Part 1: Multiple Choice (2 points each)Directions: Please circle the best answer for each of the following questions.Part 2: Short AnswerDirections: Answer each of the following questions. Be sure to use complete sentences where appropriate. For full credit be sure to show all of your work.Grossmont College Chemistry 141 Exam 3 Fall 2016Name: ___________________________________Date: ________Instructor: Diana VanceExam 3Part 1: Multiple Choice (2 points each) Directions: Please circle the best answer for each of the following questions. 1. Bonding orbitals are __________ and antibonding orbitals are __________ in energy than atomic orbitals. a. lower, lowerb. lower, higherc. higher, lowerd. higher, highere. same, same2. What is the atomic number of the as yet undiscovered element in which the 8s and 8p electron energy levelsfill? a. 118b. 6c. 168d. 8e. 8883. Which atom would be expected to have the largest third ionization energy? Consider the following electronconfigurations for neutral atoms:i. I: 1s22s22p63s2ii. II: 1s22s22p63s23p4iii. III: 1s22s22p63s23p6a. atom Ib. atom IIc. atom IIId. atom I and IIe. all of the above4. How many electrons does a dsp3 orbital hold? a. 10b. 8c. 6d. 4e. 25. Which of the following ionic compounds would be expected to have the highest lattice energy?a. Cesium fluorideb. Cesium chloridec. Cesium bromided. Cesium iodidee. All have the same lattice energyPage 1 of 9Grossmont College Chemistry 141 Exam 3 Fall 2016Name: ___________________________________Date: ________Instructor: Diana Vance6. Which ion does not have a noble gas configuration in its ground state?a. Sc3+b. Al3+c. Ga3+d. As3-e. none of the above7. The phosphorus atom in phosphorous trichloride would be expected to havea. a partial positive (δ+) chargeb. a partial negative (δ-) chargec. a 3+ charged. a 3- chargee. a 0 charge8. What are the F-S-F bond angles in sulfur hexafluoride? a. 90° and 120°b. 90°c. 109.5°d. 120°e. 180°9. Which experiment did not use a balance? a. Propagation of Errorb. Alloy Analysisc. Titrationd. Calorimetrye. none of the above10. In general, in what order should the following steps be taken when responding to a chemical spill? a. Communicate, isolate, mitigate, evacuate. b. Isolate, evacuate, mitigate, communicate.c. Evacuate, communicate, isolate, mitigate.d. Evaluate, isolate, mitigate, communicate.e. all of the above Page 2 of 9Grossmont College Chemistry 141 Exam 3 Fall 2016Name: ___________________________________Date: ________Instructor: Diana VancePart 2: Short Answer Directions: Answer each of the following questions. Be sure to use complete sentences where appropriate. Forfull credit be sure to show all of your work. 1. Given the following data (12 points): Mg (s) → Mg (g)∆H° = 148 kJF2 (g) → 2 F (g)∆H° = 159 kJMg (g) → Mg+ (g) + e-∆H° = 738 kJMg+ (g) → Mg2+ (g) + e-∆H° = 1450 kJF (g) + e- → F- (g)∆H° = -328 kJMg (s) + F2 (g) → MgF2 (s)∆H° = -1123 kJ a. Use the following to calculate the ∆H°lattice of MgF2:∆H°lattice = -1123 kJ – 148 kJ – 159 kJ – 738 kJ – 1450 kJ – 2(-328 kJ) = -2962 kJ b. Compared with the lattice energy of LiF (1050 kJ/mol) or the lattice energy ofNaCl (788 kJ), does the relative magnitude of the value for MgF2 surprise you? Explain. No, both of these compounds have +1 and -1 charges, whereas MgF2 has a +2 and -1charge, because E α q1q2 and the magnitude of the charges is higher the lattice energyshould also be higher. 2. Use the concepts of effective nuclear charge, shielding, and n value of the valence orbitalto explain the trend in atomic radius as you move across a period in the periodic table fromleft to right (5 points). As you move across a row in the periodic table, the n level stays the same. However, thenuclear charge increases and the amount of shielding stays about the same since the numberof inner electrons stays about the same. So, the effective nuclear charge experienced by theelectrons in the outermost principal energy level increases, resulting in a stronger attractionbetween the outermost electrons and the nucleus and therefore, a smaller atomic radii.3. How are electron affinity and electronegativity different (5 points)?Electron affinity is the process of a single atom gaining an electron. Electronegativity is thestrength of the attraction of a nucleus to a pair of shared (bonded) electrons within a covalentbond. Electronegativity is only important when looking at covalent bonds and electronaffinity is only important when considering single atoms gaining electrons to form anions.Page 3 of 9Grossmont College Chemistry 141 Exam 3 Fall 2016Name: ___________________________________Date: ________Instructor: Diana Vance4. Draw Lewis electron dot structure for POCl3 (P is central). Draw a structure that obeys the octet rule and then show any other structures which may better represent the molecule. Include formal charges and explain which structure is preferred and why (6 points).Cl ClClPOClClCl POObeys octet rule minimizes charges – better structure5. Two structures may be drawn for C4H5N2OBr (8 points):BrHHCHN C CHNOC HBrHHC N NOC HCHCHStructure a Structure ba. Are these two resonance structures of the same molecule? Explain.These are different molecules because they have different skeleton structures; thereforethey are isomers. Resonance structures must have the same skeleton structure!b. How many sigma bonds are in structure a? __12______c. How many pi bonds?____3____d. Which bonds are longer, the CC bonds in structure a or b? Explain.The CC bonds in b are longer because single bonds are longer than triple bonds.e. Which bonds are stronger, the CN bonds in a or b? Explain. The CN bonds in b are stronger because double bonds are stronger than single bonds.6. Acetylene reacts with bromine according to the followingequation (10 points):C2H2 (g) + 2 Br2 (l)  C2H2Br4 (g)From bond energies, calculate the standard enthalpy change forthe reaction.C CBrBrHBrBrHC C HHBr BrBr Br+∆ Hrxn=∑bondsbroken−∑bonds formedPage 4 of 9Bond Energy (kJ/mol)C-C 346C=C 610C≡C 812Br-Br 193C-H 346C-Br 285H-Br 366Grossmont College Chemistry 141 Exam 3 Fall 2016Name: ___________________________________Date: ________Instructor: Diana Vance∆ H∆ H(¿¿C−H )+(2 mol)(∆ HBr−Br)∆ H∆ H(¿¿C−H)+(1mol)(∆ HC−C)(¿¿C−Br)+(2 mol)¿(4 mol)¿(¿¿C ≡C)+(2 mol)¿−¿(1 mol)¿∆ Hrxn=¿∆ Hrxn=[(1mol)(812kJmol)+(2