AP Chemistry Compounds 2015-09-14 www.njctl.org Slide 3 / 163 - - PDF document

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AP Chemistry

Compounds

2015-09-14 www.njctl.org

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Table of Contents: Compounds Pt. A

· Ionic Compounds · Chemical Formulas

Click on the topic to go to that section

· Covalent Compounds · Metals & Alloys

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Chemical Formulas

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Slide 5 / 163 The Mole

Recall that 1 mole is defined as 6.022 x 1023 units of a given substance. 1 mol of electrons = 6.022 x 1023 electrons 1 mol of H2O molecules = 6.022 x 1023 molecules of water 1 mol of NaCl formula units = 6.022 x 1023 formula units NaCl 1 mol of K atoms = 6.022 x 1023 atoms of K

Slide 6 / 163 The Mole

Within 1 mole of a compound, there are often differing moles of each element In 1 mole of aluminum nitrate, Al(NO3)3

= 1 mol of Al3+ ions

= 3 mol of NO3- ions = 3 mol of N atoms = 9 mol of O atoms

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1 How many moles of oxygen atoms are present in 2.0 moles of aluminum nitrate Al(NO3)3?

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2 How many moles of oxygen atoms are in 2.0 moles of sodium sulfate, Na2SO4?

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3 How many hydroxide ions are present in 1.2 x 1024 formula units of magnesium hydroxide: Mg(OH)2?

Slide 10 / 163 Molar Mass and Volume

Recall that the mass of 1 mol of a substance is called the molar mass and is measured in g/mol. This can be found on the periodic table. Molar mass of CaCl2 = 110 g/mol Molar Mass of Ag = 108 g/mol Recall also that 1 mol of any gaseous substance will occupy 22.4 L of space at STP. 1 mol of Ar(g) = 22.4 L @STP 1 mol of H2(g) = 22.4 L @STP

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4 What is the volume occupied @STP by 88 grams of carbon dioxide?

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5 How many mL of methane gas (CH4) are present @STP in a 100 gram sample of a gas that is 32% methane by mass?

Natural gas (methane) pipeline.

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6 Which of the following contains the most atoms of H? A 2 grams of H2 gas B 16 grams of methane (CH4) C 22.4 L of H2 gas D 9 grams of water (H2O)

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7 How many moles of fluoride ions (F-) are present in a 79 gram sample of SnF2? A 0.5 moles B 78.5 moles C 1 mole D 1.5 moles

Slide 15 / 163 Chemical Formulas

A chemical formula provides the ratio of atoms or moles of each element in a compound. H2O = 2 atoms H or 2 mol H 1 atom O 1 mol O Al(NO3)3 = 1 Al3+ ion

• r 1 mol Al3+ ions

3 NO3- ions 3 mol NO3- ions

Slide 16 / 163 Empirical and Molecular Formulas

An empirical formula provides the simplest whole number ratio of atoms or moles of each element in a compound. Examples: H2O, NaCl, C3H5O A molecular formula represents the actual number of atoms or moles of each element in a compound. Examples: H2O, C3H5O, C6H12O6

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Calculating an Empirical Formula

To find an empirical formula:

• 1. Determine the moles of each element within the compound

then..... Compound "X" consists of 1.2 g C, 0.2 g H, and 1.6 g O = 0.1 mol C, 0.2 mol H, and 0.1 mol O

• 2. Find the whole number ratio of these moles by dividing by

smallest mole value! 0.1 mol C = 1 C 0.2 mol H = 2 H 0.1 mol O = 1 O 0.1 mol 0.1 mol 0.1 mol Empirical formula = CH2O

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Calculating a Molecular Formula

Determining the molecular formula of a compound is easy once the empirical formula and the molecular weight of the compound are known.

• 1. Determine the ratio of the molecular weight to the empirical

formula weight. MW of Compound "X" = 60 u Empirical formula weight of CH2O = 30 u Ratio = 60/30 = 2/1. The molecule is twice as heavy as the empirical formula.

• 2. Multiply each subscript of empirical formula by the ratio

determined in step 1 CH2O x 2 = C2H4O2 = Molecular Formula

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8

Students type their answers here

Practice

Given the following data, calculate the empirical formula of phosphine gas. Phosphine gas is created by reacting solid phosphorus with H2(g). Mass of P(s) initial Mass of P(s) unreacted 1.45 g 1.03 g Mass of H2(g) initial Mass of H2(g) unreacted 0.041 g 0.000 g

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9

Students type their answers herePractice

Black iron oxide (aka magnetite) is used as a contrast agent in MRI scans of human soft tissue. To determine the empirical formula, a student reacted solid iron with O2(g). Fe(s) reacted Mass of iron oxide obtained. 3.05 g 4.22 g What is the empirical formula?

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10

Students type their answers herePractice

Butane gas can be produced when solid carbon is reacted with hydrogen gas. If 0.45 grams of carbon were found to react with 1.05 L of H2 gas @STP, what is the molecular formula of butane given it has a molar mass of 58 g/mol.

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11 Which of the following is NOT an empirical formula? A Fe2O3 B H2NNH2 C CH3OH D CH3CH2Cl

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12 A compound used in airbags degrades into sodium and nitrogen gas (N2) when ignited. If 3.36 L of N2(g) was produced @STP from an initial mass of the compound of 6.50 grams, what is the empirical formula? A Na2N3 B Na3N C NaN3 D NaN

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13 A compound containing carbon, hydrogen, and chlorine is 14.1% carbon by mass, 83.5% Cl, with the rest being

• hydrogen. What is the empirical formula?

A C2H5Cl B CH2Cl2 C C2H6Cl D CH3Cl

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14 Hydrazine is a component of rocket fuel. It consists of 87.5% N with the rest being hydrogen by mass. If the molecular weight of the compound is 32 grams/mol, what is the molecular formula? A NH2 B NH3 C N2H6 D N2H4

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If a compound is pure, it will always consist of the same composition no matter where the sample was taken or the size of the sample.

Law of Definite Composition

Example: calcium carbonate If it's calcium carbonate, it's guaranteed to be 40% calcium, 48%

• xygen, and 12% carbon by mass.

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Sample Location Size Analysis Composition 1 Eastern Pennsylvania 50.0 g 20.0 g Ca 24.0 g O 6.0 g C 40% calcium 48% oxygen 12% carbon 2 Wyoming 250.0 g 100.0 g Ca 120.0 g O 30.0 g C 40 % calcium 48% oxygen 12% carbon Example: calcium carbonate

Law of Definite Composition Slide 28 / 163

Some substances are not pure and do not obey the law of definite

• composition. These are called mixtures.

Law of Definite Composition

Sample Size Sample location % mass composition 1 500.0 g Atlantic Ocean 85.3% O 10.7 % H 1.6% Na 2.4 % Cl 2 330.0 g Indian Ocean 79.5 % O 10.0 % H 4.2 % Na 6.3 % Cl Sea water

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15

Students type their answers here

Water is known to be 88.9%

• xygen and 11.1 % hydrogen by
• mass. How many grams of oxygen

would be present in a 400 gram sample of pure water?

Practice

The law of definite composition can be used mathematically to see how much of a given substance is present in a sample.

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16

Students type their answers here

If a sample of water was found to contain 34.1 grams of oxygen, how many grams of hydrogen and water would be present?

Practice

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17 Hydrogen peroxide is known to be 94.1 % oxygen by mass with the rest being hydrogen. How many grams of hydrogen would be present in a pure 230. gram sample

• f hydrogen peroxide?

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18 Calcium oxide is 71.4% calcium by mass with the rest being oxygen. How many grams of calcium would be present if a sample of calcium oxide sample was found to contain 12.3 grams of oxygen?

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19 Calcium carbonate is known to be 40% Ca, 48% O, and 12% carbon by mass. When a 200 gram sample of what is thought to be pure calcium carbonate is decomposed, 18 grams of carbon are found in the sample. Is this substance pure? Yes No

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This law allows us to classify the different types of matter

MATTER Does the material obey the Law of Definite composition? Yes No Pure Substance Mixture Compounds Elements Can the material be broken down into different elements with distinct properties? Yes No

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20 Two samples of a material are taken and the composition

• f each sample is given below. Is this material a pure

substance? Yes No

Sample A Sample B 45 % Cu, 12% Si, 43 % O 34% Cu, 19% Si, 47 % O Answer

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21 A certain material is found to vary in composition by

• mass. What kind of matter is this?

A Element B Compound C Mixture D Pure Substance

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22 Which of the following is TRUE regarding a pure compound? A It will not obey the law of definite composition B It can be broken down into different elements C The amounts of each element by mass in the compound will not vary D It must contain the same equal mass % of each element in the compound

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Metals & Alloys

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Slide 39 / 163 Introduction to Chemical Bonds

There are three basic types of bonds: metallic, ionic, and covalent. A chemical bond is an attraction between atoms to form a compound that contain two or more atoms. The bond forms due to the electrostatic attraction between opposite charges.

Slide 40 / 163 Sea of Electrons

Recall that metallic atoms lose electrons easily. In a metallic compound, valence electrons become delocalized. Instead of orbiting their specific nuclei, the electrons flow freely in a "sea of electrons" between the positively charged nuclei.

Slide 41 / 163 Properties of Metals

The free flow of electrons and strong electrostatic force between metallic atoms give metals some unique properties: · They are good conductors of heat and electricity · They have typically have high melting points · They are shiny, malleable and ductile · They are readily alloyed (able to form mixtures with other elements)

click here for a tutorial on properties of metals

Slide 42 / 163 Conduction

Metals are good conductors of both heat and electricity because the electrons are delocalized and relatively free to move.

Slide 43 / 163 Melting Point

Metals have a wide range of melting points but due to high Coulombic attraction between the sea of electrons and the metallic nuclei, these melting points are typically high. Metal Melting Point (oC) Aluminum 660 Copper 1084 Lead 327.5 Magnesium 650 Phosphorus 44 Potassium 63.3 Platium 1770 Silver 961 By comparison:

Compound/Type MP (oC) CH4/molecule

• 182

H2O/molecule diamond/covalent network solid 3550 NaCl/ionic compound 801 RbBr/ionic compound 682

Slide 44 / 163 Structural Properties

Metals are malleable (able to be flatten into sheets) and ductile (able to be pulled into wires) because deforming the solid does not change the environment surrounding the metallic nuclei. The low electronegativity of these atoms allow the electrons to move in response to the change. Metals are shiny because the free flowing electrons are excited by interaction with photons, causing the electrons to vibrate and reflect the light.

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23 Electrons are stationary in a metallic compound. True False

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24 Copper is often used to make wires. This is because.... (you may select more than one response) A it is ductile B it has a high electronegativity C it is conductive D its is malleable

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25 Most metals are ___________ at room temperature. A gas B liquid C molten D solid

Slide 48 / 163 Alloys

An alloy is a mixture of two or more metals, or a metal and another element. There exists three main variants of alloys: · Heterogeneous alloys · Intermetallic alloys · Homogeneous/ Metal solution alloys

Slide 49 / 163 Heterogeneous Alloys

Heterogeneous alloys lack a regular crystal structure throughout the solid. Solder (50% Pb and 50% Sn), used in fusing together different metal work pieces, is an example of a heterogeneous alloy.

Slide 50 / 163 Intermetallic Alloys

Intermetallic alloys: · Have definite proportions of constituent elements · Have a crystal structure that is different from any of the constituent metals · Result in solids with properties often different from constituent metals

Examples: MgZn2 Na5Zn21 Cu3Zn

Slide 51 / 163 Homogeneous Alloys

There are two types of homogeneous alloys: substitutional alloys and interstitial alloys. Homogeneous alloys are alloys with the same crystal lattice structure as one of the constituent elements.

Slide 52 / 163 Substitutional Alloys

Substitutional alloys consist of atoms of similar sizes. The atoms of the base metal are replaced by atoms of the added element in the matrix. Examples: · Brass - copper and zinc · Bronze - copper and tin

Slide 53 / 163 Interstitial Alloys

Alloys containing more than 2 elements can be a combination of these types of alloys. Example: Stainless steel - iron, carbon, nickel and chromium Interstitial alloys occur when the added element is much smaller than the base metal. The smaller atoms occupy spaces between the base metal atoms in the crystal matrix. Example: Steel - iron and carbon

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26 Sterling silver is composed of silver and copper. What type of alloy is this? A interstitial alloy B substitutional alloy C substitutional-institial alloy D it's not an alloy, its an ionic compound

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27 An alloy is formed using a transition metal and carbon. It retains the crystalline structure of the transition metal. What type of alloy is created? A Hetergeneous B Homogeneous substitional C Homogenous interstitial D Intermetallic

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28 A homogenous alloy is formed by 3 elements. Which of the following categories of alloy might in fall into? A interstitial B substitutional C interstitial/substitutional D any of the above

Slide 57 / 163 Alloys

Metals must be molten in order to be alloyed. Although this requires a large energy input, the resulting alloy has enhanced properties compared to its base metal(s). Alloys are created to increase strength and reduce malleability (ex. steel and brass) or resist corrosion (ex. stainless steel and sterling silver). Note that the electrical and thermal conductivity of alloys is usually lower than that of its base metal.

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29 In general, alloys enhance the properties of metals. Which of the following is a true statement? A Alloys have increased malleability B Alloys have increased strength C Alloys have increase conductivity D Alloys have increased ductility

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30 Which of the following would have the highest melting point? A steel B carbon C iron D they have the same melting point

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Ionic Compounds

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simple cubic unit cell body-centered cubic unit cell face-centered

• rthorhombic unit

cell

Ionic Compounds

Ionic compounds form crystal lattice structures of repeating patterns called unit cells. Unit cells are form in an arrangement that maximizes attractive forces between ions and minimizes repulsion. Some examples:

click here to review ionic formulas and naming

Slide 62 / 163 Properties of Ionic Compounds

The strong electrostatic attraction between charges and their crystalline structure gives ionic compounds some unique properties: · Conductive only when molten or dissolved in water (aqueous) · High melting points · Brittle

click here for a tutorial on properties of ionic compounds

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31 Most ionic compounds are ___________ at room temperature. A gas B liquid C molten D solid

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32 The brittleness of ionic compounds can be attributed to... A the electrostatic attraction between the ions B the strength of the ionic bonds C the repeating pattern of its structure D its ability to conduct electricity

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33 The high melting point of ionic compounds can be attributed to... A the electrostatic attraction between the ions B their lack of malleability C the repeating pattern of its structure D its ability to conduct electricity

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Once a positive and negative ion are formed, they will be attracted to each other via the electrostatic force.

Ionic Bonding

r2 F = k q1 q2

Neutral atom 7 valence electrons High Electronegativity

+ Cation

• Anion

Neutral atom 1 valence electron Low Electronegativity

Slide 68 / 163 Ionic Bond Formation

Formation of LiF(s) from it's elements in their standard state. Li(s) + 1/2F2(g) --> LiF(s)

Event Reaction Energy Change Sublimation of Li(s) Li(s) --> Li(g) +180 kJ/mol Ionization of Li(g) Li(g) --> Li+(g) + e- +520 kJ/mol Breaking of existing F-F bond 1/2F2(g) --> F(g) +157 kJ/mol Ionization of F(g) F(g) + e- --> F-(g)

• 328 kJ/mol

Bond formation Li+(g) + F-(g) --> LiF(s)

• 1036 kJ/mol

Overall energy change = -505 kJ/mol Notice that breaking bonds is endothermic (requires energy input) while making bonds is exothermic (releases energy) Ionic bonding occurs in a series of steps, most of which require energy but occur because they are coupled to the highly exothermic formation of the bond.

Slide 69 / 163 Ionic Bond Formation

The energy released when the gaseous ions combine is known as the Lattice Energy Hf represents the overall energy change of the process. The thermochemical steps for ionic bond formation are often summarized in a Born-Haber Cycle Diagram

Slide 70 / 163 Lattice Energy

The higher the charges, the greater the Coulombic attraction and the higher the lattice energy.

Substance Charges Lattice Energy NaF(s) +1 and -1

• 923 kJ/mol

MgO(s) +2 and -2

• 3791 kJ/mol

The smaller the ionic radii, the greater the coulombic attraction and the higher the lattice energy

Substance Ionic radii Lattice Energy NaF(s) F- = 117 pm

• 923 kJ/mol

NaCl(s) Cl- = 167 pm

• 786 kJ/mol

The magnitude of the lattice energy is influenced by the charge and size of the ions involved.

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34 Which of the following would have the highest lattice energy? A BeO B MgS C MgCl2 D MgI2

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35 Which of the following BEST explains why the lattice energy of CaS is lower than that of MgO? A CaO has lower ionic charges than MgO B The calcium ion has more shielding than the magnesium ion C The calcium ion has a smaller nuclear charge than magnesium ion D CaO has higher ionic charges than MgO

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36 Which of the following BEST explains why the lattice energy of MgF2 is lower than that of MgO? A The oxide ion is smaller than the flouride ion B The charge of the cation is higher in MgO C The charge density of the anion is less in MgF2 D The charge density of the anion is less in MgO

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37 Rank the following from lowest to highest melting point. A I < II < III B I < III < II C II < I < III D III < II < I

• I. NaBr
• II. LiBr
• III. LiF

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Covalent Compounds

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Slide 76 / 163 Types of Covalent Compounds

There are two types of compounds created by the covalent bonding of atoms: Molecules - smaller compounds of a one or more elements bonded together, such as water (H2O) or oxygen gas (O2) Covalent networks - larger compounds consisting of repeating elemental or molecular units all covalently bonded together, such as diamond (Cn) or quartz [(SiO2)n]

Slide 77 / 163 Properties of Covalent Network Compounds

Like ionic and metallic substances, covalent network solids are giant molecules with many strong electrostatic attractions. This gives them some unique properties: · Non-conductive - because they are made of non-metallic atoms, covalent network solids conduct little or no electricity · Hardness - because all atoms in the structure are bonded together, these materials tend to be very hard · High melting points - covalent bonds are strong and covalent networks have many bonds giving them high melting points

Slide 78 / 163 Diamond vs. Graphite

Diamond and graphite are examples of covalent network solids composed of a single element, carbon. They are allotropes - different molecules made of the same element. In graphite only 3 covalent bonds are formed, creating 2-dimensional sheets of carbon atoms. The fourth electron wanders freely over the surface

• f the sheet, allowing graphite to be somewhat

conductive. Graphite is soft because the 2-D sheets are held together by weak intermolecular forces; very little force is required to break these interactions. Carbon atoms have 4 valence electrons. In diamonds, 4 covalent bonds are formed creating a very strong 3-dimensional shape.

Slide 79 / 163 Silicon

Silicon is a covalent network solid and a semiconductor. Semiconductors allow us to control the flow of electricity through a circuit and provide the basis for modern electronics. Silicon form a 3-dimensional crystalline structure similar to diamond. It is a good thermal conductor and its ability to conduct electricity increases with temperature.

Slide 80 / 163 Doping

Impurities are introduced to silicon to alter its conductive behavior; this is referred to as doping. When an element with more valence electrons, such as phosphorus or arsenic, is added to silicon the additional electrons increase the semiconductor's electron carrier concentration. This makes the silicon an n-type (negative charge carrying) semiconductor. When an element with less valence electrons, such as boron or aluminum, is added to silicon the element accepts electrons from the silicon. This makes the silicon a p-type (positive charge carrying) semiconductor.

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38 Diamonds are harder than graphite because... A they are made from different elements B they form different numbers of bonds C they are different types of compounds D they have different bond lengths

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39 Covalent network solids do not conduct electricity. True False

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40 What effect would added gallium to silicon have? A reduce its conductivity B make it an n-type semiconductor C make it a p-type semiconductor D it would have no effect

Slide 84 / 163 Properties of Molecular Compounds

Since these substances are made up of lots of small molecules they behave differently than covalent network compounds. · Non-conductive - because they are made of non-metallic atoms molecular compounds are excellent insulators · Low melting points - molecules are small and held together by weak intermolecular forces (not bonds), these are easy to break giving molecular solids low melting and boiling points.

weak inter-molecular forces between molecules

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41 Butter melts on a hot day. What type of compound is it? A metallic B ionic C covalent network D molecular

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42 You are given a substance that has a high melting point and does not conduct electricity, even when you put it in

• water. What is it?

A a metal B an ionic compound C a covalent network D a molecule

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43 An unknown white powder is found to melt at 186oC. Identify the compound. A sugar B table salt C sodium bicarbonate D pewter

Slide 88 / 163 Chemical Bonds

How ionic or covalent a bond is depends on the difference in

• electronegativity. The smaller the difference, the more likely electrons

are "shared" and the bond is considered covalent, the greater the difference, the more likely electrons have been transferred and the atoms are ionized resulting in an ionic bond. Li Be B C N O F

Electronegativity 1.0 1.6 2.0 2.5 3.0 3.5 4.0 Bond Li-F Be-F B-F C-F N-F O-O F-F Electronegativity 3 2.4 2.0 1.5 1 0.5 0

Increasing Covalent Character

Slide 89 / 163 Chemical Bonds

While bond character (between ionic and covalent) is a spectrum, we can make a few simplifications... · Ionic bonds occur when the difference in electronegativity between two atoms is 1.7 or greater. Na ---- F electronegativity = 3 · If the difference of electronegativity is less than 1.7, it is a covalent

• bond. Neither atom

takes electrons from the other; they share

• electrons. This type of bonding

typically takes place between two non- metals. H ---- Cl electronegativity = 1.1

Click here to review naming covalent compounds

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Slide 91 / 163 Slide 92 / 163 Slide 93 / 163 Covalent Bond Formation

Consider F2 [Ne] __ __ __ __ __ __ __ __ [Ne] + + e- e- Shared pair of electrons provides both fluorine atoms with a full valence shell (electrons are of opposite spin to minimize repulsions) Click here to view an interactive The sharing of electrons allows atoms to lower their potential energy by achieving a complete valence shell.

Slide 94 / 163 Covalent Bond Formation

Non-polar covalent bond

H-H

There is no difference in electronegativity so the electrons are shared equally. Polar covalent bond

H-Cl partially -

partially +

There is a significant difference in electronegativity between the atoms (3.2 - 2.1) so the bond is polar, meaning opposite charges develop across the bond The electronegativity difference between the atoms involved determines how equally the electrons are shared.

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47 Which of the following would be considered a polar covalent bond? A H-H B H-S C Cl-Cl D C-S E C-O

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48 Which of the following elements, if bonded to S would produce the most polar covalent bond? A H B P C Cl D F E C

Slide 97 / 163 Covalent Bond Strength

Effect of Atomic Radii Smaller atoms result in smaller distances between charges thereby increasing the Coulombic attractions Bond Atomic Radii Bond Enthalpy H-H H = 53 pm 436 kJ/mol Cl-Cl Cl = 79 pm 243 kJ/mol The strength of a covalent bond is influenced by the radii of the atoms, the polarity of the bond, and by the # of electron pairs being shared.

Slide 98 / 163 Covalent Bond Strength

Effect of Polarity The more polar the bond, the stronger the Coulombic attractions Bond EN difference Bond Enthalpy H-Cl 0.9 431 kJ/mol H-S 0.5 344 kJ/mol

Note: The sizes are not a constant in this comparison, however, the projected enthalpy of an H-Cl bond would be the average of an H-H and a Cl-Cl bond or 340 kJ/mol. The observed enthalpy is much higher and is explained by the polarity of the bond.

Slide 99 / 163 Covalent Bond Strength

Effect of Multiple Shared Pairs Atoms often share more than one pair of electrons to realize a full valence shell. As we will learn later, these pairs do not all form the same kind of bonds but the net effect is to increase the Coulombic attractions between the nuclei. Bond # of shared pairs Bond Enthalpy O-O 1 142 kJ/mol O=O 2 498 kJ/mol

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49 As the number of bonds between a pair of

atoms increases, the strength of the bond between the atoms:

A increases

B decreases

C remains unchanged

D

varies, depending on the atoms

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50 Which of the following bonds would be expected to have the smallest bond enthalpy? A F-F B Cl-Cl C C-C D C-O

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51 Which of the following bonds would be expected to have the highest bond enthalpy? A F-F B O-H C C-H D C-O

Slide 103 / 163 Covalent Bond Length

A high bond enthalpy is an indication of a strong coulombic attraction between nuclei, thereby indicating a small bond length between nuclei. Bond Enthalpy Length H-H 436 kJ/mol 74 pm H-Cl 431 kJ/mol 127 pm C-C 347 kJ/mol 154 pm C=C 611 kJ/mol 121 pm

Note: The C=C is longer than the H-H bond despite having the higher bond enthalpy due to the increased radii of the C atoms over the H atoms in the H-H bond.

The length of a covalent bond is influenced by enthalpy of the bond.

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52 Which of the following would be expected to have the longest bond length? A H-O B H-S C H-Cl D H-C

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53 All else being equal, the more polar the bond, the shorter the bond length. True False

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54 As the number of bonds between a pair of atoms

increases, the distance between the atoms:

A increases

B decreases

C

remains unchanged

D

varies, depending on the atoms

Slide 107 / 163 Lewis Structures

Guidelines for writing Lewis structures Guideline One: Determine the ordering of atoms in the molecule A proper Lewis structure distributes the valence electrons in the molecule so each atom has a full valence shell. Typically, the least electronegative atom is the central atom. CCl4 Cl C Cl SO2 O S O But not always... often it's the less abundant atom H2O H O H NH3 H N H Cl Cl H

Slide 108 / 163 Lewis Structures

Guidelines for writing Lewis structures Guideline Two: Determine the number of valence electrons in the molecule NH3 = 8 CCl3H = 26 If the molecules is an ion, one must either subtract or add electrons to the valence electron count. NO3- = 23 +1 = 24 NH4+ = 9-1 = 8

Slide 109 / 163 Lewis Structures

Guidelines for writing Lewis structures Guideline Three: Form a single bond (2 shared electrons) between all elements and then distribute electrons such that all atoms have a full valence shell, saving the central atom for last. Example: H2O (8 ve) Example: CO2 (16 ve) H - O - H O - C - O Notice H needs only 2 electrons for a full valence shell. Notice that C does not have a full valence shell and therefore adjustments will need to be made to this structure

Slide 110 / 163 Lewis Structures: Octet Rule

H = 2 Be = 4 B = 6 In addition, elements in period 3 or below can have expanded

• ctets or more than 8 valence electrons.

The "Octet Rule" refers to the fact that a full valence shell for most elements is a full outer s and p orbital or 8 electrons. Some elements do not follow this as shown below.

Slide 111 / 163 Lewis Structures

Guidelines for writing Lewis structures Guideline Four: If an atom is short of an octet, additional electrons must be shared between the nuclei forming pi bonds. O - C - O O C O Pi bonds Note: Pi bonds are formed from valence electrons in "p"

• rbitals.

Slide 112 / 163 Lewis Structures

Guidelines for writing Lewis structures Guideline Five: If all atoms have a full valence shell but valence electrons remain, they are to be added to the central atom in pairs. S F F F F 34 valence electrons Extra pair of un-bonded electrons is added to central atom.

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55

Students type their answers here

Diatomic Elements

Seven elements in the periodic table are always diatomic. In their elemental state, they are always seen as two atoms covalently bonded together. H2 , O2, N2 , Cl2 , Br2 , I2 , F2

H

O N Cl Br I F Which of these diatomic elements contains pi bonds?

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I

CH 4 Practice

Elements Bonds & Electrons

Cl H C

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I

N2 Practice

Elements Bonds & Electrons

Cl H C O N

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I

CO Practice

Elements Bonds & Electrons

Cl H C O N

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I

SO 42- Practice

Elements Bonds & Electrons

Cl H C O N S

For Ions

+ -

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I

H3O+ Practice

Elements Bonds & Electrons

Cl H C O N S

For Ions

+ -

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I

SF 6 Practice

Elements Bonds & Electrons

Cl H C O N S F

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I

XeF 4 Practice

Elements Bonds & Electrons

Cl H C O N S F Xe

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56 Which of the following elements is not diatomic in its elemental state? A Br2 B S2 C I2 D O2

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57 Which of the following is the correct Lewis Structure for

H2O?

A B C D

H O H H H O H H O H H O Slide 123 / 163

58 Which of the following is the correct Lewis Structure for

PH3?

A B C D

H P H H H P H H H H P H H P H Slide 124 / 163

59

A B C

H H H H H H C C C C H H H H H H

Which of the following is the correct Lewis Structure for C2H6?

C C H H H H H H C C H H H H H H

D

Slide 125 / 163

60 Which of the following molecules would have 10 valence electrons in the Lewis structure? A NH4+ B CN- C H2O D NO2-

Slide 126 / 163

61 How many valence electrons can be used in the Lewis structure for NO+?

Slide 127 / 163

62 Which of the following molecules has a central atom with an expanded octet? A SO2 B SCl2 C PF3 D XeF2

Slide 128 / 163

63 Which of the following molecules would require pi bonds in the Lewis structure? A I only B II only C III only D I, II, and III

• I. NO3-
• II. CO32-
• III. HCN

Slide 129 / 163

64 How many unbounded pairs of electrons are on the central atom in ClO3-?

Slide 130 / 163

65 Which of the following molecules would have a Lewis structure most similar to CO2? A SO2 B CS2 C NO2- D CO32-

Slide 131 / 163

66 Below is a skeleton for the Lewis structure for alphaketoglutarate, a Kreb's cycle intermediate. After finishing the lewis structure, how many pi bonds are needed to complete the structure?

C - C - C - C - C H H H H O O O O O

Slide 132 / 163

67 Which of the following would contain the largest number

• f pi bonds?

A CH4 B CO32- C C2H2 D SF6

Slide 133 / 163

68 Which of the following is the correct Lewis structure for the ammonium ion? A B C D

N H H H H + N H H H H N H H H H N H H H

Slide 134 / 163 Resonance Structures

N O O O One pi bond is needed but could be formed from electrons shared by any of three oxygens. Resonance structures N O O O N O O O N O O O When pi bonds can be formed in more than one location, the electrons are thought to be shared across all of the possible

• locations. This is shown by writing resonance structures.

Slide 135 / 163 Resonance Structures

In essence, the pi bond electrons are shared across all of the bonds in which we find resonance.

N O O O N O O O N O O O

EQUALS N O O O Pi bond electrons shared across all three bonds. The bonds involved in resonance are equivalent in strength and in length.

Slide 136 / 163

I

SO 3 Practice

Elements Bonds & Electrons

Cl H C O N S

Slide 137 / 163

I

NO 3- Practice

Elements Bonds & Electrons

Cl H C O N S

For Ions

+ -

Slide 138 / 163

69 Which of the following show the correct Lewis structure for O3? A B C D Both A & C O O O O O O O O O

Slide 139 / 163

70 Which of the following molecules demonstrate resonance structures? A I only B II only C III only D I and II only

• I. NO2-
• II. CH3COO- (both O attached to C)
• III. CH3CH2OH

Slide 140 / 163

71 How many resonance structures would be needed to represent SO3?

Slide 141 / 163

72 All bonds that demonstrate resonance are equal in length but not in strength. True False

Slide 142 / 163

73 The C-O bonds in the carbonate ion (CO32-) would consist of … A 3 single bonds B 2 single bonds of longer length and 1 double bond

• f shorter length

C 3 double bonds D 3 bonds equal in length but shorter than a single bond

Slide 143 / 163

74 Which of the following require no resonance structures to represent? A NO+ B SO2 C CH3COOH (both O attached to C) D NO3-

Slide 144 / 163 Formal Charge

The formal charge tells us how the electrons are distributed within a

• molecule. For example, depending on how the electrons are shared,

some atoms may have more electrons than others resulting in a semi- charged state for that atom. O P O O O FC for P: 5 - 4= +1 (count each bond as one) FC for each O: 6 -7= -1 (count each bond as one)

Note: The charges must add to the charge of the molecule. So for PO4

3-

1 P atom x +1 = +1 + 4 O atoms x -1 = -4 +1 + -4 = -3

Formal Charge = # of valence electrons - # of electrons atom possesses within the lewis structure.

Slide 145 / 163 Formal Charge

The best Lewis structure will have the formal charge = 0 on each

• atom. However, if the molecule carries a charge, the more

electronegative atoms should carry a charge as they have the greater attraction for electrons. Each bond is counted as one in a formal charge calculation as each atom forming part of the bond contributes just one electron to that bond. [ O - H ]-1 FC on O = 6-7 = -1 FC on H = 1-1 = 0

O H

The oxygen is more electronegative so it makes sense that it carries the negative charge.

Slide 146 / 163

Example: Below are two possible Lewis structures for the phosphate ion, PO43-. Which Lewis structure is considered to more closely represent the actual molecule based on formal charge calculations?

O P O O O O P O O O

Structure 2 is superior because the formal charges on each atom = 0 whereas in structure 1, the P carries a +1 charge and each oxygen carries a -1 charge Structure 1 Structure 2 (exhibits resonance)

slide for answer

Formal Charge

Slide 147 / 163

75 What is the formal charge on the S atom in the molecule depicted below?

Slide 148 / 163

76 What is the formal charge on the O atom in the molecule depicted below?

Slide 149 / 163

77 What is the formal charge on the indicated N atom in the molecule below?

N N N Note: When multiple resonance structures can be written, the structure with the formal charges closest to zero is most stable.

Slide 150 / 163 Bond Order

N O O O Bond order of N-O bonds = 4/3 = 1.33 N N Bond order of N-N bond = 3/1 = 3 The bond order refers to the number of bonds between two atoms in a molecule. It is calculated by adding up the bonds attached to the atom divided by the number of atoms attached to that atom. The higher the bond order, the stronger and shorter the bond.

Slide 151 / 163

78 Which of the following contains bonds of the lowest order? A N2 B SO2 C SO3 D CF4

Slide 152 / 163

79 Which of the following would have a bond order of 1.5? A CO32- B NO2- C CO2 D CS2

Slide 153 / 163

80 Which of the following is true regarding bond order? A The higher the bond order the longer and weaker the bond B The higher the bond order the longer and stronger the bond C The higher the bond order the shorter and stronger the bond D The higher the bond order the shorter and weaker the bond

Slide 154 / 163

81 Which of the following carbon molecules would have the shortest C-O bond lengths? A CO2 B CO32- C CH3OH D CO

Slide 155 / 163 Hybridization

Carbon is known in nature to form compounds in which it must form 4 bonds. However, it's electron configuration suggests it could only share 2 electrons resulting in just 2 bonds. [Ne] __ ___ ___ ___ 2s 2p If the s and p orbitals were hybridized, four degenerate orbitals would be formed each with an electron that could be shared. [Ne] __ __ __ __ (sp3 hybrid orbitals) In order to explain observations in molecular bonding, it has been proposed that atoms will hybridize s and p orbitals to create new orbitals of equal energy which are then involved in bonding.

Slide 156 / 163 Hybridization

N H H H 4 orbitals required (3 for shared pairs, 1 for unshared pair) Requires "s" and all 3 "p" orbitals for hybridization. [Ne] __ __ __ __ N = unbonded pair form bonds with hydrogen sp3 hybridized The number of p orbitals that must be hybridized with the s

• rbital depends on the number of orbitals needed for all

shared and unshared pairs of electrons.

Slide 157 / 163 Hybridization, Sigma, and Pi Bonds

N O O O Here, there are only 3 sigma bonds so

• nly 2 "p" orbitals are needed to

hybridize with the "s" orbital so this N atom is sp2 hybridized. H - C C - H Here, each carbon has only 2 sigma bonds attached to each, so only 1 "p"

• rbital is needed to hybridize with the

"s" orbital so each C is sp hybridized. Bonds made from hybridized orbitals are called sigma bonds while those of un-hybridized p orbitals are called pi bonds.

Slide 158 / 163

82 Which of the following would the central atom be sp hybridized? A I only B II only C III only D I and II only

• I. CO2
• II. HCN
• III. PH3

Slide 159 / 163

83 Which of the following molecules would have an sp2 hybridized atom? A C2H2 B CO32- C BeCl2 D CO2

Slide 160 / 163

84 Which of the following is correct regarding the number of sigma and pi bonds in the molecule below? (Note: only the skeleton is written - you must finish the Lewis structure) A 4 sigma 5 pi B 7 sigma 2 pi C 6 sigma 3 pi D 9 sigma 1 pi

C - C - N - H O O O H H H

Slide 161 / 163

85 What kind of hybridizations are found on the C atoms in the molecule below? (Note: only the skeleton is written - you must finish the Lewis structure) A sp and sp3 B sp2 and sp3 C sp and sp2 D sp, sp2, and sp3

C - C - N - H O O O H H H

Slide 162 / 163

86 What is the hybridization of the oxygen atom in water? A sp B sp2 C sp3 D It is not hybridized

Slide 163 / 163