Answers to 1^st Semester Exam Review Questions

MC Questions:

1.  d    2. c    3. a    4. b    5. d    6. d    7. b    8. d    9. d    10. c

FREE RESPONSE QUESTIONS

Part A: 1 Writing Equations

(a)  Al + Cu²⁺ --> Al³⁺  +  Cu

(b)  K₂O  +  H₂O  -->  K⁺  +  OH^-

(c)  Cl₂  +  Fe  -->  FeCl₃  (FeCl₂ also acceptable)

(d)  PO₄^3-  +  Zn²⁺  -->  Zn₃(PO₄)₂

(e)  C₂H₅OH  +  O₂  -->  CO₂  +  H₂O

(f)  Li  +  N₂  -->  Li₃N

(g)  H⁺  +  SO₃^2-  -->  H₂O  +  SO₂  (H₂SO₃ also possible answer)

(h)  S^2-  +  Zn²⁺  -->  ZnS      (Not the best question since S^2- does not exist as such in an aqueous solution; it would exist as HS^-)

Part B Problem 2

(a) HF is the gas
(b) %F in solid product = 33.3%;  %F in gaseous product = 66.7%
(c) UF₂O₂
(d)  UF₆  +  2 H₂O  -->  UF₂O₂  +  4 HF

Part C

Problem 3.
(a)  C₃H₈  +  5 O₂  -->  3 CO₂  +  4 H₂O
(b)  135 L air
(c)  DH^o_f C₃H₈ = -101.7 kJ/mol
(d)  DT = 45.3 K

Problem 4
(a)  C₇H₁₂O₂
(b)  molar mass = 257 g/mol
(c)  molar mass = 128 g/mol
(d) This molecule must form a dimer in solution.  This means that two of the basic molecules must stick together in solution.  This could happen through hydrogen bonding with itself.

Problem 5
(a) (1) I would subtract the mass of the clean copper strip from mass of the copper strip and compound (1.2874g - 1.2789g), then divide this answer by the gram atomic mass of I, 126.91g.
  (2) I would subtract the mass of the copper strip after washing from the mass of the clean copper strip (1.2789g - 1.2748g).  Then I would divide this answer by the molar mass of copper, 63.55g.
(b)  I would find the least whole number ratio of moles of Cu and I by dividing the larger of the two by the smaller and finding the whole number ratio equal to that ratio.
(c)  (1) If some unreacted iodine vapor condensed on the strip then the mass and moles of I calculated would be too high thereby resulting in an empirical formula with too much I in it.
  (2) If some of the white copper iodide compound flaked off before weighing then the mass and moles of I calculated would be too small resulting in an empirical formula with too little I in it.

Problem 6.
(a)  In order to form NaCl(s) from its elements in their natural state one would have to consider the following processes:  Sublimation of Na(s) into a gas (endothermic).  Breaking of the Cl-Cl bond to form Cl atoms (endothermic).  First ionization energy of Na (endothermic).  Electron affinity of Cl (exothermic).  Lattice energy of NaCl (very exothermic).  It is the very exothermic lattice energy that tips the process to that point from endothermic to exothermic.
(b)  When NaCl dissolves in water, three basic processes occur, the spreading or separating of the ions from each other in the salt is very endothermic.  Likewise the spreading of the hydrogen bonded, polar water molecules from each other is also endothermic.  When the ions and the polar water mix this process is very exothermic and makes up most of the energy required in the first two processes.  Therefore, NaCl is able to dissolve in water with the ambient heat available in the environment.

Problem 7
(a)  The two carbons in C₂H₄ are held together by a double bond which means 4 electrons hold the two C nuclei together.  In C₂H₆ only a single bond, two electrons, hold the C nuclei together.  Therefore, the C=C bond is stronger and shorter.
(b)  In NH₃ there are four electron groups around the central nitrogen,3 bonding pair and one lone pair.  These four groups are arranged in a tetrahedral electron group arrangement, giving a basic angle of 109.5 degrees between electron groups.  However, the lone pair of electrons is not as localized between two nuclei as the bonding electrons are.  Because of this delocalization of the lone pair of electrons, they exert a greater repulsion on the bonding electrons forcing the H-N-H bond angles to be 107.5 degrees.
(c)  The lewis structure of SO₃ reveals that there are three resonance structures for SO₃ which include two single bonds and one double bond. These resonance structures indicate that there are actually 4 pairs of electrons that hold the three O atoms to the central S.  This indicates that the bond order for each S-O bond would be 4/3 (4bonds/3bonded atoms).  This makes each S-O bond stronger than a simple single bond and shorter as well.
(d)  I₃^- lewis structure has 5 groups of electrons around the central I.  The basic electron geometry would be trigonal bipyramidal.  With three lone pairs and two bonding pairs the lone pairs would orient themselves in the trigonal plane with the bonded I's above and below the plane.  This would minimize the repulsion between the delocalized lone pairs and give a linear geometry of atoms.

Problem 8
(a)  Across a period, from left to right, the first ionization energy of the elements increases.  This is the result of the fact that the atomic radii decrease in the same direction.  If the atomic radii are smaller that indicates that the electrons are held progressively more tightly.  Thus if the electrons are held tighter it would take more energy to remove one.
(b)  The first ionization energy of B is lower than the first IE of Be contrary to the expected trend.  The reason for this can be found in the electron configuration.  Since Boron's 5th electron is first electron in the 2p sublevel it is easier to remove than Beryllium's 4th electron.  Be's valence electrons are in the 2s sublevel.  Since a 2p orbital has higher energy than a 2s orbital it is slightly easier to remove a valence electron from B than from Be.
(c)  The first ionization energy of O is lower than the first IE of N contrary to the expected trend.  The reason for this can be found in the orbital diagram.  Nitrogen's three electron's in the 2p sublevel are evenly distributed one electron to each of the three 2p orbitals as predicted by Hund's Rule.  Oxygen however has 4 electrons in the 2p sublevel.  Therefore, two of O's valence electrons must be paired up in a 2p orbital.  The result of this is greater electron repulsion between O's valence electrons.  This repulsion makes it easier for O's valence electron to be removed.
(d)  Na's first ionization energy will be smaller than both Li's and Ne's.  It will be smaller than Li's because Na's valence electron is in 3s orbital whereas Li's is in 2s.  3s is in a higher energy level and this one valence electron has 10 electrons between it and the nucleus.  These 10 electrons shield the +11 nuclear charge.  Therefore Na's valence electron is farther from the nucleus and less tightly held than Li's valence electron.  Thus Na has a lower first IE than Li.  Na's 3s valence electron is in a higher level than Ne's 6 2p electrons.  Ne's 2 p electrons are shielded by only two electrons whereas Na's one 3s electron is shielded by 10 electrons.  Therefore the nuclear charge felt by the valence in Ne is greater than the electron in Na.  So Na's valence electron is less tightly held and its first IE is much lower.