MT2+Solutions_spring 2010 - Chemistry 1A, Spring 2010...

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Unformatted text preview: Chemistry 1A, Spring 2010 Midterm 2 Solutions Mar 8, 2010 Problems 1. Which atom has the lowest ionization energy? C) Li (1s2 4p). This will have the lowest ionization energy, because the 1s electrons will shield most of the nuclear charge, so the 4p electron only sees about +1 effective nuclear charge. The 3p electron in He(1s3p) also will see about +1 effective nuclear charge, but it is in a lower energy orbital, so it will require more energy to ionize. 2. [Ar]4s2 3d10 4p3 is the electron configuration of a(n) A) As 3. Which one of the following configuration depicts an excited oxygen atom? B) 1s2 2s2 2p2 3s2 . atom. ￿ ￿ This is the only oxygen configuration (8 electrons), other than the ground state 1s2 2s2 2p4 . 4. Which two elements have the same ground-state electron configuration? E) No neutral elements have the same ground state electron configuration 5. Which of the following will become more paramagnetic upon ionization to form a +1 ion? Mark all that apply A) O, C) Br, D) Pd All of these have paired electrons in their ground state configuration. Therefore, they will have a greater number of unpaired electrons upon ionization, making them more paramagnetic. The next three questions refer to three reactions shown below. The energy diagram has one column which shows the relative energies of products and reactants for each reaction. Ar → Ar+ + e− Cs+ + e− → Cs Ge → Ge+ + e− 1 IV: Ar+ + eGe+ + eE II: Cs+ + eI: Cs III: Ge Ar These are all ionization energies. Therefore, the highest energy difference is for Ar, because it has the highest ionization energy, followed by Ge, and finally Cs. 6. Which is the appropriate label for level I in the diagram? A) Cs 7. Which is the appropriate label for level III in the diagram? E) Ge 8. Which is the appropriate label for level IV in the diagram? D) Ar+ + e− 9. What is the P-O bond order in PO3− ? 4 D) 1 1 . 4 .. O .. .. O P 3.. O .. .. O .. O P 3.. O .. .. .. .. O .. .. O P 3.. O .. .. .. .. O .. .. O P 3.. O .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. B) antimony .. .. .. .. The Lewis structure for PO3− will have 1 P-O double bond and 3 P-O single bonds, since phosphorus 4 can expand its octet. This will give it 4 resonance structures, and thus average bond order is 1 1 4 10. Which of the following elements has the largest electronegativity? .. .. O .. O .. O .. O .. .. .. .. .. .. .. .. .. .. .. .. .. .. 2 11. Of the following, which gives the correct order for atomic radius for Mg, Na, P, Si and Ar? D) Na > Mg > Si > P > Ar 12. Which isoelectronic series is correctly arranged in order of increasing radius? D) Ca2+ < K+ < Ar < Cl− These all have the same electron configuration, therefore the atom with the largest nuclear charge will have the smallest radius, and the atom with the lowest nuclear charge will have the largest radius. 13. According to Lewis theory, what is wrong with this structure for hydrogen cyanide, HCN? H C N E) All of the above (hydrogen cannot accommodate more than 2 electrons, nitrogen does not have an octet, carbon does not have an octet, there are not enough electrons in the structure). 14. Given the electronegativity below, which covalent single bond is most polar? H = 2.1, C = 2.5, N = 3.0, O = 3.5 C) O-H O-H has the largest electronegativity difference, and will therefore be the most polar bond. 15. The Lewis structure of AsH3 shows nonbonding electron pair(s) on As. H .. As H H B) 1 3 16. What is the formal charge of the central atom in SO2− in its most stable resonance structure? 4 .. O .. C) 0 .. O 2- Sulfur can expand its octet to minimize formal charge. Therefore, it will form two double bonds and two single bonds. This gives it a total of 6 bonds and zero lone pairs, making the formal charge 0. 17. Which of the following molecules are polar? Mark all that apply .. O .. .. .. .. S O .. .. .. .. .. .. O .. .... .. O .. Cl P H H .. S .. O .. Cl .. Cl .. .. .. .. .. Cl .. .. Cl .. .. Cl .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. B) H2 O, C) SO2 Cl2 , E) PCl− 4 H2 O and PCl− will both have lone pairs causing the bond dipoles to give a net molecular dipole. 4 SO2 Cl2 has two dissimilar types of bonds, which will also lead to a net dipoles since it is tetrahedral. For the next two questions, assume the ionization energy of K is 417 kJ/mol and the electron affinity of F is -325 kJ/mol 18. What is the approximate net energy change in producing K+ and F− from K and F atoms (kJ/mol) D) 92 19. What is the net energy change in making the KF molecule? C) -241 Making bonds always releases energy, and C) is the only negative value. 4 20. The oxidation number of phosphorus in PF3 is C) 3 . PF3 will have three bonds and one lone pair in the Lewis structure. However, for oxidation number, the bonding electrons are assigned to the more electronegative atom. Therefore, the Phosphorus atom has only 2 electrons, compared to 5 in the neutral atom. This gives the oxidation number of +3. 21. The Cl-Si-Cl bond angle in the SiCl2 F2 molecule is approximately B) 109.5◦ SiCl2 F2 is tetrahedral, so the bond angles will be approximately 109.5◦ The following 2 questions are about the carbon atoms in this compound: H C H C H CH3 . 22. The molecular geometry at each of the double bonded carbons is B) trigonal planar . (credit was also given for C), or both B) and C), as there was a tetrahedral carbon attached to the double bond) 23. The hybridization at each of double bonded carbons is B) sp2 (credit was also given for C), or both B) and C), as there was an sp3 carbon attached to the double bond) 24. Which of the following is a chiral isomer of the following heptane? . D) 5 25. Valence bond theory does not address the issue of A) excited states of molecules 26. The hybridization of nitrogen in the HCN molecule is A) sp . . Consider the molecules and molecular ions of F2 , O2 and N2 to answer the following three questions (The relative energies of the molecular orbitals for each is shown. You need to fill in the electrons by yourself). 27. Which of the following, according to molecular orbital theory, has the strongest bond? E) N2 N2 has the highest bond order (3) of all these species, so it will be the strongest bond. 28. Which of the following bonds gets weaker when the species shown is ionized? D) N2 Every other species has its highest energy electrons in an antibonding MO, therefore the bond order will increase upon ionization. N2 has its highest energy electrons in a bonding MO (σ2p ), so ionization will decrease the bond order, weakening the bond. N2 2p * 2p * 2p 2p 2s * 2s 2p 2p * 2p * 2p O2 2p * 2p * 2p 2p 2s * 2s 2p * 2p F2 2p * 2p * 2p 2p 2s * 2s 29. Which of the following is the most paramagnetic? B) F2+ 2 F2+ has two unpaired electrons, making it the most paramagnetic of the species listed. 2 6 30. The hybrid orbitals used for bonding by the sulfur atoms in the SF4 molecule are D) sp3 d orbitals. SF4 has 34 valence electrons, so sulfur will have 4 bonds and one lone pair, giving is a steric number of 5, requiring sp3 d hybrid orbitals. 31. There are C) 2 sp hybrids require 1 s orbital and 1 p orbital. Since there are 3 p orbitals for a given value of n, there are 2 unhybridized p orbitals remaining. 32. How many equivalent resonance forms can be drawn for CO2− ? 3 unhybridized p atomic orbitals in an sp-hybridized carbon atom. .. O .. 2O .. O .. 2O .. .. .. O .. 2O C) 3 There are 3 C-O bonds, but only one double bond. Therefore, there are three different positions to place the double bond, making 3 resonance structures. 33. The volume of a sample of gas (2.49 g) was 752 mL at 1.98 atm and 62◦ C. the gas is . To solve this problem, we first must find the number of moles of gas from the ideal gas law, and use this to get the molecular mass of the gas. 1L ✘ = 0.752 L 1000✘✘ mL ◦ 62 C = (273 + 62) K = 336 K ✘ 752✘✘ × mL .. .. C O .. .. .. .. .. .. .. .. .. PV .. .. .. .. C O .. .. .. .. .. C O .. .. .. .. = nRT PV n= RT ✘ ✚ (1.98✘✘ (0.752✚ atm) L) =￿ ✘ ✘￿￿ L atm 336.15 ✚￿ ✘✘ 0.8206 K ✚ mol￿ K = 0.0540 mol 7 .. .. .. .. .. .. .. .. MW 2.49 g 0.05398 mol g = 46.1 mol = Therefore, the gas is D) NO2 34. A balloon originally had a volume of 4.39 L at 44◦ C and a pressure of 729 torr. The balloon must ◦ C to reduce its volume to 3.78 L (at constant pressure). be cooled to 1 atm ✘ = 0.955 atm 760✘✘ torr 44◦ C = (273 + 44) K = 317 K ✘ 726✘✘ × torr For this problem, we know that volume and the number of moles gas are kept constant (since the balloon is sealed, no gas can enter/escape), therefore: P V = nRT P T = constant = nR V therefore, we can compare the initial and final cases: Tinitial = constant = Vinitial Tf inal Tf inal Vf inal Tinitial = · Vf inal Vinitial 317.15 K ✚ = · 3.78✚ L ✚ 4.39✚ L = 273.08 K = (273.08 − 273.15) K ≈ 0 K (B) This matches what we would expect, as the decreasing temperature causes a decrease in the volume of the balloon. For example, the demo in lecture where a balloon was placed in liquid nitrogen, causing it to shrink. 35. A sample of gas (1.9 mol) is in a flask at 21◦ C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new pressure is 795 mm Hg and the temperature is now 26◦ C. There is now mole of gas in the flask. 21◦ C = (273 + 21) K = 294 K 8 26◦ C = (273 + 26) K = 299 K 1 atm ✘ 697✘✘✘✘ × mm Hg ✘ = 0.917 atm 760✘✘✘✘ mm Hg 1 atm ✘ 795✘✘✘✘ × mm Hg ✘ = 1.05 atm 760✘✘✘✘ mm Hg This problem has a fixed volume (since the flask does not change). Therefore: P V = nRT V nT = constant = R P Comparing the initial and final cases gives: ni T i = constant = Pi nf nf T f Pf n f T f Pi = · Pf Ti ✘ ✚ 1.9 mol · 294.15✚ 1.046✘✘ K atm = · ✘ ✚ 0.917✘✘ atm 299.15✚ K = 2.1 mol (B) note: we must convert Celsius to Kelvin, as Celsius can have negative values. However, if we leave the pressure in mm Hg instead of atm, we get the same answer, as the units for pressure will cancel in both cases. nf 1.9 mol · 294.15 K · 697 mm Hg 1.9 mol · 294.15 K = · 697 mm Hg ✚ 1.9 mol · 294.15✚ K = · ✘ ✘ 697✘✘✘ mm Hg = 2.1 mol = 795 mm Hg 299.15 K 795 mm Hg 760 mm✘✘ Hg 1 atm ✟ ✟ · ✘✘✘✘ · ✟✟ Hg 299.15 K 1 atm 760 mm ✟✟ ✘ ✘✘ 795✘✘ Hg mm ✚ 299.15✚ K 36. Which of the following statements about gases is false? E) All gases are colorless and odorless at room temperature. There are many gases which are colored (eg. NO2 is brown) and have odors (eg. H2 S smells like rotten eggs). 37. Calcium hydride (CaH2 ) reacts with water to form hydrogen gas: CaH2 (s) + 2 H2 O(l) → Ca(OH)2 (aq) + 2 H2 (g) 9 How many grams of CaH2 are needed to generate 48.0 L of H2 gas at a pressure of 0.888 atm and a temperature of 32◦ C? First, we need to solve for the number of moles of hydrogen gas, using the ideal gas law. Then, we can use the balanced chemical equation to find the moles of Calcium Hydride. 32◦ C = (273 + 32) K = 305 K PV n H2 = nRT PV = RT ✘ ✚ 0.888✘✘ · 48.0✚ atm L ✚ = K ✘ ✘✘ · 305.15✚ L 0.08206 ✘ atm mol ￿ K = 1.07 mol H2 ✭ 1.072 mol H2 × 1✭✭✭✭✭ mol CaH2 42.094 g CaH2 × ✭ = 35.8 g CaH2 (D) 2✘✘✘✘ mol H2 1✭✭✭✭✭ mol CaH2 38. An absorbance of 0.234 at 520 nm is measured for a 0.0100 M solution of compound X. The cuvette has a path length of 1.00 cm. What is the absorbance of the compound X in a 4.82×10−3 M solution? For this problem, the molar absorptivity and cuvette path length are constant. Therefore, using Beer’s Law A520 = ￿520 · b · C A520 ￿520 · b = constant = C Af Ai = Ci Cf Ai Af = · Cf Ci 0.234 −3 ✚ = M ✟ · 4.82 × 10 ✚ 0.001✟ M = .113(B ) We can also calculate the molar absorptivity: A520 = ￿520 · b · C A520 ￿520 = b·C 0.234 = 1.00 cm · 0.01 M = 23.4 M−1 cm−1 10 And use this to find the absorbance at 4.82×10−3 M A520 = ￿520 · b · C = .113 ✚ = 23.4✭✭✭✭−1 · 1.00✘✘ · 4.82 × 10−3 ✚ M−1 cm cm M ✭ ✭ 39. C2 H2 is a compound you investigated in the experiment ”How the nose knows”. Each molecule of this compound contains . C) 3 σ bonds and 2 π bonds. C2 H2 , or acetylene, has a one C-C triple bond (1 σ , 2 π ) and two C-H bonds (1 σ each), giving 3 σ bonds and 2 π bonds. 40. In the experiment ”Determination of the molarity of a strong acid”, if you titrated passed the end point until the indicator turned yellow, how would that affect calculated concentration of the HCl? A) The concentration would be lower than the true value. If you titrated past the end point, then you would have put in more HCl solution than necessary to reach the endpoint. However, the amount of TRIS in the flask would remain constant. We calculate the HCl concentration as follows: [HCl] = mol TRIS vol HCl from buret Since the volume from the buret will be larger than required, the [HCl] will be lower than the true value 11 ...
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This note was uploaded on 05/01/2011 for the course CHEM 1A taught by Professor Nitsche during the Spring '08 term at University of California, Berkeley.

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