Chemical Bonds Worksheet

Chem Worksheet - Week 11 - Solutions Chemical Bonds Key Questions 1. Using the electronegativity values in the table above, calculate δχ for the pairs of elements forming bonds in each of the following species, and describe the type of bonding as either pure covalent, polar covalent, or ionic: H2, HCl, PbBr2, NaCl, NCl3. H2 HCl PbBr2 NBr3 NaCl δχ = *2.1 - 2.1* = 0 δχ = *2.1 - 3.0* = 0.9 δχ = *1.8 - 2.8* = 1.0 δχ = *3.0 - 2.8* = 0.2 δχ = *0.9 - 3.0* = 2.1 pure covalent polar covalent polar covalent polar covalent (slightly) ionic 2. Calculate δχ for the bonds in HF(g) and NCl3(l) and describe them. HF δχ = *2.1 - 4.0* = 1.9 very polar covalent Although δχ suggests an ionic bond, HF is a gas and exists as a molecular species. Its bonds are very polar covalent, but HF does not contain discrete H+ and F– ions like a typical ionic compound. NCl3 δχ = *3.0 - 3.0* = 0 virtually pure covalent The zero value of difference in electronegativity suggests that there is very little polarity to the N–Cl bonds in NCl3. But because N and Cl are two different elements, there is actually some difference in their abilities to attract electrons in the bond. Pauling electronegativity values are simply too imprecise to indicate the polarity. 3. Explain the following trends in lattice energy. a. KF < CaO < ScN The ion charges are +1/–1 for KCl, +2/–2 for CaO, and +3/–3 for ScN. Lattice energy rises sharply with increasing ion charge. b. NaCl > RbBr > CsBr All of these are +1/–1 ion pairings, so size is the dominating factor. As ion size increases, the separation between oppositely charged ions increases, causing the forces of attraction to diminish. c. MgO > MgCl2 The Cl– ion is larger and has a lesser charge than the O2– ion. Both factors result in MgCl2 having a lower lattice energy than MgO. 4. Following the steps given above, draw Lewis dot structures for the following molecular species: NF3, H2O, NH4+, SO42–, H2CO, C2H2. N + 3F = 5 +(3)(7) = 26 NF3 F N F F H2O O + 2H = 6 +(2)(1) = 8 H O NH4+ H N + 4H + (+) = 5 + (4)(1) – 1 + H H N H H SO42– S + 4O + (2–) = 6 + 4(6) + 2 = 32 2- O O S O O H2CO 2H + C + O = (2)(1) + 4 + 6 = 12 O H C2H2 C H 2C + 2H = (2)(4) + (2)(1) = 10 H C C H 5. Following the steps above, draw Lewis dot structures for the following molecular species, which may be electron deficient or hypervalent: AlCl3, XeF4, XeF2, BeCl2 AlCl3 Cl Al + 3Cl = 3 + (3)(7) = 24 Al Al is electron deficient (3 pairs). Cl Cl XeF4 Xe + 4F = 8 + (4)(7) = 36 Xe is hypervalent (6 pairs). F F Xe F F XeF2 Xe + 2F = 8 + (2)(7) = 22 F Xe BeCl2 Cl F Xe is hypervalent (5 pairs). Be + 2Cl = 2 + (2)(7) = 16 Be Cl Be is electron deficient (2 pairs) 6. Show the formal charges on all atoms in the sulfate ion, SO42–, whose Lewis dot structure you developed in Key Question 4. 2O O S O 2+ O 7. The nitrate ion, NO3–, can be represented by a series of three equivalent resonance forms. Show the resonance forms of NO3–, including formal charges, and estimate the bond order of the N–O bonds in this ion. NO3– N + 3O + (–) = 5 + (3)(6) + 1 = 24 _ O O N O O O N O O O N O As with ozone, the single and double bonds in these structures do not really exist in the nitrate ion. All N-O bonds are equivalent. The three N-O bonds in any one of the three equivalent resonance forms involve four pairs of electrons. Therefore the approximate bond order for a N-O bond in the nitrate ion is 4/3 = 1a. 8. FNO2 can be represented by three non-equivalent resonance forms. Draw the three canonical forms of FNO2 and assess their relative contributions to the resonance hybrid description of the molecule. What is the approximate N–O bond order? FNO2 F + N + 2O = 7 + 5 + (2)(6) = 24 O F N O O F N O O F N O The third form is insignificant (adjacent like formal charges). In the remaining forms, the two N–O bonds are approximately equivalent. The two N–O bonds use 3 pairs, so the bond order of each is about 3/2 = 1½.