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Chapter 12 Chemical Bonding

Chapter 12 Chemical Bonding. QUESTION. Most chemical bonds consist of electrostatic attractive forces and are called ____ bonds, or of shared electrons and are called _____bonds. electric; shared ionic; covalent ionic; molecular electronic; coordinate. ANSWER.

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Chapter 12 Chemical Bonding

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  1. Chapter 12Chemical Bonding

  2. QUESTION • Most chemical bonds consist of electrostatic attractive forces and are called ____ bonds, or of shared electrons and are called _____bonds. • electric; shared • ionic; covalent • ionic; molecular • electronic; coordinate

  3. ANSWER • Choice #2 properly identifies the bonds as ionic and covalent. • Section 12.1: Types of Chemical Bonds

  4. QUESTION • When electrons in a covalent bond are shared equally the bond is _________, but if the electrons are not shared equally the bond is ________, which means that it has a positive side and a negative side. • ionic; covalent • strong; weak • nonpolar; polar • stable; unstable

  5. ANSWER • Choice #3 is the correct answer. • Section 12.1: Types of Chemical Bonds

  6. QUESTION • When considering a bond between two atoms, the greater the difference in ____________, the more __________is the bond. • polarity; divided • atomic weight; nonpolar • electronegativity; polar • electronegativity; nonpolar

  7. ANSWER • Choice #3 correctly expresses the idea that bonds become more polar as the difference in the electronegativity of the atoms increases. • Section 12.2: Electronegativity

  8. QUESTION • Rank the following bonds from least polar to most polar: • Si-Cl P-Cl Mg-Cl S-Cl • 1. P-Cl, S-Cl, Si-Cl, Mg-Cl • 2. Mg-Cl, Si-Cl, P-Cl, S-Cl • 3. Mg-Cl, S-Cl, P-Cl, Si-Cl • 4. S-Cl, P-Cl, Si-Cl, Mg-Cl

  9. ANSWER • Choice #3 is the correct answer. Bonds become more polar as the difference in the electronegativity of the atoms increases. • Section 12.2: Electronegativity

  10. QUESTION • Choose the bond that is the most polar. • Sr–O • C–O • N–N • Fe–P

  11. ANSWER • Choice #1 is the correct answer. Bonds become more polar as the difference in the electronegativity of the atoms increases. • Section 12.2: Electronegativity

  12. QUESTION • Which of the following bonds would be the least polar yet still be considered polar covalent? • Mg–O • N–O • Si–O • O–O

  13. ANSWER • Choice #2 is the correct answer. To be considered polar covalent, unequal sharing of electrons must still occur. Choose the bond with the least difference in electronegativity yet there is still some unequal sharing of electrons. • Section 12.2: Electronegativity

  14. QUESTION • Which of the following bonds would be the most polar without being considered ionic? • Mg–O • N–O • Si–O • O–O

  15. ANSWER • Choice #3 is the correct answer. To not be considered ionic, generally the bond needs to be between two nonmetals. The most polar bond between the nonmetals occurs with the bond that has the greatest difference in electronegativity. • Section 12.2: Electronegativity

  16. QUESTION • The difference in electronegativity between hydrogen and iodine is (4.0 – 2.5) = 1.5, which means that hydroiodic acid has a(n) ________________. • low boiling point • low acidity • unstable gas phase • dipole moment

  17. ANSWER • Choice #4 associates the bond polarity of a diatomic molecule with having a dipole moment. • Section 12.3: Bond Polarity and Dipole Moments

  18. QUESTION • If we consider the electron configuration of strontium, [Kr]5s2, and that of oxygen, [He]2s22p4,both atoms will attain stable noble gas electron configurations by the transfer of ____ electron(s). This will give Sr a charge of ___ and O a charge of ___. Hence the ionic compound formed has the formula ___________ and is named strontium oxide. • 1 ; 1+ ; 1– ; SrO • 2 ; 2+ ; 2– ; SrO • 2 ; 2+ ; 1– ; SrO • 2 ; 2– ; 2+ ; SrO

  19. ANSWER • Choice #2 correctly indicates the transfer of two electrons from strontium to oxygen to form the ionic compound known as strontium oxide. • Section 12.4: Stable Electron Configurations and Charges on Ions

  20. QUESTION • What is the expected ground state electron configuration for Te2–? • 1. [Kr]5s24d105p4 • 2. [Kr]5s24d104f145p6 • 3. [Kr]5s24d105p6 • 4. [Ar]5s24d105p2

  21. ANSWER • Choice #3 is the correct answer. Te2–contains 54 electrons and has the noble gas configuration of Xe. • Section 12.4: Stable Electron Configurations and Charges on Ions

  22. QUESTION • What is the correct electron configuration for the most stable form of the sulfur ion in an ionic compound? • 1. 1s22s22p63s2 • 2. 1s22s22p63s23p2 • 3. 1s22s22p63s23p4 • 4. 1s22s22p63s23p6

  23. ANSWER • Choice #4 is the correct answer. S2–is the most stable form of the sulfur ion in an ionic compound. It contains 18 electrons and has the noble gas configuration of Ar. • Section 12.4: Stable Electron Configurations and Charges on Ions

  24. QUESTION • The structures of ionic compounds are usually described as the packing of ____________ with smaller ____________ fitting into the interstices. • anions; cations • anions; electrons • cations; anions • cations; electrons

  25. ANSWER • Choice #1 correctly describes the packing of anions (which tend to be larger) into specific patterns known as crystal lattices, while cations fit into the spaces or interstices between the packed anions. • Section 12.5: Ionic Bonding and Structures of Ionic Compounds

  26. QUESTION • Rank the following from smallest to largest atomic radius: • Ar, S2–, Ca2+, K+, Cl– • 1. Ar < K+ < Ca2+ < S2– < Cl– • 2. Ca2+ < K+ < Ar < Cl– < S2– • 3. Ar < Cl– < S2– < Ca2+ < K+ • 4. S2– < Cl– < Ar < K+ < Ca2+

  27. ANSWER • Choice #2 is the correct answer. They all have the same electron configuration of the noble gas Ar. Therefore, nuclear charge becomes very important in determining the sizes relative to each other. The higher the nuclear charge, the smaller the ion/atom. • Section 12.5: Ionic Bonding and Structures of Ionic Compounds

  28. QUESTION • Which atom or ion has the smallest radius? • O2+ • O+ • O • O2–

  29. ANSWER • Choice #1 is the correct answer. As electrons are removed, the effective nuclear charge has a stronger effect and thus makes the ion, O2+, the smallest. • Section 12.5: Ionic Bonding and Structures of Ionic Compounds

  30. QUESTION • Which of the following has the lowest ionization energy? • Se2– • Br– • Sr2+ • Rb+

  31. ANSWER • Choice #1 is the correct answer. Se2– has the smallest effective nuclear charge and thus does not bind the electrons as strongly as the others. • Section 12.5: Ionic Bonding and Structures of Ionic Compounds

  32. QUESTION • Lewis structures show the arrangement of ________electrons in an atom or ion. • all • core • valence • missing

  33. ANSWER • Choice #3 correctly indicates that only valence electrons are included in Lewis structures. • Section 12.6: Lewis Structures

  34. QUESTION • In the Lewis structure for H2S there are a total of ______ electrons and ____ pair(s) of nonbonding electrons. • 9 ; 2 • 9 ; 1 • 8 ; 2 • 8 ; 1

  35. ANSWER • Choice #3 correctly describes the Lewis structure for hydrosulfuric acid: (6 electrons from S) + (2 × 1 = 2 electrons from H) = 8 total electrons. The Lewis structure is the same as that for H2O, with S in place of O. • Section 12.6: Lewis Structures

  36. QUESTION • The Lewis structure for SO2 contains ____ total electrons and____ nonbonding pairs of electrons, as well as one ______ bond and one _______ bond between the central sulfur and the oxygen atoms. • 16 ; 4 ; single ; double • 16 ; 6 ; single ; double • 18 ; 4 ; single ; triple • 18 ; 6 ; single ; double

  37. ANSWER • Choice #4 correctly describes the SO2 molecule with (3 × 6 =) 18 electrons and S located between the two O’s with one single and one double bond. • Section 12.7: Lewis Structures of Molecules with Multiple Bonds

  38. QUESTION • Consider the following compounds: • CO2 N2 CCl4 • Which compound exhibits resonance? • 1. CO2 • 2. N2 • 3. CCl4 • 4. At least two of the above compounds exhibit resonance.

  39. ANSWER • Choice #1 is the correct answer. Only CO2 exhibits resonance. • Section 12.7: Lewis Structures of Molecules with Multiple Bonds

  40. QUESTION • Which of the following supports why Lewis structures are not a completely accurate way to draw molecules? • 1. We cannot say for certain where an electron is located yet when drawing Lewis structures, we assume the electrons are right where we place them. • 2. When adding up the number of valence electrons for a molecule, it is possible to get an odd number which would make it impossible to satisfy the octet rule for all atoms. • 3. Both statements 1 and 2 above support why Lewis structures are not a completely accurate way to draw molecules. • 4. Lewis structures are the most accurate way to draw molecules and are completely correct.

  41. ANSWER • Choice #3 is the correct answer. • Section 12.7: Lewis Structures of Molecules with Multiple Bonds

  42. QUESTION • It is important to fully understand that Lewis structures are useful in determining the bonding relationships between atoms in a molecule, but that they do not directly provide a true picture of molecular shape. While the Lewis structure for methane, CH4, an important greenhouse gas, suggests a flat structure with 4 hydrogens arranged around a central carbon, the methane molecule is actually ________. • square planar • trigonal • tetrahedral • octahedral

  43. ANSWER • Choice #3 correctly describes the methane molecule as a tetrahedron, with the carbon centrally located in the geometric center and the 4 hydrogens forming the corners of the tetrahedron. • Section 12.8: Molecular Structure

  44. QUESTION • While the electron pair geometry of NH3 is _______, VSEPR predicts the molecular shape as _________, due to the pair of nonbonding electrons on the central N. • tetrahedral; trigonal pyramidal • trigonal planar; tetrahedral • tetrahedral; trigonal planar • Both are trigonal planar.

  45. ANSWER • Choice #1 answers both geometry questions correctly. The central N has four pairs of electrons around it, giving it a tetrahedral electron pair geometry. But because one of the corners of the tetrahedron is an electron pair, the molecule is a trigonal pyramid with the N forming the apex and the three H’s forming the pyramidal base. • Section 12.9: Molecular Structure: The VSEPR Model

  46. QUESTION • Draw the Lewis structures for the following compounds: • CBr2H2 BH3 HCl • Which compound has bond angles of 109.5˚ around the central atom? • BH3 • CBr2H2 • HCl • At least two of the above compounds have bond angles of 109.5˚.

  47. ANSWER • Choice #2 is the correct answer. CBr2H2 is tetrahedral, BH3 is trigonal planar, and HCl is linear. • Section 12.9: Molecular Structure: The VSEPR Model

  48. QUESTION • In determining the shape of the SO2 molecule we examine the • Lewis structure and find the central S atom attached to O’s via • one single and one double bond. The electron pair geometry is • _______ and the molecular geometry is _________, with bond • angles of ____ degrees. • trigonal planar ; linear ; 180 • trigonal planar ; bent ; 120 • tetrahedral ; linear ; 120 • tetrahedral ; trigonal planar ; 120

  49. ANSWER • Choice #2 correctly describes the electron pair about the central S as trigonal planar. S has one nonbonding electron pair, a single bond, and a double bond (counts as a single bond for shape). Because one corner of the triangle is occupied by an electron pair, the molecule has a bent shape with bond angles of 120o. • Section 12.10: Molecular Structure: Molecules with Double Bonds

  50. QUESTION • Draw the Lewis structures for the following compounds: • HCN NH4+ NO2– • Which compound has bond angles of 120˚ around the central atom? • HCN • NH4+ • NO2– • At least two of the above compounds have bond angles of 120˚.

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