Bonding Metals versus non-metals AP Question 1984 Discuss some differences in physical and chemical properties of metals and nonmetals. What characteristic of the electronic configurations of atoms distinguishes metals from nonmetals? On the basis of these characteristics, explain why there are many more metals than nonmetals. Answer: Physical properties: Melting points Electricity Luster (shine) Physical state solids metals relatively high conductivity (due to mobile electrons) high most solids*** Ductile yes non-metals relatively low good insulators (due to a lack of mobile electrons) little or none gases, liq. or (mostly gases) no *** Most metals are solids due to the strong attraction of the positive metal ions being so very attracted to the free mobile electrons Chemical properties: Oxides Reacts with metals reducing electropositive basic or amphoteric nonmetals Electronegativity generally low Redox agents non-metals oxid. or reducing electronegative acidic metals & nonmetals generally high Electron configurations: Metals: Outermost electrons filling in s or d sublevels of their atoms. (A few heavy elements have atoms with a couple of electrons in p sublevels.) Nonmetals: Valence electrons in the s and p sublevels of their atoms. There are more metals than nonmetals because filling d orbitals in a given energy level involves the atoms of ten elements and filling the f orbitals involves the atoms of 14 elements. In the same energy levels, the maximum number of elements with atoms receiving p electrons is six. Types of Bonding Atoms form molecules because atoms want to have a full outer valence shell. This means that they usually will have eight electrons in their outer shells (s and p sublevels). If atoms gain or lose electrons, then they will form ionic bonds. If atoms share electrons, then they have covalent bonds. The electronegativity difference determines the bond character. A difference of 00.3 is designated as a NON-POLAR covalent bond while a difference of 0.311.67 is considered POLAR covalent and a difference greater than 1.67 is IONIC since the electrons are spending most of their time around only one of the nuclei. Ionic bonds occur between metals and non-metals (high difference between electronegative values). Metal loses electrons (cation) and nonmetals gain electrons (anion). Covalent bonds occur when two atoms are sharing electrons. Each atom counts the shared pair of electrons in its valence shell, thereby fulfilling the octet rule. THE NUMBER OF COVALENT BONDS THAT AN ATOM CAN FORM IS THE SAME AS THE NUMBER OF ELECTRONS IN ITS VALENCE SHELL. The first covalent bond formed between two atoms is a sigma (σ) bond and the second and/or third is a pi (π) bond. Pi’s are formed from the side to side overlap of unhybridized p orbitals. SINGLE AND MULTIPLE BONDS: • • • • single bond–one pair of electrons shared = sigma (σ) bond • sigma’s form from the overlap of an s and a p, or an s and an s, OR a p and a p that are UNHYBRIDIZED MULTIPLE BONDS ARE MOST OFTEN FORMED BY C,N,O, P and S ATOMS [pronounced “SEE-NOPS”] double bond–two pairs of electrons shared. One σ bond and one π bond triple bond–three pairs of electrons shared. One σ bond and two π bonds Coordinate covalent bond: When one atom provides both electrons to form a sigma bond. Some atoms such as N and P tend to share a lone pair with another atom that is short of electrons, leading to the formation of a coordinate covalent bond. All complex “coordinated” ions contain a coordinate covalent bond - this is also where the “coordination” number concept comes from. THINK: Lewis Acid-Base reactions – see the definition of Lewis acids and bases. “Non-octets” • Fewer than eight – H has, at most, only 2 electrons! BeH2, only has 4 valence electrons around Be! Boron compounds, only have 6 valence electrons! Hydrogen = 2 electrons Beryllium = 4 electrons Boron = 6 electrons • Expanded Valence–3rd or higher period can be surrounded by more than four valence pairs in certain compounds. The number of bonds depends on the balance between the ability of the nucleus to attract electrons and the repulsion between the pairs. • Odd-electron compounds–A few stable compounds contain an odd number of valence electrons and thus cannot obey the octet rule: NO, NO2 , and ClO2 Hybridization: Hybridization occurs when an atom moves electrons around within the SAME energy level to provide more electrons for bonding; this may include a central atom expanding its octet to include the d sublevel. We can think of this in terms of “sites” of electron density surrounding an atom. A site can be a lone pair, or a bond of any variety. Double and triple bonds count as ONE site—they are just more electron dense than singles or lone pairs. THIS IS WHY I ALWAYS SAY, “MULTIPLE BONDS COUNT AS A SINGLE BOND FOR GEOMETRY”. We can also think of this as pairs of electrons around the central atoms. Pairs of electrons = “sites” and multiple bonds count as a single bond i.e. only one site. Hybridization— use as many letters as there are sites starting with s, then p’s then d’s sp sp2 sp3 sp3d # of “Sites” of Electron Density -or# of pairs of electrons around central atom Bond Angle 2 3 4 5 sp3d2 6 180 120 109.5 3 @ 120 & 2 @ 90 all 90 Example Carbon in CO2 Carbon in CO32Carbon in CH4 Phosphorous in PF5 Sulfur in SF6 To determine the hybridization: 1. Draw the Lewis structure then count the pairs of electrons around the central atom (or sites). 2. Use as many letters as there are pairs of electrons (remember multiple bonds count as a single bond for this) or count the “sites”. 4 sites, 4 letters therefore sp3 tetrahedral. 3. Only expanded octet elements exhibit hybrids involving d orbitals. This is WHY they had to be from at least the 3rd principal energy level If you see d2sp3 it just means the d’s that hybridized were from the LOWER energy level than the s and/or p’s. Lewis Dot Diagrams Rules for Lewis Dot Diagrams 1) 2) 3) 4) 5) 6) 7) 8) 9) Determine the central atom. The central atom is: a) The least electronegative element. b) The atom with the smallest number of valence electrons. c) The “oddball” element. d) The central atom can never be hydrogen. Count the number of valence electrons in each atom. Include the number of electrons lost or gained due to charge. Total the number of valence electrons. Draw the skeletal structure with the central atom in the middle. Draw the first sigma bonds connecting all atoms to the central atom. Subtract the total number of electrons involved in the sigma bonds (two electrons for each bond) from the total number of electrons. Finish drawing lines around the atoms up to the number of electrons left. Work from the outside in. Make sure that all atoms satisfy the octet rule. If they don’t, move electrons around to form multiple bonds (double or triple) so that everything does have a full octet. See-NOPS only. 10) If it is an ion, put [square brackets] around the diagram and put the charge outside. Almost always atoms will have an EVEN number of electrons, but compounds of nitrogen may not – nitrogen may have an odd number of electrons around it. Resonance forms: When a molecule has more than one way to draw its structure it will have resonance structures. This occurs ONLY with multiple bonds. A double arrow (⇔) separates the pictures in the resonance diagrams. Valence Shell Electron Pair Repulsion Theory (VSEPR Theory) VSEPR is a theory that allows students to determine the shape of the molecule and the bond angles in the molecule. This is done by: 1. Drawing the Lewis Dot Diagram. 2. Determining the number of bonding pairs of electrons around the central atom. 3. Determining the number of unbonded pairs of electrons around the central atom. Reference chart on next page → V alence S hell E lectron P air R epulsion Theory sp2 sp Structural Bonded Lone Pairs Pairs (σ) Pairs . Geometry 2 0 Linear 180 3 3 0 120 2 1 Triangular planar Bent < 120 4 0 Tetrahedral 109.5 3 1 < 109.5 2 2 Triangular pyramidal Bent 5 0 4 sp3 sp3d1 5 6 sp3d2 Bond angle 2 4 Hybridization Molecular << 109.5 1 Triangular bipyramidal See-saw 120 & 90 < 120 & 90 3 2 T-shape 90 2 3 Linear 180 6 0 Octahedral 90 5 1 90 & < 90 4 2 Square pyramidal Square planar 3 3 T-shape < 90 2 4 Linear 180 90 Practice Draw the Lewis Dot Diagrams for CCl4, NH3, PCl5, PCl3, SF6, SF4, and SF2. Explain each of their shapes and their hybridization. Typical AP question: Account for the fact that PCl3, NCl3 and PCl5 exist, but NCl5 does not. Answer: Draw the arrow diagram for nitrogen and for phosphorus and use hybridization. Other Terms Bond order = The number of bonding electron pairs shared by two atoms in a molecule. Easiest way to find this is use the formula: Bond Order = # of bonds in a " site" # of resonance structures Bond length = The distance between the nuclei of two bonded atoms. Bond energy = The energy supplied to a gaseous molecule to separate two of its atoms. This is also called the bond dissociation energy. The process of breaking bonds in a molecule is endothermic, as energy must be added. THIS IS ANOTHER WAY TO DETERMINE ΔHrxn : ΔHrxn = Σ bonds broken – Σ bonds formed Bond polarity – Determines if the electrons of a bond are being shared equally or unevenly. If a bond does not share its electrons equally it is said to be polar. One end of the bond must be positive and the other must be negative. If the electrons in the bonds are shared evenly, then the bond is said to be non-polar. Determines the Intermolecular Forces involved in the molecule and explains the behavior of the molecule. Determining the Electronegativity Values For Elements H = 2.1 Li = 1.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O = 3.5 F = 4.0 Na = 0.9 Mg = 1.2 Al = 1.5 Si = 1.8 P = 2.1 S = 2.5 Cl =3.0 K = 0.8 Ca = 1.0 Ga = 1.6 Ge = 1.8 As = 2.0 Se = 2.4 Br =2.8 Rb = 0.8 Sr = 1.0 Y = 1.2 Sn = 1.8 Sb = 1.9 Te = 2.1 I = 2.5 Cs = 0.7 Ba = 0.9 ****** Pb = 1.9 Bi = 1.9Po = 2.0 At = 2.2 ****** La Lu = 1.0 1.2 Another way of looking at the type of bond that you have is to look at the difference in the electronegativity values. There are two types of covalent bonds: non-polar and polar. Although different textbooks classify them differently, we will use the point breakdown below. Non-polar covalent 0.0 0.3 Polar covalent weakly IONIC 1.67 4.0 strongly 100% covalent 0.0 % ionic 0.0 % covalent 100% ionic The differences in the electronegativity values determine the type of bond. The shape of the molecule and the type of bond determine if the molecule is polar or non-polar. • • A molecule can be non-polar with symmetrical polar bonds. This can only be in a molecule that has ZERO unbonded pairs of electrons around the central atom. Molecular polarity = if a molecule has dipoles on its surface, due to its shape and the polarity of the bonds that make it up. Dipole Moment = a measurement of the polarity of a bond or molecule. Generally, the more polar the bond, the greater the dipole moment. Intermolecular Forces Polar bonds = When electrons are not shared equally, one end of the bond has a positive dipole and the other end has a negative dipole. Must use electronegativity values to determine the type of bond. Polar molecules = molecules are polar if one end of the molecule is positive and the other is negative. Molecules that are symmetrical are non-polar. The only shapes of molecules that COULD be non-polar are linear (all three forms), triangular planar (trigonal planar), tetrahedral, triangular bipyramidal (trigonal bipyrimidal), square planar and octahedral. Just because a molecule is one of these shapes does not mean that it will be non-polar. It must be symmetrical. Take for instance, CCl4 and CCl3F – both are tetrahedral, but only CCl4 is non-polar because it is symmetrical down each one of the axes (the bond between the central atom (carbon) and the attached atoms (either chlorine or fluorine). CCl4 Symmetrical Non-polar CCl3F Non-symmetrical Polar Intermolecular Attractions The weak attraction between molecules (within a molecule it is called “intramolecular”) due to arrangement of electrons. Dipole to dipole interactions only has about 1% the strength of a covalent bond. Named after Johannes van der Waals, these are called van der Waals forces. Weakest to the strongest: • • • • London Dispersion • Induced dipole to induced dipole Induce dipole to dipole Dipole to dipole Hydrogen bonding Dipole interactions • London Dispersion Forces = caused by the motion of the electrons. As electrons move around the nucleus, they can create an asymmetrical distribution that results in temporary dipoles. If the atom or molecule is moving slowly enough (i.e. very low temperature) then these dipoles can cause a neighboring atom/molecule to form a dipole. It is this interaction that causes the noble gases to liquefy. Named for Fritz London and not the city in England. • • • Strength increases with increased number of electrons (more electrons means that there are more polarizable electrons and the stronger the resulting dipoles are) • Strength increases with more mass. explains behavior of noble gases explains behavior of the halogens • Dipole interactions occur when polar molecules are involved • Hydrogen bonding = attractive forces in which a hydrogen atom that is bonded to a highly electronegative atom has a very strong attraction to an unbonded pair of electrons. IMF allows gases to become a liquid. IMF is involved with the Ideal Gas Law The Ideal Gas Law says gas particles have no volume and they don’t interact with each other (intermolecular forces). This is an oversimplification of the following equation. You can notice the terms for the intermolecular forces and the volume of the particles. When using the Ideal Gas Law, the terms “a” and “b” are both equal to zero and then the equation just becomes PV = nRT [P + an2][V- bn] = nRT v2 Correction for Intermolecular forces Correction for molecular volume PHASES OF MATTER AND THE KINETIC MOLECULAR THEORY (a) The volume of the molecules themselves [when in the gas phase] is negligible (b) No attractive forces between molecules (c) Molecules are in constant random motion (d) KE is determined by temperature only • • compressibility—due to (a) above—not appreciable in the liquid or solid phase intermolecular forces—KT assumes weak forces • expansion—occurs when KE overcomes attractive IMF’s; gases that resist have strong IMF’s • • pouring—due to IMF’s in solids and liquids condensation—molecules lose their KE [especially upon collision with each other and with cooler surfaces] and the attractive forces “kick” in. INTERMOLECULAR FORCES Collectively called van der Waals forces. These are the attractive forces between molecules, between ions, or between ions and molecules • ION-DIPOLE forces—weaker than ion/ion interactions; occurs when a polar molecule is attracted to an ion • • • Distance—the larger the interacting species, the weaker this force between them Charge—follow Coulomb’s Law; the greater the charge the stronger the force of attraction Magnitude—the greater the magnitude of the dipole [larger dipole moment], the stronger the attraction • • • hydrated—metals bond to several water molecules; porcelain also holds water heat or energy of hydration—when the water sets up an IMF with the metal, energy is released in the process • energy of solvation—when “hydration” occurs with something other than water • acids—H3O+ is a hydrated proton DIPOLE-DIPOLE forces—Energy is released upon interaction; Energy is required to separate interacting dipoles [high MP, high BP, high ΔHvap] • • Energy released when polar molecules interact with one another, and this is one of the reasons you must cool a gas consisting of polar molecules to convert it into a liquid, also reduces KE. Hydrogen bonding—really, it’s not a bond, it’s an attractive force that is NEVER to be confused with a bonded hydrogen!! • • • • Two requirements for this IMF. It’s a dipole-dipole interaction that involves an H that is bonded to a highly electronegative atom • • N, O, or F—very highly electronegative atoms • That THEN interacts with a pair of unbonded electrons on a highly electronegative atom on an adjacent molecule. Cl or S—weaker, yet still high in electronegative, however, size of the atoms (since period 3 elements) can offset the effects. • It’s a really big deal in Biology! That highly electronegative atom requirement keeps lots of hydrocarbons from exhibiting H-bonding • About 10% the strength of a covalent bond. melting point vs. intermolecular forces—the stronger the IMF the higher the MP Interactions Involving Induced Dipoles—without this, Br2 wouldn’t be a liquid nor would I2 be solid! • Dipole-Induced Dipole forces—a polar molecule induces a temporary situation where a NON-polar molecule becomes slightly polar since the electron cloud is mushy • Induced Dipole-Induced Dipole forces—since the electron cloud is mushy and electrons move all the time, a molecule may build up a concentration of electrons on one side of the surface leaving a slight positive charge on the opposite side—this sets off a chain reaction! Now very nonpolar molecules are behaving as if they were slightly polar—happens in iodine all the time! The larger the molecule, the more likely this will happen. • London or dispersion forces—another name just to confuse you! Melting points vs. dispersion forces—again, the stronger the IMF, the higher the MP since the attractive forces must be overcome by increasing the KE [temperature] Although the term “van der Waals forces” usually refers to ALL IMF, other terms may be used interchangeably with it. Those terms are: • • • dispersion forces London forces dipole-induced dipole forces AP Multiple Choice Questions: 1984 Questions 8 - 9 (A) (B) (C) (D) (E) A network solid with covalent bonding A molecular solid with zero dipole moment A molecular solid with hydrogen bonding An ionic solid A metallic solid 8. Solid ethyl alcohol, C2H5OH 9. Silicon dioxide, SiO2 18. Hydrogen Halide HF HCl HBr HI Normal Boiling Point, °C +19 - 85 - 67 - 35 The liquefied hydrogen halides have the normal boiling points given above. The relatively high boiling point of HF can be correctly explained by which of the following? (A) (B) (C) (D) (E) HF gas is more ideal. HF is the strongest acid. HF molecules have a smaller dipole moment. HF is much less soluble in water. HF molecules tend to form hydrogen bonds. 40. The geometry of the SO3 molecule is best described as (A) (B) (C) (D) (E) 41. trigonal planar triangular pyramidal square pyramidal bent tetrahedral Which of the following molecules has the shortest bond length? (A) N2 51. (D) Br2 (E) I2 (B) C2H4 (C) CN-1 (D) C6H6 (E)CH4 Which of the following has no dipole moment? (A) HCN 80. (C) Cl2 Pi bonding occurs in each of the following species EXCEPT (A) CO2 60. (B) O2 (B) NH3 (C) SO2 (D) NO2 (E) PF5 For which of the following molecules are resonance structures necessary to describe the bonding satisfactorily? (A) H2S (B) SO2 (C) CO2 (D) OF2 (E) PF3 1989 a. b. c. d. e. hydrogen bonding hybridization ionic bonding resonance van der Waals forces (London dispersion forces) 11. Is used to explain why iodine molecules are held together in the solid state. 12. Is used to explain why the boiling point of HF is greater than the boiling point of HBr. 13. Is used to explain the fact that the four bonds in methane are equivalent. 14. Is used to explain the fact that the carbon-to-carbon bonds in benzene, C6H6, are identical. 17. The Lewis dot structure of which of the following molecules shows only one unshared pair of valence electron? (A) Cl2 47. (B) N2 CCl4, CO2, (C) NH3 PCl3, (D) CCl4 PCl5, (E) H2O2 SF6 Which of the following does not describe any of the molecules above? (A) Linear (B) Octahedral (C) Square planar (D) Tetrahedral (E) Triangular pyramidal 57. Which of the following compounds is ionic and contains both sigma and pi covalent bonds? (A) (D) Fe(OH)3 NO2 (B) (E) HClO NaCN (C) H2 S 1994 Questions 8-10 refer to the following diatomic species. (A) (B) (C) (D) (E) Li2 B2 N2 O2 F2 8. Has the largest bond-dissociation energy 9. Has a bond order of 2 10. Contains 1 sigma and 2 pi bonds 15. In a molecule in which the central atom exhibits sp3d2 hybrid orbitals, the electron pairs are directed toward the corners of (A) (B) (C) (D) (E) a tetrahedron a square-based pyramid a trigonal bipyramid a square an octahedron 57. Which of the following molecules have planar configurations? I. BCl3 II. CHCl3 III. NCl3 (A) (B) (C) (D) (E) I only III only I and II only II and III only I, II, and III I2(g) + 3 Cl2(g) 60. 2 ICl3(g) According to the data in the table below, what is the value of ΔHO for the reaction represented above? Bond I–I Cl – Cl I – Cl Average Bond Energy (Kilojoules/mole) 149 239 208 (A) -860 kJ (B) -382 kJ (C) +180 kJ (D) +450 kJ (E) +1,248 kJ 62. The electron-dot structure (Lewis structure) for which of the following molecules would have two unshared pairs of electrons on the central atom? (A) (B) (C) (D) (E) H2 S NH3 CH4 HCN CO2 68. Which of the following molecules has a dipole moment of zero? (A) (B) (C) (D) (E) C6H6 (benzene) NO SO2 NH3 H2 S 1999 Questions 13-16 refer to the following descriptions of bonding in different types of solids. (A) (E) Lattice of positive and negative ions held together by electrostatic forces Closely packed lattice with delocalized electrons throughout Strong single covalent bonds with weak intermolecular forces Strong multiple covalent bonds (including π-bonds) with weak intermolecular forces Macromolecules held together with strong polar bonds. 13. Cesium chloride, CsCl(s) 14. Gold, Au(s) 15. Carbon dioxide, CO2(s) 16. Methane, CH4(s) 32. Types of hybridization exhibited by the C atoms in propene, CH3CHCH2, include which of the following? (B) (C) (D) I. sp II. sp2 III. sp3 (A) (B) (C) (D) (E) I only III only I and II only II and III only I, II, and III 40. Of the following molecules, which has the largest dipole moment? (A) CO (B) CO2 (C) O2 (D) HF (E) F2 68. In which of the following processes are covalent bonds broken? (A) (B) (C) (D) (E) I2(s) I2(g) CO2(s) CO2(g) NaCl(s) NaCl(l) C(diamond) C(g) Fe(S) Fe(l) 74. Which of the following gases deviates most from ideal behavior? (A) SO2 (B) Ne (C) CH4 (D) N2 (E) H2 AP Multiple Choice Answers Question Answer % correct 1984 8 9 18 40 41 51 60 80 C A E A A E E B 66 56 65 54 49 56 34 45 1989 11 12 13 14 17 47 57 E A B D C C E 57 72 44 55 76 50 38 1994 8 9 10 15 57 60 62 68 C D C E A B A A 21 47 57 50 46 47 64 41 1999 13 14 15 16 32 40 68 74 A B D C D D D A 60 64 65 61 38 54 54 48 AP Free Response Questions 1990 D – Lewis Dot Diagrams Use simple structure and bonding models to account for each of the following. (a) The bond length between the two carbon atoms is shorter in C2H4 than in C2H6. (b) The H-N-H bond angle is 107.5o‚ in NH3. (c) The bond lengths in SO3 are all identical and are shorter than a sulfur-oxygen single bond. (d) The I3-1 ion is linear. (e) Molecules of NF3 are polar, but those of BF3 are not. 1989 D – Lewis Dot Diagrams CF4 XeF4 ClF3 (a) Draw a Lewis electron–dot structure for each of the molecules above and identify the shape of each. (b) Use the valence shell electron–pair repulsion (VSEPR) model to explain the geometry of each of these molecules. 1982 D – Lewis Dot Diagrams (a) Draw the Lewis electron–dot structures for CO32-, CO2, and CO, including resonance structures where appropriate. (b) Which of the three species has the shortest C-O bond length? Explain the reason for your answer. (c) Predict the molecular shapes for the three species. Explain how you arrived at your predictions. 1979 D – Lewis Dot Diagrams Draw Lewis structures for CO2, H2S, SO3 and SO32- and predict the shape of each species. 1990 – Lewis Dot Diagrams and Bond Energy ______________________________________________ Average Bond Dissociation Energies at 298 K ______________________________________________ Bond C-H C-C C-Cl Cl-Cl H-Cl Energy, kJ mol-1 414 347 377 243 431 ______________________________________________ The tables above contain information for determining thermodynamic properties of the reaction below. C2H5Cl(g) + Cl2(g) C2H4Cl2(g) + HCl(g) (a) Calculate the ΔHrxn for the reaction above, using the table of average bond dissociation energies. (b) Calculate the ΔSrxn for the reaction at 298 K, using data from either table as needed. (c) Calculate the value of Keq for the reaction at 298 K. (d) What is the effect of an increase in temperature on the value of the equilibrium constant? Explain your answer. 1992 D – Lewis Dot Diagrams and IMF Explain each of the following in terms of atomic and molecular structures and/or intermolecular forces. (a) Solid K conducts an electric current, whereas solid KNO3 does not. (b) SbCl3 has measurable dipole moment, whereas SbCl5 does not. (c) The normal boiling point of CCl4 is 77oC, whereas that of CBr4 is 190oC. (d) NaI(s) is very soluble in water, whereas I2(s) has a solubility of only 0.03 gram per 100 grams of water. 1996 – Lewis Dot Diagram and IMF Explain each of the following observations in terms of the electronic structure and/or bonding of the compounds involved. (a) (b) (c) (d) At ordinary conditions, HF (Boiling point = 20oC) is a liquid, where as HCl (B. Pt. = 114oC) is a gas. Molecules of AsF3 are polar, whereas molecules of AsF5 are non-polar. The N-O bonds in NO2-1 ion are equal in length, whereas they are unequal in HNO2. For sulfur, the fluorides SF2, SF4, and SF6 are known to exist, whereas for oxygen only OF2 is know to exist. AP Free Response Answer 1992 Answer: (a) K conducts because of its metallic bonding - or - “sea” of mobile electrons (or free electrons). KNO3 does not conduct because it is ionically bonded and has immobile ions (or immobile electrons). (b) SbCl3 has a measurable dipole moment because it is asymmetrical due to the lone pair of electrons which causes a dipole - or - its dipoles do not cancel - or - it has a trigonal pyramidal structure - or - a clear diagram illustrating any of the above. (c) CBr4 boils at a higher temperature than CCl4 because it has stronger intermolecular forces (or van der Waal or dispersion). These stronger forces occur because CBr4 has more polarizable electrons than CCl4. (d) NaI has greater aqueous solubility than I2 because NaI is ionic (or polar), whereas I2 is nonpolar (or covalent). Water, being polar, interacts with the ions of NaI but not with I2. (Like dissolves like accepted if polarity of water is clearly indicated.) 1996 (a) hydrogen bonding is much larger in HF than in HCl (hydrogen bonding). (b) AsF3 has triangular pyramidal shape and a lone pair of electrons, while AsF5 is triangular bipyramidal and is symmetrical. (c) The N-O bond in nitrite ion has resonance structure (bond order = 1½) while that of HNO2 is prevented to have resonance structures by the addition of the hydrogen. (d) Hybridization of sulfur can go to sp3d2 while that of oxygen is only sp3. . 1979 Answer: or or bent with sp3 hybridization Linear or Trigonal pyramidal or triangular planar 1982 Answer: (a) (b) CO has the shortest bond because there is a triple bond. OR because there is the greatest number of electrons between C and O in CO. (c) CO32- trigonal planar (planar and triangular). C bonding is sp2 hybrid - or - C has three bonding pairs and no lone pair. CO2 linear. C bonding is sp hybrid - or - C has two bonding pairs and no lone pairs - or CO2 is nonpolar and must be linear. CO linear. Two atoms determine a straight line. 1989 D Answer: (a) Tetrahedral square planar T shape (b) CF4 - 4 bonding pairs around C at corners of regular tetrahedron to minimize repulsion (maximize bond angles). XeF4 - 4 bonding pairs and 2 lone pairs give octahedral shape with lone pairs on opposite sides of Xe atom (along the same axis) ClF3 - 3 bonding pairs and 2 lone pairs give trigonal bipyramid with one pairs in equatorial positions 120O apart. 1990 Answer: (a) C2H4 has a multiple bond; C2H6 has a single bond. Multiple bonds are stronger and, therefore, shorter than single bonds. (b) NH3 has 3 bonding pairs of electrons and 1 lone pair. Bonding pairs are forced together because repulsion between lone pair and bonding pairs is greater than between bonding pairs. (c) The bonding in SO3 can be described as a combination of 3 resonance forms of 1 double and 2 single bonds. The actual structure is intermediate among the 3 resonance forms, having 3 bonds that are equal and stronger (therefore, shorter) than an S-O single bond. (d) The central I atom has 3 lone pairs and 2 bonding pairs around it. To minimize repulsion, the 3 lone pairs on the central atom are arranged as a triangle in a plane and are right angles to the I-I-I- axis (equatorial positions) 1990 Answer: (a) ΔH = energy of bonds broken - energy of bonds formed C2H5Cl + Cl2 C2H4Cl2 + HCl ΔH = (2794 + 243) - (2757 + 431) kJ = -151 kJ OR CH + Cl-Cl C-Cl + HCl (representing the changes) ΔH = (414) + 243) - (377 + 431) = -151 kJ (b) ΔG = ∑ ΔG°(products) − ∑ ΔG °(reactants) = [-80.3 + (-95.3)] - [-60.5 + 0] = -115 kJ ΔS° = ΔH ° − ΔG ° T = −151 − ( −115 )kJ 298 K = −0 .120 kJ K (c) Keq = e-G/RT = e-(-115100/(8.3143)(298)) = 1.50 x1020 (d) Keq will decrease with an increase in T because the reverse (endothermic) reaction will be favored with the addition of heat. OR ΔG will be less negative with an increase in temperature (from ΔG = ΔH - TΔS), which will cause Keq to decrease.