FINAL EXAM REVIEW CHM IB DR. DIXON
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1 DISCLAIMER: SOME OF THIS MATERIAL WAS TAKEN FROM OTHER MATERIALS CREATED BY OTHER SI LEADERS. THERE IS A POSSIBILITY THAT THIS REVIEW CONTAINS ERRORS. PLEASE REFER TO YOUR TEXTBOOK, CLASS SLIDES OR YOUR PROFESSOR TO STUDY AND ADDRESS ERRORS. ADDITIONALLY, THIS REVIEW SUMMARIZES KEY POINTS BUT DOES NOT INCLUDE ALL POSSIBLE MATERIAL GOING FOR THE FINAL EXAM. MAKE SURE YOU PREPARE FOR YOUR FINAL EXAM USING YOUR OTHER RESOURCES AS WELL. FINAL EXAM REVIEW CHM IB DR. DIXON
2 MAIN ENERGY EQUATIONS E = hv (MAKE SURE YOU KNOW) v = c/λ (MAKE SURE YOU KNOW) 1 λ = R ( 1 nf 2 1 ni 2) h = J s (Planck s constant) R = m^-1 c=2.998 x 10^8 m/s Electromagnetic Radiation 3.00 x 108 m/s = speed of light Frequency = # of waves that pass a point during certain period of time Greater wavelength = smaller frequency ROY G BIV FOR VISIBLE LIGHT 1. An energy of 2.0 x 10^2 kilojoules/ mol is required to cause a Cs atom on a metal surface to lose an electron. Calculate the longest possible wavelength of light in nanometers that can ionize a Cs atom
3 Quantum numbers n= l = ml = Heisenberg s Uncertainty Principle Q: Which of the following represent valid sets of quantum numbers? For a set that is invalid, explain briefly why it is invalid. a) n = 3, l = 2, ml: -3 b) n = 2, l = 1, m l : +1 c) n = 6, l = 5, ml: 3 d) n = 4, l = 4, ml: +3
4 Orbital Filling Electrons occupy the lowest energy orbital available (i.e. 1s before 2s) Fill in the following order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s... Pauli exclusion principle: no two electrons in an atom can have the same four quantum numbers Hund s Rule: when filling degenerate orbitals, electrons filled them singly first, with parallel spins Aufbau principle: lower energy levels are filled first Orbitals can only hold 2 electrons each
5 Be able to distinguish valence electrons from core electrons Valence electrons: o Main group elements: o Transition metals: 1. Determine the number of valence electrons in each element a. Ba b. Cs c. Ni d. S Electron Configuration Writing all the energy levels that are filled with electrons (and how many in each) Shorthand = using nearest noble gas preceding the element in question to shorten the electron configuration Li = [He] 2s 1 2. Choose the correct electron configuration for Se. a. 1s 2 2s 2 2p 6 3s 2 3p 4 b. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4
6 c. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 4p 4 d. 1s 2 2s 2 2p 6 3s 2 3d 4 3. What is the electron configuration for Fe 2+? Indicate whether it is paramagnetic or diamagnetic. a. [Ar]4s 2 3d 6 b. [Ar]4s 2 3d 4 c. [Ar]4s 0 3d 6 d. [Ar]4s 2 3d 8 Periodic trends 4. Arrange these atoms and ions in order of increasing radius: Cs +, Xe, I - a) I - < Xe < Cs + b) Cs + < Xe < I - c) Xe < Cs + < I - d) I - < Cs + < Xe
7 5. The ionization energies of an unknown third period element are shown below. Identify the element. 1 st IE = 786 kj/mol 2 nd IE = 1580 kj/mol 3 rd IE = 3230 kj/mol 4 th IE = 4360 kj/mol 5 th IE = 16,100 kj/mol a) Mg b) Al c) Si d) P 6. For which element is the gaining of an electron the most exothermic? a) Li b) N c) F d) B 7. Choose the larger atom from each pair a) Al or In b) Si or N c) P or Pb d) C or F 8. Arrange this isoelectronic series in order of decreasing radius: F -, Ne, O -2, Mg +2, Na + 9. Draw the full orbital diagram for each element a. N b. F c. Mg d. Al 10. Which pair of elements is most likely to form an ionic bond? a. Nitrogen and Oxygen b. Carbon and Hydrogen c. Sulfur and Oxygen d. Calcium and Oxygen 11. Which set of elements is arranged in order of decreasing electronegativity? a. O < S < As < Ge
8 b. Ge < As < S < O c. S < O < As < Ge d. As < O < Ge < S IONIC BONDING Transfer of electrons Occurs between a metal and a nonmetal High melting and boiling points High polarity Solid at room temperature Lattice Energy & Bohr-Haber Cycle Examples: NaCl, MgCl 2 COVALENT BONDING Sharing of electrons Occurs between two or more nonmetals Low melting and boiling points Low polarity Liquid or gaseous at room temperature Examples: CH 4, N 2 Lattice Energy Lattice energy becomes less exothermic as ionic radius increases Lattice energy becomes more exothermic as magnitude of ionic charge increases Lewis Structures [1] count valence electrons [2] determine central atom [3] distribute bonds between atoms, then add lone pairs to terminal atoms [4] any remaining electrons, after terminal atoms have their octet, go to the central atom Some can have incomplete octets (BF3) or expanded octets (SF6)
9 Bond Lengths Distance between nuclei of bonded atoms: Single bond > double bond > triple bond o Single bond = sigma o Double bond = sigma and pi o Triple bond = sigma and 2 pi Length and strength are inversely proportional relationships, i.e. the shorter the length, the stronger the bond Polarity Means that one atom has a stronger pull on the electrons than the other atom Differences in electronegativities to determine polarity Non-polar or Pure Covalent Examples: Cl-Cl (0) C-H (0.4) Polar bond Examples: H-Cl (0.9) Ionic bonds Examples: NaCl Na + and Cl -
10 2 conditions must be true for a molecule to be polar: 1. There must be at least one polar bond (ClF) or, one lone pair of electrons present (NH3) 2. The polar bonds, if there are more than one, and lone pairs, must be arranged so that their dipole moments do NOT cancel each other Hybridization [1] Count up sigma bonds and lone electron pairs [2] Determine the principal energy level of atom s valence electrons [3] Take s-orbital to hybridize [4] Take as many p-orbitals and d-orbitals (if applicable) needed for same number of orbitals hybridized as sigma bonds/lone pairs Valence Shell Electron-Pair Repulsion (VSEPR) Theory Regions of high electron density (electron pairs) around the CENTRAL ATOM are arranged as far apart as possible to minimize repulsions Lone pairs (LP) of electrons require more volume than shared pairs LP to LP is the strongest repulsion LP to BP is intermediate repulsion
11 BP to BP is the weakest repulsion
12 State the hybridization and shape for each of the non-hydrogen atoms in the following molecule. Also state the number of sigma and pi bonds.
13 Naming hydrocarbons Bond Order Bonding order = (# of electrons in bonding MOs) (# of electrons in antibonding MOs) / 2 Be able to fill in an energy level diagram (10.8)
14 Draw the molecular orbital diagram of C 2-2 Gases Pressure: atm is standard unit o 1 atm = 760 mmhg a.k.a. torr Temperature: Kelvin is standard unit o Kelvin = C Density = mass/volume o Can find mass from using PV=nRT to find moles, then convert o STP = standard temperature and pressure o Temperature = 273 K o o Pressure = 1 atm (760 mmhg)
15 Boyle s Law: Pressure and Volume o Pressure of a gas is inversely proportional to its volume o P1V1=P2V2 Charles s Law: Volume and Temperature o Volume is directly proportional to temperature o V1/T1=V2/T2 Avogadro s Law: Volume and Number of Particles o Volume is directly proportional to the number of particles o V1/n1=V2/n2 Standard Conditions o 1atm, K (0 C) o Used because volume of gases varies with temperature and pressure Mole Fraction o PA/PTotal = na/ntotal o Mole fraction = na/ntotal o PA = xa(ptotal) Collecting Gases via Water Displacement o Can use water displacement to collect a gas o Vapor pressure of water is known (table) o Subtract partial pressure of water vapor from total pressure to find collected gas pressure Molecular Velocities of Gases o Use root-mean-squared method to find average velocity of gas sample o Urms = 3RT MM Real Gases o In real life, intermolecular interactions occur o Volume of particles actually has impact o Use Van der Waals constants to account for these factors
16 Van der Waals Equation for real gases Diffusion vs. Effusion Diffusion = Effusion = 1. What volume of O2 measured at 17.7 degrees Celsius and atm reacts with 15.1 g C4H10 to produce CO2 and H2O? 2C4H O2 8CO H2O
17 2. Calculate the density of Kr at 308 K and atm. 3. A sample of oxygen was collected by displacement of water at 25 degrees Celsius. The atmospheric pressure was 700 torr. What is the mole fraction of O2? Vapor pressure of H2O is 24 torr at 25 degrees Celsius.
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