chem - Chemistry 101 ANSWER KEY REVIEW QUESTIONS Chapter 9...

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Unformatted text preview: Chemistry 101 ANSWER KEY REVIEW QUESTIONS Chapter 9 1. For each set shown below, select the atoms or ions that are isoelectronic with each other, and write their electron configuration: + + 2+ a) K , Rb , Ca + 2+ 2 6 K and Ca are isoelectronic [Ne] 3s 3p 2– 2– b) S , Ar , Se 2– S and Ar are isoelectronic 2 6 [Ne] 3s 3p 2+ 3+ Mg and Al are isoelectronic 2 2 6 1s 2s 2p 2+ – 3+ c) Mg , Cl , Al 2. Explain each of the following trends in ionic radii: – + a) I > I > I – All species have 5 energy levels. I has the largest number of electrons (54) and has the + largest inter­electron repulsion, making it the largest. I has the lowest number of electrons (52) and has the lowest inter­electron repulsion, making it the smallest. 2+ 2+ 2+ b) Ca > Mg > Be 2+ 2+ 2+ Ca has 3 energy levels, Mg has 2 energy levels, while Be has only one energy level. 2+ 2+ Therefore Ca is the largest and Be is the smallest. 3. Arrange the atoms or ions in each of the following sets in order of increasing radius: – + 2+ a) Br , Na , Mg 2+ + – Mg < Na < Br – 2– b) Ar, Cl , S – 2– Ar < Cl < S 3+ 2+ 3+ c) Co , Fe , Fe 3+ 3+ 2+ Co < Fe < Fe + – 2+ 3– d) K , Cl , Ca , P 2+ + – 3– Ca < K < Cl < P In an isoelectronic series size increases as the Zeff decrease (charge becomes less positive) 1 4. Draw Lewis structures for each of the following structures and assign formal charges to each atom: a) SF2 20 electrons 0 0 0 b) NH2OH (N and O are bonded to one another) 14 electrons 0 0 0 0 0 c) PO4 3– 32 electrons –1 –1 +1 3– –1 –1 d) I3 – 22 electrons –1 0 1– 0 5. Draw two possible resonance structures for the isocyanate ion (NCO – ) and using formal charges determine which structure is more plausible. 16 electrons 1– –1 0 1– 0 0 0 (A) –1 (B) Both structures possess the same magnitude of formal charges. However, structure B is more plausible since the negative charge is carried by oxygen which is more electronegative than nitrogen. 2 6. Arrange the bonds in each of the following sets in order if increasing polarity: a) C ¾ S B ¾ F N ¾ O 2.0 0.5 S ¾ Br C ¾ P DEN 0 b) O ¾ Cl C ¾S < N ¾ O < B ¾F DEN 0.5 0.3 S ¾ Br < C ¾ P < O ¾ Cl 0.4 7. Classify each of the following bonds as ionic, polar covalent or non­polar covalent: a) B ¾ Cl polar covalent DEN= 1.0 b) Mg ¾ Br polar covalent DEN= 1.6 c) Cl ¾ Cl non­polar covalent DEN= 0 d) Na ¾ Br ionic DEN= 1.9 8. Use bond energies listed in Table 9.5 in your textbook to find DH for the reactions shown below: ¾¾ ® a) Bond Breaking (BB) C=C 1 x 602 = 602 kJ Bond Forming (BF) C¾C 1 x 346 = 346 kJ O¾O 1 x 142 = 142 kJ C¾O 2 x 358 = 716 kJ DH = S BB – S BF = (602+142) – (346+716) = –318 kJ b) CH2NH + H2O ¾¾ ® CH2O + NH3 ¾¾ ® Bond Breaking (BB) C=N 1 x 615 = 615 kJ O¾H 2 x 459 = 918 kJ Bond Forming (BF) C=O 1 x 745 = 745 kJ N¾H 2 x 386 = 772 kJ DH = S BB – S BF = (615+918) – (745+772) = +16 kJ 3 9. Use the data provided below to calculate the lattice energy of RbCl. Is this value greater or less than the lattice energy of NaCl? Explain. Electron affinity of Cl = –349 kJ/mol st 1 ionization energy of Rb = 403 kJ/mol Bond energy of Cl2 = 242 kJ/mol Sublimation energy of Rb= 86.5 kJ/mol DHf [RbCl (s)] = –430.5 kJ/mol Rb (s) + ½ Cl2 (g) ® RbCl (s) This equation can be written as the sum of the following: Rb (s) ® Rb (g) sublimation DH1= +86.5 kJ/mol + – Rb (g) ® Rb + e st 1 ionization energy DH2= +403 kJ/mol ½ Cl2 (g) ® Cl DH3= +121 kJ/mol bond energy of Cl2 Cl + e ® Cl electron affinity of Cl DH4= –349 kJ/mol Rb + Cl ® RbCl lattice energy DH5= ??? – + – – Rb (s) + ½ Cl2 (g) ® RbCl (s) DHf = –430.5 kJ/mol DHf = DH1 + DH2 + DH3 + DH4 + DH5 Lattice energy = DH5 = DHf – (DH1 + DH2 + DH3 + DH4) = –430.5 – (86.5 + 403 + 121 – 349) = –692 kJ This value would be expected to be smaller than NaCl (–786 from notes 9A). This is because Rb is a larger ion than Na and would be further apart from the anion. Lattice energy is inversely proportional to the distance between the ions. 10. Arrange the following compounds in order of increasing lattice energy: NaF CaO CsI CsI < NaF < CaO Calcium has a +2 ion and oxygen has –2 ion, while both NaF and CsI possess +1 and –1 charges. Since lattice energy is directly proportional to the charges, CaO would have the largest value. Sodium ion and fluoride ions are smaller than cesium and iodide ions. Since lattice energy is inversely proportional to the size of the ions, CsI would have the lowest value. 4 ...
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