Intro to VSEPR Geometry and Bonding February 28, 2018

Intro to VSEPR Geometry and Bonding February 28, 2018
Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

DO NOW (4 MIN) February 28, 2018 Learning Target: I can draw the Lewis Dot structures and determine the molecular and electronic geometries of molecules based on VSEPR theory Do Now: (on your DO Now Sheet) Compare and contrast the following molecules based on shape: NH NH3 NH2- Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

REVIEW: DO NOW (4 MIN) NH4+ NH3 NH2- February 28, 2018

Lewis Dot Structures February 28, 2018 Lewis structures tell us which atoms are bonded together, but we will now explore the geometric shapes of these molecules. - Shows how atoms are connected - Shows where electron pairs are located - Shows whether bonds are single, double, or triple BUT, only 2 dimensional Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Video: VSEPR February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Ch. 9 Molecular Geometry & Bonding Theories
February 28, 2018 Geometry: Overall shape is determined by bond angles. Bond angles are determined by the VSEPR theory. Electrons repel & will try to get as far away from each other as possible Nonbonded electron pairs take up more space than bonded electrons. We will first explore the molecular form ABn. This represents one central atom bonded to 1-6 other atoms. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Ch. 9 Molecular Geometry & Bonding Theories
February 28, 2018 First you must determine the # of electron domains on the central atom. An electron domain is a region of electrons that are either bonded or non-bonded (lone pairs). A double or triple bond only counts as one domain. TRY IT: How many electron domains are around each central atom: CO2 CO32- CH4 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Electron Domain Geometry
February 28, 2018 The arrangement of electron domains about the central atom of an ABn molecule is its electron-domain geometry. There are five different electron-domain geometries: linear --(2 electron domains) trigonal planar --(3 domains) tetrahedral --(4 domains) trigonal bipyramidal --(5 domains) octahedral --(6 domains) We can imagine all the electron domains as taking up space Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

February 28, 2018 You gotta know the angles! Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

February 28, 2018 For example… :O=C=O: There are 2 electron domains on carbon…Its shape must therefore be linear. H–O –H There are 4 electron domains on oxygen….Its shape is based on the tetrahedral. TRY IT: What is the electron geometry for: CO32- CH4 H2O NH3 .. .. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

H-O-H Oxygen has 2 bonding and 2 nonbonding domains
Molecular Geometry February 28, 2018 **In reality, electrons are not visible. We can only really see atoms. So we need to understand what the molecular geometry is too.** The molecular geometry is the arrangement of the atoms in space To determine the shape of a molecule we will distinguish between bonding pairs and lone pairs. Count the # of bonding domains vs. nonbonding domains. H-O-H Oxygen has 2 bonding and 2 nonbonding domains With this information, we can determine the molecular geometry…bent .. .. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Molecular Geometry February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Molecular Geometry February 28, 2018 TRY IT: Identify the electronic and molecular geometries: NH3 XeF5 NO3- PCl5 H2O Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Molecular Geometry—Most Common Shapes
February 28, 2018 The most common shapes we deal with are as follows: Tetrahedral, pyramidal, bent, linear, and trigonal planar. (It is to your advantage to know some common examples of each of these shapes!!) The “ideal” bond angle between the central atom and the other atoms should be noted… Linear= 180º Tetrahedral = 109.5º Trigonal Planar =120º Due to the lone pairs of electrons on pyramidal and bent shapes, the ideal bond angles will be less than 109.5º Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Molecular Geometry— e- repulsion
February 28, 2018 In general, multiple bonds repel more as do lone pairs. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Molecular Geometry— e- repulsion
February 28, 2018 In general, multiple bonds repel more as do lone pairs. As we are looking at angles, we can consider the optimal angle and then modify based upon lone pairs and double/triple bonds Example: H2O NH3 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Shapes of Larger Molecules
February 28, 2018 In acetic acid, CH3COOH, there are three interior atoms: two C’s and one O. •We assign the molecular (and electron-domain) geometry about each interior atom separately: -The geometry around the first C is tetrahedral. -The geometry around the second C is trigonal planar. -The geometry around the O is bent (tetrahedral). Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Finish the Geometries Worksheet
February 28, 2018 We haven’t yet covered hybridization, but we can now answer the other parts of the worksheet Work in the same groups to complete those columns. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Exit Ticket (3 min) Complete on a separate sheet of paper:
February 28, 2018 Complete on a separate sheet of paper: Identify the lewis dot structure, electronic and molecular geometry, and angle for the following: O || H3CCCH3 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Polarity, Hybridization, MO, Bond Order
February 28, 2018 Polarity, Hybridization, MO, Bond Order Geometry and Bonding Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

DO NOW (7 MIN) February 28, 2018 Learning Target: I can determine the polarity, hybridization, the molecular orbitals, and bond order of a molecule from the molecular and electronic geometry using VSEPR Do Now: (on your DO Now Sheet) Complete Question 3 on your MC DO NOW SHEET Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

REVIEW: DO NOW February 28, 2018 The photoelectron spectra above show the energy required to remove a 1s electron from a nitrogen atom and from an oxygen atom. Which of the following statements best accounts for the peak in the upper spectrum being to the right of the peak in the lower spectrum? Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

REVIEW: DO NOW February 28, 2018 The photoelectron spectra above show the energy required to remove a 1s electron from a nitrogen atom and from an oxygen atom. Which of the following statements best accounts for the peak in the upper spectrum being to the right of the peak in the lower spectrum? Options Why is each option partially right? Choose the most correct answer, and why Nitrogen atoms have a half-filled p subshell There are more electron-electron repulsions in oxygen atoms than in nitrogen atoms. Electrons in the p subshell of oxygen atoms provide more shielding than electrons in the p subshell of nitrogen atoms. Nitrogen atoms have a smaller nuclear charge than oxygen atoms. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Electronegativity February 28, 2018 The tendency for an atom to attract electrons to itself Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Polarity: Of bonds - A bond can be polar
February 28, 2018 - A bond can be polar When there is a difference in electronegativity between two atoms, then the bond between them is polar. δ = partial charge δ- = partial negative charge (more EN) δ+ = partial positive charge (less EN) Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Polarity: Of bonds February 28, 2018 TRY IT: Determine whether the bond is polar or nonpolar, and draw the δ’s and the dipole arrow C-O N-N C-H H-Cl Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Polarity: Of molecules
February 28, 2018 - A molecule can be polar If the bonds are polar, and the molecule is not symmetric (ASYMMETRIC), then the molecule is POLAR H----F O H H Overall Dipole Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

February 28, 2018 - A molecule can be polar It is possible for a molecule to contain polar bonds, but not be polar. -For example, the bond dipoles in CO2 cancel each other because CO2 is linear. LOOK! They cancel!! Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

February 28, 2018 In water, the molecule is not linear and the bond dipoles do not cancel each other. Therefore, water is a polar molecule. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

February 28, 2018 The overall polarity of a molecule depends on its molecular geometry. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

February 28, 2018 TRY IT: Are the molecules polar or nonpolar? H2 N2 HI NH4+ NH NH2- H2O2 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

February 28, 2018 TRY IT: Are the molecules polar or nonpolar? CH2O CO CO CH3Cl CH2Cl2 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Why do bonds form? Bonds form when orbitals on atoms overlap.
February 28, 2018 Bonds form when orbitals on atoms overlap. There are two electrons of opposite spin in the overlapping orbitals. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Why do bonds form? February 28, 2018 The overlapping of the orbitals will lower the overall energy of the 2 atoms, therefore it is more stable. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

When you mix n atomic orbitals we must get n hybrid orbitals.
February 28, 2018 A hybrid orbital is simply a mixing of different orbitals together to form a new “hybridized orbital”. We need the concept of hybrid orbitals to explain molecular shapes. (Let’s try to keep it simple…) When you mix n atomic orbitals we must get n hybrid orbitals. Example: If you mix one “s” orbital and three “p” orbitals you will get four “sp3” hybrid orbitals that all have exactly the same energies. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Hybrid Orbitals February 28, 2018 The # of electron domains on the atom will indicate the hybridization needed. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Hybrid Orbitals CH4 1. Draw the Lewis Dot Structures
February 28, 2018 The # of electron domains on the atom will indicate the hybridization needed. CH Draw the Lewis Dot Structures 2. Count the number of e- domains 3. Match to the number of hybrid orbitals Options: Hybrid Orbital Explanation # hybrid orbitals # e- domains sp 1 s orbital and 1 p orbital 2 sp2 1 s orbital and 2 p orbitals 3 sp3 1 s orbital and 3 p orbitals 4 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Hybrid Orbitals TRY IT: Identify the hybrid orbitals for each molecule
February 28, 2018 TRY IT: Identify the hybrid orbitals for each molecule H2C=CH2 CH4 CO2 NH NO3- Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Sigma and Pi Bonds Overlapping orbitals come in 2 varieties…
-Bonds: electron density lies on the axis between the nuclei. - All single bonds are -bonds. February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Sigma and Pi Bonds -Bonds: electron density lies above and below the plane of the nuclei. -A double bond consists of one -bond and one -bond. -A triple bond has one -bond and two - bonds. Often, the p-orbitals involved in -bonding come from unhybridized orbitals. A total of 5 -bonds are formed from the overlapping sp2 hybrid orbitals of carbon, and the -bond is from the unhybridized overlapping p-orbitals on each carbon. H2C=CH2 February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Sigma and Pi Bonds H2C=O H–C≡C–H
C and O both have sp2 hybridization and each has an unhybridized p-orbital available to make the -bond portion of the double bond. H–C≡C–H In this case, C has sp hybridization. One -bond and two -bonds form the triple bond between the carbon atoms. February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Sigma and Pi Bonds February 28, 2018 TRY IT: Identify the number of sigma (σ) and pi (π) bonds CH4 CO2 NH3 NO3- H2CO Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Delocalized Pi Bonds Simply put, if there are resonance structures, the -bond is delocalized or “smeared” between the 2 resonance structures. (By the way, -bonds are never delocalized!) Example: Benzene (C6H6) February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Molecular Orbitals February 28, 2018 Some aspects of bonding are not explained by Lewis structures, VSEPR theory or hybridization. For example, why does O2 interact with a magnetic field? For these molecules, we use Molecular Orbital (MO) Theory. Just as electrons in atoms are found in atomic orbitals, electrons in molecules are found in molecular orbitals. FOR AP: You need to know that MO better explains than VSEPR or Hybridization. That’s it. If you want more, please see your textbook or pull up this ppt on weebly Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Confused yet? Let’s look at a diagram!
Molecular Orbitals Let’s look at the formation of H2… 1s (H) + 1s (H) must result in two MO’s for H2: - One MO has electron density between the nuclei. This is called a bonding MO. The other MO has little or no electron density between nuclei. This is called an antibonding MO. MO’s resulting from s-orbitals are  MOs.  (bonding) MO is lower energy than * (antibonding) MO… Confused yet? Let’s look at a diagram! February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Molecular Orbitals (caused by destructive interference)
(caused by constructive interference) February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Molecular Orbitals Energy level diagrams or MO diagrams show the energies and electrons in an orbital. The total number of electrons in all atoms are placed in the MO starting from lowest energy (1s) and ending when you run out of electrons. Note that electrons in MOs have opposite spins. (Sound familiar?!) H2 has a total of two bonding electrons. February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Molecular Orbitals Let’s look at another example… He2
He2 has two bonding electrons and two antibonding electrons. That is not a stable situation and therefore He2 will not form. In order to predict stability of a molecule from a MO diagram, we need to learn about the concept of “BOND ORDER”… February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Bond Order Here’s another example…O2 ↑ ↑
How many total # of e-’s will we fill in? _________ ↑↓ ↑↓ 8+8 = 16 ↑↓ What is the bond order for O2? __________________________ ↑↓ ½ (10-6) = 2…double bond ↑↓ Notice that there are 2 unpaired e- in the * 2p orbitals. This leads to our last topic of discussion. ↑↓ ↑↓ February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Bond Order Shows strength of Bonds - Bond order = 1 for single bond.
February 28, 2018 Shows strength of Bonds - Bond order = 1 for single bond. - Bond order = 2 for double bond. - Bond order = 3 for triple bond. Fractional bond orders are possible… This indicates that resonance structures are likely. This is when we look at resonance structures! - Draw NO3- Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Bond Order This is when we look at resonance structures! NO3-
February 28, 2018 This is when we look at resonance structures! NO3- Bond order = (total # of bonds) / (# of resonance structures) Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Electron Configurations and Molecular Properties
Molecules have two types of magnetic behavior: - paramagnetism (unpaired electrons in molecule): strong attraction between magnetic field and molecule - diamagnetism (no unpaired electrons in molecule): weak repulsion between magnetic field and molecule. Magnetic behavior is detected by determining the mass of a sample in the presence and absence of a magnetic field. (diamagnetic) small decrease in mass (paramagnetic) large increase in mass February 28, 2018 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Practice!! In PAIRS: Complete the worksheet
February 28, 2018 In PAIRS: Complete the worksheet Make sure you show work!!!! (It is really hard to solve Lewis Dot Structures in your head) Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Exit Ticket (3 min) Complete on a separate sheet of paper:
February 28, 2018 Complete on a separate sheet of paper: Determine whether the following is polar and the hybridization of the central atom: HCF3 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Review Final Review February 28, 2018

DO NOW (4 MIN) February 28, 2018 Learning Target: I can demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies Do Now: (on your DO Now Sheet) What are some strategies that you employ to successfully complete a test (ANY TEST)? Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

REVIEW: DO NOW (4 MIN) February 28, 2018 What are some strategies that you employ to successfully complete a test (ANY TEST)? Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Write THIS…Not THAT When responding to AP Chemistry Free Response Questions Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

General Stuff Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Generally Write This Not That Rational
February 28, 2018 Write This Not That The language used in the question when asked to make a choice examples: “increases”, “decreases”, etc. Other words that may mean the same thing but are likely more ambiguous examples: “goes up”, “goes down”, etc. Rational Make it easy to give you points, and be sure the reader can understand what you saying Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Generally Write This Not That Rational Make it easy to give you points
February 28, 2018 Write This Not That Answer the specific question first, then “justify”, “explain” etc. Burying the answer in the text of the response Rational Make it easy to give you points Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Generally Write This Not That Rational Ambiguous
February 28, 2018 Write This Not That Names of specific elements and compounds, “reactants”, “products”, etc. it Rational Ambiguous Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Generally Write This Not That Rational Be formal in language Species
February 28, 2018 Generally Write This Not That Species It Stuff Rational Be formal in language Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

February 28, 2018 Generally Write This Not That A justification or explanation when it is part of the question Only the answer without supporting it Rational Justification/explanation required to earn point Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Generally Write This Not That Rational Be specific mass, volume, etc.
February 28, 2018 Generally Write This Not That mass, volume, etc. size Rational Be specific Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Generally Write This Not That Rational Required to earn point
References to specific data or graphs when prompted to “explain how the data…” or something similar Make generalizations about the data without specifically citing provided data or trials Rational Required to earn point Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Net ionic equations only containing species that change Aqueous ionic compounds in their undissociated form Spectator ions Rational Including these is not a net ionic, it’s a molecular or complete ionic Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Particle view diagrams with ions and polar molecules orientated in the correct direction relative to each other Incorrectly oriented dipoles Rational Drawings must demonstrate understanding of interactions at the molecular level (ref #4) Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

An answer with units if “include units” is stated in the problem An answer without units Rational If “include units” is written in the prompt, a unit is required to earn full points Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Show all work used to derive an answer An answer without supporting work shown Rational Work is often what earns some/all of the points Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Answers expressed to the correct number of significant figures Answers with an incorrect number of significant figures Rational 1 pt traditionally is assessed somewhere in the FR for significant figures – usually the LAB question Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Gases Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Gases Write This Not That Rational
Components of the Kinetic Molecular Theory as justifications for changes at the molecular level Ideal gas law for molecular level justification Rational arguments based on PV = nRT are at the bulk level and not the molecular level (ref #5) Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Thermodynamics Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Thermodynamics Write This Not That Rational
Values with correct signs Values with incorrect signs Rational Necessary for correct calculations and determinations – watch signs based on bonds breaking/forming, heat flow in calorimetry indicated by temperature changes, signs that may change in application of Hess’ Law, etc. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Thermodynamics Write This Not That Thermodynamically favorable
Spontaneous Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Thermodynamics Write This Not That Rational
Overcome intermolecular forces Break up a solid/liquid Like dissolves like Rational State the actual reason not the memory aid Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Thermodynamics Write This Not That Effective nuclear charge increases
It wants to have a full octet It’s close to having a full octet It’s on the right side of the periodic table Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Thermodynamics Write This Not That
It has more polarizable cloud of electrons It has more electrons More mass More surface area Bigger More protons Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Kinetics Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Kinetics Write This Not That Rational Units required to earn point
Value of k with units Value of k without units Rational Units required to earn point Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Kinetics Write This Not That Rational AP wants more specific answer
Specific parts of the molecules that must collide in order for the reaction to occur Collision must occur in the correct orientation Rational AP wants more specific answer Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Kinetics Write This Not That Rational
A rate law that includes the rate constant k as part of it A rate law without k being included Rational Incomplete rate law if k is not included Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Kinetics Write This Not That Rational
A rate law based only on reactants A rate law that includes products Rational Rate laws are based only on reactants Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Equilibrium Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Equilibrium Write This Not That Rational Preferred AP language
Discussion of Q vs. K Due to Le Châtelier’s Principle Reduce the stress Rational Preferred AP language Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Equilibrium Write This Not That Rational
Proceeds Shift – if equilibrium has not yet been established (i.e. a precipitate has not yet been formed when evaluating Ksp) Rational If equilibrium is not yet established, then it cannot “shift” – rxn will proceed in a certain direction until equilibrium is established Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Ksp expressions that only contain the ions Ksp expressions that contain or imply a species in the denominator Rational Solids and liquids are not included in equilibrium expressions Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Correct formulas (including charges!) for all species in equilibrium expressions Substitutions Abbreviations Chargeless ions Other shorthand that may work out in calculations but does not represent the correct species Rational Equilibrium expressions must be written formally when requested Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

In Kp expressions: Pspecies In Kp expressions: [species] Rational Concentration is not used in Kp, partial pressures are Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

“x” has been assumed to be so small relative to the original concentrations that it can be ignored Nothing about why you ignore x to avoid quadratics Rational Show you understand why you are making the decision Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Acids and Bases Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Acids and Bases Write This Not That Rational
The pH > 7 because the salt produced in the neutralization behaves as a base A- + H2O ⇄ HA + OH- The pH > 7 because it’s a battle between weak acid and strong base and strong base wins Rational State the actual reason not the memory aid Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

The solution is neutral when [H3O+] = [OH-] The solution is neutral when pH=7 Rational True definition of neutral – neutral is only pH of 7 when Kw = 1.0 x (at 298 K) Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Kw = Ka x Kb for a conjugate pair Kw = Ka x Kb for an unrelated acid/base pair Rational This equation only holds true for conjugate acid-base pairs Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

pH = pKa because it is at ½ the equivalence point of a titration of a weak acid with a strong base pH = pKa Rational Explains the reason behind this, and shows you understand this is only true at this point Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Atomic Structure Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Atomic Structure Write This Not That Rational
Effective nuclear charge increases It wants to have a full octet It’s close to having a full octet Rational State the actual reason not the memory aid Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

It has a more polarizable cloud of electrons It has more electrons it has more mass it has more surface area it is bigger it has more protons Rational This is the shortest way to show the reason – simply mentioning “more” of something is probably not enough to demonstrate without further explanation of why that is the case Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Period Shell when referring to elements and their location on the Periodic Table Rational Elements are in a period, electrons are in a shell Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Reference reasons for periodic trends (i.e. effective nuclear charge, coulomb’s law, polarizability, etc.) Stating the trend as the reason (because it is to the left, because it is further down the periodic table, etc.) Rational State the actual reason not the memory aid Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Electrons in higher energy levels are farther from the nucleus, resulting in a larger atom/ion More electrons/more energy levels make the atom/ion bigger Rational Explanation of reason, not just statement of fact, required for point (Ref 2016 #1) Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Bonding and IMF Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Bonding and IMFs Write This Not That Rational
Overcome intermolecular forces Break up a solid/liquid Rational IMFs should be used to justify Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Ion interactions LDF’s when discussing ionic compounds Rational Ionic compounds have ions with whole charges, which dominate interactions Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Coulombic attraction Opposites attract Rational State the actual reason not the memory aid Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Describe the process of overcoming intermolecular forces/polarity Like dissolves like Rational State the actual reason not the memory aid Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Has hydrogen bonds between the molecules Has hydrogen bonds Rational Shows that you understand hydrogen bonds are not actually bonds Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Ionic compound Molecule when discussing an ionic compound Rational A molecule is a covalent compound Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Ions Atoms when discussing ionic compounds Rational Ionic compounds contain ions Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Atoms Ions when discussing covalent compounds Rational Covalent compounds do not contain ions Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Lewis structures that are complete with necessary lone pairs and/or resonance Lewis structures that are missing lone pairs and/or resonance (if needed for correct structures) Rational Lewis structures are incorrect without necessary lone pairs Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Identify specific intermolecular forces at play Stronger intermolecular forces Rational Shows your understanding of the chemistry at play Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Dissolve when discussing interactions between molecular substances in solution Ionize Dissociate Bond React Break up Rational Molecular substances do not dissociate into ions, dissolving is not reacting, and otherwise be formal in usage Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Electrochemistry Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Electrochemistry Write This Not That Rational
Loss of mass of electrode is due to atoms of electrode going into solution as ions Loss of mass of electrode is due to loss of electrons Rational Electrons have extremely small (negligible in this case) mass (ref #3) Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Electrochemistry Write This Not That Rational
Discussion of Q vs. K for changes in cell potential after a change, or qualitative discussion of Nernst Equation Discussion of Le Châtelier’s principle Rational Preferred AP language (ref #3) Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Speed Dating Instructions for Starting the Date
February 28, 2018 Speed Dating Instructions for Starting the Date We need all groups of two. Every group needs a whiteboard and a marker. One person from each group, raise your hand. Just one person from each group. Okay, after a few minutes, you guys will go clockwise. Everyone else will go counterclockwise. So, everyone will move places. Everyone is going to work on the same problem. So, let’s spend 5 minutes now working on problem 9. Work straight on the whiteboard. We’ll have time at the end for you to copy down anything you want to have in your packet. No pencils. Just straight on the whiteboard. Okay, go! Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies

Na2O (molar mass = 62 g/mol) Li2O (molar mass = 30 g/mol)
A 4.5 gram sample of which of the following would have the greatest mass percent of oxygen? Na2O (molar mass = 62 g/mol) Li2O (molar mass = 30 g/mol) MgO (molar mass = 40 g/mol) SrO (molar mass = 104 g/mol) Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

A 4.5 gram sample of which of the following would have the greatest mass percent of oxygen?
Answer: 16/62 x 100 = 26 % 16/30 x 100 = 53% 40 x 100 = 40% 16/104 x 100 = 15% Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

A compound is determined to contain 14g nitrogen and 32g oxygen
A compound is determined to contain 14g nitrogen and 32g oxygen. The empirical formula of the compound is NO N2O NO2 NO3 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

A compound is determined to contain 14g nitrogen and 32g oxygen
A compound is determined to contain 14g nitrogen and 32g oxygen. The empirical formula of the compound is Answer: NO N2O NO2 NO3 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

The mass percent of oxygen in pure glucose, C6H12O6 is 53. 3 percent
The mass percent of oxygen in pure glucose, C6H12O6 is 53.3 percent. A chemist analyzes a sample of glucose that contains impurities and determines that the mass percent of oxygen is percent. Which of the follow impurities could account for the low mass percent of oxygen in the sample? n-eicosane (C20H42) ribose C5H10O5 fructose, C6H12O6 sucrose C12H22O11 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

How many moles of carbon are in 88 grams of propane, C3H8 ?
2.0 16.0 6.0 96.0 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Of the elements below, __________ reacts the most quickly with water.
A) sodium B) barium C) calcium D) cesium E) magnesium Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Rank the six solutions above in order of increasing molarity
Rank the six solutions above in order of increasing molarity. Pay attention to volume, Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

C,D, and E (tied); A and F (tied); most concentrated is B

The +2 and -2 ions attract each other more strongly than +1 attracts -1.
The ions Mg+2 and O-2 are smaller than Na+1 and Cl-1, therefore the ions can get closer together, increasing their electrostatic attractions. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Which of the following are broken when water boils?
a. Covalent bonds b. Hydrogen bonds c. Dipole-dipole interactions d. London Dispersion Forces Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Al2S3 + 6 H2O ---> 2Al(OH)3 + 3 H2S
15.00 g aluminum sulfide and g water react Identify the Limiting Reactant What is the maximum mass of H2S which can be formed from these reagents? How much excess reactant is left in the container? Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

a) Limiting Reactant 15.00g Al2S3 x (1mol/ g) x (6 mol H2O/1mol Al2S3) x (18g/mol H20 ) = g H20 needed 10g H20 x (1mol/ g) x (1 mol Al2S3 / 6mol H2O) x ( g/mol) = g Al2S3 needed H20 is limiting, because we need more than we were given b) Maximum yield Theoretical Yield 10.00 g H20 x (1mol/ g) x (3/6) x ( g/mol ) = g H2S produced c) Excess 15.00 g – g = 1.11g Al2S3 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

X2 + Y2  X2Y2 rate = k[X2] A reaction and its experimentally determined rate law are represented above. A chemist proposes two different possible mechanisms for the reaction, which are given below. Mechanism Mechanism 2 X2  2X (slow) X2  2X (slow) X + Y2  XY2 (fast) X + Y2  XY + Y (fast) X + XY2  X2Y2 (fast) X + XY  X2Y (fast) X2Y + Y  X2Y2 (fast) Based on the information above, which of the mechanisms is/are consistent with the rate law? List the intermediates in each mechanism: Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Answer: Both are consistent
Answer: Both are consistent. In both mechanisms, the molecularity of the slow, rate determining step is consistent with the rate law. Furthermore, the sum of the elementary steps for both mechanisms gives the overall balanced equation for the reaction. Intermediates in mechanism 1: X, XY2. Intermediates in mechanism 2: X, XY, Y, X2Y Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

The rate law for a reaction is found to be Rate = k[A]2[B]
The rate law for a reaction is found to be Rate = k[A]2[B]. What is the intermediate? Which of the following mechanisms gives this rate law? I. A + B ⇄ E (fast) E + B  C + D (slow) II. A + B ⇄ E (fast) E + A  C + D (slow) III. A + A  E (slow) E + B  C + D (fast) A. I B. II C. III D. Two of these Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Answer: E is the intermediate
Answer: E is the intermediate. Only Mechanism II is consistent with the rate law. Whenever a fast equilibrium step producing an intermediate precedes the slow rate determining step and we want to remove the intermediate from the rate law, we can solve for the concentration of the intermediate by assuming that an equilibrium is established in the fast step. The concentration of the intermediate in the rate determining slow step can be replaced with an expression derived from the equilibrium constant [E] =Keq[A][B]. This substitution gives us the desired rate law: rate = k’[A]2[B] Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

Reaction Mechanisms and Energy Profiles – Practice Problem
Draw and label axes for the energy profiles below. Match the curves with the appropriate description. exothermic reaction with a 2 step mechanism where the first step is slow. endothermic reaction with a 2 step mechanism where the first step is slow. endothermic reaction with a 2 step mechanism where the second step is slow. exothermic reaction with a 1 step mechanism. exothermic reaction with a 2 step mechanism where the second step is slow. endothermic reaction with a 1 step mechanism. Reaction pathway Potential Energy Reaction pathway Potential Energy Reaction pathway Potential Energy D C F Reaction pathway Potential Energy Reaction pathway Potential Energy Reaction pathway Potential Energy E B A LO 4.7 Cont Dena K. Leggett, PhD Advanced Chemistry Teacher Allen High School Copyright 2015

A 25 mL sample of hydrofluoric acid (HF) is titrated with 25 mL of 0
A 25 mL sample of hydrofluoric acid (HF) is titrated with 25 mL of 0.30M sodium hydroxide (NaOH). At the equivalence point of the titration, what would the pH of the solution be? Justify with a reaction. pH < 7 pH = 7 pH > 7 pH = pKa Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

A 25 mL sample of hydrofluoric acid (HF) is titrated with 25 mL of 0
A 25 mL sample of hydrofluoric acid (HF) is titrated with 25 mL of 0.30M sodium hydroxide (NaOH). At the equivalence point of the titration, what would the pH of the solution be? Justify with a reaction. pH < 7 pH = 7 pH > 7 pH = pKa The correct answer is “c” pH>7. At the equivalence point, the moles of acid equal the moles of base. The remaining species would be Na+ and F-. The conjugate base, F- will hydrolyze with water, producing OH- in solution: F-(aq) + H2O(l)  HF(aq) + OH-(aq) Select the Source link to see more calculations in this titration. Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

a. [HC2H3O2] > [C2H3O2 1-] > [ H+ ] > [ OH- ]
A 50.0 mL sample of 0.50 M HC2H3O2 is titrated to the half equivalence point with 25.0 mL of 0.50 M NaOH. Which of the following options shows the correct ranking of the molarities of the species in solution? (pKa for HC2H3O2 is 4.7) a. [HC2H3O2] > [C2H3O2 1-] > [ H+ ] > [ OH- ] b. [HC2H3O2] = [C2H3O2 1-] > [ H+ ] > [ OH- ] c. [HC2H3O2] > [C2H3O2 1-] = [ H+ ] > [ OH- ] d. [C2H3O2 1-] > [HC2H3O2] > [ OH- ] > [ H+ ] Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

a. [HC2H3O2] > [C2H3O2 1-] > [ H+ ] > [ OH- ]
(pKa for HC2H3O2 is 4.7) a. [HC2H3O2] > [C2H3O2 1-] > [ H+ ] > [ OH- ] b. [HC2H3O2] = [C2H3O2 1-] > [ H+ ] > [ OH- ] c. [HC2H3O2] > [C2H3O2 1-] = [ H+ ] > [ OH- ] d. [C2H3O2 1-] > [HC2H3O2] > [ OH- ] > [ H+ ] Option B is correct. At the half equivalence point, the concentrations of the weak acid and its conjugate base are equal because the OH- ion has reacted with half of the original acetic acid. The pH of the buffer is equal to the pKa at that point, so the [H+] is which is far lower than the concentrations of the conjugates but far higher than the [OH-] which has a value of Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

What is the maximum number of moles of AgBr that will fully dissolve in 1.0 L of water, if the Ksp value of silver bromide is 4.0 x 10-12? a. 4.0 x b. 2.0 x c x 10-6 d. 2.0 x 10-6 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

What is the maximum number of moles of AgBr that will fully dissolve in 1.0 L of water, if the Ksp value of silver bromide is 4.0 x 10-12? 4.0 x b. 2.0 x c x 10-6 d. 2.0 x 10-6 Answer: The correct answer is “d” 2.0 x To determine the solubility of AgBr we need to determine the maximum concentrations of Ag+ and Br- that will equal the equilibrium constant. The Ksp equation is Ksp = [Ag+][Br-] with [Ag+] = [Br-] = X, so to solve for X we need to take the square root of 4.0 x 10-12 Learning Target: I CAN demonstrate understanding of Big Ideas 1-6 through practice and test-taking strategies February 28, 2018

2Fe(OH)3 (s) + 3H2SO4 (aq) à Fe2(SO4)3 (aq) + 6H2O (l)
a. Ron Weasley had to perform the above reaction in Professor Snape’s potions class. Ron mixed 45.0 g iron III hydroxide with 45.0 ml 2.00 M sulfuric acid in a cauldron, leaving the lid off. The heat capacity of his cauldron was 645 J/K. i. What other information would Ron need to know or measure experimentally to determine the molar change in enthalpy of the reaction, ∆H°rxn? ii. Unbeknownst to Ron, Draco Malfoy dropped a 100 g sneakoscope (assume this is unreactive and at room temperature) into Ron’s cauldron before Ron measured his temperature but after the reaction was initiated. Explain in which direction this would alter Ron’s experimental value of ∆H°rxn. b. Hermione Granger also performed this reaction. She mixed 45.0 g iron III hydroxide with 450 ml 2.00 M sulfuric acid in her cauldron, and kept the lid on, using a spell to enable her to measure temperature (revelio thermo). i. How would you expect her temperature change to differ from Ron’s, if at all? For this comparison, assume that Ron’s reaction was performed with the lid on and with no sneakoscope. ii. Hermione repeated the experiment using all the same measurements, but used aluminum hydroxide instead of iron III hydroxide. Assuming that both reactions have the same ∆H°rxn, would her temperature change be smaller, the same, or greater in her second experiment. c. Given the following values for absolute entropies, what would ∆S°rxn be?

Just for Fun… Answers c. ∆S°rxn = sum product S° - sum reactant S°
ai. qrxn = -(Ccal∆T + c(mixture)m(mixture)∆T); ∆H°rxn = qrxn / moles rxn Ron would need to measure the starting and final temperatures, know (or measure) the specific heat capacity and mass of the reaction mixture, and the number of moles of reaction. In this case, 45.0 g Fe(OH)3 / g/mole = moles; L x 2 M = 0.09 moles H2SO4 (limiting), so moles reaction (to those who dislike this concept, I apologize). bii. 45.0 g Al(OH)3 / 78 g/mole = mole. This is still limiting, but now there are about moles of reaction, so more heat is produced. As the total mass doesn’t change, and the specific heat capacity is unlikely to change dramatically, it is reasonable to predict that ∆T would be larger. aii. Because the sneakoscope would absorb heat, ∆T would be lower, so the heat of the reaction would appear to be lower as would ∆H°rxn. c. ∆S°rxn = sum product S° - sum reactant S° ∆S°rxn = 6(69.91) + 2( ) + 3(20.1) – 2(107) – 3(20.08) = J/mole*K. bi. For Hermione, the Fe(OH)3 is now limiting, as there are now 0.9 moles H2SO4. So, there are more moles of reaction, about 0.21, a seven-fold increase. This would increase the temperature change. However, there is also a significant increase in mass of the reaction mixture, from about 90 g to 500 g. This would decrease the temperature change. So the exact ∆T outcome, increase or decrease, depends on the specific heat capacity of the reaction mixture.

2016 Free Response Question #1
A student investigates the enthalpy of solution, ∆Hsoln, for two alkali metal halides, LiCl and NaCl. In addition to the salts, the student has access to a calorimeter, a balance with a precision of +0.1 g, and a thermometer with a precision of +0.1 °C. a). To measure ∆Hsoln for LICl, the student adds g of water initially at 15.0 °C to a calorimeter and adds 10.0 g of LiCl (s) stirring to dissolve. After the LiCl dissolves completely, the maximum temperature reached by the solution is 35.6 °C. i) Calculate the magnitude of the heat absorbed by the solution during the dissolution process, assuming that the specific heat capacity of the solution is 4.18 J/g°C. Include units with your answer. Released answer: 2 pts; 1 for correct setup, 1 for answer with units. q = mc∆T = (110.0)(4.18)( ) = 9,470 J = 9.47 kJ Additional comments: Had to use 110 g, not 100 g. If 100 g was used, correct calculation could earn the second pt. Inserting a negative sign or wrong/unshown units cost one pt.

aii) Determine the value of ∆Hsoln for LiCl in kJ/molrxn. Released answer: 2 pts g x 1 mol/42.39 g = mol LiCl -9.47 kJ/0.236 mol = kJ/molrxn. 1 pt for correct number of moles, 1 pt for correct ∆H with the correct sign. Additional comments: correct setup of mole calculation earned the point even if the answer wasn’t shown. Again, carrying over number from first part correctly would earn credit. Many students did not give the negative sign in the answer, losing credit.

To explain why ∆Hsoln for NaCl is different than that for LiCl, the student investigates factors that affect ∆Hsoln and finds that ionic radius and lattice enthalpy (which can be defined as the ∆H associated with the separation of a solid crystal into gaseous ions) contribute to the process. The student consults references and collects the data shown: Li+ 76 pm ionic radius; Na+ 102 pm ionic radius. b. Write the complete electron configuration for the Na+ ion in the ground state. Released answer: 1 pt for complete correct configuration. 1s22s22p6 Additional information: Had to be complete (no noble gas symbol); could not be an orbital diagram unless the configuration information was also present. Way too many students gave us the configuration for the sodium atom, not ion.

c) Using principles of atomic structure, explain why the Na+ ion is larger than the Li+ ion. Released answer: 1 pt. The valence electrons in the Na+ ion are in a higher principal energy level than the valence electrons in the Li+ ion. Electrons in higher principal energy levels are, on average, farther from the nucleus. Additional information: we wanted to see some sense of occupied principal energy levels being different (saying ‘more’ was no good – no sense of occupancy here, but ‘one extra’ was) (levels or shells, not orbitals or sublevels; 2p is higher energy than 2s, but closer to nucleus on average). For group trends, the best explanation is that higher principal energy levels are occupied, which are further from the nucleus. Many students wanted to talk about shielding, which is more useful for periodic trend reasoning. There is an effect of shielding, but it serves largely to cancel out the increased nuclear charge, so that Zeff (which was also frequently referred to) is largely unchanged.

d) Which salt, LiCl or NaCl, has the greater lattice enthalpy? Justify your answer. Released answer: 1 pt. LiCl. Because the Li+ ion is smaller than the Na+ ion, the columbic attractions between ions in the LiCl are stronger than in NaCl, resulting in greater lattice energy. Additional information: We wanted some discussion about since Li+ is smaller, it and Cl- can get closer, and therefore have a stronger attraction. Many students got into Coulombic attraction conversation here, which could work, but it had to be between Li+ and Cl- as opposed to between the lithium nucleus and its electrons. Students also frequently referred to electronegativity and ionization energies, which did not earn credit.

e) In the representation of a portion of a crystal of LiCl, identify the ions by writing in the appropriate formulas (Li+ or Cl-) in the boxes. Released answer: 1 pt. The larger ion is Cl-, the smaller is Li+. Additional information: Not much to say here, there were only two choices, but many students apparently didn't realize that losing or gaining electrons alters the size of the entity.

f) The lattice enthalpy of LiCl is positive, indicating that it takes energy to break the ions apart in LiCl. However, the dissolution of LiCl in water is an exothermic process. Identify all particle-particle interactions that contribute significantly to the dissolution process being exothermic. For each interaction, include the particles that interact and the specific type of intermolecular force between those particles. Released answer: There are interactions between Li+ ions and polar water molecules and between Cl- ions and polar water molecules. These are ion-dipole interactions. 1 pt for identifying the particles that interact, 1 pt for identifying the type of interaction. Additional information: The first point was for naming the three particles involved in the exothermic processes: H2O with Li+, and H2O with Cl-. Showing H+ or OH- or O-2 lost this point. The second point was for ion-dipole attraction. We did accept a thorough description of an ion-dipole attraction in the absence of that specific phrase which could’ve included a diagram. Again, any reference to water ions (such as ‘the H+ of water) lost the point; however, partial charge was good. Many students invoked LDF, which as they were quick to point out, are always present. However, it was also irrelevant to the question. Mentioning dipole-dipole lost the point.

Common Problems and Misconceptions
Going from mass to empirical formula – often switch the coefficients Transition metals lose the s electrons first s electrons are further on average from nucleus than p for same energy level Units: kJ vs J, °C vs K, per mole or per gram; don’t lose track of which unit you’re using; not always at STP for a gas Explaining is more than just an observation: a lone pair on a central atom is not sufficient for shape, the pair must act (repel the other electrons) What occurs in the process of dissolving? The solute is not disappearing; it is mixing Show combustion analysis slide for Big Idea #1 Van der Waal’s is not LDF; mass is not an explanation for LDF, # electrons, polarizability, size, volume have been accepted Students get mixed up in limiting reactant problems – amounts of whatever is left over Don’t make H+ from a strong base, don’t make OH- from a strong acid

Common Problems and Misconceptions
Don’t make H+ from the addition of a strong base; don’t make OH- from the addition of a strong acid Van der Waal’s is not LDF; do not use mass as an explanation for LDF. You may use polarizability, # of electrons, size, and volume. Limiting reactant problems are confusing – which is limiting, how much of the other is used up, how much of the other remains, etc Combustion analysis to get empirical formula can be confusing, particularly if not using oxygen as the oxidizer

Intro to VSEPR Geometry and Bonding February 28, 2018

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Intro to VSEPR Geometry and Bonding February 28, 2018

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