AP* Chemistry Chemical Foundations · 2019-07-09 · AP* Chemistry Chemical Foundations Chemistry:...

AP* Chemistry Chemical Foundations Chemistry: An Overview

• Matter – takes up space, has mass, exhibits inertia - composed of atoms only 100 or so different types - Water made up of one oxygen and two hydrogen atoms - Pass an electric current through it to separate the two types of atoms and they rearrange to become two different types of molecules

- reactions are reversible

Chemistry – is defined as the study of matter and energy and more importantly, the changes between them • Why study chemistry?

- become a better problem solver in all areas of your life - safety – had the Roman’s understood lead poisoning, their civilization would not have fallen - to better understand all areas of science

The Scientific Method

• A plan of attack!

The fundamental steps of the scientific method

*AP is a registered trademark of the College Board, which was not involved in the production of this product. © 2013 by René McCormick. All rights reserved.

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To be completed by the STUDENT

• Good experimental design coupled with repetition is key! Theory – hypotheses are assembled in an attempt at explaining “why” the “what” happened.

• Model – we use many models to explain natural phenomenon – when new evidence is found, the model changes!

• Robert Boyle

o loved to experiment with air o created the first vacuum pump o coin and feather fell at the same rate due to gravity o in a vacuum there is no air resistance to impede the fall of either

object! o Boyle defined elements as anything that cannot be broken into

simpler substances. Boyle’s Gas Law: P1V1 = P2V2

• Scientific Laws – a summary of observed (measurable) behavior [a theory is an explanation of behavior] A law summarizes what happens; a theory (model) is an attempt to explain WHY it happens. - Law of Conservation of Mass – mass reactants = mass products - Law of Conservation of Energy – (a.k.a. first law of thermodynamics) Energy CANNOT be created NOR destroyed; can only change forms.

- Scientists are human and subjected to • Data misinterpretations • Emotional attachments to theories • Loss of objectivity • Politics • Ego • Profit motives • Fads • Wars • Religious beliefs

• Galileo – forced to recant his astronomical observations in the face of strong religious resistance

• Lavoisier – “father of modern chemistry”; beheaded due to political affiliations.

• The need for better explosives; (rapid change of solid or liquid to gas where molecules become ≈2,000 diameters farther apart and exert massive forces as a result) for wars have led to

-fertilizers that utilizes nitrogen - Nuclear devices

Chemical Foundations 2

Units of Measure A quantitative observation, or measurement, ALWAYS consists of two parts: a number and a unit. Two major measurements systems exist: English (US and some of Africa) and Metric (the rest of the globe!) • SI system – 1960 an international agreement was reached to set up a system of units so scientists everywhere

could better communicate measurements. Le Système International in French; all based upon or derived from the metric system

KNOW THE UNITS AND PREFIXES shown in BLUE!!! • Volume – derived from length; consider a cube 1m on each edge ∴1.0 m3 - A decimeter is 1/10 of a meter so (1m)3 = (10 dm)3 = 103 dm3 = 1,000 dm3 1dm3 = 1 liter (L) and is slightly larger than a quart also 1dm3 = 1 L = (10 cm)3 = 103cm3 = 1,000 cm3 = 1,000 mL AND 1 cm3 = 1 mL = 1 gram of H2O (at 4ºC if you want to be picky) Mass vs. Weight – chemists are quite guilty of using these terms interchangeably.

o mass (g or kg) – a measure of the resistance of an object to a change in its state of motion (i.e. exhibits inertia); the quantity of matter present

o weight (a force∴ has units of Newtons) – the response of mass to gravity; since all of our measurements will be made here on Earth, we consider the acceleration due to gravity a constant so we’ll use the terms interchangeably as well although it is technically incorrect! We “weigh” chemical quantities on a balance NOT a scale!!


Physics connection: Fw = ma Fw = mg

29.8 m

sw mF

=

∴its units are

( ) 2mkgs

N

=

Exercise 1 Precision and Accuracy To check the accuracy of a graduated cylinder, a student filled the cylinder to the 25-mL mark using water delivered from a buret and then read the volume delivered. Following are the results of five trials: Trial Volume Shown by Volume Shown Graduated Cylinder by the Buret 1 25 mL 26.54 mL 2 25 mL 26.51 mL 3 25 mL 26.60 mL 4 25 mL 26.49 mL 5 25 mL 26.57 mL Average 25 mL 26.54 mL Is the graduated cylinder accurate? Note that the average value measured using the buret is significantly different from 25 mL. Thus, this graduated cylinder is not very accurate. It produces a systematic error (in this case, the indicated result is low for each measurement).

Gravity – varies with altitude here on planet Earth • The closer you are to the center of the Earth, the stronger the gravitational field

SINCE it originates from the center of the Earth. • Every object has a gravitational field – as long as you’re on Earth, they are masked

since the Earth’s field is so HUGE compared to the object’s. • The strength of the gravitational field ∝ mass • Ever seen astronauts in space that are “weightless” since they are very far removed

from the center of Earth? Notice how they are constantly “drawn” to the sides of the ship and must push away?

• The ships’ mass is greater than the astronaut’s mass ∴ “g” is greater for the ship and the astronaut is attracted to the ship just as you are attracted to Earth! The moon has

61 the mass of the Earth ∴ you would experience 6

1 the gravitational field you experience on Earth and ∴ you’d WEIGH 6

1 of what you weigh on Earth.

Precision and Accuracy - Accuracy – correctness; agreement of a measurement with the true value - Precision – reproducibility; degree of agreement among several measurements. - Random or indeterminate error – equal probability of a measurement being high or low - Systematic or determinate error – occurs in the same direction each time

The results of several dart throws show the difference between precise and accurate. (a) Neither nor precise (large random errors). (b) Precise but not accurate (small random errors, large systematic error). (c) Bull’s-eye! Both precise and accurate (small random errors, no systematic error).


Significant Figures and Calculations

Determining the Number of Significant Figures (or Digits) in a Measurement • Nonzero digits are significant. (Easy enough to identify!) • A zero is significant IF and ONLY IF it meets one of the conditions below:

- The zero in question is “terminating AND right” of the decimal [must be both] - The zero in question is “sandwiched” between two significant figures

• Exact or counting numbers have an ∞ amount of significant figures as do fundamental constants (never to be confused with derived constants)

Reporting the Result of a Calculation to the Proper Number of Significant Figures • When × and ÷ , the term with the least number of significant figures (∴least accurate measurement)

determines the number of maximum number of significant figures in the answer. (It’s helpful to underline the digits in the least significant number as a reminder.)

4.56 × 1.4 = 6.38 →corrected 6.4

• When + and (−), the term with the least number of decimal places (∴least accurate measurement) determines the number of significant figures in the final answer.

12.11 18.0 ← limiting term (only 1 decimal place) 1.013 31.123 corrected→ 31.1 (limits the overall answer to only one decimal place)

• pH – the number of significant figures in least accurate measurement determines number decimal places on the reported pH (usually explained in the appendix of your text)

Rounding Guidelines for the AP Exam and This Course:

• Round ONLY at the end of all calculations (keep the numbers in your calculator) • Examine the significant figure one place beyond your desired number of significant figures.

IF > 5 round up; < 5 drop the remaining digits. • Don’t “double round”! Example: The number 7.348 rounded to 2 SF is reported as 7.3

In other words, DO NOT look beyond the 4 after the decimal and think that the 8 rounds the 4 up to a five which in turn makes the final answer 7.4. [Even though you may have conned a teacher into rounding your final average this way before!]

Exercise 2 Significant Figures (SF) Give the number of significant figures for each of the following experimental results. a. A student’s extraction procedure on a sample of tea yields 0.0105 g of caffeine. b. A chemist records a mass of 0.050080 g in an analysis. c. In an experiment, a span of time is determined to be 8.050 × 10‒3 s .

a. three; b. five; c. four


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Dimensional Analysis Example: Consider a straight pin measuring 2.85 cm in length. Calculate its length in inches. Start with a conversion factor such as 2.54 cm = 1 inch ∴ you can write TWO

Conversion factors: 1 in2.54 cm

or 2.54 cm1 in

. Why is this legal? Both quantities

represent the exact same “thing” so the conversion factor is actually equal to “1”. To convert the length of the pin from cm to inches, simply multiply your given quantity by a conversion factor you engineer so that it “cancels” the undesirable unit and places the desired unit where you want it. For our example, we want inches in the numerator so our numerical answer is not reported in reciprocal inches! Thus,

2.85 cm 1 in2.54 cm

× = 1.12 in

Let’s practice!

Exercise 3 A pencil is 7.00 in. long. Calculate the length in centimeters?

17.8 cm

Exercise 4 You want to order a bicycle with a 25.5-in. frame, but the sizes in the catalog are given only in centimeters. What size should you order? 64.8 in

Exercise 5 A student has entered a 10.0-km run. How long is the run in miles? We have kilometers, which we want to change to miles. We can do this by the following route:

kilometers → meters → yards → miles To proceed in this way, we need the following equivalence statements (conversion factors):

1 km = 1000 m 1 m = 1.094 yd

1760 yd = 1 mi

6.22 mi


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Temperature I suspect you are aware there are three temperature scales commonly in use today. A comparison follows:

Notice a degree of temperature change on the Celsius scale represents the same quantity of change on the Kelvin scale.

Exercise 6 The speed limit on many highways in the United States is 55 mi/h. What number would be posted if expressed in kilometers per hour?

88 km/h

Exercise 7 A Japanese car is advertised as having a fuel economy of 15 km/L. Convert this rating to miles per gallon.

35 mi /gal


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Density Classification of Matter States of Matter (mostly a vocabulary lesson) Be very, very clear that changes of state involve altering IMFs not altering actual chemical bonds!! solid – rigid; definite shape and volume; molecules close together vibrating about fixed points

∴ virtually incompressible liquid – definite volume but takes on the shape of the container; molecules still vibrate but also have

rotational and translational motion and can slide past one another BUT are still close together ∴ slightly compressible

gas – no definite volume and takes on the shape of the container; molecules vibrate, rotate and translate and are independent of each other ∴ VERY far apart ∴ highly compressible

- vapor – the gas phase of a substance that is normally a solid or liquid at room temperature - fluid – that which can flow; gases and liquids

• Mixtures – can be physically separated - homogeneous – have visibly indistinguishable parts, solutions including air - heterogeneous – have visibly distinguishable parts - means of physical separation include: filtering, fractional crystallization, distillation, chromatography

Density = volumemass

Exercise 8 Determining Density A chemist, trying to identify the main component of a compact disc cleaning fluid, determines that 25.00 cm3 of the substance has a mass of 19.625 g at 20°C. Use the information in the table below to identify which substance may serve as the main component of the cleaning fluid. Justify your answer with a calculation.

Compound Density (g/cm3) at 20°C Chloroform 1.492 Diethyl ether 0.714 Ethanol 0.789 Isopropyl alcohol 0.785 Toluene 0.867

Density = 0.7850 g / cm3∴ isopropyl alcohol


Paper Chromatography: Distillation:

Paper chromatograph of ink. (a) A line of the mixture to be separate is placed at one end of a sheet of porous paper. (b) The paper acts as a wick to draw up the liquid. (c) The component with the weakest attraction for the paper travels faster than those that cling to the paper.

• Pure substances – compounds like water, carbon dioxide etc. and elements. Compounds can be separated into elements by chemical means - electrolysis is a common chemical method for separating compounds into elements - elements can be broken down into atoms which can be further broken down into

- nuclei and electrons - p+, n0 and e- - quarks

Electrolysis is an example of a chemical change. In this apparatus, water is decomposed to hydrogen gas (filling the red balloon) and Oxygen gas (filling the blue balloon).


AP* Chemistry ATOMS, MOLECULES & IONS

*AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. © 2008 by René McCormick. All rights reserved.

The highest honor given by the American Chemical Society. He discovered oxygen. Ben Franklin got him interested in electricity and he observed graphite conducts an electric current. Politics forced him out of England and he died in the US in 1804. The back side, pictured below was given to Linus Pauling in 1984. Pauling was the only person to win Nobel Prizes in TWO Different fields: Chemistry and Peace.

2.1 THE EARLY HISTORY OF CHEMISTRY

• 1,000 B.C.—processing of ores to produce metals for weapons and ornaments; use of embalming fluids

• 400 B.C.—Greeks—proposed all matter was make up of 4 “elements” : fire, earth, water and air

• Democritus—first to use the term atomos to describe the ultimate, smallest particles of matter

• Next 2,000 years—alchemy—a pseudoscience where people thought they could turn metals into gold. Some good chemistry came from their efforts—lots of mistakes were made!

• 16th century—Georg Bauer, German , refined the process of extracting metals from ores & Paracelsus, Swiss, used minerals for medicinal applications

• Robert Boyle, English—first “chemist” to perform quantitative experiments of pressure versus volume. Developed a working definition for “elements”.

• 17th & 18th Centuries—Georg Stahl, German—suggested “phlogiston” flowed OUT of burning material. An object stopped burning in a closed container since the air was “saturated with phlogiston”

• Joseph Priestley, English—discovered oxygen which was originally called “dephlogisticated air”

2.2 FUNDAMENTAL CHEMICAL LAWS

• late 18th Century—Combustion studied extensively • CO2, N2, H2 and O2 discovered • list of elements continued to grow • Antione Lavoisier, French—explained the true

nature of combustion—published the first modern chemistry textbook AND stated the Law of Conservation of Mass. The French Revolution broke out the same year his text was published. He once collected taxes for the

government and was executed with a guillotine as an enemy of the people in 1794. He was the first to insist on quantitative experimentation.

THE LAW OF CONSERVATION OF MASS: Mass is neither created nor destroyed.Joseph Proust, French—stated the Law of Definite Proportions

Atoms, Molecules and Ions 2

THE LAW OF DEFINITE PROPORTIONS: A given compound always contains exactly the same proportions of elements by mass.

• 1808--John Dalton stated the Law of Definite proportions. He then went on

to develop the Atomic Theory of Matter. THE LAW OF MULTIPLE PROPORTIONS: When two elements combine to form a series of compounds, the ratios of the masses of the second element that combine with 1 gram of the first element can always be reduced to small whole numbers.

Dalton considered compounds of carbon and oxygen and found: Mass of Oxygen that

combines with 1 gram of C Compound I 1.33 g Compound II 2.66 g Therefore Compound I may be CO while Compound II may be CO2. Exercise 2.1 Illustrating the Law of Multiple Proportions The following data were collected for several compounds of nitrogen and oxygen: Mass of Nitrogen That Combines With 1 g of Oxygen Compound A 1.750 g Compound B 0.8750 g Compound C 0.4375 g Show how these data illustrate the law of multiple proportions.

A = 1.750 = 2 B 0.875 1

B = 0.875 = 2

C 0.4375 1

A = 1.750 = 4 C 0.4375 1

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2.3 DALTON’S ATOMIC THEORY Dalton’s ATOMIC THEORY OF MATTER: (based on knowledge at that time): 1. All matter is made of atoms. These indivisible and indestructible objects are the ultimate

chemical particles. 2. All the atoms of a given element are identical, in both weight and chemical properties.

However, atoms of different elements have different weights and different chemical properties. 3. Compounds are formed by the combination of different atoms in the ratio of small whole

numbers. 4. A chemical reaction involves only the combination, separation, or rearrangement of atoms;

atoms are neither created nor destroyed in the course of ordinary chemical reactions. **TWO MODIFICATIONS HAVE BEEN MADE TO DALTON' S THEORY 1. Subatomic particles were discovered. 2. Isotopes were discovered.

• 1809 Joseph Gay-Lussac, French—performed experiments [at constant temperature and pressure] and measured volumes of gases that reacted with each other.

• 1811 Avogardro, Italian—proposed his hypothesis regarding Gay-Lussac’s work [and you thought he was just famous for 6.02 x 1023] He was basically ignored, so 50 years of confusion followed.

AVOGADRO’S HYPOTHESIS: At the same temperature and pressure, equal volumes of different gases contain the same number of particles.


2.4 EARLY EXPERIMENTS TO CHARATERIZE THE ATOM Based on the work of Dalton, Gay-Lussac, Avogadro, & others, chemistry was beginning to make sense [even if YOU disagree!] and the concept of the atom was clearly a good idea! THE ELECTRON

• J.J. Thomson, English (1898-1903)—found that when high voltage was applied to an evacuated tube, a “ray” he called a cathode ray [since it emanated from the (-) electrode or cathode when YOU apply a voltage across it] was produced.

o The ray was produced at the (-) electrode o Repelled by the (-) pole of an applied electric field, E o He postulated the ray was a stream of NEGATIVE particles now called electrons, e-

o He then measured the deflection of beams of e- to determine the charge-to-mass ratio

o e is charge on electron in Coulombs, (C) and m is its mass. o Thomson discovered that he could repeat this deflection and

calculation using electrodes of different metals ∴ all metals contained electrons and ALL ATOMS contained electrons

o Furthermore, all atoms were neutral ∴ there must be some (+) charge within the atom and the “plum pudding” model was born. Lord Kelvin may have played a role in the development of this model. [the British call every dessert pudding—we’d call it raisin bread where the raisins were the electrons randomly distributed throughout the + bread]

• 1909 Robert Millikan, American—University of

Chicago, sprayed charged oil drops into a chamber. Next, he halted their fall due to gravity by adjusting the voltage across 2 charged plates. Now the voltage needed to halt the fall and the mass of the oil drop can be used to calculate the charge on the oil drop which is a whole number multiple of the electron charge. Mass of e- = 9.11 × 10-31 kg.

81 .7 6 1 0e Cm g

= − ×


RADIOACTIVITY

• Henri Becquerel, French—found out quite by accident [serendipity] that a piece of mineral containing uranium could produce its image on a photographic plate in the absence of light. He called this radioactivity and attributed it to a spontaneous emission of radiation by the uranium in the mineral sample.

• THREE types of radioactive emission: o alpha, α--equivalent to a helium nucleus; the largest particle radioactive particle emitted;

7300 times the mass of an electron. He42 Since these are larger that the rest, early atomic

studies often involved them. o beta, β--a high speed electron. β0

1− OR e01−

o gamma, γ--pure energy, no particles at all! Most penetrating, therefore, most dangerous. THE NUCLEAR ATOM

• 1911 Ernest Rutherford, England—A pioneer in radioactive studies, he carried out experiments to test Thomson’s plum pudding model.

o Directed α particles at a thin sheet of gold foil. He thought that if Thomson was correct, then the massive α particles would blast through the thin foil like “cannonballs through gauze”. [He actually had a pair of graduate students Geiger & Marsden do the first rounds of experiments.] He expected the α particles to pass through with minor and occasional deflections.

o The results were astounding [poor Geiger and Marsden first suffered Rutherford’s wrath and were told to try again—this couldn’t be!].

Most of the α particles did pass straight through, BUT many were deflected at LARGE angles and some even REFLECTED!

Rutherford stated that was like “shooting a howitzer at a piece of tissue paper and having the shell reflected back”.

He knew the plumb pudding model could not be correct! Those particles with large deflection angles had a “close encounter” with the

dense positive center of the atom Those that were reflected had a “direct hit” He conceived the nuclear atom; that with a dense (+) core or nucleus


Particle Mass Charge

e- 9.11 × 10-31 1-

p+ 1.67 × 10-27 1+

n0 1.67 × 10-27 None

This center contains most of the mass of the atom while the remainder of the

atom is empty space! 2.5 THE MODERN VIEW OF ATOMIC STRUCTURE: AN INTRODUCTION ELEMENTS All matter composed of only one type of atom is an element. There are 92 naturally occurring, all others are manmade. ATOMS atom--the smallest particle of an element that retains the chemical properties of that element.

• nucleus--contains the protons and the neutrons; the electrons are located outside the nucleus. Diameter = 10-13 cm. The electrons are located 10-8cm from the nucleus. A mass of nuclear material the size of a pea would weigh 250 million tons! Very dense! - proton--positive charge, responsible for the identity of the

element, defines atomic number - neutron--no charge, same size & mass as a proton,

responsible for isotopes, alters atomic mass number - electron--negative charge, same size as a proton or

neutron, BUT 1/2,000 the mass of a proton or neutron, responsible for bonding, hence reactions and ionizations, easily added or removed.

• atomic number(Z)--The number of p+ in an atom. All atoms of the same element have the same number of p+.

• mass number(A)--The sum of the number of neutrons and p+ for an atom. A different mass number does not mean a different element--just an isotope.

mass number → ← element symbol atomic numbe r→

• actual mass is not an integral number! mass defect--causes this and is related to the energy binding the particles of the nucleus together

Exercise 2.2 Writing the Symbols for Atoms Write the symbol for the atom that has an atomic number of 9 and a mass number of 19. How many electrons and how many neutrons does this atom have?

F; 9 electrons and 10 neutrons

ΧAZ

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ISOTOPES

• isotopes--atoms having the same atomic number (# of p+) but a different number of neutrons • most elements have at least two stable

isotopes, there are very few with only one stable isotope (Al, F, P)

• hydrogens isotopes are so important they have special names: - 0 neutrons hydrogen - 1 neutron deuterium - 2 neutrons tritium

2.6 MOLECULES AND IONS Electrons are responsible for bonding and chemical reactivity

• Chemical bonds—forces that hold atoms together • Covalenet bonds—atoms share electrons and make molecules [independent units]; H2, CO2,

H2O, NH3, O2, CH4 to name a few. • molecule--smallest unit of a compound that retains the chem. characteristics of the compound;

characteristics of the constituent elements are lost. • molecular formula--uses symbols and subscripts to represent the composition of the molecule.

(Strictest sense--covalently bonded) • structural formula—bonds are shown by lines [representing shared e- pairs]; may NOT

indicate shape H O H O H H

• ions--formed when electrons are lost or gained in ordinary chem. reactions; affect size of atom dramatically

• cations--(+) ions; often metals since metals lose electrons to become + charged

• anions--(-) ions; often nonmetals since nonmetals gain electrons to become - charged

• polyatomic ions--units of atoms behaving as one entity MEMORIZE formula and charge!

• ionic solids—Electrostatic forces hold ions together. Strong ∴ ions held close together

∴ solids.


Current Name Original Name Symbol Antimony Stibium Sb Copper Cuprum Cu Iron Ferrum Fe Lead Plumbum Pb Mercury Hydrargyrum Hg Potassium Kalium K Silver Argentum Ag Sodium Natrium Na Tin Stannum Sn Tungsten Wolfram W

2.7 AN INTRODUCTION TO THE PERIODIC TABLE

• Atomic number = number of protons and is

written above each symbol • metals—malleable, ductile & have luster;

most of the elements are metals—exist as cations in a “sea of electrons” which accounts for their excellent conductive properties; form oxides [tarnish] readily and form POSITIVE ions [cations]. Why must some have such goofy symbols?

• groups or families--vertical columns; have

similar physical and chemical properties (based on similar electron configurations!!) • group A—Representative elements • group B--transition elements; all metals; have numerous oxidation/valence states

• periods--horizonal rows; progress from metals to metalloids [either side of the black “stair step” line that separates metals from nonmetals] to nonmetals

• MEMORIZE: 1. ALKALI METALS—1A 2. ALKALINE EARTH METALS—2A 3. HALOGENS—7A 4. NOBLE (RARE) GASSES—8A


2.8 NAMING SIMPLE COMPOUNDS BINARY IONIC COMPOUNDS Naming + ions: usually metals

• monatomic, metal, cation → simply the name of the metal from which it is derived. Al3+ is the aluminum ion

• transition metals form more than one ion; Roman Numerals (in) follow the ion=s name. Cu2+ is copper(II) (Yeah, the no space thing between the ion’s name and (II) looks strange, but it is the correct way to do it.) Mercury (I) is an exception it is Hg2

2+ ∴ two Hg+ associated together. • NH4

+ is ammonium • NO ROMAN NUMERAL WHEN USING silver, cadmium and zinc. [Arrange

their SYMBOLS in alphabetical order—first one is 1+ and the other two are 2+] Naming - ions: monatomic and polyatomic

• MONATOMIC--add the suffix -ide to the stem of the nonmetal's name. Halogens are called the halides.

• POLYATOMIC--quite common; oxyanions are the PA's containing oxygen. - hypo--”ate” least oxygen - -ite--”ate” more oxygen than hypo- - -ate--”ate” more oxygen than -ite - hyper---ate--”ate” the most oxygen

NAMING IONIC COMPOUNDS: The + ion name is given first followed by the name of the negative ion.


Exercise 2.3 Naming Type I Binary Compounds Name each binary compound. a. CsF b. A1C13 c. LiH

a. cesium fluoride b. aluminum chloride

c. lithium hydride Exercise 2.4 Naming Type II Binary Compounds Give the systematic name of each of the following compounds. a. CuC1 b. HgO c. Fe2O3 d. MnO2 e. PbC12

a. copper(I) chloride, b. mercury(II) oxide.

c. iron(III) oxide. d. manganese(IV) oxide. e. lead(II) chloride.

TYPE II: Involve a transition metal that needs a roman numeral Mercury (I) is Hg2

+2 Exceptions: these never need a roman numeral even though transition metals. MEMORIZE Ag+, Cd2+, Zn2+

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Exercise 2.5 Naming Binary Compounds Give the systematic name of each of the following compounds. a. CoBr2 b. CaC12 c. A12O3 d. CrC13

a. Cobalt(II) bromide b. Calcium chloride c. Aluminume oxide

d. Chromium(III) chloride Exercise 2.6 Naming Compounds Containing Polyatomic Ions Give the systematic name of each of the following compounds. a. Na2SO4 b. KH2PO4 c. Fe(NO3)3 d. Mn(OH)2 e. Na2SO3 f. Na2CO3 g. NaHCO3 h. CsC1O4 i. NaOC1 j. Na2SeO4 k. KBrO3

a. Sodium sulfate b. Potassium dihydrogen phosphate

c. Iron(III) nitrate d. Manganese(II) hydroxide

e. Sodium sulfite f. Sodium carbonate

g. Sodium hydrogen carbonate h. Cesium perchlorate i. Sodium hypochlorite

j. Sodium selenate k. Potassium bromate

NAMING BINARY COVALENT COMPOUNDS : (covalently bonded)

• use prefixes!!! Don’t forget the –ide ending as well. Exercise 2.7 Naming Type III Binary Compounds Name each of the following compounds. a. PC15 b. PC13 c. SF6 d. SO3 e. SO2 f. CO2

a. Phosphorus pentachloride b. Phosphorus trichloride

c. Sulfur hexafluoride d. Sulfur trioxide e. Sulfur dioxide

f. Carbon dioxide

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ACIDS

• hydrogen, if present, is listed first - naming acids -ide → hydro [negative ion root]ic ACID -ate → -ic ACID -ite → -ous ACID


• PAINS IN THE GLUTEUS MAXIMUS: these lovely creatures have been around longer than the naming system and no one wanted to adapt!! - water - ammonia - hydrazine - phosphine - nitric oxide - nitrous oxide (“laughing gas”)

Exercise 2.8 Naming Various Types of Compounds Give the systematic name for each of the following compounds. a. P4O10 b. Nb2O5 c. Li2O2 d. Ti(NO3)4

a. Tetraphosphorus decaoxide b. Niobium(V) oxide c. Lithium peroxide

d. Titanium(IV) nitrate

Exercise 2.9 Writing Compound Formulas from Names Given the following systematic names, write the formula for each compound. a. Vanadium(V) fluoride b. Dioxygen difluoride c. Rubidium peroxide d. Gallium oxide

a. VF5 b. O2F2

c. Rb2O2 d. Ga2O3

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AP* Chemistry Stoichiometry


ATOMIC MASSES

12C—Carbon 12—In 1961 it was agreed that this isotope of carbon would serve as the standard used to determine all other atomic masses and would be defined to have a mass of EXACTLY 12 atomic mass units (amu). All other atomic masses are measured relative to this.

mass spectrometer—a device for measuring the mass of atoms or molecules

o atoms or molecules are passed into a beam of high-speed electrons o this knocks electrons OFF the atoms or molecules transforming them into cations o apply an electric field o this accelerates the cations since they are repelled from the (+) pole and attracted toward the (−)

pole o send the accelerated cations into a magnetic field o an accelerated cation creates it’s OWN magnetic field which perturbs the original magnetic field o this perturbation changes the path of the cation o the amount of deflection is proportional to the mass; heavy cations deflect little o ions hit a detector plate where measurements can be obtained.

o 13

1312

Mass C1.0836129 Mass C (1.0836129)(12 amu) 13.003355 amu

Mass C

Exact by definition

average atomic masses—atoms have masses of whole numbers, HOWEVER samples of quadrillions of atoms have a few that are heavier or lighter [isotopes] due to different numbers of neutrons present

percent abundance--percentage of atoms in a natural sample of the pure element represented by a particular isotope

percent abundance = number of atoms of a given isotope × 100% Total number of atoms of all isotopes of that element

counting by mass—when particles are small this is a matter of convenience. Just as you buy 5 lbs of sugar rather than a number of sugar crystals, or a pound of peanuts rather than counting the individual peanuts….this concept works very well if your know an average mass.

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SHOW ALL WORK to receive CREDIT!

Stoichiometry Screencasts are available at www.apchemistrynmsi.wikispaces.com

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mass spectrometer to determine isotopic composition—load in a pure sample of natural neon or other substance. The areas of the “peaks” or heights of the bars indicate the relative abundances of Ne20

10 ,

Ne2110 , and Ne22

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Exercise 1 The Average Mass of an Element When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown in the figure are obtained. Use these data to calculate the average mass of natural copper. (The mass values for 63Cu and 65Cu are 62.93 amu and 64.93 amu, respectively.)

63.55 amu/atom

THE MOLE mole—the number of C atoms in exactly 12.0 grams of 12C; also a number, 6.02 × 1023 just as the word

“dozen” means 12 and “couple” means 2. Avogadro’s number—6.02 × 1023, the number of particles in a mole of anything DIMENSIONAL ANALYSIS DISCLAIMER: I will show you some alternatives to dimensional analysis. WHY? First, some of these techniques are faster and well-suited to the multi-step problems you will face on the AP Exam. Secondly, these techniques better prepare you to work the complex equilibrium problems you will face later in this course. Lastly, I used to teach both methods. Generations of successful students have encouraged me to share these techniques with as many students as possible. They themselves did, once they got to college, and made lots of new friends once word got out they had this “easy way” to solve stoichiometry problems—not to mention their good grades! Give this a try. It doesn’t matter which method you use, I encourage you to use the method that works best for you and lets you solve problems accurately and quickly!


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ALTERNATE TECHNIQUE #1—USING THE MOLE MAP: Simply reproduce this map on your scratch paper until you no longer need to since the image will be burned into your brain!

MULTIPLY[bytheconversionfactoronthearrow]when“traveling”INTHEDIRECTIONOFTHEARROWandobviously,dividewhen“traveling”againstanarrow.


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Exercise 2 Determining the Mass of a Sample of Atoms Americium is an element that does not occur naturally. It can be made in very small amounts in a device known as a particle accelerator. Calculate the mass in grams of a sample of americium containing six atoms.

2.42 × 1021 g

Exercise 3 Determining Moles of Atoms Aluminum is a metal with a high strength-to-mass ratio and a high resistance to corrosion; thus it is often used for structural purposes. Calculate both the number of moles of atoms and the number of atoms in a 10.0-g sample of aluminum.

0.371 mol Al 2.23 × 1023 atoms

Exercise 4 Calculating the Number of Moles and Mass Cobalt (Co) is a metal that is added to steel to improve its resistance to corrosion. Calculate both the number of moles in a sample of cobalt containing 5.00 × 1020 atoms and the mass of the sample.

8.31 × 104 mol Co

4.89 × 102 g Co MOLAR MASS AND FORMULA WEIGHT

molar mass, MM--the sum of all of the atomic masses in a given chemical formula in units of g/mol. It is also equal mass in grams of Avogadro’s number of molecules; i.e. the mass of a mole

empirical formula--the ratio in the network for an ionic substance formula weight--same as molecular weight, just a language problem “molecular” implies covalent

bonding while “formula” implies ionic bonding {just consider this to be a giant conspiracy designed to keep the uneducated from ever understanding chemistry—kind of like the scoring scheme in tennis}. Just use “molar mass” for all formula masses.

A WORD ABOUT SIG. FIG.’s—It is correct to “pull” from the periodic table the least number of sig. figs for your MM’s as are in your problem—just stick with 2 decimal places for all MM’s —much simpler!


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Exercise 5 Calculating Molar Mass I Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural herbicide (weed killer) that kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants [a concept called allelopathy]. The formula for juglone is C10H6O3. (a) Calculate the molar mass of juglone. (b) A sample of 1.56 × 102 g of pure juglone was extracted from black walnut husks. Calculate the number of

moles of juglone present in this sample.

a. 174.1 g b. 8.96 × 105 mol juglone

Exercise 6 Calculating Molar Mass II Calcium carbonate (CaCO3), also called calcite, is the principal mineral found in limestone, marble, chalk, pearls, and the shells of marine animals such as clams. (a) Calculate the molar mass of calcium carbonate. (b) A certain sample of calcium carbonate contains 4.86 moles. Calculate the mass in grams of this sample.

Calculate the mass of the CO32 ions present.

a. 100 g/mol b. 486 g; 292g CO3

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Exercise 7 Molar Mass and Numbers of Molecules Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bees release about 1µg (1 × 10-6 g) of this compound when they sting. The resulting scent attracts other bees to join the attack. (a) Calculate the number of molecules of isopentyl acetate released in a typical bee sting. (b) Calculate the number of carbon atoms present.

5 × 1015 molecules 4 × 1016 carbon atoms

ELEMENTS THAT EXIST AS MOLECULES Pure hydrogen, nitrogen, oxygen and the halogens exist as DIATOMIC molecules under normal conditions. MEMORIZE!!! Be sure you compute their molar masses as diatomics. We lovingly refer to them as the “gens”, “Hydrogen, oxygen, nitrogen & the halogens!” Others to be aware of, but not memorize:

P4—tetratomic form of elemental phosphorous; an allotrope S8—sulfur’s elemental form; also an allotrope Carbon—diamond and graphite covalent networks of atoms

PERCENT COMPOSITION OF COMPOUNDS There are two common ways of describing the composition of a compound: 1) in terms of the number of its constituent atoms and 2) in terms of the percentages (by mass) of its elements. Percent Composition (by mass): The Law of Constant Composition states that any sample of a pure compound always consists of the same elements combined in the same proportions by mass. Remember, all

percent calculations are simply part

100%whole

% comp = mass of desired element × 100% total mass of compound

Consider ethanol, C2H5OH

Mass of C = 2 mol × mol

g01.12 = 24.02 g

Mass of H = 6 mol × mol

g01.1 = 6.06 g

Mass of O = 1 mol × mol

g00.16 = 16.00g

Mass of 1 mol of C2H5OH = 46.08 g


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NEXT, THE MASS PERCENT CAN BE CALCULATED: Mass percent of C = 24.02 g C × 100% = 52.14% 46.08 g Repeat for the H and O present.

Exercise 8 Calculating Mass Percent I Carvone is a substance that occurs in two forms having different arrangements of the atoms but the same molecular formula (C10H14O) and mass. One type of carvone gives caraway seeds their characteristic smell, and the other type is responsible for the smell of spearmint oil. Calculate the mass percent of each element in carvone.

C = 79.96% H = 9.394% O = 10.65%

Exercise 9 Calculating Mass Percent II Penicillin, the first of a now large number of antibiotics (antibacterial agents), was discovered accidentally by the Scottish bacteriologist Alexander Fleming in 1928, but he was never able to isolate it as a pure compound. This and similar antibiotics have saved millions of lives that might have been lost to infections. Penicillin F has the formula C14H20N2SO4. Calculate the mass percent of each element.

C = 53.82% H = 6.47% N = 8.97%

S = 10.26% O = 20.49%


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DETERMINING THE FORMULA OF A COMPOUND

When faced with a hydrocarbon compound of “unknown” formula, one of the most common techniques is to combust it with oxygen to produce oxides of the nonmetals CO2 and H2O which are then collected and weighed.

Calculating empirical and molecular formulas: empirical formulas represent the simplest or smallest ratio of elements within a compound while molecular formulas represent the actual numbers of elements within a compound. The empirical mass is the least common multiple of the molar mass. Example: CH2O is the empirical for a carbohydrate—get it? “carbon waters”. Anyway, glucose is a perfect example of a carbohydrate (a sugar to be exact) with an empirical molar mass of 12 + 2(1) + 16 = 30 g/mol and since glucose is 6 units of CH2O which is equivalent to (CH2O)6 or C6H12O6; the empirical mass of 30 is also multiplied by 6. Thus the MM of glucose is 180 g/mol.

Make your problem solving life easy and assume a 100 gram sample if given %’s—that way you can convert the percents given directly into grams and subsequently into moles in order to simplify your life!

Other twists and turns occurring when calculating molar masses involve: hydrates—waters of hydration or “dot waters”. They count in the calculation of molar masses for

hydrates and used to “cement” crystal structures together

anhydrous—means without water—just to complete the story—just calculate the molar masses of anhydrous substances as you would any other substance

Example: A compound is composed of carbon, nitrogen and hydrogen. When 0.1156 g of this compound is reacted with oxygen [a.k.a. “burned in air” or “combusted”], 0.1638 g of carbon dioxide and 0.1676 g of water are collected. Determine the empirical formula of the compound. So, Compound + O2 oxides of what is burned. In this case Compound + O2 CO2 + H2O + N2

(clearly not balanced) You can see that all of the carbon ended up in CO2 so…when in doubt, calculate THE NUMBER OF MOLES!! 0.1638 g CO2 ÷ 44.01 g/mol CO2 = 0.003781 moles of CO2 = 0.003781 moles of C (why?)


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Next, you can see that all of the hydrogen ended up in H2O, so….calculate THE NUMBER OF MOLES!! So, 0.1676 g H2O ÷ 18.02 g/mol H2O = 0 .009301 moles of H2O, BUT there are 2 moles of H for each mole of water [ Think “organ bank” one heart per body, one C per molecule of carbon dioxide while there are 2 lungs per body, 2 atoms H in water and so on…] thus, DOUBLE THE NUMBER OF MOLES of H2O GIVES THE NUMBER OF MOLES OF HYDROGEN!! moles H = 2 × 0 .009301 moles of H2O = 0.01860 moles of H Therefore, the remaining mass must be nitrogen, BUT we only have mass data for the sample so convert your moles of C and H to grams:

grams C = 0.003781 moles C × 12.01 g

mol = 0.04540 grams C

grams H = 0.01860 moles H × 1.01 g

mol = 0.01879 grams H

Total grams : 0.06419 total grams accounted for thus far What to do next? SUBTRACT! 0.1156 g sample – 0.06419 total grams accounted for thus far = grams N left = 0.05141 g N so….

0.05141 g N ÷ 14.01 g

mol = 0.003670 moles N

Next, realize that chemical formulas represent mole to mole ratios, so…divide the number of moles of each by the smallest # of moles for any one of them to get a guaranteed ONE in your ratios…multiply by 2, then 3, etc to get to a ratio of small whole numbers. Clear as mud? WATCH THE SCREENCAST!!

Element # moles ALL Divided by the smallest (0.003670 moles)

C 0.003781 1 H 0.01860 5 N 0.003670 1

Therefore, the correct EMPIRICAL formula based on the data given is CH5N. Finally (this is drumroll worthy), IF we are told that the MM of the original substance is 31.06 g/mol, then simply use this relationship: (Empirical mass) × n = MM (12.01 + 5.05 + 14.01) × n = 31.07 g/mol n = 0.999678 This is mighty close to 1.0! Thus, the empirical formula and the molecular formula are one and the same.


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Exercise 10 Determine the empirical and molecular formulas for a compound that gives the following analysis in mass percents:

71.65% C1 24.27% C 4.07% H The molar mass is known to be 98.96 g/mol. Empirical formula = CH2 C1

Molecular formula = C2H4C12 Exercise 11 A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. The compound has a molar mass of 283.88 g/mol. What are the compound’s empirical and molecular formulas? Empirical formula = P2O5

Molecular formula = (P2O5)2 or P4O10


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Exercise 12 Caffeine, a stimulant found in coffee, tea, and chocolate, contains 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of 194.2 g/mol. Determine the molecular formula of caffeine.

Molecular formula = C8H10N4O2

BALANCING CHEMICAL EQUATIONS

Chemical reactions are the result of a chemical change where atoms are reorganized into one or more new arrangements. Bonds are broken [requires energy] and new ones are formed [releases energy]. A chemical reaction transforms elements and compounds into new substances. A balanced chemical equation shows the relative amounts of reactants [on the left] and products [on the right] by molecule or by mole.

Subtle details: s, l, g, aq—state symbols that correspond to solid, liquid, gas,

aqueous solution NO ENERGY or TIME is alluded to Antoine Lavoisier (1743-1794)—The Law of Conservation of Matter: matter can be neither created nor

destroyed this means you having to “balance equations” is entirely his fault!!

BALANCING CHEMICAL EQUATIONS Begin with the most complicated-looking thing (often the scariest, too). Save the elemental thing for last. If you get stuck, double the most complicated-looking thing. MEMORIZE THE FOLLOWING: metals + halogens MaXb

CH and/or O + O2 CO2(g) + H2O(g) H2CO3 [any time formed!] CO2 + H2O; in other words, never write carbonic acid as a

product, it spontaneously decomposes [in an open container] to become carbon dioxide andwater

metal carbonates metal OXIDES + CO2


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Exercise 13 Chromium compounds exhibit a variety of bright colors. When solid ammonium dichromate, (NH4)2Cr2O7, a vivid orange compound, is ignited, a spectacular reaction occurs. Although the reaction is actually somewhat more complex, let’s assume here that the products are solid chromium(III) oxide, nitrogen gas (consisting of N2 molecules), and water vapor. Balance the equation for this reaction.

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g) (4 × 2) H (4 ×2) H

http://www.youtube.com/watch?v=CW4hN0dYnkM

Exercise 14 At 1000ºC, ammonia gas, NH3(g), reacts with oxygen gas to form gaseous nitric oxide, NO(g), and water vapor. This reaction is the first step in the commercial production of nitric acid by the Ostwald process. Balance the equation for this reaction.

4 NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)


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STOICHIOMETRIC CALCULATIONS: AMOUNTS OF REACTANTS AND PRODUCTS Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions. Stoichiometry is the most important thing you can learn as you embark upon AP Chemistry! Get good at this and you will do well all year. This NEVER goes away! It’s time to repeat my dimensional analysis disclaimer. DIMENSIONAL ANALYSIS DISCLAIMER: I will show you some alternatives to dimensional analysis. WHY? First, some of these techniques are faster and well-suited to the multi-step problems you will face on the AP Exam. Secondly, these techniques better prepare you to work the complex equilibrium problems you will face later in this course. The first problem you must solve in the free response section of the AP Exam will be an equilibrium problem and you will need to be able to work them quickly. Lastly, I used to teach both methods. Generations of successful students have encouraged me to share these techniques with as many students as possible. They did, once they got to college, and made lots of new friends once word got out they had this “cool way” to solve stoichiometry problems—not to mention their good grades! Give this a try. It doesn’t matter which method you use, I encourage you to use the method that works best for you and lets you solve problems accurately and quickly! First you have to be proficient at the following no matter which method you choose!:

Writing CORRECT formulas—this requires knowledge of your polyatomic ions and being able to use the periodic table to deduce what you have not had to memorize. Review section 2.8 in your Chapter 2 notes or your text.

Calculate CORRECT molar masses from a correctly written formula Balance a chemical equation Use the mole map to calculate the number of moles or anything else!

Remember the mole map? It will come in mighty handy as well!


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Here’s the “template” for solving the problems…you’ll create a chart. Here’s a typical example: Example: Calculate the mass of oxygen will react completely with 96.1 grams of propane? [notice all words—you supply the chemical formulas!] Molar Mass: (44.11) (32.00) (44.01) (18.02) Balanced Eq’n C3H8 + 5 O2 3 CO2 + 4 H2O

mole:mole 1 5 3 4 # moles Amount

1. Write a chemical equation paying special attention to writing correct chemical formulas! 2. Calculate the molar masses and put in parentheses above the formulas—soon you’ll figure out you don’t

have to do this for every reactant and product, just those in which you are interested. 3. Balance the equation! Examine the coefficients on the balanced equation, they ARE the mole:mole

ratios! Isolating them helps you internalize the mol:mol until you get the hang of this. 4. Next, re-read the problem and put in an amount—in this example it’s 96.1 g of propane.

Molar Mass: (44.11) (32.00) (44.01) (18.02) Balanced Eq’n

C3H8 + 5 O2 3 CO2 + 4 H2O

mole:mole 1 5 3 4 # moles 2.18 10.9 6.53 8.71 amount 96.1 grams

5. Calculate the number of moles of something, anything! Use the mole map. Start at 96.1 grams of C3H8, divide the 96.1 g [against the arrow on the mole map] by molar mass to calculate the # moles of propane.

6. USE the mole: mole to find moles of EVERYTHING! If 1 = 2.18 then oxygen is 5(2.18) etc…. [IF the

first mol amount you calculate is not a “1”, just divide appropriately to make it “1” before moving on to calculate the moles of all the rest!] Leave everything in your calculator—I only rounded to save space!

7. Re-read the problem to determine which amount was asked for…here’s the payoff….AP problems ask

for several amounts! First, we’ll find the mass of oxygen required since that’s what the problem asked. 10.9 moles × 32.00 g/mol = 349 g of oxygen


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Now, humor me…What if part (b) asked for liters of CO2 at STP [1 atm, 273K]? Use the mole map. Start in the middle with 6.53 moles × [in direction of arrow] 22.4 L/mol = 146 L

What if part (c) asked you to calculate how many water molecules are produced?

Use the mole map , start in the middle with 8.71 mol water × 6.02 × 1023 molecules

mol = 5.24 × 1024 molecules

of water. Try these two exercises with whichever method you like best! Exercise 15 Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide?

920. g

Molar Mass: (44.11) (32.00) (44.01) (18.02) Balanced Eq’n

C3H8 + 5 O2 3 CO2 + 4 H2O

mole:mole 1 5 3 4 # moles 2.18 10.9 6.53 8.71

amount 96.1 grams 349 g 146 L


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Exercise 16 Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric acid secreted by the stomach:

NaHCO3(s) + HC1(aq) → NaC1(aq) + H2O(l) + CO2(aq)

Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, is also used as an antacid: Mg(OH)2(s) + 2HC1(aq) → 2H2O(l) + MgC12(aq)

Which is the more effective antacid per gram, NaHCO3 or Mg(OH)2? Justify your answer.

Mg(OH)2

CALCULATIONS INVOLVING A LIMITING REACTANT Ever notice how hot dogs are sold in packages of 10 while the buns come in packages of 8? What’s up with that?! The bun is the limiting reactant and limits the hot dog production to 8 as well! The limiting reactant [or reagent] is the one consumed most entirely in the chemical reaction. Let’s use a famous process [meaning one the AP exam likes to ask questions about!], the Haber process. This reaction is essentially making ammonia for fertilizer production from the nitrogen in the air reacted with hydrogen gas. The hydrogen gas is obtained from the reaction of methane with water vapor. This process has saved millions from starvation!! The reaction is shown below. Exercise17 Examine the particle views and explain the differences between the two situations pictured below with regard to what is or is not reacting and total yield of ammonia. N2(g) + 3H2(g) 2NH3(g) Situation 1 Situation 2


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Plan of attack: First, you must realize that you even need a plan of attack! IF ever you are faced with TWO starting amounts of matter reacting, you have entered “The Land of Limiting Reactant”. When faced with this situation …calculate the number of moles of everything you are given. Set up your table like before, only now you’ll have TWO amounts and thus TWO # ‘s of moles to get you started. Cover one set of moles up (pretending you only had one amount to work from) and ask yourself, “What if all of these moles reacted?” “How many moles of the other reactants would I need to use up all of these moles?” Next, do the calculation of how many moles of the “other” amount(s) you would need. Do you have enough? If so, the reactant you began with IS the limiting reactant. If not repeat this process with the “other” reactant amount you were given. It doesn’t matter where you start the “What if?” game….you get there either way. Clear as mud? Read on…(and consider listening to the SCREENCAST!) Let’s revisit the Haber process:

Molar Mass: (28.02) (2.02) (17.04) Balanced Eq’n N2 + 3 H2 2 NH3

mole:mole 1 3 2 # moles amount Suppose 25.0 kg of nitrogen reacts with 5.00 kg of hydrogen to form ammonia. What mass of ammonia can be produced? Which reactant is the limiting reactant? What is the mass of the reactant that is in excess? **Insert the masses in the amount row and find the number of moles of BOTH! Molar Mass: (28.02) (2.02) (17.04) Balanced Eq’n

N2 + 3 H2 2 NH3

mole:mole 1 3 2 # moles 892 moles 2,475 moles amount 25,000 g 5,000 g WHAT IF I used up all the moles of hydrogen? I’d need 1/3 × 2,475 moles = 825 moles of nitrogen. Clearly I have EXCESS moles of nitrogen!! Therefore, hydrogen limits me.

OR WHAT IF I used up all the moles of nitrogen? I’d need 3 × 892 moles = 2,676 moles of hydrogen. Clearly I don’t have enough hydrogen, so it limits me!! Therefore nitrogen is in excess. Continued on next page.


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Either way, I’ve established that hydrogen is the limiting reactant so I modify the table:

In English, that means I’ll use up all the hydrogen but not all the nitrogen!

Molar Mass: (28.02) (2.02) (17.04) Balanced Eq’n N2 + 3 H2 2 NH3 mole:mole 1 3 2 # moles 825 mol used

892 moles

2,475 moles

1650 mol produced

amount 825 mol (28.02) = 23,116 g used

25,000 g

5,000 g

1650 mol (17.04) = 28,116 g produced

1,884 g excess!! Here’s the question again, let’s clean up any sig.fig issues: Suppose 25.0 kg of nitrogen reacts with 5.00 kg of hydrogen to form ammonia. (3 sig. fig. limit) What mass of ammonia can be produced? 28,100 g produced = 28.1 kg (It is always polite to respond in the unit given). Which reactant is the limiting reactant? Hydrogen—once that’s established, discard the nitrogen amounts and let hydrogen be your guide! What is the mass of the reactant that is in excess? 1,884 g = 1.88 kg excess nitrogen!!

Exercise 18 Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed?

CuO is limiting; 10.6 g N2


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Theoretical Yield: The amount of product formed when a limiting reactant is completely consumed. This assumes perfect conditions and gives a maximum amount!! Not likely!

Actual yield: That which is realistic.

Percent yield: The ratio of actual to theoretical yield.

Actual Yield × 100% = Percent yield Theoretical Yield

Exercise 19 Methanol (CH3OH), also called methyl alcohol, is the simplest alcohol. It is used as a fuel in race cars and is a potential replacement for gasoline. Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5 kg CO(g) is reacted with 8.60 kg H2(g). Calculate the theoretical yield of methanol. If 3.57 × 104 g CH3OH is actually produced, what is the percent yield of methanol?

Theoretical yield is 6.82 × 104 g Percent yield is 52.3%

AP* Chemistry GASES


Early barometer

THE PROPERTIES OF GASES Only 4 quantities are needed to define the state of a gas: a) the quantity of the gas, n (in moles) b) the temperature of the gas, T (in KELVINS) c) the volume of the gas, V (in liters) d) the pressure of the gas, P (in atmospheres) A gas uniformly fills any container, is easily compressed & mixes completely with any other gas. GAS PRESSURE: A measure of the force that a gas exerts on its container. Force is the physical quantity that interferes with inertia. Gravity is the force responsible for weight. Force = mass × acceleration; Newton’s 2nd Law. The units of force follow: N = kg × m/s2 Pressure-- Force/ unit area; N/m2 Barometer--invented by Evangelista Torricelli in 1643; uses the height of a column of mercury to measure gas pressure (especially atmospheric) 1 mm of Hg = 1 torr 760.00 mm Hg = 760.00 torr =1.00 atm = 101.325 kPa ≈ 105 Pa At sea level all of the above define STANDARD PRESSURE. The SI unit of pressure is the Pascal (Blaise Pascal); 1 Pa = 1 N / m2

The manometer—a device for measuring the pressure of a gas in a container. The pressure of the gas is given by h [the difference in mercury levels] in units of torr (equivalent to mm Hg).

a) Gas pressure = atmospheric pressure – h

b) Gas pressure = atmospheric pressure + h

Exercise 1 Pressure Conversions The pressure of a gas is measured as 49 torr. Represent this pressure in both atmospheres and pascals.

6.4 × 10-2 atm 6.5 × 103 Pa

holsteint

Text Box


Gases 2

Exercise Rank the following pressures in decreasing order of magnitude (largest first, smallest last): 75 kPa, 300. torr, 0.60 atm and 350. mm Hg. GAS LAWS: THE EXPERIMENTAL BASIS • BOYLE’S LAW: “father of chemistry”--the volume of a confined gas is

inversely proportional to the pressure exerted on the gas. ALL GASES BEHAVE IN THIS MANNER!

• Robert Boyle was an Irish chemist. He studied PV relationships using a J-tube set up in the multi-story entryway of his home. o P ∝ 1/V plot = straight line o pressure and volume are inversely proportional o Volume ↑ pressure ↓ at constant temperature, the converse is also

true o for a given quantity of a gas at constant temperature, the product of

pressure and volume is a constant. PV = k

Therefore, P

kPkV 1

==

which is the equation for a straight line of the type

y = mx + b where m = slope, and b the y-intercept

In this case, y = V, x = 1/P and b = 0. Check out the plot on the right (b). The data Boyle collected is graphed on (a) above.

o P1V1 = P2V2 is the easiest form of Boyle’s law to memorize o Boyle’s Law has been tested for over three centuries. It holds true

only at low pressures. A plot of PV versus P for several gases at pressures below 1atm is pictured at left. An ideal gas is expected to have a constant value of PV, as shown by the dotted line. CO2 shows the largest change in PV, and this change is actually quite small: PV changes from about 22.39 L·atm at 0.25 atm to 22.26 L·atm at 1.00 atm. Thus Boyle’s Law is a good approximation at these relatively low pressures.

Gases 3

Exercise 2 Boyle’s Law I Sulfur dioxide (SO2), a gas that plays a central role in the formation of acid rain, is found in the exhaust of automobiles and power plants. Consider a 1.53- L sample of gaseous SO2 at a pressure of 5.6 × 103 Pa. If the pressure is changed to 1.5 × 104 Pa at a constant temperature, what will be the new volume of the gas ?

0.57 L Exercise 3 Boyle’s Law II In a study to see how closely gaseous ammonia obeys Boyle’s law, several volume measurements were made at various pressures, using 1.0 mol NH3 gas at a temperature of 0ºC. Using the results listed below, calculate the Boyle’s law constant for NH3 at the various pressures.

Experiment Pressure (atm) Volume (L) 1 0.1300 172.1 2 0.2500 89.28 3 0.3000 74.35 4 0.5000 44.49 5 0.7500 29.55 6 1.000 22.08

experiment 1 is 22.37 experiment 2 is 22.32 experiment 3 is 22.31 experiment 4 is 22.25 experiment 5 is 22.16 experiment 6 is 22.08

PLOT the values of PV for the six experiments above. Extrapolate it back to see what PV equals at 0.00 atm pressure. Compare it to the PV vs. P graph on page 2 of these notes. What is the y-intercept for all of these gases? Remember, gases behave most ideally at low pressures. You can’t get a pressure lower than 0.00 atm!

Gases 4

These balloons each hold 1.0 L of gas at 25°C and 1 atm. Each balloon contains 0.041 mol of gas, or 2.5 × 1022 molecules.

• CHARLES’ LAW: If a given quantity of gas is held at a constant pressure, then its volume is directly proportional to the absolute temperature. Must use KELVIN

• Jacques Charles was a French physicist and the first person to fill a hot “air” balloon with hydrogen gas and made the first solo balloon flight! o V ∝ T plot = straight line o V1T2 = V2T1 o Temperature ∝ Volume at constant pressure o This figure shows the plots of V vs. T (Celsius) for several

gases. The solid lines represent experimental measurements on gases. The dashed lines represent extrapolation of the data into regions where these gases would become liquids or solids. Note that the samples of the various gases contain different numbers of moles.

o What is the temperature when the Volume extrapolates to zero?

Exercise 4 Charles’s Law A sample of gas at 15ºC and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38ºC and 1 atm ?

2.79 L

• GAY-LUSSAC’S LAW of combining volumes: volumes of gases always combine with one another in the ratio of small whole numbers, as long as volumes are measured at the same T and P.

- P1T2 = P2T1 - Avogadro=s hypothesis: equal volumes of gases under the

same conditions of temperature and pressure contain equal numbers of molecules.

• AVOGADRO’S LAW: The volume of a gas, at a given temperature and pressure, is directly proportional to the quantity of gas.

- V ∝ n

- n ∝ Volume at constant T & P HERE’S AN EASY WAY TO MEMORIZE ALL OF THIS! Start with the combined gas law: P1V1T2 = P2V2T1 Memorize it.

Next, put the fellas’ names in alphabetical order. Boyle’s uses the first 2 variables, Charles’ the second 2 variables & Gay-Lussac’s the remaining combination of variables. What ever doesn’t appear in the formula, is being held CONSTANT!

Gases 5

Exercise 5 Avogadro’s Law Suppose we have a 12.2-L sample containing 0.50 mol oxygen gas (O2) at a pressure of 1 atm and a temperature of 25ºC. If all this O2 were converted to ozone (O3) at the same temperature and pressure, what would be the volume of the ozone ?

8.1 L THE IDEAL GAS LAW Four quantities describe the state of a gas: pressure, volume, temperature, and # of moles (quantity). Combine all 3 laws: V ∝ nT P Replace the ∝ with a constant, R, and you get: PV = nRT The ideal gas law! It is an equation of state. R = 0.08206 L• atm/mol • K also expressed as 0.08206 L atm mol−1 K−1

Useful only at low Pressures and high temperatures! Guaranteed points on the AP Exam! These next exercises can all be solved with the ideal gas law, BUT, you can use another if you like! Exercise 6 Ideal Gas Law I A sample of hydrogen gas (H2) has a volume of 8.56 L at a temperature of 0ºC and a pressure of 1.5 atm. Calculate the moles of H2 molecules present in this gas sample.

0.57 mol Exercise 7 Ideal Gas Law II Suppose we have a sample of ammonia gas with a volume of 3.5 L at a pressure of 1.68 atm. The gas is compressed to a volume of 1.35 L at a constant temperature. Use the ideal gas law to calculate the final pressure.

4.4 atm

Gases 6

Exercise 8 Ideal Gas Law III A sample of methane gas that has a volume of 3.8 L at 5ºC is heated to 86ºC at constant pressure. Calculate its new volume.

4.9 L

Exercise 9 Ideal Gas Law IV A sample of diborane gas (B2H6), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a temperature of -15ºC and a volume of 3.48 L. If conditions are changed so that the temperature is 36ºC and the pressure is 468 torr, what will be the volume of the sample?

3.07 L Exercise 10 Ideal Gas Law V A sample containing 0.35 mol argon gas at a temperature of 13ºC and a pressure of 568 torr is heated to 56ºC and a pressure of 897 torr. Calculate the change in volume that occurs.

decreases by 3 L GAS STOICHIOMETRY Use PV = nRT to solve for the volume of one mole of gas at STP: Look familiar? This is the molar volume of a gas at STP. Work stoichiometry problems using your favorite method, dimensional analysis, mole map, the table way…just work FAST! Use the ideal gas law to convert quantities that are NOT at STP.

Gases 7

Exercise 11 Gas Stoichiometry I A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present?

7.81 × 10-2 mol N2 Exercise 12 Gas Stoichiometry II Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3). Calculate the volume of CO2 at STP produced from the decomposition of 152 g CaCO3 by the reaction

CaCO3(s) → CaO(s) + CO2(g)

34.1 L CO2 at STP Exercise 13 Gas Stoichiometry III A sample of methane gas having a volume of 2.80 L at 25ºC and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35.0 L at 31ºC and 1.25 atm. The mixture was then ignited to form carbon dioxide and water. Calculate the volume of CO2 formed at a pressure of 2.50 atm and a temperature of 125ºC.

2.47 L

Gases 8

THE DENSITY OF GASES:

d = m = P(MM) {for ONE mole of gas}= MM AND Molar Mass = MM = dRT V RT 22.4 L P “Molecular Mass kitty cat”—all good cats put dirt [dRT] over their pee [P]. Corny? Yep! But, you’ll thank me later! Just remember that densities of gases are reported in g/L NOT g/mL. What is the approximate molar mass of air? _________ The density of air is approx. _______ g/L. List 3 gases that float in air: List 3 gases that sink in air: Exercise 14 Gas Density/Molar Mass The density of a gas was measured at 1.50 atm and 27ºC and found to be 1.95 g/L. Calculate the molar mass of the gas.

32.0 g/mol GAS MIXTURES AND PARTIAL PRESSURES The pressure of a mixture of gases is the sum of the pressures of the different components of the mixture: Ptotal = P1 + P2 + . . . Pn John Dalton’s Law of Partial Pressures also uses the concept of mole fraction, χ χ A = moles of A . moles A + moles B + moles C + . . . so now, PA = χ A Ptotal The partial pressure of each gas in a mixture of gases in a container depends on the number of moles of that gas. The total pressure is the SUM of the partial pressures and depends on the total moles of gas particles present, no matter what they are!

Gases 9

Exercise 15 Dalton’s Law I Mixtures of helium and oxygen are used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46 L He at 25ºC and 1.0 atm and 12 L O2 at 25ºC and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25ºC.

PHe = 9.3 atm PO2 = 2.4 atm

PTOTAL = 11.7 atm Exercise 16 Dalton’s Law II The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present.

0.210 Exercise 17 Dalton’s Law III The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760. torr.

593 torr WATER DISPLACEMENT It is common to collect a gas by water displacement which means some of the pressure is due to water vapor collected as the gas was passing through! You must correct for this. You look up the partial pressure due to water vapor by knowing the temperature.

Gases 10

Exercise 8 Gas Collection over Water A sample of solid potassium chlorate (KClO3) was heated in a test tube (see the figure above) and decomposed by the following reaction:

2 KClO3(s) → 2 KCl(s) + 3 O2(g) The oxygen produced was collected by displacement of water at 22ºC at a total pressure of 754 torr. The volume of the gas collected was 0.650 L, and the vapor pressure of water at 22ºC is 21 torr. Calculate the partial pressure of O2 in the gas collected and the mass of KClO3 in the sample that was decomposed.

Partial pressure of O2 = 733 torr 2.12 g KClO3

KINETIC MOLECULAR THEORY OF GASES Assumptions of the MODEL: 1. All particles are in constant, random, motion. 2. All collisions between particles are perfectly elastic. 3. The volume of the particles in a gas is negligible 4. The average kinetic energy of the molecules is its Kelvin temperature. This neglects any intermolecular forces as well. Gases expand to fill their container, solids/liquids do not. Gases are compressible, solids/liquids are not appreciably compressible. This helps explain Boyle’s Law: If the volume is decreased that means that the gas particles will hit the wall more often, thus increasing pressure

V

nRTP 1)(=

Constant

Gases 11

And helps explain Charles’ Law When a gas is heated, the speed of its particles increase and thus hit the walls more often and with more force. The only way to keep the P constant is to increase the volume of the container.

TP

nRV ⎟⎠⎞

⎜⎝⎛=

And also helps explain Gay-Lussac’s Law When the temperature of a gas increases, the speeds of its particles increase, the particles are hitting the wall with greater force and greater frequency. Since the volume remains the same this would result in increased gas pressure.

TVnRP ⎟

⎠⎞

⎜⎝⎛=

And even helps explain Avogadro’s Law An increase in the number of particles at the same temperature would cause the pressure to increase if the volume were held constant. The only way to keep constant P is to vary the V.

nP

RTV ⎟⎠⎞

⎜⎝⎛=

What about Dalton’s Law? The P exerted by a mixture of gases is the SUM of the partial pressures since gas particles are acting independent of each other and the volumes of the individual particles DO NOT matter.

Constant

Constant

Constant

Gases 12

DISTRIBUTION OF MOLECULAR SPEEDS Plot # of gas molecules having various speeds vs. the speed and you get a curve. Changing the temperature affects the shape of the curve NOT the area beneath it. Change the # of molecules and all bets are off! Maxwell’s equation:

2 3rms

RTu uMM

= =

Use the “energy R” or 8.314510 J/K• mol for this equation since kinetic energy is involved.

Exercise 19 Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of helium gas at 25ºC.

1.36 × 103 m/s

If we could monitor the path of a single molecule it would be very erratic. Mean free path—the average distance a particle travels between collisions. It’s on the order of a tenth of a micrometer. WAAAAY SMALL!

Examine the effect of temperature on the numbers of molecules with a given velocity as it relates to temperature. HEAT ‘EM UP, SPEED ‘EM UP!! Drop a vertical line from the peak of each of the three bell shaped curves—that point on the x-axis represents the AVERAGE velocity of the sample at that temperature. Note how the bells are “squashed” as the temperature increases. You may see graphs like this on the AP exam where you have to identify the highest temperature based on the shape of the graph!

GRAHAM’S LAW OF DIFFUSION AND EFFUSION Effusion is closely related to diffusion. Diffusion is the term used to describe the mixing of gases. The rate of diffusion is the rate of the mixing. Effusion is the term used to describe the passage of a gas through a tiny orifice into an evacuated chamber as shown on the right. The rate of effusion measures the speed at which the gas is transferred into the chamber.

Gases 13

The rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure.

REMEMBER rate is a change in a quantity over time, NOT just the time! Exercise 20 Effusion Rates Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6), a gas used in the enrichment process to produce fuel for nuclear reactors.

13.2 Exercise A pure sample of methane is found to effuse through a porous barrier in 1.50 minutes. Under the same conditions, an equal number of molecules of an unknown gas effuses through the barrier in 4.73 minutes. What is the molar mass of the unknown gas? Diffusion This is a classic!

Distance traveled by NH3 = urmsfor NH3 = Distance traveled by HCl urms for HCl

2

1

Rate of effusion of gas 1 = Rate of effusion of gas 2

MMMM

3

36.5 1.517

HCl

NH

MMMM

= =

Gases 14

The observed ratio is LESS than a 1.5 distance ratio—why? This diffusion is slow considering the molecular velocities are 450 and 660 meters per second—which one is which? This tube contains air and all those collisions slow the process down in the real world. Speaking of real world…. REAL, thus NONIDEAL GASES Most gases behave ideally until you reach high pressure and low temperature. (By the way, either of these can cause a gas to liquefy, go figure!) van der Waals Equation--corrects for negligible volume of molecules and accounts for inelastic collisions leading to intermolecular forces (his real claim to fame).

a and b are van der Waals constants; no need to work problems, it’s the concepts that are important! Notice pressure is increased (intermolecular forces lower real pressure, you’re correcting for this) and volume is decreased (corrects the container to a smaller “free” volume). These graphs are classics and make great multiple choice questions on the AP exam.

When PV/nRT = 1.0, the gas is ideal This graph is just for nitrogen gas. All of these are at 200K. Note that although nonideal behavior Note the P’s where the curves is evident at each temperature, the cross the dashed line [ideality]. deviations are smaller at the higher Ts. Don’t underestimate the power of understanding these graphs. We love to ask question comparing the behavior of ideal and real gases. It’s not likely you’ll be asked an entire free-response gas problem on the real exam in May. Gas Laws are tested extensively in the multiple choice since it’s easy to write questions involving them! You will most likely see PV = nRT as one part of a problem in the free-response, just not a whole problem! GO FORTH AND RACK UP THOSE MULTIPLE CHOICE POINTS!!

[ ][ ]2nP + a( V bn = nRT)V

−

AP* Chemistry THERMOCHEMISTRY

*AP is a registered trademark of the College Board, which was not involved in the production of this product. Special thanks to the contributions of Lisa McGaw and David Wentz. © 2013 by René McCormick. All rights reserved.

Let’s begin with terms for you to master: Energy (E) – the ability to do work or produce heat ; the sum of all potential and kinetic energy in a system

is known as the internal energy of the system

Potential energy – energy by virtue of position. In chemistry this is usually the energy stored in bonds (i.e., when gasoline burns there are differences in the attractive forces between the nuclei and the electrons in the reactants and the products) When bonded atoms are separated, the PE is raised because energy must be added to overcome the coulombic attraction between each nucleus and the shared electrons.

When atoms bond, the above mentioned coulombic attraction results in energy being released and a subsequently lower PE.

Kinetic energy – energy of motion (translational, rotational & vibrational motion of particles in our case), proportional to Kelvin temperature; kinetic energy depends on the mass and the velocity of the object: KE = ½ mv2

Check out this simulation regarding the Boltzmann distribution: http://phet.colorado.edu/en/simulation/atomic-interactions

And this one regarding the effect of temperature on molecular motion: http://phet.colorado.edu/en/simulation/states-of-matter-basics

Law of Conservation of Energy – You may know it as “energy is never created nor destroyed” which implies that any change in energy of a system must be balanced by the transfer of energy either into or out of the system.

AKA energy of the universe is constant & the First Law of Thermodynamics

Heat (q) – Two systems with different temperatures that are in thermal contact will exchange thermal energy, the quantity of which is call heat. This transfer of energy in a process (flows from a warmer object to a cooler one, transfers heat because of temperature difference but, remember, temperature is not a measure of energy—it just reflects the motion of particles)

Temperature (T)—is proportional to the average kinetic energy of the molecules, KEave . “Heat ‘em up and speed ‘em up” as you saw in the states of matter animation.

Enthalpy (H)– flow of energy (heat exchange) at constant pressure when two systems are in contact. Enthalpy of reaction (∆Hrxn) – amount of heat released (negative values) or absorbed (positive values)

by a chemical reaction at constant pressure in kJ/molrxn Enthalpy of combustion (∆Hcomb)—heat absorbed or released by burning (usually with O2) in

kJ/molrxn; note that combustion reactions yield oxides of that which is combusted Enthalpy of formation (∆Hf) – heat absorbed or released when ONE mole of compound is formed

from elements in their standard states in kJ/molrxn Enthalpy of fusion (∆Hfus)—heat absorbed to melt (overcome IMFs) 1 mole of solid to liquid @ MP

expressed in kJ/molrxn Enthalpy of vaporization (∆Hvap)—heat absorbed to vaporize or boil (overcome IMFs) 1 mole liquid to

vapor @BP in kJ/molrxn

http://phet.colorado.edu/en/simulation/atomic-interactions

http://phet.colorado.edu/en/simulation/states-of-matter-basics

holsteint

Text Box


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System – area of the universe we are focusing on (i.e., the experiment) Surroundings – everything outside of the system Endothermic – net absorption of energy (heat exchange) by the system; energy is a reactant; (i.e., baking

soda and vinegar when mixed get very cold to the touch) ; +ΔH Exothermic – net release of energy (heat exchange) by the system; energy is a product; (i.e., burning

methane gas in the lab burner produces heat; light sticks give off light which is also energy); −ΔH Entropy (s) – measure of the dispersal of matter and energy; increase dispersal +ΔS; decrease dispersal −ΔS Gibbs Free Energy (G)– criteria for determining thermodynamic favorability and calculating the

theoretical amount of energy to do work Thermodynamics – study of energy and its interconversions Work – force acting over distance in physics often expressed as work = −PΔV where gases are involved;

expressed in Joules or kJ Standard Conditions—you already know about STP, but recall that the T is STP is 0°C and humans are not

happy lab workers when it is that cold! So, think of standard conditions as standard lab conditions which are 1 atm of pressure, 25°C (much more comfy!) and if solutions are involved, their concentration is 1.0 M. All of this information is communicated by adding the symbol ° to G, H or S. So, if you see ΔH°, then you automatically know the pressure, temperature and conditions that apply to that value!

There has recently been a change in how enthalpy, entropy and free energy units are expressed. For example, you may see ΔH° values expressed as kJ in older printed material. Currently, they should be expressed in kJ/molrxn where the “molrxn” is “moles of reaction”. See Jim Spencer’s article on AP Central for additional information.

ENERGY AND WORK Energy is often defined as

the “ability to do work”. ∆E = q (heat) + w

(work) Signs of q +q if heat absorbed –q if heat released

Algebraic sign of w as it relates to work done by or work done on gases

• + w if work done on the system (i.e., compression) • −w if work done by the system (i.e., expansion)

When related to gases, work is a function of pressure

Pressure is defined as force per unit of area, so when the volume is changed work was either done on the gas or by the gas. work = −P∆V

Exercise 1 Internal Energy Calculate ∆E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

17.0 kJ

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Exercise 2 PV Work Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.

‒270 L•atm

Exercise 3 Internal Energy, Heat, and Work A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 × 106 L to 4.50 × 106 L by the addition of 1.3 × 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ∆E for the process. (To convert between L ⋅ atm and J, use 1 L ⋅ atm = 101.3 J.)

8.0 × 107 J ENTHALPY Measure only the change in enthalpy, ∆H ( the difference between the potential energies of the products

and the reactants) ∆H is a state function

∆H = q at constant pressure (i.e. atmospheric

pressure)

Enthalpy can be calculated from several sources including: Stoichiometry Calorimetry From tables of standard values Hess’s Law Bond energies

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Stoichiometrically:

Calorimetry:

The process of measuring heat based on observing the temperature change when a body absorbs or discharges energy as heat.

“Coffee Cup” calorimetry:

Coffee-cup calorimetry – in the lab this is how we experiment to find energy of a particular system. We use a Styrofoam® cup, reactants that begin at the same temperature and look for change in temperature. After all data is collected (mass or volume; initial and final temperatures) we can use the specific formula to find the energy released or absorbed. We refer to this process as constant pressure calorimetry. ** q = ∆H @ these conditions**

Terms to know: Heat capacity – energy required to raise temp. by 1 degree (Joules/ °C)

Specific heat capacity (Cp) – same as above but specific to 1 gram of substance and the experiment is

carried out at constant pressure. Constant pressure is achieved using open containers, so you will be doing experiments of that kind in lab.

quantity of heat transferredspecific heat ( of material) (degrees of temperature change)g

=

Molar heat capacity—same as above but specific to one mole of substance (J/mol K or J/mol °C )

Exercise 4 Upon adding solid potassium hydroxide pellets to water the following reaction takes place:

KOH(s) → KOH(aq) + 43 kJ/mol Answer the following questions regarding the addition of 14.0 g of KOH to water:

Does the beaker get warmer or colder? Is the reaction endothermic or exothermic? What is the enthalpy change for the dissolution of the 14.0 grams of KOH?

Answers: (a) warmer (b) exothermic (c) −10.7 kJ/molrxn

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Energy (q) released or gained at constant pressure: q = mCp∆T q = quantity of heat (Joules or calories) m = mass in grams ΔT = Tf −Ti (final – initial)

Cp = specific heat capacity ( J/g°C)

Specific heat of water (liquid state) = 4.184 J/g°C ( or 1.00 cal/g°C) Water has one of the highest specific heats known! This property makes life on earth possible and regulates earth’s temperature year round!

Heat lost by substance = heat gained by water

Units of Energy:

calorie--amount of heat needed to raise the temp. of 1.00 gram of water 1.00 °C kilocalorie—1,000 calories AND the food label calorie with a capital C.

212

KE mv= units are 2

kg×ms

joule--SI unit of energy; 1 cal = 4.184 J

Exercise 6 Enthalpy When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ/mol of energy is released as heat. Calculate ∆H for a process in which a 5.8 gram sample of methane is burned at constant pressure.

∆H = heat flow = ‒320 kJ/molrxn

Exercise 5 In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions were originally at 24.6°C. After the reaction, the final temperature is 31.3°C. Assuming that all solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/g°C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter.

‒5.6 kJ/molrxn

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Exercise 7 Constant-Pressure Calorimetry When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25°C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, and that the specific heat capacity of the solution is 4.18 J/°C ⋅ g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.

‒26 kJ/molrxn

Tables:

♦ ∆Hf° = enthalpy of formation

= Production of ONE mole of compound FROM

its ELEMENTS in their standard states (°)

= ZERO (0.00) for ELEMENTS in standard states

♦ Standard States: 25°C (298 K), 1 atm, 1M

The “Big Mamma” Equation: ∆Hrxn = Σ ∆Hf (products) - Σ ∆Hf (reactants)

(also known as Hess’s Law)

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Exercise 8 Substance ∆Hf

° (kJ/mol) NH4ClO4(s) −295 Al2O3(s) −1676 AlCl3(s) −704 NO(g) 90.0 H2O(g) −242 Given the information above, calculate the ∆H°rxn for the following chemical reaction. 3 Al(s) + 3 NH4ClO4(s) → Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)

‒2,677 kJ/molrxn (exo) Exercise 9 C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O() + 2800 kJ Occasionally, not all values are found in the table of thermodynamic data. For most substances it is impossible to go into a lab and directly synthesize a compound from its free elements. The heat of formation for the substance must be calculated by working backwards from its heat of combustion. Calculate the ∆Hf of C6H12O6(s) given the combustion reaction above along with the following information. Substance ∆Hf

° (kJ/mol) CO2(g) −393.5 H2O() −285.8

∆Hf ° for glucose = ‒1276 kJ/mol

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Exercise 10 The thermite reaction occurs when a mixture of powdered aluminum and iron(III) oxide is ignited with a magnesium fuse. Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction: 2Al (s) + Fe2O3(s) → A12O3(s) + 2Fe(s)

‒850. kJ/molrxn

Enthalpy is independent of the reaction pathway. If you can find a combination of chemical equations that add up to give you the desired overall equation, you can also sum up the ∆H’s for the individual reactions to get the overall ∆Hrxn. ♦ First, decide how to rearrange equations such that the reactants and products are on appropriate sides of the

arrows in the chemical equation. It is often helpful to begin by working backwards from the final or summary chemical equation.

♦ If an equation had to be reversed, also reverse the sign of ∆Hrxn ♦ If an equation had to be multiplied by a given factor to obtain correct coefficients, also multiply the

∆H rxn by this factor since ∆Hrxn’s are in kJ/MOLErxn (division applies similarly) ♦ Double check to ensure that everything cancels out to give you the exact summary chemical equation you

desire.

Exercise 11 Calculate the ∆H for this overall reaction 2 H3BO3(aq) → B2O3(s) + 3 H2O() given the following equations: H3BO3(aq) → HBO2(aq) + H2O() ∆H = −0.02 kJ/molrxn

H2B4O7(aq) + H2O() → 4 HBO2(aq) ∆ H = −11.3 kJ/molrxn

H2B4O7(aq) → 2 B2O3(s) + H2O() ∆ H = 17.5 kJ/molrxn

14.4 kJ/molrxn endothermic

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Exercise 12 Calculate the change in energy that accompanies the following reaction given the data below.

H2(g) + F2(g) → 2 HF(g) Bond Type Bond Energy H−H 432 kJ/mol F−F 154 kJ/mol H−F 565 kJ/mol

−544 kJ/molrxn

Bond Energies

• Energy must be added/absorbed to BREAK bonds (endothermic) in order to overcome the coulombic attraction between each nuclei and the shared electrons. Energy is released when bonds are FORMED (exothermic) because the resultant coulombic attraction between the bonded atoms lowers potential energy causing a release. This is a giant misconception among students! Once again, it “takes” energy to break bonds and energy is released when a bond forms.

• ∆H = sum (Σ) of the energies required to break old bonds (positive signs since energy is added to the system) plus the sum of the energies released in the formation of new bonds (negative signs since energy is lost from the system).

∆H = Σ Bond Energies broken – Σ Bond Energies formed

SUMMARY FOR ENTHALPY: What does it really tell you about the changes in energy regarding a

chemical reaction? ∆H = + reaction is endothermic and heat energy is added into the system ∆H = − reaction is exothermic and heat energy is lost from the system

(Nature tends toward the lowest energy state!)

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Speaking of bond energies, allow us to clear up some common misconceptions AND make some dazzling connections. Let’s start with the vocabulary used to describe phase changes. First, you must realize that the vocabulary is “directional” (hence the arrows on this diagram) as well as very specific. You’ll have to mean what you say and say what you mean when answering a free response question!

Phase transitions involving overcoming intermolecular attractions or IMFs which should never be confused with ionic or covalent chemical bonds. Fusion (melting), vaporization, and sublimation require an input of energy to overcome the attractive forces between the particles of the substance. NOTICE we did not speak of “breaking bonds”. Freezing, condensation, and deposition (opposite of sublimation) release energy as IMFs form since the particles achieve a lower energy state mainly due to a decrease in temperature.

Is there a difference between a vapor and a gas? Yes, it’s primarily semantics. A gas is a gas at room temperature, we don’t speak of “oxygen vapor”. However, we do use the term “vapor” when the substance is normally a liquid or solid at room temperature. We say “water vapor”, “carbon dioxide vapor”, “iodine vapor”, etc. Be very, very clear that changes in the phases of matter involve altering IMFs, not altering chemical bonds.* The strength of the intermolecular attractions between molecules, and therefore the amount of energy required to overcome these attractive forces (as well as the amount of energy released when the attractions are formed) depends on the molecular properties of the substance, ionic, polar, nonpolar, etc. Generally, the more polar a molecule is, the stronger the attractive forces between molecules are. Hence, more polar molecules typically require more energy to overcome the intermolecular attractions in an endothermic phase transition, and release more energy by forming intermolecular attractions during an exothermic phase transition.

Phase transitions involve the “breaking” or forming of intermolecular forces (attractive interactions between molecules). Hence, as with other chemical reactions, it is necessary to discuss the energy that is absorbed or given off during the breaking or forming of intermolecular interactions in a phase transition.

*UNLESS you have a network solid or covalent network solid which is a chemical compound in which the atoms are bonded by covalent bonds in a continuous network. In a network solid there are no individual molecules and the entire crystal may be considered a macromolecule. Examples of network solids include diamond with a continuous network of carbon atoms and silicon dioxide or quartz with a continuous three dimensional network of SiO2 units. Graphite and the mica group of silicate minerals structurally consist of continuous two-dimensional layers covalently bonded within the layer with other bond types holding the layers together. That means they essentially slide in sheets the way your pencil “lead” (actually graphite) glides across the page leaving a trail.

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You have probably seen a graph like this before arriving in AP Chemistry. Let’s make some additional dazzling connections!

Suppose this heating curve is for water (it isn’t because the horizontal line for boiling condensing is too short. Why? (It takes WAY more energy to boil water than what is shown here due to having to overcome all those IMFs called hydrogen bonds).

Take a moment and ponder the differences in molecular structure and molecular motion among the different states of water represented above, the kinetic energy changes, and the potential energy changes.

1. Does the process above represent a collection of chemical changes, physical changes or both? Physical changes.

2. What type of force is involved in the changes you identified in question 1? Intermolecular forces as opposed to intramolecular forces (chemical bonds). Students may also specify exact IMFs such as H-bonding.

3. Define temperature: Temperature is defined as the average kinetic energy of the molecules.

4. Which conversions involve temperature changes? warming and cooling

5. Which mathematical formula is appropriate for calculating the energy associated with the processes you identified in question 4? q = mcΔT

6. Which conversions involve potential energy changes? freezing, melting, condensing, boiling, vaporizing

7. Which mathematical formula is appropriate for calculating the energy associated with the processes you identified in question 4? q = mΔHvap

8. How do you calculate q for the processes identified in number 6 if you are given only the mass of the water sample? Simply divide the mass given by the molar mass of water to determine the number of moles involved, then multiply by the enthalpy of vaporization or fusion.

9. Which portions of this graph represent equilibrium conditions? Phase changes: ice water, water steam, the plateaus where temperature is not changing ∴only potential energy is changing as system is absorbing or releasing heat. PE is added to overcome IMFs (melting or boiling) or released to when IMFs form (condensing or freezing).

AP* Chemistry ATOMIC STRUCTURE


ELECTROMAGNETIC RADIATION James Maxwell developed an elegant mathematical theory in 1864 to describe all forms of radiation in terms of oscillating or wave-like electric and magnetic fields in space. • electromagnetic radiation—UV, visible light, IR, microwaves,

television and radio signals, and X-rays • wavelength (λ)—(lambda) length between 2 successive crests. • frequency (υ)—(nu in chemistry; f in physics—either is OK), number

of cycles per second that pass a certain point in space (Hz-cycles per second)

• amplitude—maximum height of a wave as measured from the axis of propagation

• nodes—points of zero amplitude (equilibrium position); always occur

at λ/2 for sinusoidal waves velocity—speed of the wave

velocity = λ υ “c”—the speed of light; 2.99792458 [just call it 3] × 108 m/s; ALL EM RADIATION TRAVELS AT THIS SPEED! I call it easy, you’ll call it a trick! If “light” is involved, it travels at 3 × 108 m/s

• Notice that λ and υ are inversely proportional. When one is large, the other is small.

axis of propagation

holsteint

Text Box

SHOW ALL WORK to receive CREDIT! Refer to KEY WITH ANSWERS for problems with no answers given on last page.

Atomic Structure 2

Exercise 1 Frequency of Electromagnetic Radiation The brilliant red colors seen in fireworks are due to the emission of light with wavelengths around 650 nm when strontium salts such as Sr(NO3)2 and SrCO3 are heated. (This can be easily demonstrated in the lab by dissolving one of these salts in methanol that contains a little water and igniting the mixture in an evaporating dish.) Calculate the frequency of red light of wavelength 6.50 × 102 nm.

υ = 4.61 × 1014 Hz THE NATURE OF MATTER At the end of the 19th century, physicists were feeling rather smug. All of physics had been explained [or so they thought]. Students were being discouraged from pursuing physics as a career since all of the major problems had been solved! Matter and Energy were distinct: Matter was a collection of particles and Energy was a collection of waves. Enter Max Planck stage left… THE QUANTIZATION OF ENERGY "Ultraviolet catastrophe"—defined as the fact that a glowing hot object did not emit UV light as predicted. • 1900--Max Planck solved the problem. He made an incredible assumption: There is a minimum

amount of energy that can be gained or lost by an atom, and all energy gained or lost must be some integer multiple, n, of that minimum. (As opposed to just any old value of energy being gained or lost.)

ΔEnergy = n(hυ) • where h is a proportionality constant, Planck's constant, h = 6.6260755 × 10-34 joule • seconds.

This υ is the lowest frequency that can be absorbed or emitted by the atom, and the minimum energy change, hυ, is called a quantum of energy. Think of it as a “packet” of E equal to hυ.

• No such thing as a transfer of E in fractions of quanta, only in whole numbers of quanta. • Planck was able to calculate a spectrum for a glowing body that reproduced the experimental

spectrum. • His hypothesis applies to all phenomena on the atomic and molecular scale. Exercise 2 The Energy of a Photon The blue color in fireworks is often achieved by heating copper(I) chloride (CuCl) to about 1200°C. Then the compound emits blue light having a wavelength of 450 nm. What is the increment of energy (the quantum) that is emitted at 4.50 × 102 nm by CuCl?

= 4.41 × 10-19 J

Atomic Structure 3

THE PHOTOELECTRIC EFFECT AND ALBERT EINSTEIN In 1900 Albert Einstein was working as a clerk in the patent office in Bern, Switzerland. This left him time to work on Physics. He proposed that EM radiation itself was quantized; he was a great fan of Planck’s work! He proposed that EM radiation could be viewed as a stream of “particles” called photons. • photoelectric effect--light bombards the surface of a metal and electrons are ejected. • frequency—a minimum υ must be met or alas, no action! Once minimum is met, intensity increases

the rate of ejection. • photon--massless particles of light. Ephoton = hυ = hc λ You know Einstein for the famous E = mc2 from his second “work” as the special theory of relativity published in 1905. Such blasphemy, energy has mass?! That would mean: m = E c2 therefore,

m = E = hc/λ = h c2 c2 λc Does a photon have mass? Yep! In 1922 American physicist Arthur Compton performed experiments involving collisions of X-rays and electrons that showed photons do exhibit the apparent mass calculated above.

SUMMARY: Energy is quantized. It can occur only in discrete units called quanta [hυ]. EM radiation [light, etc.] exhibits wave and particle properties. This phenomenon is known as the dual nature of light.

Since light which was thought to be wavelike now has certain characteristics of particulate matter, is the converse also true? Enter Louis de Broglie [French physicist, 1923] stage right…

IF m = h λc substitute v [velocity] for c for any object NOT traveling at the speed of light, then rearrange and solve for lambda

Atomic Structure 4

this is called the de Broglie equation: λ = h mv Exercise 3 Calculations of Wavelength Compare the wavelength for an electron (mass = 9.11 × 10-31 kg) traveling at a speed of 1.0 × 107 m/s with that for a ball (mass = 0.10 kg) traveling at 35 m/s.

λe = 7.27 × 10-11 m λ b = 1.9 × 10-34 m

• The more massive the object, the smaller its associated wavelength and vice

versa! • Davisson and Germer @ Bell labs found that a beam of electrons was

diffracted like light waves by the atoms of a thin sheet of metal foil and that de Broglie's relation was followed quantitatively.

• ANY moving particle has an associated wavelength. • Silly physicists! We now know that E is really a form of matter, and ALL

matter shows the same types of properties. That is, all matter exhibits both particulate and wave properties.

HYDROGEN’S ATOMIC LINE SPECTRA AND NIELS BOHR • emission spectrum—the spectrum of bright lines, bands, or

continuous radiation that is provided by a specific emitting substance as it loses energy and returns to its ground state OR the collection of frequencies of light given off by an "excited" electron

• absorption spectrum—a graph or display relating how a substance absorbs electromagnetic radiation as a function of wavelength

• line spectrum--isolate a thin beam by passing through a slit then a prism or a diffraction grating which sorts into discrete frequencies or lines

• Johann Balmer--worked out a mathematical relationship that accounted for the 3 lines of longest wavelength in the visible emission spectrum of H. (red, green and blue lines)

Atomic Structure 5

• Niels Bohr connected spectra, and the quantum ideas of Einstein and Planck: the single electron of the hydrogen atom could occupy only certain energy states, stationary states

"The Mother of all Assumptions" An electron in an atom would remain in its lowest E state unless otherwise disturbed. • Energy is absorbed or emitted by a change from this ground state • an electron with n = 1 has the most negative energy and is thus the most strongly

attracted to the positive nucleus. [Higher states have less negative values and are not as strongly attracted to the positive nucleus.]

• ground state--n = 1 for hydrogen

photon = = hchE υλ

• To move from ground to n = 2 the electron/atom must absorb no more or no less than 0.75 Rhc. [that’s a collection of constants]

• So, a move of n = 2 to n = 1 emits 985 kJ of energy. • What goes up must come down. Energy absorbed must eventually be emitted. • The origin or atomic line spectra is the movement of electrons between quantized energy states. • IF an electron moves from higher to lower E states, a photon is emitted and an emission line is

observed.

Atomic Structure 6

• Bohr’s equation for calculating the energy of the E levels available to the electron in the

hydrogen atom:

• where n is an integer [larger n means larger orbit radius, farther from nucleus], Z is the nuclear charge

• The NEGATIVE sign simply means that the E of the electron bound to the nucleus is lower that it world be if the electron were at an infinite distance [n = ∞] from the nucleus where there is NO interaction and the energy is zero.

• ΔE is simply the subtraction of calculating the energy of two different levels, say n=6 and n=1. If the difference is negative, E was lost. If the difference is positive, E was gained.

• TWO Major defects in Bohr's theory: 1) Only works for elements with ONE electron. 2) The one, lonely electron DOES NOT orbit the nucleus in a fixed path!!

Exercise 4 Energy Quantization in Hydrogen Calculate the energy required to excite the hydrogen electron from level n = 1 to level n = 2. Also calculate the wavelength of light that must be absorbed by a hydrogen atom in its ground state to reach this excited state.

∆E = 1.633 × 10-18 J λ = 1.216 × 10-7 m

Exercise 5 Electron Energies Calculate the energy required to remove the electron from a hydrogen atom in its ground state.

∆E = 2.178 × 10-18 J

218

22.178 10 J ZEn

− ⎛ ⎞= − × ⎜ ⎟

⎝ ⎠

Atomic Structure 7

THE WAVE PROPERTIES OF THE ELECTRON SCHRODINGER, HEISENBERG, AND QUANTUM NUMBERS • After World War I--Niels Bohr assembled a group of physicists in Copenhagen hoping to derive

a comprehensive theory for the behavior of electrons in atoms from the viewpoint of the electron as a particle

• Erwin Schrodinger--independently tried to accomplish the same thing but focused on de Broglie's eqn' and the electron as a wave. Schrodinger's approach was better; explained more than Bohr's and met with more success. Quantum mechanics was born!

• de Broglie opened a can of worms among physicists by suggesting the electron had wave properties

• the electron has dual properties • Werner Heisenberg and Max Born provided the uncertainty principle

- if you want to define the momentum of an electron, then you have to forego knowledge of its exact position at the time of the measurement

• Max Born on the basis of Heisenberg's work suggested: if we choose to know the energy of an electron in an atom with only a small uncertainty, then we must accept a correspondingly large uncertainty about its position in the space about the atom's nucleus. So What? We can only calculate the probability of finding an electron within a given space.

THE WAVE MECHANICAL VIEW OF THE ATOM • Schrodinger equation: solutions are called wave functions--chemically

important. The electron is characterized as a matter-wave • sort of standing waves--only certain allowed wave functions (symbolized by

the Greek letter, ψ, pronounced “sigh”) • Each ψ for the electron in the H atom corresponds to an allowed energy

(−Rhc/n2). For each integer, n, there is an atomic state characterized by its own ψ and energy En.

• In English? Points 1 & 2 above say that the energy of electrons is quantized.

Notice in the figure to the right, that only whole numbers of standing waves can “fit” in the proposed orbits.

The hydrogen electron is visualized as a standing wave around the nucleus [above right]. The circumference of a particular circular orbit would have to correspond to a whole number of wavelengths, as shown in (a) and (b) above, OR else destructive interference occurs, as shown in (c). This is consistent with the fact that only certain electron energies are allowed; the atom is quantized. (Although this idea encouraged scientists to use a wave theory, it does not mean that the electron really travels in circular orbits.)

Atomic Structure 8

• the square of ψ gives the intensity of the electron wave or the probability of finding the electron at the point P in space about the nucleus—the intensity of color in (a) above represents the probability of finding the electron in that space, the darker the color—the more probable it is we would find the electron in that region.

• electron density map, electron density and electron probability ALL mean the same thing! When we say “orbital” this image at right is what we picture in our minds.

• matter-waves for allowed energy states are also called (drum roll please…) orbitals.

• To solve Schrodinger's equation in a 3-dimensional world we need the quantum numbers n, , and mℓ

• The amplitude of the electron wave at a point depends on the distance of the point from the nucleus.

• Imagine that the space around a H nucleus is made up of a series of thin “shells” like the layers of an onion, but these “shells” as squishy (Very scientific word, that “squishy”!)

• Plot the total probability of finding the electron in each shell versus the distance from the nucleus and you get the radial probability graph you see in (b) above

• The maximum in the curve occurs because of two opposing effects. 1) the probability of finding an electron is greatest near the nucleus [electrons just can’t resist the attraction of a proton!], BUT 2) the volume of the spherical shell increases with distance from the nucleus, SO we are summing more positions of possibility, so the TOTAL probability increases to a certain radius and then decreases as the electron probability at EACH position becomes very small.

Try not to stress over this introduction stuff! Learn to do the calculations in the practice exercises & do the best you can with the “theory” stuff. It’s my moral obligation to TRY to explain it to you.

Stress over quantum numbers and electron configurations and periodicity if you must—that’s the important stuff in this chapter!

I just feel the need to give you a complete education!

Atomic Structure 9

-1 0 +1 -1 0 +1 0 0 0

QUANTUM NUMBERS & ATOMIC ORBITALS There are 4 quantum numbers that describe the “address” and “spin” of an electron. The value of n limits the possible values of ℓ, which in turn limit the values of mℓ. • n—principal energy level--1 to infinity. Determines the total energy of the electron. It indicates the

most probable [within 90%] distance of the electron from the nucleus. A measure of the orbital size or diameter. 2n2 electrons may be assigned to a shell. It’s simply the Energy level that electron is in. If it’s a 3s electron, n = 3, if it’s a 4d electron, n = 4, etc.

• ℓ--angular momentum--0,1,2,....(n-1) electrons w/in shell may be grouped into subshells [or sublevels, same thing!], each characterized by its certain wave shape. Each ℓ is a different orbital shape or orbital type. - n limits the values of ℓ to no larger than n-1. Thus, the number of possibilities for ℓ is equal to n.

(English translation: 3 sublevels for 3rd E level, 4 for 4th E level, etc.) - s,p,d,f sublevels 0,1,2,3 ℓ-values respectively (So, what do spdf stand for? Sharp, principle,

diffuse, fundamental--early days of atomic spectroscopy)

• mℓ--magnetic—assign the “blanks” in orbital notation with zero on the middle blank and then -ℓ through zero to +ℓ. I’ll bet this looks familiar for Sulfur from Chem. I!

1s 2s 2p 3s 3p

- that means that the range of orbitals is from +ℓ to - ℓ - it describes the orientation of an orbital in a given subshell - The down arrow in the 3p, -1 slot is the last electron placed [valence electron], so far it’s set

of quantum numbers is 3, 1, -1

You can keep going from g…

DON’T try to make this hard! Embrace the elegant patterns!

Atomic Structure 10

Exercise 6 Electron Subshells For principal quantum level n = 5, determine the number of allowed subshells (different values of ℓ), and give the designation of each.

ℓ = 0; 5s ℓ = 1; 5p ℓ = 2; 5d ℓ = 3; 5f ℓ = 4; 5g

THE SHAPES OF ATOMIC ORBITALS There is no sharp boundary beyond which the electrons are never found!!

• s--spherical; the size increases with n. The nodes you see at left represent ZERO probability of finding the electron in that region of space. The number of nodes equals n-1 for s orbitals.

• p--have one plane that slices through the nucleus and divides the region of electron density into 2 halves. Nodal plane--the electron can never be found there!!

• 3 orientations: px, py, and pz.

• d--2 nodal planes slicing through the nucleus to create four sections; 5 orbitals.

- The dz2 orbital is really strange!

Atomic Structure 11

• f--3 nodal planes slicing through the nucleus; eight

sections; 7 orbitals ELECTRON CONFIGURATIONS Chemical properties depend on the number and arrangement of electrons in an atom. Usually only the valence electrons play the reaction game. ELECTRON SPIN • 1920--chemists realized that since electrons interact with a magnetic field,

there must be one more concept to explain the behavior of electrons in atoms.

• ms--the 4th quantum number; accounts for the reaction of electrons in a magnetic field

MAGNETISM • magnetite--Fe3O4, natural magnetic oxide of iron • 1600--William Gilbert concluded the earth is also a large spherical magnet

with magnetic south at the geographic North Pole (Santa's habitat). • NEVER FORGET: opposites attract & likes repel whether it’s opposite

poles of a magnet or opposite electrical charges

PARAMAGNETISM AND UNPAIRED ELECTRONS • diamagnetic--not magnetic [magnetism dies]; in fact they are slightly repelled by a magnetic field.

Occurs when all electrons are PAIRED. • paramagnetic--attracted to a magnetic field; lose their magnetism when removed from the magnetic

field; HAS ONE OR MORE UNPAIRED ELECTRONS • ferromagnetic--retain magnetism upon introduction to, then removal from a magnetic field • All of these are explained by electron spins

o Each electron has a magnetic field with N & S poles o electron spin is quantized such that, in an external magnetic field, only two orientations of the

electron magnet and its spin are possible o +1/2 and −1/2 o H is paramagnetic; He is diamagnetic, WHY?

- H has one unpaired electron - He has NO unpaired electrons; all spins offset and cancel each other out

Atomic Structure 12

What about ferromagnetic? clusters of atoms have their unpaired electrons aligned within a

cluster, clusters are more or less aligned and substance acts as a magnet. Don't drop it!!

When all of the domains, represented by these arrows are aligned, it behaves as a magnet.

This is what happens if you drop it: The domains go in all different directions and it no longer operates as a magnet.

THE PAULI EXCLUSION PRINCIPLE In 1925 Wolfgang Pauli stated: no two electrons in an atom can have the same set of four quantum numbers. This means no atomic orbital can contain more than 2 electrons & for two electrons to occupy the same orbital, they must be of opposite spin!! ATOMIC ORBITAL ENERGIES AND ELECTRON ASSIGNMENTS

• Order of orbital energies--the value of n (E = −Rhc/n2) determines the energy of an electron • “many-electron” atoms depend on both n & ℓ • Use the diagonal rule or Aufbau series • Orbital radius changes slightly with ℓ as well as with n • subshell orbitals contract toward the nucleus as the value of ℓ

increases • contraction is partially into the volume of space occupied by

the core electrons • the energy of the electrons in these subshells is raised due to

the repulsions between the subshell electrons and the core electrons.

- a subshell's energy rises as its ℓ quantum number increases when inner electrons are present

• Order of Orbital Assignments • each electron is lazy and occupies the lowest energy space available • based on the assumption inner electrons have no effect on which orbitals are assigned to outer or

valence electrons - not exactly true (use diagonal rule)—note that the one pictured shows the lowest

energy state (1s) at the top of the diagram—some variations show 1s at the bottom of the diagram

• electron configurations (spectroscopic notation) • clump the 1's, 2's, etc. TOGETHER rather than written in the order the orbitals fill

Atomic Structure 13

HUNDS RULE The most stable arrangement of electrons is that with the maximum number of unpaired electrons; WHY?? it minimizes electron-electron repulsions (everyone gets their own room)

• all single electrons also have parallel spins to reduce e-/e- repulsions (aligns micromagnets) • When 2 electron occupy the same orbital they must have opposite spins (Pauli exclusion

principle); this also helps to minimize e-/e- repulsions Personally, I think this whole quantum number thing is easiest when we start with the electron configurations THEN write the quantum numbers. Allow me to recap: Don’t try to make this hard! It just isn’t. Energy Level 1 2 3 4 5 6 7…. # of sublevels 1 2 3 4 5 6 7…. Names of sublevels s s, p s,p,d s,p,d,f s,p,d,f,g s,p,d,f,g,h s,p,d,f,g,h,i n, principal quantum number

1 2 3 4 5 6 7….

Name of sublevel

s p d f g h i…

ℓ, angular momentum quantum number [= n-1]

0 1 2 3 4 5 6

# of orbitals [= -ℓ to +ℓ]

1 3 5 7 9 11 13

mℓ for each orbital within a sublevel

0

-1 0 +1

-2 -1 0 +1 +2

And so on, just pretend you are in second grade and make a number line with negative

numbers on the left, ZERO in the middle, and positive numbers on the right.

The first electron placed in an orbital is assigned the +1/2 and the second one is assigned the -1/2.

Atomic Structure 14

-1 0 +1

Let’s practice: Give the electron configurations for the elements within this figure. I’ll get you started! S 1s2 2s2 2p6 3s2 3p4

Cd La Hf Ra Ac

And their Orbital Notation: Sulfur 1 2 3 4 5 8 6 9 7 10 11 12 13 16 14 15 1s 2s 2p 3s 3p Cd La Hf Ra Ac There is a super cool animation that illustrates this concept. The website is from the Chief Reader of the AP Exam. This site is http://intro.chem.okstate.edu/APnew/Default.html Click on Electron Configuration Animation. You’ll need the shockwave plug-in. Once the animation comes up, click on the screen to advance from Hydrogen on up by atomic number.

As electrons enter these sublevels their wave functions interfere with each other causing the energy of these to change and separate. Do not be misled by this diagram, there ARE INDEED energy differences between all of these sublevels. See below.

Atomic Structure 15

mighty close to the nucleus. “E is ZAPPED”

What are the quantum numbers for the outermost [valence] electron? S--since the last electron put in is #16 and it is in the 3p sublevel, n = 3 and ℓ = 2. It’s in the -1 slot, the mℓ = -1 and since it’s the second arrow placed [down arrow] its ms = -1/2. So the set of quantum numbers for the 16th electron in sulfur is: 3,2,-1, -1/2. You’ll find this really easy IF you draw the orbital notation for the last sublevel! Cd La Hf Ra Ac • You accepted that the sublevels had differences in energies long ago. • You even know the increasing order of energies: s < p < d < f < g…

• Now you have to be able to EXPLAIN it on the AP test. Throughout this discussion keep some fundamental scientific principles close at hand

o electrons repel each other o electrons are negative and thus, attracted by the

positive nucleus o forces (attractive & repulsive) dissipate with

increasing distance. • We need to examine the graph at the right, radial

probabilities again. o See the small hump near the origin? That’s the

distance from the nucleus that a 2s electron occupies a small but significant amount of the time—we say the 2s electron “penetrates to the nucleus” more than for the 2p orbital. This causes a 2s electron to be ATTRACTED to the nucleus more than a 2p electron making the 2s orbital LOWER in E than the 2p orbital.

o Think of the nucleus as “zapping” the energy of the electrons that penetrate closer to it. [Just don’t write that!] Imagine a hyper child—it’s on its best behavior, sitting still, being quiet, etc. when it’s close to Mom. The closer to the Mother Nucleus the hyper electron is, the less hyper or energetic it is. Don’t EVER write this as an answer to an essay question! It’s just a model to help you get your teeth into this concept!

“penetrates” closest to the nucleus

Atomic Structure 16

• Same song second verse—the last hump represents the greatest probability for predicting the distance of an electron from the nucleus, BUT the first humps determine the order of the energy.

o The top graph is for 3s—note it has 2 humps close to the nucleus

o The bottom graph is for 3s, 3p and note that 3d only has one hump.

o 3s penetrates most [has least energy], then 3p [higher E than 3s, lower E than 3d] then 3d penetrates least [so it has the highest energy].

o MORAL: The greater the penetration, the less energy that orbital has. Since you already knew the order with respect to energy, s < p < d < f , the degree of penetration is s’s penetrate most and f’s penetrate least.

ION ORBITAL ENERGIES AND ELECTRON CONFIGURATIONS The dfs overlay [that thing that happens when the configurations don’t fit the pattern in transition metals and rare earth metals] does not occur in ion configurations since the valence (outermost n) electrons are the first to go! THE SHELL ENERGY RANGES SEPARATE MORE WIDELY AS ELECTRONS ARE REMOVED.

• Atoms and ions with unpaired electrons are paramagnetic (attracted to a magnetic field) • Transition metals with +2 or higher have no ns electrons • Fe+2 is paramagnetic to the extent of 4 unpaired electrons and Fe+3 is paramagnetic to the extent

of 5 unpaired electrons THE HISTORY OF THE PERIODIC TABLE

1800ish—Johann Dobereiner, triads; 1864 John Newlands octaves; 1870--Dmitrii Mendeleev & Julius Lothar Meyer--by mass; 1913 Mosley--by number of protons.

Atomic Structure 17

o Group IA(1A or 1)--alkali metals o Group IIA(2A or 2)--alkaline earth metals o Group VIA(6A or 16)--Chalcogens o Group VIIA(7A or 17)—Halogens o Group VIIIA (8A or 18)—Noble gas

SOME PROPERTIES OF COMMON GROUPS:

• Alkali metals—the most reactive metal family; must be stored under oil; react with water violently!

• Alkaline-earth metals-- except for Be(OH)2, the metal hydroxides formed by this group provide basic solutions in water ; pastes of these used in batteries

• Chalcogen family – many found combined with metal ores • Halogen family – known as the “salt-formers” ; used in modern lighting • Noble Gas family – known for their disinterest in other elements; once thought to

never react; neon used to make bright RED signs • Transition metals—fill the d orbitals.

o Anomalies occur at Chromium and Copper to minimize electron/electron repulsions. If you learned that there is special stability associated with a half-filled sub-level, IT’S A LIE!! No such stability exists! NEVER, EVER write that! It’s all about lowering energy by minimizing electron/electron repulsions.

Atomic Structure 18

PERIODIC TRENDS: A trend is NOT an

EXPLANATION!

• Rare Earth metals—fill the f sublevels.

Lanthanides and Actinides. These sometimes put an electron in d [just one or two electrons] before filling f. This is that dsf overlay referred to earlier—the energies of the sublevels are very similar.

This is an important section—there is almost always an essay involving this topic on the AP exam. There are several arguments you will evoke to EXPLAIN a periodic trend. Remember opposites attract and likes repel. The trick is learning which argument to use when explaining a certain trend! THE ARGUMENTS:

1. Effective nuclear charge, Zeff—essentially equal to the group number. Think of the IA’s having a Zeff of one while the VII A’s have a Zeff of 7! The idea is that the higher the Zeff, the more positive the nucleus, the more attractive force there is emanating from the nucleus drawing electrons in or holding them in place. Relate this to ENERGY whenever possible.

2. Distance—attractive forces dissipate with increased distance. Distant electrons are held loosely and thus easily removed. Relate this to ENERGY whenever possible.

3. Shielding—electrons in the “core” effectively shield the nucleus’ attractive force for the valence electrons. Use this ONLY when going up and down the table, NOT across. There is ineffective shielding within a sublevel or energy level. Relate this to ENERGY whenever possible. {I can’t help but think of dodge ball, such a barbaric ritual—since you’re the smart kids, you figured out in elementary school to stay behind the bigger kids to keep from getting hit! The electrons in the first or second energy level, shield the outer valence electrons from the Mother Nucleus’ attractive force. }

4. Minimize electron/electron repulsions—this puts the atom at a lower energy state and makes it more stable. Relate this to ENERGY whenever possible.

Here we go!

Atomic Structure 19

1. ATOMIC RADIUS--No sharp boundary beyond which the electron never strays!! • Use diatomic molecules and determine radius then react with other atoms to

determine the radius of those atoms • ATOMIC radii decreases (↓) moving across a period AND increases (↑)

moving down a row (family)

WHY ↓ across? The effective nuclear charge (Zeff) increases (more protons for the same number of energy levels) as we move from left to right across the periodic table, so the nucleus has a greater positive charge, thus the entire electron cloud is more strongly attracted and “shrinks”.

o This shrinks the electron cloud until… …the point at which electron/electron repulsions overcome the nuclear attraction and stop the contraction of the electron cloud.

WHY ↑ down? The principal level, n, determines the size of an atom—add another principal level and the atoms get MUCH larger radii

As we move down a family, the attractive force the nucleus exerts on the valence electrons dissipates.

Shielding is only a valid argument when comparing elements from period to period (up and down the table) since shielding is incomplete within a period—use this argument with extreme caution! It should NOT be your favorite!

2. IONIZATION ENERGY--energy required to remove an electron from the atom IN THE GAS PHASE. Costs Energy. • Removing each subsequent electron requires more energy:

Second IE, Third IE, etc. Some subsequent IEs cost more E than others!! A

HUGE energy price is paid if the subsequent removal of electrons is from another sublevel or, Heaven forbid, another principal E level (core).

- ↓ down a family—increased distance from the nucleus and increased shielding by full principal E levels means it requires less E to remove an electron

- ↑across a period—due to increasing Zeff. The higher the Zeff, the stronger the nucleus attracts valence electrons, the more energy required to remove a valence electron. Let’s talk EXCEPTIONS!!

Atomic Structure 20

• an anomaly occurs at “messing up a half-filled or filled sublevel” There’s nothing magical about this and electrons are not “happier” as a result. The simple truth is that when electron pairing first occurs within an orbital, there is an increase in electron/electron repulsions which makes it require less energy [easier] to remove an electron thus the IE drops. NEVER, EVER write about the fictitious stability of a ½-filled shell—even though you may see it in books!

• Look at oxygen vs. nitrogen—it requires less energy to remove an electron from oxygen’s valence

IN SPITE OF AN INCREASING Zeff because oxygen’s p4 electron is the first to pair within the orbital thus experiencing increased repulsion. The increased repulsion lowers the amount of energy required to remove the newly paired electron!

• Also, look at the drop in IE from any s2 to p1. This is also IN SPITE OF AN INCREASING Zeff. This

drop in the energy required to remove a valence electron is due to the fact that you are removing a p electron rather than an s electron. The p’s are less tightly held BECAUSE they do not penetrate the electron cloud toward the nucleus as well as an s electron. The general trend is that s is held most tightly since it penetrates more then p, d and f…

3. ELECTRON AFFINITY—an affinity or “liking” for electrons—force feeding an element an electron—Energy associated with the addition of an electron to a gaseous atom: X (g) + e− X− (g) If the addition of an electron is exothermic, then you’ll see a negative sign on the energy change and the converse is also true. The more negative the quantity, the more E is released. This matches our sign convention in thermodynamics. • ↓ down a family [that means becomes less negative a.k.a. more positive, giving off less energy]—due to increased distance from the nucleus with each increasing principal E level. The nucleus is farther from the valence level and more shielded. • ↑ across a period [that means become more negative, giving off more energy]—Again the increasing Zeff more strongly attracts the electron. The interactions of electron/electron repulsions wreaks havoc with this generalization as we shall soon see!

Atomic Structure

21

• Let’s talk EXCEPTIONS!! First the lines on the diagram below connect adjacent elements. The absence of a line indicates missing elements whose atoms do not add an electron exothermically and thus do not form stable isolated anions [remember these are all –1 ions at this point].

- an anomaly—No N− yet there is a C−--this is due to their electron/electron repulsions compared to their electron configurations. N is p3 while C is p2. C adds an electron WITHOUT PAIRING which does NOT increase the e-/e- repulsions and therefore, carbon forms a stable -1 ion while N would have to pair p electrons and the increased e-/e- repulsions overcome the increasing attractive force due to the increase in Zeff and no -1 N ion forms! [horrid sentence structure!]

- O2- doesn’t exist in isolated form [gasp] for the same reason. It’s p4, so adding the first electron causes a subsequent pairing BUT it has a greater Zeff than N, so it can form O-. BUT adding the second electron fills the p’s and that increased e-/e- repulsion overpowers the Zeff of oxygen. Never fear, oxide ions exist in plenty of compounds so we haven’t exactly been lying to you!

- F is weird, it has really strong e-/e- repulsion since the p orbitals are really small in the second level therefore repulsions are high. In subsequent halogen orbitals, it’s not as noticeable.

4. IONIC RADII • Cations—shrink big time since the nucleus is now attracting

fewer electrons • Anions—expand since the nucleus is now attracting MORE

electrons than there are protons AND there is enhanced electron/electron repulsion to boot.

• Isoelectronic—ions containing the same number of electrons - consider the # of protons to determine size. Oxide vs. Fluoride.

Fluoride has one more proton which further attracts the electron cloud, so it is smaller.

Atomic Structure

22

5. ELECTRONEGATIVITY (En)—The ability of an atom IN A MOLECULE [meaning it’s participating in a BOND] to attract shared electrons to itself. Think “tug of war”. Now you know why they teach you such games in elementary school! o Linus Pauling’s scale—Nobel Prize for Chemistry & Peace

o Fluorine is the most En and Francium is the least En o Why is F the most? Highest Zeff and smallest so that the nucleus is closest to the valence

“action”. o Why is Fr the least? Lowest Zeff and largest so that the nucleus is farthest from the

“action”. o We’ll use this concept a great deal in our discussions about bonding since this atomic

trend is only used when atoms form molecules.

Exercise 8 Trends in Ionization Energies The first ionization energy for phosphorus is 1060 kJ/mol, and that for sulfur is 1005 kJ/mol. Why?

Atomic Structure

23

Exercise 9 Ionization Energies Consider atoms with the following electron configurations: a. 1s22s22p6 b. 1s22s22p63s1 c. 1s22s22p63s2 Identify each atom. Which atom has the largest first ionization energy, and which one has the smallest second ionization energy? Explain your choices.

A: Ne; largest IE B: Na

C: Mg; smallest IE2

Exercise 10 Trends in Radii Predict the trend in radius for the following ions: Be2+, Mg2+, Ca2+, and Sr2+.

Be2+ < Mg2+ < Ca2+ < Sr2+

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The Do’s and Don’ts of Teaching Periodic Trends The AP® Chemistry test typically addresses periodic trends in the free response questions. Students are often given a statement such as “In terms of atomic structure, explain why the first ionization energy of selenium is less than that of bromine.” Often these questions are not well answered because students simply state the trend, such as “ionization energy increases as you move left in a period,” and fail to address the forces at work that created the trend in the first place. Your first order of business when teaching this unit is to emphasize that students must address the forces that create the trend and avoid the impulse to use the location on the table as an explanation of the observed property. If needed, chant over and over again, “A trend is an observation, not an explanation!” It is fine to state the trend in their answer, but they must also go further by explaining what causes the observed trend. In addition to discussing forces, train your students to mention BOTH of the atoms or ions in the question when stating their answer. Addressing one and leaving the other implied does not usually earn the point for this type of question. Almost all of the properties that are asked about in exam questions rely on the Coulombic attraction between the outer electrons and the nucleus. Answers to these questions should always include a statement about how this attraction is affected. The concept of Coulomb’s Law should be taught and students should be encouraged to mention it in their explanations. Coulomb’s Law shows that the force of attraction between two oppositely charged particles is directly proportional to the magnitude of the charges and inversely proportional to the distance between those charges.

2attractionFd

( )( )q q+ −

∝

Chemistry is often defined as “the study of matter and energy”. Emphasize the energy of attractions and repulsions throughout this unit of study. Trends in the Periodic Table Justifying all of the trends on the periodic table can be simplified using these two generalizations:

1) Use Zeff to justify trends across a period. 2) Use increased distance (greater value of n) to justify trends down a group.

Atomic radius refers to the distance between the nucleus and the outer edge of the electron cloud. It is influenced by the nuclear pull and the number of energy levels.

Atomic radii decrease as atomic numbers increase in any given period DO DON’T

Teach students that the effective nuclear charge, Zeff, increases the attraction of the nucleus and therefore pulls the electron cloud closer to the nucleus resulting in a smaller atomic radius.

Don’t’ let students get away with simply stating that atomic radii decrease from left to right across a period.

© René McCormick & Mary Payton 2008. All rights reserved.

Atomic radii increase as atomic number increases down a column or group

DO DON’T Teach students that the increased number of energy levels (n) increases the distance over which the nucleus must pull and therefore reduces the attraction for electrons.

Don’t let students get away with simply saying that radii increase down a column.

Teach students that full energy levels provide some shielding between the nucleus and valence electrons.

Don’t let students use shielding for explanations across a period. Only full energy levels, not full sublevels, are of concern in a shielding argument.

Ionization energy refers to the energy needed to remove an electron from a gaseous atom or ion, i.e. an isolated one, not part of a solid, liquid or a molecule. It is always endothermic.

Ionization energy increases as atomic number increases in any given period DO DON’T

Teach students that the effective nuclear charge, Zeff, increases the attraction of the nucleus and therefore holds the electrons more tightly.

Don’t’ let them get away with simply stating that ionization energy increases from left to right across a period.

Teach students the exceptions that occur between groups II and III and V and VI.

Don’t let them think that the trend is unwavering.

1) A drop in IE occurs between groups II and III because the p electrons do not penetrate the nuclear region as greatly as s electrons do and are therefore not as tightly held.

1) Don’t let them state that p electrons are farther away from the nucleus.

2) A drop in IE occurs between groups V and VI because the increased repulsion created by the first pairing of electrons outweighs the increase in Zeff and thus less energy is required to remove the electron.

2) Don’t let them state that the atoms in group V are more stable because they have a half filled sublevel. This is wrong, wrong, wrong!

Ionization energy decreases as atomic number increases down a column or group


Don’t let students get away with simply saying that IE decreases down a column.




Electron affinity is NOT the opposite of ionization energy, but involves the addition of an electron to a gaseous atom or ion, which can be exothermic or endothermic. The exothermic values can be confusing for students since − 500 kJ represents a higher electron affinity than −100 kJ. You can tell students to consider the absolute value of the energy term since the negative sign is simply indicating the direction of energy flow (out of the system). Electronegativity is a property (there are several scales) which measures the attraction of an atom for the pair of outer shell electrons in a covalent bond with another atom. Electronegativity patterns are the same as electron affinity patterns for the same reasons. Both of these properties focus on the attraction that the nucleus has for electrons.

Electronegativity increases as atomic numbers increase in any given period DO DON’T

Teach students that the effective nuclear charge, Zeff, increases the attraction of the nucleus and therefore it strengthens the attraction for the electrons.

Don’t’ let them get away with simply stating that electronegativity increases from left to right across a period.

Electronegativity decreases as atomic number increases down a column or group


Don’t let students get away with simply saying that electronegativity decreases down a column.



Ionic radius is the distance from the nucleus to the outer edge of the electron cloud in a charged ion. The same radii trends apply once you divide the table into the metal and non-metal sections. Within the metal section the positive ionic radii decrease from left to right with only minor changes in the transition metals. Once you get to the nonmetal section and the ions are now negative and larger they will again decrease in radii from left to right. Ionic radii increase going down all columns because of the additional energy levels present (n).

Positive ions are smaller than their respective neutral atoms DO DON’T

Teach students that positive metal ions result from the loss of valence electrons. In many cases this means the farthest electrons are now in a smaller principal energy level (n) than the original neutral atom.

Don’t let students stop at saying that the positive ion is smaller because it lost electrons. The mention of energy levels (n) is essential to earning the point on this type of question.

Teach students to address the ratio of protons to electrons. As electrons are lost the ratio of p+/e− increases and thus the electrons are held closer and with more strength.

Don’t let students neglect this important effect. This is especially useful when comparing ionic radii that do not involve a complete loss of a valence energy level.



Negative ions are smaller than their respective neutral atoms

DO DON’T Teach students that negative nonmetal ions result from the addition of valence electrons. The primary explanation is the change in the proton to electron ratio. As electrons are added the p+/e− ratio decreases and the electrons are not as closely held.

Don’t let students say that the ion is bigger simply because it has more electrons. They must address the p+/e− ratio and the e−/e− repulsions to earn maximum credit.

Teach students that increased electron/electron repulsions also play a role in expanding the electron cloud.

Don’t let students neglect this important effect. Electron repulsions are a powerful force within the atom.

Reactivity depends on whether the element reacts by losing electrons (metals) or gaining electrons (non-metals).

Metals are more reactive as you move down a column DO DON’T

Teach students that because metals react by losing electrons, a loosely held electron will result in a more reactive metal. This is directly tied to ionization energy. With an increased number of energy levels (n) comes increased distance from the nuclear attraction and thus a more loosely held electron available for reacting.

Don’t let students simply say that metals are more reactive at the bottom left corner of the table.

Non-metals are more reactive as you move up a column DO DON’T

Teach students that because nonmetals tend to gain electrons, a strong nuclear attraction will result in a more reactive non-metal. This means that an atom with the highest Zeff

and the least number of energy levels should be the most reactive nonmetal (F) because its nucleus exerts the strongest pull.

Don’t let students simply say that nonmetals are most reactive at the top right corner of the table.

Final thoughts Students often have trouble immediately recognizing the difference between the two species given. Teach them to follow these three steps EVERY time they answer a periodicity question and their scores are sure to increase. 1) Locate both elements on the periodic table and state the principal energy level (n) and the sublevel

containing the valence electrons for each element. 2) Do they have the same or different n values? 3) If same n: argue with Zeff ; if different n: argue with n vs. n distances.

AP* Chemistry CHEMICAL BONDING & MOLECULAR STRUCTURE


WHAT IS A CHEMICAL BOND? It will take us the next two chapters to answer this question! Bonds are attractive forces that hold groups of atoms together and make them function as a unit. [Defined in 25 words or less, but leaves out LOTS of details!] Bonding relates to physical properties such as melting point, hardness and electrical and thermal conductivity as well as solubility characteristics. The system is achieving the lowest possible energy state by bonding. If you think about it, most of the chemical substances you can name or identify are NOT elements. They are compounds. That means being bound requires less energy than existing in the elemental form. It also means that energy was released from the system. This is a HUGE misconception most students have—it takes energy to break a bond, not make a bond! Energy is RELEASED when a bond is formed, therefore, it REQUIRES energy to break a bond. Bond energy—energy required to break the bond TYPES OF CHEMICAL BONDS • ionic bonds—an electrostatic attraction between ions; usually produced by the reaction between a metal and

nonmetal. Cause very high melting points and usually a solid state since the attraction is SO strong that the ions are VERY close together in a crystal formation.

• covalent bonds—electrons are shared by nuclei [careful, sharing is hardly ever 50-50!] • Coulomb’s Law--used to calculate the Energy of an ionic bond.

r is the distance between the ion centers in nanometers [size matters!] J is the energy in Joules Q1 and Q2 are the numerical ion charges; don’t neglect their signs There will be a negative sign on the Energy once calculated—it indicates an attractive force so that the ion pair has lower energy than the separated ions. You can also use Coulomb’s Law to calculate the repulsive forces between like charges. What sign will that calculation have?

VALENCE ELECTRONS • valence electrons--outermost electrons; focus on ns, np and d electrons of transition elements. • Lewis dot structures--(usually main group elements) G.N. Lewis, 1916 • Emphasizes rare gas configurations, s2p6, as a stable state. All rare gasses except He have 8 valence electrons

and follow the octet rule. CHEMICAL BOND FORMATION Level 1—When 2 hydrogen atoms approach each other 2 “bad” E things happen: electron/electron repulsion and proton/proton repulsion. One “good” E thing happens: proton/electron attraction. When the attractive forces offset the repulsive forces, the energy of the two atoms decreases and a bond is formed. Remember, nature is always striving for a LOWER ENERGY STATE. • bond length—the distance between the 2 nuclei where the

energy is minimum between the two nuclei. • energy decrease is small--van der Waals IMFs [another chapter!] • energy decrease is larger--chemical bonds

19 1 22.31 10 J nm Q QEr

− ⎛ ⎞= × • ⎜ ⎟⎝ ⎠

too FAR

too CLOSE

Just right!

getting better

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Chemical Bonding and Molecular Structure 2

Level 2--Orbital Theory − electrons and + nucleus of one atom strongly perturb or change the spatial distribution of the other atom’s valence electrons. A new orbital (wave function) is needed to describe the distribution of the bonding electrons bond orbital • bond orbital--describes the motion of the 2 electrons of opposite spin • lone pair orbital--the orbitals of electrons on a bonded atom that are distorted away from the bond region also

have new descriptions (wave functions) • The new bond orbital is “built” from the atomic orbitals of the two bonded atoms. Looks a lot like the original

BUT, the bond orbital is concentrated in the region between the bonded nuclei. • The energy of the electrons in a bond orbital, where the electrons are attracted by two nuclei, is lower than

their energy in valence electron orbitals where the electrons are attracted to only one nucleus. [ZAPPED!!] • ionic bond--the bonding orbital is strongly displaced toward one nuclei (metal from the left side of table +

nonmetal from right side of the periodic table) • covalent bond--bond orbital is more or less (polar or non-polar) evenly distributed and the electrons are

shared by two nuclei. (elements lie close to one another on the periodic table) • Most chemical bonds are in fact somewhere between purely ionic and purely covalent.

Recall the information you’ve already learned about electronegativity: ELECTRONEGATIVITY (En)—The ability of an atom IN A MOLECULE [meaning it’s participating in a BOND] to attract shared electrons to itself. Think “tug of war”. Now you know why they teach you such games in elementary school!

o Linus Pauling’s scale—Nobel Prize for Chemistry & Peace o Fluorine is the most En and Francium is the least En

o Why is F the most? Highest Zeff and smallest radius so that the nucleus is closest to the “action”.

o Why is Fr the least? Lowest Zeff and largest radius so that the nucleus is farthest from the “action”.

o We’ll use this concept a great deal in our discussions about bonding since this atomic trend is only used when atoms form molecules.

Use the difference in En to determine the type of bond formed.

o ionic--Electronegativity difference > 1.67

o covalent--Electronegativity difference < 1.67

o NONpolar En difference < 0.4


Exercise 1 Relative Bond Polarities Order the following bonds according to polarity: H—H, O—H, Cl—H, S—H, and F—H.

• bond polarity and electronegativity--En (χ) determines polarity since it measures a

nucleus’ attraction or “pull” on the bonded electron pair. En ranges from 0--4.0. When 2 nuclei are the same, the sharing is equal ∴ the bond is described as NONPOLAR (a). When the 2 nuclei are different the electrons are not shared equally, setting up slight +/− poles ∴ POLAR (b). When the electrons are shared

unequally to a greater extent ∴ IONIC (c). • The polarity of a bond can be estimated from Δχ/Σχ. Range is 0 for pure covalent bonds to 1

for completely ionic bonds. IONIC BONDING The final result of ionic bonding is a solid, regular array of cations and anions called a crystal lattice. At right, you can see the energy changes involved in forming LiF from the elements Li and F2 • Enthalpy of dissociation--energy required to decompose an ion pair (from a

lattice) into ions ∴ a measure of the strength of the ionic bond • from Coulomb’s law:

where q+ is the charge on the positive ion and q− is the charge on the negative ion and r is the distance between the ion centers in the crystal lattice.

• energy of attraction depends directly on the magnitude of the charges (higher the charges the greater the attractive energy) and inversely on the distance between them (greater the distance, the smaller the attractive energy).

• the larger the ion the smaller the ΔHdissociation (it’s a distance thing) - ion-ion attractions have a profound effect on melting points and

solubilities. • Water must overcome the ion-ion attractions to dissolve an ionic substance.

Size affects this as does charge HOW?? • The crystal lattice for LiF is shown at the left. • Lattice energy can be represented by a modified form of Coulomb’s Law: k is a

proportionality constant that depends on the structure of the solid and the electron configurations of the ions.

+dissociation q qE H r

−= Δ ∝

1 2Lattice Energy Q Qkr

⎛ ⎞= ⎜ ⎟⎝ ⎠


N H+ N : + H+ →

COVALENT BONDING Most compounds are covalently bonded, especially carbon compounds. We have 3 major bonding theories to discuss. Only one for this chapter though!

• Localized Electron [LE] Bonding Model—assumes that a molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms. Electron pairs are assumed to be localized on a particular atom [lone pairs] or in the space between two atoms [bonding pairs]. 1. Lewis Structures describe the valence electron arrangement 2. Geometry of the molecule is predicted with VSEPR 3. Description of the type of atomic orbitals “blended” by the atoms to share electrons or hold lone

pairs [hybrids—next chapter]. • Number of Bond Pairs: The Octet Rule—“noble is good”

- predict # of bonds by counting the number of unpaired electrons in a Lewis structure - a dash is used to represent a pair of shared electrons, : is used to represent a lone pair

SINGLE AND MULTIPLE BONDS:

• single bond--one pair of electrons shared sigma (σ) bond • MULTIPLE BONDS ARE MOST OFTEN FORMED by C,N,O,P and S ATOMS—say “C-NOPS” • double bond--two pairs of electrons shared one σ bond and one π bond • triple bond--three pairs of electrons shared one σ bond and two π bonds • obviously, combinations of σ and π are stronger than σ alone. Pi bonds are weaker than sigma but never

exist alone • Multiple bonds increase the electron density between two nuclei and therefore decrease the nuclear

repulsions while enhancing the nucleus to electron density attractions—either way, the nuclei move closer together and the bond length is shorter for a double than a single and triple is shortest of all!

COORDINATE COVALENT BONDS: Some atoms such as N and P, tend to share a lone pair with another atom that is short of electrons, leading to the formation of a coordinate covalent bond: These bonds are in all coordination compounds and Lewis Acids/Bases ammonium ion formation: We show that N is sharing the lone PAIR of electrons by drawing an arrow from it to the H+, remember H+ has NO electrons to contribute to the bond. Note that all four bonds are actually identical in length and strength. EXCEPTIONS TO THE OCTET RULE:

• Fewer than eight--H at most only 2 electrons (one bond)! BeH2, only 4 valence electrons around Be (only two bonds)! Boron compounds, only 6 valence electrons (three bonds)! ammonia + boron trifluoride is a classic Lewis A/B reaction.

• Expanded Valence—can only happen if the central element has d-orbitals which means it is from the 3rd period or greater [periods 4, 5, 6…] and can thus be surrounded by more than four valence pairs in certain compounds. The number of bonds depends on the balance between the ability of the nucleus to attract electrons and the repulsion between the pairs.

• odd-electron compounds--A few stable cmpds. contain an odd number of valence electrons and thus cannot obey the octet rule. NO, NO2 , and ClO2.

H

H

H

H

H

H

HUGE concept

CRITICAL stuff!


Drawing Lewis Structures: (VERY USEFUL when predicting molecular shape)

To predict arrangement of atoms within the molecule use the following rules: 1. H is always a terminal atom. ALWAYS connected to only one other

atom!! 2. LOWEST En is central atom in molecule [not just the oddball element] 3. Find the total # of valence electrons by adding up group #’s of the

elements. FOR IONS add for negative and subtract for positive charge. Divide by two to get the number of electron PAIRS.

4. Place one pair of electrons, a σ bond, between each pair of bonded atoms. 5. Subtract from the total the number of bonds you just used. 6. Place lone pairs about each terminal atom (EXCEPT H) to satisfy the octet

rule. Left over pairs are assigned to the central atom. If the central atom is from the 3rd or higher period, it can accommodate more than four electron pairs, up to six pairs.

7. If the central atom is not yet surrounded by four electron pairs, convert one or more terminal atom lone pairs to pi bonds pairs. NOT ALL ELEMENTS FORM pi BONDS!! only C, N, O, P, and S!!

Exercise 6 Writing Lewis Structures Give the Lewis structure for each of the following: a. HF d. CH4 b. N2 e. CF4 c. NH3 f. NO+

Exercise 7 Lewis Structures for Molecules That Violate the Octet Rule I Write the Lewis structure for PCl5.

Exercise 8 Lewis Structures for Molecules That Violate the Octet Rule II Write the Lewis structure for each molecule or ion. a. ClF3 b. XeO3 c. RnCl2 d. BeCl2 e. ICl4

-


RESONANCE STRUCTURES: Experiments show that ozone, O3 has equal bond lengths and equal bond strengths, implying that there is an equal number of bond pairs on each side of the central O atom. When you draw the Lewis structure, that situation is NOT what you draw! So, we draw a resonance structure:

We draw it as having a double bond and a single bond [the dashes are another way of representing lone pairs] BUT since there are equal bond lengths and strengths, they are clearly NOT as pictured above. The bonds are more equivalent to a “bond and ½” in terms of length and strength. We use the double edged arrows to indicate resonance. We also bracket the structures just as we do for polyatomic ions. In an attempt to improve the drawing, we sometimes use a single composite picture. The drawing at right better shows the bond angle. Focus on the central atom in either of the resonance structures above. The central atom is surrounded by three sites of electron density: a lone pair, a single σ bond and a double bond (consisting of one σ bond and one π bond). Three electron-dense sites surrounding a central point in a 3-dimensional space will orient themselves at 120° so that electron/electron repulsions are minimized. Carbonate ion: NOTE: These all 3 need brackets and the charge shown in the upper right corner [like the composite at right] to gain full credit on the AP Exam!!!!

Notice: 1) resonance structures differ only in the assignment of electron pair positions, NEVER atom positions. 2) resonance structures differ in the number of bond pairs between a given pair of atoms Exercise 9 Resonance Structures Describe the electron arrangement in the nitrite anion (NO2

-) using the localized electron model.

O..

O

O..

:

:_

C:_

.. 2/3_

2/3

2/3

2_

..

:

_ ..

O C

O

O

..

....:

:

: :

_

_

..

O

: :O C

O....

..

:

:

__

O C

O

O

..

..

: :

: :_

CO32-

THREE EQUIVALENT STRUCTURES

RESONANCE HYBRID


BOND PROPERTIES • bond order—simply the number of bonding electron pairs shared by two atoms in a molecule.

- 1--only a sigma bond between the 2 bonded atoms - 2--2 shared pairs between two atoms; one sigma and one pi (CO2 and ethylene) - 3--3 shared pairs between two atoms; one sigma and two pi (acetylene and CO and cyanide) - fractional--resonance; ozone 3/2; carbonate 2/3

bond order = # of shared pairs linking X and Y

number of X−Y links

• bond length--distance between the nuclei of two bonded atoms C−N< C−C < C−P The effect of bond order is evident when you compare bonds between the same two atoms

Bond C−O C=O C≡O Bond order 1 2 3 Bond length (pm) 143 122 113

The bond length is reduced by adding multiple bonds since it increases the electron density between the two nuclei. Variations in neighboring parts of a molecule can affect the length of a particular bond as much as 10%

• bond energy—the greater the number of electron pairs between a pair of atoms, the shorter the bond. This implies that atoms are held together more tightly when there are multiple bonds, so there is a relation between bond order and the energy required to separate them.

• bond dissociation energy(D)--gaseous atom; E supplied to break a chemical bond (bond energy for short!) o D is + and breaking bonds is endothermic. The converse is also true.

- D in table 10.4 is + - D is an average with a +/- 10% - Gaseous atoms - What is the connection between bond energy and bond order???

• Calculating reaction energies from bond energies: Bonds in reactants are broken while bonds in products are formed. Energy released is greater than energy absorbed in EXOthermic reactions. The converse is also true.

oreaction = mD (bonds broken) nD(bonds made)HΔ Σ − Σ


ΔHoreaction = reactants(E cost) − products(E payoff)

NOTE THIS IS “BACKWARDS” FROM THE THERMODYNAMICS “BIG MAMMA” EQUATION. We’re back to that misconception if you are confused by this. It takes energy to break bonds NOT make bonds! First we must break the bonds of the reactants [costs energy] then subtract the energy gained by forming new bonds in the products.

Exercise 5 ∆H from Bond Energies Using the bond energies listed in Table 8.4, calculate ∆H° for the reaction of methane with chlorine and fluorine to give Freon-12 (CF2Cl2).

CH4(g) + 2Cl2(g) + 2F2(g) → CF2Cl2(g) + 2HF(g) + 2HCl(g)

∆H = -1194 kJ/mol

FORMAL CHARGE Often, many nonequivalent Lewis structures may be obtained which all follow the rules. Use the idea of formal charge to determine the most favored structure. Physicists tell us that oxidation states of more than +/− two are pure fantasy and that formal charges are much more realistic. • formal charge--The difference between the number of valence electrons on the free element and the number of

electrons assigned to the atom in the molecule.

Atom’s formal charge = group number − [# of lone electrons − 2 (# of bonding electrons)] THE SUM OF THE FORMAL CHARGES MUST EQUAL AN ION’S CHARGE!! • Use formal charges along with the following to determine resonance structure

- Atoms in molecules (or ions) should have formal charges as small as possible—as close to zero as possible [principle of electroneutrality]

- A molecule (or ion) is most stable when any negative formal charge resides on the most electronegative atom.


Example: Draw all possible structures for the sulfate ion. Decide which is the most plausible using formal charges. ** Caution** Although formal charges are considered closer to the atomic charges than the oxidation states, they are still only estimates and should not be taken as the actual atomic charges. Second, using formal charges can often lead to erroneous structures, so tests based on experiments must be used to make the final decisions on the correct description of bonding. Exercise 10 Formal Charges Give possible Lewis structures for XeO3, an explosive compound of xenon. Which Lewis structure or structures are most appropriate according to the formal charges?

Structures with the lower values of formal charge would be most appropriate

MOLECULAR SHAPE—MINIMIZE ELECTRON PAIR REPULSIONS!!! VSEPR--valence shell electron pair repulsion theory • Molecular shape changes with the numbers of σ bonds plus lone pairs about the central atom

σ bonds + lone pairs

on central atom STRUCTURAL PAIR or ELECTRONIC GOMETRY

(NOT to be confused with molecular geometry!)

2 Linear

3 Trigonal planar

4 Tetrahedral (or pyramidal)

5 Trigonal bipyramidal

6 Octahedral THE VSPER MODEL AND MOLECULAR SHAPE

• molecular geometry--the arrangement in space of the atoms bonded to a central atom not necessarily the same as the structural pair geometry lone pairs have a different repulsion since they are experiencing an attraction or “pull” from only one nucleus as opposed to two nuclei. They also take up more space around an atom as you can see on the left.

• Each lone pair or bond pair repels all other lone pairs and bond pairs--they try to avoid each other making as wide an angle as possible. - works well for elements of the s and p-blocks - VSEPR does not apply to transition element compounds (exceptions)


• MOLECULAR SHAPES FOR CENTRAL ATOMS WITH NORMAL

VALENCE: no more than 4 structural pairs if the atom obeys the octet rule. Since no lone pairs are present, the molecular and structural pair [or electronic] geometry is the same. 109.5° bond angle.

- Ignore lone pairs AFTER you’ve determined the angles ∴ only the relative positions of the atoms are important in molecular geometry

- Below is ammonia—3 sigma bonds and one lone pair—when lone pairs are present, the structural pair or electronic geometry are different!!

- “electronic geometry”—tetrahedral while molecular geometry—trigonal pyramidal, 107.5° bond angle

- Water has 2 lone pairs and still obeys the octet rule. Again, the electronic and molecular geometries will be different. The electronic is still tetrahedral since the octet rule is obeyed BUT the molecular geometry is described as “bent” or V-shaped, 104.5° bond angle.

" Now, let’s discuss the warping of the 109.5° bond angle. The lone pairs have more repulsive

force than shared pairs and are “space hogs”. They force the shared pairs to squeeze together.

- To determine the geometry: 1. Sketch the Lewis dot structure [DO NOT SKIP THIS STEP.]

2. Describe the structural pair or electronic geometry 3. Focus on the bond locations (ignore lone pairs) and assign a molecular geometry based on their locations

ANY time lone pairs are present, the

electronic and molecular geometries will be

different.

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• MOLECULAR SHAPES FOR CENTRAL ATOMS WITH

EXPANDED VALENCE—only elements with a principal energy level of 3 or higher can expand their valence and violate the octet rule on the high side. Why? d orbitals are needed for the expansion to a 5th or 6th bonding location—the combination of 1 s and 3 p’s provides the four bonding sites that make up the octet rule.

- seems to be a limit of 3 lone pairs about the central atom - XeF4; There are 2 lone pairs and 4 shared pairs. Two possible

arrangements exist for this situation. - equatorial—shared pairs surround Xe. In both molecular

arrangements the electronic geometry is octahedral with 90° angles. (a) has a molecular geometry known as “see saw”. This puts the lone pairs only 90° apart.

- axial—lone pairs are located “top and bottom” This is preferred for this molecule since the lone pairs are 180° apart which minimizes their repulsion. Lone pairs prefer maximum separation—use this in your determinations! (b at upper right) has a molecular geometry that is more stable—square planar

Which of these arrangements do you predict to be the most stable for I3¯ ? Remember to look at the lone pair-lone pair angles to make your determination. Exercise 11 Prediction of Molecular Structure I Describe the molecular structure of the water molecule.

Two bonding and two non-bonding pairs of electrons Forming a V shape molecule

Exercise 12 Prediction of Molecular Structure II When phosphorus reacts with excess chlorine gas, the compound phosphorus pentachloride (PCl5) is formed. In the gaseous and liquid states, this substance consists of PCl5 molecules, but in the solid state it consists of a 1:1 mixture of PCl4

+ and PCl6- ions.

Predict the geometric structures of PCl5, PCl4+, and PCl6

-.


• RECAP including bond angles for all • structural pairs--σ bond (π bond pairs occupy the same space) pairs about an atom

- 2 @ 180o ∴ linear [and of course, planar]

- 3 @ 120o ∴ trigonal planar " “

" “

- 4 @ 109.5o ∴ tetrahedral NOT SQUARE PLANAR! " “

" “

" “

- 3 @ 120 & 2 @ 90o with each other through central atom ∴ trigonal bipyramidal

- 6 @ 90o ∴ octahedral

The presence of lone pairs alters the six basic MOLECULAR geometries, but the electronic or structural pair geometry remains one of these six basic types.


MOLECULAR POLARITY polar--bonds can be polar while the entire molecule isn’t and vice versa. dipole moment--separation of the charge in a molecule; product of the size of the charge and the distance of separation • align themselves with an electric field (figure b at right) • align with each other as well in the absence of an electric field • water—2 lone pairs establish a strong negative pole • ammonia—has a lone pair which establishes a neg. pole • note that the direction of the “arrow” indicating the dipole moment always points to the negative pole with the cross

hatch on the arrow (looks sort of like we’re trying to make a + sign) is at the positive pole. Logical?

• IF octet rule is obeyed [which it is in both water and ammonia] AND all the surrounding bonds are the same [even if very polar] then the molecule is NONpolar since all the dipole moments cancel each other out.

• carbon dioxide, above is nonpolar since the dipole moments cancel. For those of you in physics, the dipole moments

are vector quantities.

• Methane is a great example. Replace one H with a halogen and it becomes polar. Replace all and it’s nonpolar again!

Draw CH4 , CH3Cl, CH2Cl2, CHCl3, CCl4 Indicate dipole moment(s) where necessary.


Exercise 2 Bond Polarity and Dipole Moment For each of the following molecules, show the direction of the bond polarities and indicate which ones have a dipole moment: HCL, Cl2, SO3 (a planar molecule with the oxygen atoms spaced evenly around the central sulfur atom), CH4 [tetrahedral (see Table 8.2) with the carbon atom at the center], and H2S (V-shaped with the sulfur atom at the point).

Exercise 13 Prediction of Molecular Structure III Because the noble gases have filled s and p valence orbitals, they were not expected to be chemically reactive. In fact, for many years these elements were called inert gases because of this supposed inability to form any compounds. However, in the early 1960s several compounds of krypton, xenon, and radon were synthesized. For example, a team at the Argonne National Laboratory produced the stable colorless compound xenon tetrafluoride (XeF4). Predict its structure and whether it has a dipole moment. Exercise 14 Structures of Molecules with Multiple Bonds Predict the molecular structure of the sulfur dioxide molecule. Is this molecule expected to have a dipole moment?


ONCE MORE WITH FEELING… CHEMICAL BONDS—Forces of attraction that hold groups of atoms together within a molecule or crystal lattice and make them function as a unit.

IIOONNIICC Characteristics of ionic substances usually include: • electrons that are transferred between atoms having high differences in electronegativity (greater than

1.67) • compounds containing a metal and a nonmetal (Remember that metals are located on the left side of the

periodic table and nonmetals are to the right of the “stairs”.) • strong electrostatic attractions between positive and negative ions • formulas given in the simplest ratio of elements (empirical formula; NaCl, MgCl2) • crystalline structures that are solids at room temperature • ions that form a crystal lattice structure as pictured in Figure 1 • compounds that melt at high temperatures • substances that are good conductors of electricity in the molten or dissolved state

Fig. 1

CCOOVVAALLEENNTT Characteristics of covalent substances usually include: • the sharing of electrons between atoms having small differences in electronegativities (less than 1.67) • nonmetals attracted to other nonmetals • formulas that are given in the true ratios of atoms (molecular formulas; C6H12O6) • substances that may exist in any state of matter at room temperature (solid, liquid, or gas) • compounds that melt at low temperatures • substances that are nonconductors of electricity

MMEETTAALLLLIICC Characteristics of metallic substances usually include: • substances that are metals • a “sea” of mobile or delocalized electrons surrounding a positively charged metal center • an attraction between metal ions and surrounding electrons • formulas written as a neutral atom (Mg, Pb) • solids with a crystalline structure at room temperature • a range of melting points—usually depending on the number of valence electrons • substances that are excellent conductors of electricity since the electrons in the “sea” are free to move


Most chemical bonds are in fact somewhere between purely ionic and purely covalent.

DRAW THE DANG LEWIS STRUCTURE when answering bonding multiple choice or free-response questions.

AP* Chemistry COVALENT BONDING: ORBITALS


The localized electron model views a molecule as a collection of atoms bound together by sharing electrons between their atomic orbitals. The arrangement of valence electrons is represented by the Lewis structure and the molecular geometry can be predicted from the VSEPR model. There are 2 problems with this.

1) Using the 2p and the 2s orbitals from carbon in methane would result in 2 different types of bonds when they overlap with the 1s from Hydrogen. [three 2p/1s bonds and one 2s/2p bond] However, experiments show that methane has FOUR IDENTICAL bonds. Uh, oh….we need to modify the model!

2) Since the 3p orbitals occupy the x, y and z-axes, you would expect those overlaps of atomic orbitals to be at bond angles of 90°. Darn those experiments! All 4 angles are 109.5°.

It’s not that the localized electron model is wrong; it’s just that carbon adopts a set of orbitals rather than its “native” 2s & 2p. THIS IS WHY THESE ARE MODELS/THEORIES rather than LAWS!! VALENCE BOND THEORY—an extension of the LE model—it’s all about hybridization! Two atoms form a bond when both of the following conditions occur: 1. There is orbital overlap between two atoms. 2. A maximum of two electrons, of opposite spin, can be present in the overlapping orbitals. • Because of orbital overlap, the pair of electrons is found within a region influenced by both nuclei. Both

electrons are attracted to both atomic nuclei and this leads to bonding. • As the extent of overlap increases, the strength of the bond increases. The electronic energy drops as the

atoms approach each other but, increases when they become too close. This means there is an optimum distance, the observed bond distance, at which the total energy is at a minimum.

• sigma (σ) bond--overlap of two s orbitals or an s and a p orbital or head-to-head p orbitals. Electron density of a sigma bond is greatest along the axis of the bond.

• maximum overlap: forms the strongest possible bond, two atoms are arranged to give the greatest possible orbital overlap. Tricky with p orbitals since they are directional.

• hybrid orbitals-- a blending of atomic orbitals to create orbitals of intermediate energy. Methane: all of the C−H bonds are 109.5° apart while p orbitals are only 90° apart. Pauling explained:

The orbitals on the left are for a carbon atom [no bonding] Once the carbon atom begins to bond with say, H to keep it simple, the atomic orbitals HYBRIDIZE which changes their shape considerably! There’s an energy payoff, else they wouldn’t behave this way! leads to

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SHOW ALL WORK Only do problems through Page 6 #5. Molecular Orbitals are outside of our curriculum

Covalent Bonding: Orbitals

2

Ammonia also has sp3 hybridization even though it has a lone pair. I find it helpful to think of electron pairs as “sites” of electron density that can be occupied by either a lone pair or a shared pair. If there are 4 “sites” then 4 orbitals need to hybridize so use one s and three p’s to make FOUR equivalent orbitals named sppp or sp3 orbitals . [1 s + 3 p = 4 sp3 orbitals]

MULTIPLE BONDING—lowers the number of hybridizing orbitals since Unhybridized orbitals are necessary to form the pi bonds Pi (π) bonds--come from the sideways overlap of p atomic orbitals; the region above and below the internuclear axis. NEVER occur without a sigma bond first! • may form only if unhybridized p orbitals remain on the bonded atoms • occur when sp or sp2 hybridization is present on central atom NOT sp3 hybridization. • Carbon often does this with N, O, P, and S since they have p orbitals (remember CNOPS?) • This is the formation of an sp2 set of orbitals [3 orbitals formed, 3 sites, 3 letters!]. This molecule would

contain a double bond like ethene. The molecule reserves a set of p’s to form the π bond.

leads to

The set of p’s that are UNhybridized are not pictured here at left, they are hovering above and below this very page.

A different view, without the hydrogens, centering on the C atoms shows the unhybridized p orbitals that are making the sideways overlap necessary to create the double (π) bond. Here’s the whole mess altogether: Carbon #1 Carbon #2

OVERLAPPING UNhybridized p-orbitals that form pi bond.


3

• This is the formation of an sp set of orbitals [2 orbitals formed, 2 sites, 2 letters!]. This molecule

would contain a triple bond like ethyne or the double-double arrangement in carbon dioxide. leads to

The molecule reserves TWO sets of UNhybridized p’s to form the 2 π -bonds. At right, is a picture of the 2 unhybridized p’s on the C atom that is about to make a triple bond. The one labeled at the top is in the plane of this page, the other plain p is in a plane perpendicular to this page. (No typo—just appropriate use of homonyms; plain vs. plane!) Look at the CO2 Lewis diagram. The carbon has 2 sites of electron density, each occupied by a double bond, and is therefore sp [2 sites, 2 letters] hybridized while the oxygens have 3 sites [2 lone pairs and a double bond]. The oxygen’s have sp2 hybridization [3 sites, 3 letters].

This should help:

HYBRIDIZATION # OF HYBRID ORBITALS

GEOMETRY EXAMPLE

sp 2 Linear

sp2 3 Trigonal planar

sp3 4 Tetrahedral

dsp3 5 Trigonal bipyramidal

d2sp3 6 Octahedral


4

Check out Benzene The sigma bond formations The pi bond formations

Draw the Lewis structure for benzene:


5

Exercise 1 The Localized Electron Model I Describe the bonding in the ammonia molecule using the localized electron model.

tetrahedral, sp3 hybrid

Exercise 2 The Localized Electron Model II Describe the bonding in the N2 molecule.

linear, sp hybrid

Exercise 3 The Localized Electron Model III Describe the bonding in the triiodide ion (I3

-).

trigonal bipyramidal arrangement, e- pair geometry, linear molecular geometry

central iodine is dsp3 hybridized

Exercise 9 The Localized Electron Model IV How is the xenon atom in XeF4 hybridized?

d2sp3 hybridized

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6

Exercise 5 The Localized Electron Model V For each of the following molecules or ions, predict the hybridization of each atom, and describe the molecular structure. a. CO b. BF4

- c. XeF2

A: linear, sp hybridized

B: tetrahedral, sp3 hybridizied C: trigonal bipyramidal e- pair,

Xe dsp3, linear THE MOLECULAR ORBITAL MODEL

Though the molecular orbital model will not be covered on the AP exam, I feel that students should be exposed to a little of this theory for several reasons.

1. Electrons are not always localized as in the VSEPR theory; therefore resonance must be added and explained as best possible.

2. Molecules containing unpaired electrons are not easily dealt with using the localized model.

3. Magnetism is easily described for molecules using the MO theory. (Oxygen is paramagnetic which is unexplained by the localized electron model.) 4. Bond energies are not easily related using the localized model.

TERMS TO KNOW: Bonding molecular orbital - an orbital lower in energy than the atomic orbitals from which it is composed.(favors formation of molecule) Antibonding molecular orbital - an orbital higher in energy than the atomic orbitals from which it is composed. (favors separated atoms) represented by a * [The diagram at the top of the next page uses A for antibonding and B for bonding—I prefer the *, it has more personality!] Bond order - the difference between the number of bonding electrons and the number of antibonding electrons divided by two. Indicates bond strength. Homonuclear diatomic molecules - those composed of two identical atoms. Heteronuclear diatomic molecules - those composed of two different atoms Paramagnetism - causes the substance to be drawn into a magnetic field; associated with unpaired electrons. Diamagnetic - causes the substance to be repelled by the magnetic field; associated with paired electrons.

General Energy Level Sequence for Filling Orbitals Using the MO Theory

σ1s2 σ1s2* σ2s2 σ2s2* ( π2px2 π2py

2)σ2p2(π2px2* π2py

2*) σ2p2*

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7

Let’s start simple. 2 Hydrogen atoms.

If we assume that the molecular orbitals can be constructed from the atomic orbitals, the quantum mechanical equations result in two molecular orbitals

MO1 = 1sA + 1sB and MO2 = 1sA − 1sB

Where 1sA and 1sB represent the 1s orbitals from the two separated hydrogen atoms.

This is actually a simple model to follow. Look at the diagram on the right, each H entered with its lone 1s electron. As they approach each other, their atomic orbitals [two of them] blend to form molecular orbitals [two of them—no magic here]. One MO is of high energy and one MO is of low energy. Which will the electrons choose? The LOW, of course! The electrons occupy the lower energy level and thus a bond is formed. The diagram at left uses the symbols we want to use. Try this again with He:

Since 4 electrons are involved, the first 2 get to be lazy and go to the low E state, the other 2 must occupy the higher energy state and thus cancel out the bond, therefore He2

DOES NOT EXIST!! Now bond order can be redefined in this theory: Bond order = number of bonding electrons – number of antibonding electrons 2 If the bond order is zero no bond!


8

Shall we predict if Li2 is possible? Li has its valence electrons in the 2s sublevel. Yes! It may also exist. What is its bond order? Can Be2 exist? Things get slightly more complicated when we leave Be and move to 2p…

General Energy Level Sequence for Filling Orbitals Using the MO Theory

σ1s2 σ1s2* σ2s2 σ2s2* ( π2px2 π2py

2)σ2p2(π2px2* π2py

2*) σ2p2* The filling order for p’s is pi, pi, sigma all bonding, followed by pi, pi, sigma all antibonding.

HUND’S RULE AND PAULI EXCULSION PRINCIPLES APPLY!!

Try to predict the configuration of B2


9

One of the most useful parts of this model is its ability to accurately predict para- and diamagnetism as well as bond order. This device is used to test the paramagnetism of homonuclear samples. When the electromagnet is on, a paramagnetic substance is drawn down into it and appears heavier on the balance. B2 is paramagnetic! That means that the pi orbitals are of LOWER energy than the sigma’s and Hund’s rule demands that the 2 electrons fill the 2 bonding pi orbitals singly first before paring.

Will C2 exist? Will it be para- or diamagnetic?

Exercise: Write the appropriate energy diagram using the MO theory for the nitrogen molecule. Find the bond order for the molecule and indicate whether this substance is paramagnetic or diamagnetic.


10

If you use the usual models to examine the paramagnetism of oxygen, you’d say it was diamagnetic. The truth is that it is paramagnetic. If you pour liquid oxygen into the space between the poles of a strong horseshoe magnet, it says there until it boils away in the warm room!


11

Exercise 6 For the species O2, O2

+, O2-, give the electron configuration and the bond order for each. Which has the

strongest bond? Exercise 7 Use the molecular orbital model to predict the bond order and magnetism of each of the following molecules. a) Ne2 b) P2

This model also works in heteronuclear molecules.

Exercise 8 Use the MO Model to predict the magnetism and bond order of the NO+ and CN- ions.

AP* Chemistry INTERMOLECULAR FORCES LIQUIDS & SOLIDS


Now it is time to consider the forces that condense matter. The forces that hold one molecule to another molecule are referred to as intermolecular forces (IMFs). These forces arise from unequal distribution of the electrons in the molecule and the electrostatic attraction between oppositely charged portions of molecules. We briefly visited the IMFs earlier when discussing the nonideal behavior of gases. These forces cause changes of state by causing changes among the molecules NOT within them. Physical properties such as melting points, boiling points, vapor pressures, etc. can be attributed to the

strength of the intermolecular attractions present between molecules. It works like this: the lower the boiling point (or vapor pressure or melting point), the weaker the intermolecular attractions; the higher the boiling point, the stronger the intermolecular attractions. For example, gasoline evaporates much more quickly than water. Therefore, the intermolecular attractive forces that hold one gasoline molecule to another are much weaker than the forces of attraction that hold one water molecule to another water molecule. In fact, water molecules are held together by the strongest of the intermolecular attractive forces, hydrogen bonds. Hydrogen bonds are not true bonds—they are just forces of attraction that exist between a hydrogen atom on one molecule and the unshared electron pair on fluorine, oxygen or nitrogen atoms of a neighboring molecule. The strands of DNA that make up our genetic code are held together by this type of intermolecular attraction. THE TYPES OF INTERMOLECULAR FORCES IN ORDER OF DECREASING STRENGTH: � Dipole-dipole—the force of attraction that enables two polar molecules to

attract one another. Polar molecules are those which have an uneven charge distribution since their dipole moments do not cancel. Compounds exhibiting this type of IMF have higher melting and boiling points than those exhibiting weaker IMFs.

o Hydrochloric acid molecules are held to each other by this type of force. HCl—the chlorine pulls the electrons in the bond with greater force than hydrogen so the molecule is polar in terms of electron distribution. Two neighboring HCl molecules will align their oppositely charged ends and attract one another.

Intermolecular Forces, Liquids & Solids 2

H “bond” bonded H o Hydrogen bonding—the force of attraction between the

hydrogen atom of one molecule and an unshared electron pair on F, O, or N of a neighboring molecule (a special case of dipole-dipole). This is the strongest IMF. Never confuse hydrogen bonding with a bonded hydrogen. The unique physical properties of water are due to the fact that it exhibits hydrogen bonding. As a result of these attractions, water has a high boiling point, high specific heat, and many other unusual properties.

o WHY is there such variation in the boiling point among the covalent hydrides of groups IV through VII? One would expect that BP would increase with increasing molecular mass [since the more electrons in a molecule, the more polarizable the cloud {more about that in the next section}, the stronger the IMFs, therefore the more E needed to overcome those increased attractions and vaporize, thus the higher the boiling point. That’s how it is supposed to work!]. Hydrogen bonding, that’s why!

o TWO reasons: both enhance the IMF we refer to as hydrogen bonding. 1. The lighter hydrides have the highest En values which lead to especially polar H�X

bonds. Increased polarity means increased attraction which makes for a stronger attraction.

2. The small size of each dipole allows for a closer approach of the dipoles, further strengthening the attractions. Remember, attractive forces dissipate with increased distance. � Ion-induced dipole—the force of attraction between a charged ion and a

nonpolar molecule. The ion greatly perturbs the electron cloud of the nonpolar molecule and polarizes it transforming it into a temporary dipole which enhances the ion’s attraction for it.


� Dipole-induced dipole—the force of attraction between a polar molecule and a

nonpolar molecule. The polar molecule induces a temporary dipole in the nonpolar molecule. Larger molecules are more polarizable than smaller molecules since they contain more electrons. Larger molecules are more likely to form induced dipoles.

� Induced dipole-induced dipole or London dispersion force—the force

of attraction between two non polar molecules due to the fact that they can form temporary dipoles. Nonpolar molecules have no natural attraction for each other. This IMF is known by both names! Without these forces, we could not liquefy covalent gases or solidify covalent liquids.

o These forces are a function of the number of electrons in a given molecule and how

tightly those electrons are held. o Let us assume that the molecule involved is nonpolar.

A good example would be H2. Pretend that the molecule is all alone in the universe. If that were the case, the electrons in the molecule would be perfectly symmetrical. However, the molecule is not really alone. It is surrounded by other molecules that are constantly colliding with it. When these collisions occur, the electron cloud around the molecule is distorted. This produces a momentary induced dipole within the molecule. The amount of distortion of the electron cloud is referred to as polarizability. Since the molecule now has a positive side and a negative side, it can be attracted to other molecules. You’ll want to write about polarizability when explaining these concepts.

o Since all molecules have electrons, all molecules have these forces. These forces range from 5—40 kJ/mol. The strength of this force increases as the number of electrons increases due to increasing polarizability.

o To better understand the induced dipole-induced dipole IMF, examine the halogens. Halogens exist as diatomic molecules at room temperature and atmospheric conditions. F2 and Cl2 are gases, Br2 is a liquid and I2 is a solid. Why?

� All of these molecules are completely nonpolar and according to theory, not attracted to each other, so one might predict they would all be gases at room temperature.


� Bromine exists as a liquid at room temperature simply because there is a greater attractive force between its molecules than between those of fluorine or chlorine. Why?

� Bromine is larger than fluorine or chlorine; it has more electrons and is thus more polarizable. Electrons are in constant motion so it is reasonable that they may occasionally “pile up” on one side of the molecule making a temporary negative pole on that end, leaving a temporary positive pole on the other end.

� This sets off a chain-reaction of sorts and this temporary dipole induces a dipole in its neighbors which induces a dipole in its neighbors and so on.

� Iodine is a solid since it is larger still, has even more electrons, is thus even more polarizable and the attractive forces are thus even greater. Note the two spheres representing I2 the diagram at right. Each iodine atom experiences a nonpolar covalent bond within the molecule. Be very clear that the IMF is between molecules of iodine, NOT atoms of iodine!

Clear as mud? This flow chart is worth studying. NOTE that dipole-dipole (and its special case of Hydrogen bonding), dipole-induced dipole, induced dipole-induced dipole (a.k.a. London dispersion forces) are collectively called van der Waals forces. It’s a simple conspiracy designed to keep you confused—much like the scoring scheme in tennis!

AP* Chemistry PROPERTIES OF SOLUTIONS


IMPORTANT TERMS

� Solution—a homogeneous mixture of two or more substances in a single phase. Does not have to involve liquids. Air is a solution of nitrogen, oxygen, carbon dioxide etc.; solder is a solution of lead, tin etc.

� solute—component in lesser concentration;

dissolvee � solvent—component in greater

concentration; dissolver

� solubility—maximum amount of material that will dissolve in a given amount of solvent at a given temperature to produce a stable solution. In other words, the solution is saturated. Study the solubility rules!!

� molar solubility—the number of moles of solute that dissolves in exactly 1.0 L of solvent, expressed in units of molarity, M or the use of square brackets.

� Saturated solution— a solution containing the maximum

amount of solute that will dissolve under a given set of pressure and temperature conditions. Saturated solutions are at dynamic equilibrium with any excess undissolved solute present. Solute particles dissolve and recrystallize at equal rates.

� Unsaturated solution— a solution containing less than the maximum amount of solute that will dissolve under a given set of conditions. (more solute can dissolve)

� Supersaturated solution—oxymoron—a solution that has

been prepared at an elevated temperature and then slowly cooled. It contains more than the usual maximum amount of

solution dissolved. A supersaturated solution is very unstable and agitation (stirring, pouring, etc.) or the addition of a “seed crystal’ will cause all excess solute to crystallize out of solution leaving the remaining solvent saturated. (rock candy is made this way as are those liquid hand warmer packets of sodium acetate solution, pictured at left, folks use hunting, at football games, skiing, etc.)

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Properties of Solutions 2

� miscible—When two or more liquids mix (ex. Water and food coloring)

� immiscible—When two or more liquids DON’T mix.—they usually layer if allowed to set for a while. (ex. Water and oil)

Concentration Units

� Molarity (M) = # of moles of solute per liter of solution; IS temperature dependent. The liquid solvent can expand and contract with changes in temperature. Thus, not a constant ratio of solute:solvent particles. Most molar solutions are made at 25� C so this point is subtle and picky, but important none the less!!

solutionofliterssolute of molesM �

� Mass percent (weight percent) = percent by mass of the solute in the solution

grams of soluteMass percent 100%grams of solution

� �

� Mole fraction (�) = ratio of the number of moles of a given component to the total number

of moles of present.

Mole fractiona = �a Mole Fraction a aa

a b

nn n

��

� Molality (m) = # of moles of solute per kilogram of solvent; NOT temperature dependent.

Represents a ratio of solute:solvent molecules at all times.

moles of solute

kilograms of solventm �


Exercise 1 Various Methods for Describing Solution Composition A solution is prepared by mixing 1.00g ethanol (C2H5OH) with 100.0g water to give a final volume of 101 mL. Calculate the molarity, mass percent, mole fraction, and molality of ethanol in this solution.

molarity = 0.215 M mass percent = 0.990% C2H5OH

mole fraction = 0.00389 molality = 0.217 m

Exercise 2 Calculating Various Methods of Solution Composition from the Molarity The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent, molality, and normality of the sulfuric acid.

mass percent = 29.9% H2SO4 molality = 4.35 m

normality is 7.50 N

THE SOLUTION PROCESS

� Energies involved in solution formation

When a solute is dissolved in a solvent, the attractive forces between solute and solvent particles are great enough to overcome the attractive forces within the pure solvent and within the pure solute. The solute becomes solvated (usually by dipole—dipole or ion—dipole forces). When the solvent is water the solute is hydrated.

� Substances with similar types of intermolecular forces dissolve in each other. “Like

dissolves like.” � Polar solvents dissolve polar or ionic solutes. � Nonpolar solvents dissolve nonpolar solutes.


� Water dissolves many salts because the stronger ion—dipole attractions water forms with

the ions of the salt are very similar to the strong attractions between the ions themselves. The same salts are insoluble in hexane (C6H14) because the weaker London dispersion forces their ions could form with this nonpolar solvent are much weaker than the attraction between the ions of the salt.

� Oil does not dissolve in water. Oil is immiscible in water due to the fact that any weak dipole-induced dipole attractions that form between oil and water cannot overcome the stronger dipole-dipole hydrogen bonding that water molecules have for each other.

� Solubilities of alcohols in water: As the hydrocarbon portion of the alcohol increases in

length, the alcohol becomes less soluble. (More of the molecule is nonpolar; the dipole moment is diminished.)

� Solubilities of alcohol in nonpolar solvents: As the hydrocarbon portion of the alcohol increases in length, the alcohol becomes more soluble in a nonpolar solvent such as hexane.

� Enthalpy of solution (�Hsoln) = the enthalpy change associated with the formation of a

solution (just the sum of all of the steps involved!)

3 steps: �Hsoln = �H1 + �H2 + �H3

� �Hsoln can be positive (endothermic) or negative (exothermic).

Step 1 (�H1) � Separating the solute into individual components of the solute (expanding the solute).

This requires E be added to the system, therefore endothermic. The magnitude of the value is high in ionic and polar solutes, low in nonpolar solutes.

�Hsolute = �Hlattice energy

Step 2(�H2) � Overcoming IMFs in solvent to make room for the solute (expanding the solvent).

Requires that E be added to the system, therefore endothermic. The magnitude of the value is high in polar solvents, low in nonpolar solvents.


Step 3 (�H3) � Interaction of solute and solvent to form the solution. Energy must be released here, else

the solution would never form since nature always tends toward a lower energy state, therefore exothermic. The magnitude of this value is high in polar solute—polar solvent interactions, low in other types of interactions.

�H2 + �H3 = enthalpy of hydration (�Hhyd) Enthalpy of hydration is more negative for small ions and highly charged ions.

� Some heats of solution are positive (endothermic). The reason that the solute dissolves is that the solution process greatly increases the entropy (disorder) which overrides the cost of the small positive heat of solution. This makes the process spontaneous. The solution process involves two factors; the change in heat and the change in entropy, and the relative magnitude of these two factors determine whether a solute dissolves in a solvent.

� Hot and cold packs:

These often consist of a heavy outer pouch containing water and a thin inner pouch containing a salt. A squeeze on the outer pouch breaks the inner pouch and the salt dissolves. Some hot packs use anhydrous CaCl2 (�Hsoln = �82.8 kJ/mol) whereas many cold packs use NH4NO3 (�Hsoln = +25.7 kJ/mol). We discussed other hot packs earlier that function on the principle of a supersaturated solution crystallizing, releasing the heat of crystallization. Other hot packs contain iron filings and the process of rusting is sped up thus, producing heat energy.

Exercise 3 Differentiating Solvent Properties Decide whether liquid hexane (C6H14) or liquid methanol (CH3OH) is the more appropriate solvent for the substances grease (C20H42) and potassium iodide (KI).

hexane � grease methanol � KI

Explain why foranswers below.


FACTORS AFFECTING SOLUBILITY

� Molecular Structure:

� Fat soluble vitamins, (A,D,E,K) –nonpolar (can be stored in fatty body tissues) � Water soluble vitamins, (B&C) –polar (are not stored, wash away and must be consumed

regularly) � Hydrophobic— water fearing (nonpolar) � Hydrophilic – water loving (polar)

� The Effect of Increasing Pressure

� The solubility of a gas increases with increasing pressure. Increasing pressure has very little effect on the solubility of liquids and solids. (carbonated beverages must be bottled at high pressures to ensure a high concentration of carbon dioxide in the liquid)

� Henry’s Law— the amount of a gas dissolved in a solution is directly proportional to the

pressure of the gas above the solution. Henry’s Law is obeyed best for dilute solutions of gases that don’t dissociate or react with the solvent.

Henry’s Law: C = kP

P = partial pressure of the gaseous solute above the solution k = constant (depends on the solution) C = concentration of dissolved gas


Exercise 4 Calculations Using Henry’s Law A certain soft drink is bottled so that a bottle at 25°C contains CO2 gas at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO2 in the atmosphere is 4.0 × 10�4 atm, calculate the equilibrium concentrations of CO2 in the soda both before and after the bottle is opened. The Henry’s law constant for CO2 in aqueous solution is 0.031 mol/L • atm at 25°C.

before = 0.16 mol/L after = 1.2 × 10—5 mol/L

� The Effect of Increasing Temperature

� The amount of solute that will dissolve usually increases with increasing temperature

since most solution formation is endothermic. Solubility generally increases with temperature if the solution process is endothermic (�Hsoln > 0). Solubility generally decreases with temperature if the solution process is exothermic (�Hsoln < 0). Potassium hydroxide, sodium hydroxide and sodium sulfate are three compounds that become less soluble as the temperature rises. This can be explained by LeChatelier’s Principle.

� Remember, the dissolving of a solid occurs more rapidly with an increase in temperature, but the amount of solid may increase or decrease with an increase in temperature. It is very difficult to predict what this solubility may be—experimental evidence is the only sure way.

� The solubility of a gas in water always

decreases with increasing temperature.

� There are all types of environmental issues involved with the solubility of a gas at higher temperatures. Thermal pollution – water being returned to its natural source at a higher ambient temperature has killed much wildlife as less oxygen is dissolved in the water. Boiler scale is another problem. This is where a coating builds up on the walls of containers such as industrial boilers and pipes causing inefficient heat transfer and blockage.

(1) Copyright © 2008 by René McCormick. All rights reserved. (2) AP® is a registered trademark of the College Entrance Examination Board and was not involved in the production of and does not endorse this product. Permission is granted for individual classroom teachers to reproduce the activity sheets and illustrations for their own classroom use. Any other type of reproduction of these materials is strictly prohibited.

Very Last Minute

AP® Net Ionic Equation Attack Strategies

These probably won’t get you a 15 on Question 4, BUT they WILL HELP YOU BEAT THE NATIONAL AVERAGE. The national average on this question is usually between 7—8. If you can consistently score 11—12, that puts you WAY ahead of the game! First, the minimum knowledge required to survive this question… Solubility rules:

1. Big Mamma: All nitrates are soluble.

2. Big Daddy: All IA metals and ammonium salts are soluble. 3. Halides: All are soluble except silver, mercury or lead. 4. Strong acids: hydrochloric, hydrobromic, hydroiodic, nitric, perchloric, sulfuric—

WRITE THESE DISSOCIATED except concentrated sulfuric, it really is 97% H2SO4 and 3% water in the jug, so water is way outnumbered and the molecules don’t dissociate completely. If carbonic acid is formed as a product CO2 + H2O.

5. Strong bases: hydroxides [and oxides] of IA and IIA* metals—write these bases dissociated. WRITE ALL WEAK ACIDS AND BASES AS MOLECULES—be on the look out for BF3 and its cousins BCl3, etc. They are classic Lewis acids and when reacting with ammonia (a classic weak Lewis base), the product is F3BNH3 (just smash everything together) since nitrogen donated its unshared electron pair to boron in an act of extreme generosity and formed a coordinate covalent bond. Lewis Acids Accept an electron pair. * The “little guys” in the IIA’s have solubility issues, write Be and Mg UNdissociated—calcium can go either way, the big guys are soluble. HF is not a strong acid since it’s the little guy as well in the halogen series. The little guys make a stronger bond with OH− or H+ and do not dissociate as much in water. Also remember that the IA metals are named the alkali metals and the IIA’s are the alkaline earth metals. What does “alkaline” mean? BASIC, so put them in water as metals, they dissolve and you make OH−. Put IA metals in water and KABOOM! KABOOM = formation of explosive hydrogen gas, H2.

TURN THIS PUPPY OVER FOR THE ATTACK STRATEGIES!

Attack Strategies

Before trying to figure out the “answers”, scan the words on question 4 (a) thru (c) and do the following: [don’t write any products until you’ve done all SIX things!]

1. Cross out the word nitrate any time it appears on the page. 2. Circle any word that implies solid or gas. (powdered, turnings, chunk, vapor, etc.) 3. Cross out any IA metal that you see UNLESS it is associated with a circled solid or gas

word. 4. Underline halides then ask yourself if silver, mercury or lead is present—if not you can

cross the halide off as well such as with hydrochloric acid. The H+ is the reacting species. (Bring the halide back as a reacting ion IF you need to oxidize something halide-1 → halogen2.)

5. Circle “burned in air” or “combines with oxygen” or anything that implies combustion

and celebrate! 6. Circle the word concentrated. Get very excited if you see excess concentrated. It means

you have entered the land of complex coordinated ions (excess is not necessary, but often appears). Sounds scary, but VERY easy. LOSE THE FEAR!

Now, focus on the reactions you just marked. WRITE THE REACTANT SETS FOR ALL THREE. Spend 3 minutes writing products using the solubility rules and strong acid-base guidelines listed on the other side of this page. To get the easy three points involved with step SIX above do the following:

Write the reactants. On the product side, open a set of brackets [ ]. Put the metal ion in the brackets first then open a set of parentheses [M ion( )]. Next put a subscript on the parentheses that is twice the charge on the metal—I’m not proud of this, but it will earn credit. For a +2 metal it becomes [M2+( )4]. Finally, plop the ligand inside the parentheses and do the math to get the charge. If the ligand is ammonia or water, the ligand is neutral, so our example carries a +2 overall charge, [M2+( )4]2+, if the ligand is hydroxide or a halide, which are both negative one, then our example becomes [M2+(OH)4]2−. Other ligands are possible, like SCN−, the thiocyanate ion and other polyatomic ions you should recognize.

Additional knowledge that contributes to survival: • metal oxides + water → bases (ask yourself strong or weak? Dissociate the strong) • nonmetal oxides + water → acids (ask yourself strong or weak? Dissociate the strong) • metal carbonate heated → CO2 + metal oxide • Redox, “acidified”? H+ is a reactant and water is a product. • React a metal with oxygen → metal oxide • React a nonmetal with oxygen [combustion] → make oxides of the nonmetal(s), NOT always

CO2 & H2O!

AP* Chemistry Chemical Foundations · 2019-07-09 · AP* Chemistry Chemical Foundations Chemistry:...

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Transcript of AP* Chemistry Chemical Foundations · 2019-07-09 · AP* Chemistry Chemical Foundations Chemistry:...