root Mean Square Velocity: urms=(3RT/M)

M=mass of one mole of gas particles in kg.

R=8.345 j/k*mol

Equation gives you the average velocity of a gas particle.

Chapter 7 – 9 Part I

Relationships: As energy increases, frequency also increases, but wavelength decreases.
Periodic Trends
-Ionization Energy: energy required to remove an electron from an atom. The further the electron is from the nucleus, the easier it is to remove it.
*Increases across a period
*Decreases with increasing atomic # within a group
-First ionization energy:energy it takes to remove the first electron from an atom
*increases across the period
*decreases down a group
*there are some exceptions-p has a higher 1st IE then S, and copper is the other exception
-Electron Affinity: energy associates with the addition of an electron
*Increases across a period
*Decreases down a group
-Atomic Radius: half of the distance between radius in a covalently bonded diatomic molecule
*Decreases across a period
*Increases down a group
-Alkali Metals
Easily lose valence electrons
*react with halogens to form salts
*react violently with water
Large hydration energy
*positive ionic charge makes ions attractive to polar water molecules
WHY?:Because as you go across the period, one more proton is added which creates a stronger attraction between the nucleus and the electrons. As you go down a period, more electron orbitals are added, which weakens the attraction between the electrons and the nucleus.

Ionic Bonds

An ion has a different size than its parent atom
Anion is larger than its parent ion
A cation is smaller than its parent atom
Lattice energy: the change in energy when ions are packed together to form an ionic solid
Lattice energy=k(Q1 Q2/r)
K= constant
Q1, Q2 = charges on the ions
r= distance between ion center

Covalent Bonds
A covalent bond is a form of chemical bonding that is characterized by the sharing of pairs of electrons betweenatoms.
Electrons are shared equally
Atoms end up with fractional charges (positive side and negative side)
Can model covalent bonds using lewis dot structures.
Ex)
H, Li, Be, B forms stable molecules when they share two electrons
Elements Carbon and beyond stable when they are surrounded by eight molecules.
Second row elements (C, N, O, and F) should always obey the octet rule
B and Be often have fewer than 8 electrons around them - highly reactive molecules
Second row elements NEVER exceed octet rule, third row often satisfies or exceeds octet rule

Multiple Bonds: single bond, double bond, triple bond (based on number of shared electrons)
Bond energy: ΔH = bonds broken - bonds formed
Chapter 7-9 Part II

Shapes and Angles of molecules

Bonding electron pairs / Lone pairs / Electron domains (Steric #) / Shape / Ideal bond angle (example's bond angle) / Example / Image
2 / 0 / 2 / linear / 180° / CO2 /
3 / 0 / 3 / trigonal planar / 120° / BF3 /
2 / 1 / 3 / bent / 120° (119°) / SO2 /
4 / 0 / 4 / tetrahedral / 109.5° / CH4 /
3 / 1 / 4 / trigonal pyramidal / 107.5° / NH3 /
2 / 2 / 4 / angular / 104.5° / H2O /
5 / 0 / 5 / trigonal bipyramidal / 90°, 120°, 180° / PCl5 /
4 / 1 / 5 / seesaw / 180°, 120°, 90° (173.1°, 101.6°) / SF4 /
3 / 2 / 5 / T-shaped / 90°, 180° (87.5°, < 180°) / ClF3 /
2 / 3 / 5 / linear / 180° / XeF2 /
6 / 0 / 6 / octahedral / 90°, 180° / SF6 /
5 / 1 / 6 / square pyramidal / 90° (84.8°), 180° / BrF5 /
4 / 2 / 6 / square planar / 90° 180° / XeF4 /
7 / 0 / 7 / pentagonal bipyramidal / 90°, 72°, 180° / IF7 /

Hybridization

Steric
No. / Basic geometry
0 lone pair / 1 lone pair / 2 lone pairs / 3 lone pairs / Hybridization
2 /
Linear / SP
3 /
Trigonal planar /
Bent / SP2
4 /
Tetrahedral /
Trigonal pyramid /
Bent / SP3
5 /
Trigonal bipyramid /
Seesaw /
T-shaped /
Linear / SP3d
6 /
Octahedral /
Square pyramid /
Square planar / SP3d2

Sigma bonds (bond)

a. Bond in which the electron pair is shared in an area centered on a line running between the atoms

b. Lobes of bonding orbital point toward each other

c. All bonds in methane are sigma bonds

Pi bonds (bonds)

a. Electron pair above and below the bond

b. Created by overlapping of nonhybridized 2p orbitals on each carbon

Double bonds

a. Double bonds always consist of one 

bond and one bond

Chapter 10 Part 1

1.Inter-molecular forces

a.Dipole - Dipole: (+) …... (-) interactions

b.Hydrogen Bonding: Hydrogen covalently bonded to N, O, F

c.London Dispersion Forces: random movement of electrons, exist between all molecules

d.Polarizability increases with the number of electrons in a molecule Ex: CCl4 experiences greater LDF than CH4 Higher LDF = larger size * Higher Dipole - Dipole = smaller sizeGreater IMFs: Higher BP , Higher FP, Lower VPWeaker IMFs: Lower BP, Lower FP, Lower VP

Chapter 10 Part II

1.Metallic Bonding :

a.The chemical bond characteristic of metals, in which mobile valence electrons are shared among atoms in a usually stable crystalline structure

2.Covalent network:

a.Solid is a chemical compound in which the atoms are bonded by covalent bonds in a continuous network. For example, diamond, graphite, silica, ceramics and semiconductors.

3.Triple Point Graph

a.Triple Point

i.Solid and liquid have identical vapor pressure

ii.All three phases exist together in equilibrium

b.Critical temperature:

i.The temperature above which the substance cannot exist as a liquid, regardless of how great the pressure

c.Critical pressure:

i.The pressure required to produce liquefaction at the critical temperature

d.Critical point

i.Point defined by the critical temperature and critical pressure

Chapter 11 Part 1Molality: Moles of solute Kilograms of SolventMolarity: Moles of soluteLiters of SolutionMole Fraction: Moles of Part AMoles of EVERYTHINGMass Percent: Mass of SoluteMass of Solution X 100!!!Van’t Hoff Factor (i) : Moles of Particles in SolutionMoles of Solute DissolvedFor ionic compounds, the expected value of i is an integer greater than 1a. NaCl, i = 2b. BaCl2, i = 3c. Al2(SO4)3, i = 5Solution ConductivityIf the the solute is a molecular solid, there will be no metals so it will not conduct.Covalently bonded compounds are weak or nonelectrolytes which resist conduction.Example-(C6H12O6 + H2O...Nonconductive)NO METAL, NO CONDUCTIf the Solute is a Ionic Compound, the lattice will break apart and individual ions will retain their own properties and allow the solution to conduct.Ionic Compounds are strong electrolytes which allow compounds to conduct.Example-(CaCO3 . + H2O.....conductive)METAL! Disassociate Conductive Properties! CONDUCT!! Solute: is dissolved by a solvent (kool-aid powder)Solvent: dissolves the solute (water)Solution: solute + solvent (kool-aid)

Chapter 11 Part II

Vapor Pressure (
Pressure of a gas of a substance that is in equilibrium
Nonvolatile solutes lower vapor pressure
Since they won’t evaporate, they block the surface and prevent molecules of the solvent from escaping
Raoult’s Law
Psol = Xsolvent * Psolvent
Just Like PVNRT
Ionic Solid dissociate
Creates more particles that clog up surface.

Boiling Point (
Nonvolatile “get in the way” of things that are trying to evaporate.
Need more energy to get them to evaporate.
Therefore boiling point rises.
Change in Temp.=Kb*msolute
Delta T is the boiling point elevation
Kb is the molal boiling point elevation constant of the solvent (given)
msolute is the molality of the solute in the solution

Freezing Point (
Solutes get in the way of the freezing lattice
More energy must be taken out to “freeze” around the solutes
Freezing Point Decreases
Change in Temp.=Kf*msolute
Delta T is the boiling point elevation
Kf is the molal freezing point depression constant of the solvent (given)
msolute is the molality of the solute in the solution

Chapter 12
Differential Rate Law

Represents how rate depends on concentration
Delta [A] / Delta t
Where [A] is concentration of reactant or product in a reaction

Reaction Mechanisms

Most chemical reactions occur by a series of steps, not one reaction
This series of steps is called the reaction mechanism
Two ways to determine reaction mechanism fits a reaction.
Use rate determining step to see if the reaction occurs at the same rate as the proposed mechanism
Cancel intermediaries to see if reaction mechanism matches reaction
Example: 2NO2 + F2 →2NO2F
Rate law = k[NO2]2[F2]
A suggested mechanism for this reaction is:
NO2+F2 →NO2F+F SLOW
F + NO2→NO2F FAST
Is this reaction mechanism acceptable?
To determine, first cancel intermediaries
The equation ends up looking like
2NO2 + F2 →2NO2F
Then, find if the slow step is equal to the rate law of the reaction
The rate for the slow step is
k[NO2][F2]
Since the rate law for the slow step doesn’t match the rate of the actual reaction, the reaction mechanism that was proposed is not an acceptable mechanism.

Collision Theory

Molecules must collide to react
Collide enough to lead to activation energy to induce the reaction

Reaction profile

Exothermic reaction
Products have less energy than reactants
Endothermic reaction
Reactants have less energy than products
Change in energy from the initial reactants to the transition state is called the activation energy

Chapter 13

When the rate of the forward reaction is equivalent to the rate of the reverse reaction the system is at equilibrium.
jA+kB<-->lC+mD
j, j, l, and m are coefficient.
A and B are the concentrations of the reactants
C and D are the concentrations of the products

k= [C]^l x [D]^m
[A]^j x [B]^k

K is the equilibrium constant and varies depending on temperature and the coefficients of the balanced equation. Concentration does NOT effect k.
kp=kc(RT)^change in n
kc is the constant for concentration
kp is the constant for pressure
R=0.0821
T=temperature in Kelvin
change in n= difference in sums of coefficients for the products and the reactants
The effect of change in pressure
When gaseous products or reactants are added pressure shifts away from the added gas.
When gaseous products or reactants are removed pressure shifts towards the removed gas.
Solids, liquids, and inert gases are not included in the equilibrium equation.
When the volume of the container holding a gaseous system is reduced the system responds by reducing its own volume.
When the container volume is increased the system will shift as to increase its own volume.
Change in Temperature
When temperature increases the system will shift in the direction that consumes the energy.
For an exothermic reaction energy is a product. The reaction will shift to the left to use up the excess energy.
For an endothermic reaction energy is a reactant. The reaction will shift to the right to use up the energy.
A decrease in temperature causes a system to shift in the direction that replaces the lost energy.
Examples:
At a certain temperature, 4.0 mol NH3 is introduced onto a 2.0-L container, and the NH3 partially dissociates by the reaction
2NH3(g)<----->N2(g) + 3H2(g)
At equilibrium, 2.0 mol NH3 remains. What is the value of K for the reaction?
(Answer: K=1.7)
At 25 C, K=0.090 for the reaction
H2O(g) + Cl2O(g)<------>2HOCl(g)
Calculate the concentrations of all species at equilibrium for each of the following cases.
1.0 mol pure HOCl is placed in a 1.0-L flask.
(Answer: 122 H2O+Cl2O; .06 HOCl)