Nuggets: Forming a Bond; Electronegativity; Types of Bonds; Lewis Dot Structures; Resonance; Bond Order; Bond Polarity; Formal Charge; Bond Enthalpy

Transcription

1 EMITRY 103 elp heet #14 hapter 6 (Part I): 6.1, 6.2, Do the topics appropriate for your lecture Prepared by Dr. Tony Jacob (Resource page) uggets: orming a ond; Electronegativity; Types of onds; Lewis Dot tructures; Resonance; ond rder; ond Polarity; ormal harge; ond Enthalpy orm a ond: hown below is the energy change as 2 forms a covalent bond. ovalent ond: Two orbitals overlap and contain two shared electrons of opposite spin in the overlap region. Potential Energy (kj/mol) uclei repel one another as they approach and energy increases 3. ond formed; energy at lowest point; internuclear distance at smallest value Energy released when bond is formed 2. Atoms approach one another; bond starts to form and energy drops Internuclear Distance (pm) 1. Atoms far apart; not interacting ELETREGATIVITY (χ): The degree to which an atom can attract an e in a bond. The greater the electronegativity (E), the greater the "pull" an atom has on an e pair. Trend: increasing Increases toward (up and to the right). TYPE D onpolar ovalent: equal sharing of e between two identical nonmetal elements (e.g., 2, and 2 ). Electronegativity difference (ΔE) is zero. Polar ovalent: unequal sharing of e between two nonmetals (e.g., and ). Electronegativity difference (ΔE) is small. Ionic: metal + nonmetal, e are transferred from the metal to the nonmetal (e.g., a). Electronegativity difference (ΔE) is large. D PLARITY: Greater ΔE the more polar, the more ionic, and the less covalent the bond is. ΔE bond polarity ; bond ionic character ; bond covalent character Polar bond: A bond in which e are pulled to one side of the molecule with a slight positive charge residing on one side (δ + ) and a slight negative charge (δ ) residing on the other side of the molecule. The arrow points toward the negative side of the bond. ( ). A dipole (µ) is a measure of bond polarity and occurs when there is charge separation; measured in debyes (D). 1D = 3.34 x m (coulombmeters) µ = Qr where r is the distance between the charges and Q is the charge (this equation is not always covered) Valence e : Add up the valence e from each atom, only s and p e past the last noble gas (no d electrons).

2 LEWI DT TRUTURE (LD): Using valence e to create covalent bonding schemes. This is one of many methods. 1. ount valence e in molecule. 2. Draw a skeleton structure. Keep symmetric if possible. Usually (not always) the central atom in the Lewis structure is the first atom in the chemical formula. 3. uild octets (8 e ) around each atom except (2 e around ). 4. ount #e drawn. If it equals the #valence e in tep 1, you're done. 5. If #e is too large: Go to tep 2 and redraw the skeleton structure, add 1 double bond, and continue with the steps. If #e is still too large, repeat by adding another double bond or create a triple bond. If #e is too small: Add e pairs to the central atom. ther info:,, r, and I can have only 1 bond unless they're the central atom. always has only 1 bond. LEWI DT TRUTURE: ategories of compounds to draw LD for (see the next page for examples): 1. "ormal" compounds: follows the octet rule; e.g., 3, nothing tricky 2. harged species: add charge to #valence e for anions, e.g., for 4 2 add 2 to valence e count (1(6) + 4(6) + 2 = 32e ); subtract charge from #valence e for cations, e.g., 4 + (1(5) + 4(1) 1 = 8e ) 3. Multiple bonds: double or triple bonds; e.g., 2, 2, etc. 4. Resonance structures: typically, a double bond that moves to an equivalent location, e.g., 3 2, entral atom with less than 8e : and e compounds commonly; e.g., 3 ; does not have to have 6e it can have 8e, e.g., 4 6. entral atom with more than 8e : central atom must be in the 3rd period or later of Periodic Table, e.g., 6 7. xyacids = n m : skeleton structure: in middle; s bonded to s, e.g., 2 4, 3 8. Individual atoms a + ) 9. Ionic compounds ( 10. Radicals: have an odd #e, e.g., (1(5) +1(6) = 11e ) 11. rganic molecules (e.g., 4, 2 2, 2 4, 2 6, 3 8, 3 ; 2 ; 2 () 2, etc.) REAE: ometimes multiple bonds can be placed in several equivalent locations. If this occurs, you must draw all structures (called resonance structures) for the complete and correct LD (e.g., there are three LD for carbonate, 3 2, and all must be drawn; for benzene, 6 6 there are two LD). D RDER () (not in text): ingle bond = sharing of 1 e pair; = 1 Double bond = sharing of 2 e pairs; = 2 Triple bond = sharing of 3 e pairs; = 3 Multiple onds = Double or Triple bonds. #bonds With resonance structures (one method of many): ond rder = #locations bonds are distributed 2 D RDER or Atomic Radii bond strength bond length bond dissociation energy

3 Example 1. 3 ( simple ) (charged) 3. 2 (multiple bond) 4. 3 (resonance) 5. 3 (less than an octet) 6. 4 (more than an octet) 7. 3 P 4 (oxyacid) tep 1: count valence electrons 1() + 3() = 1(5) + 3(7) = 26e tep 2: draw skeleton structure 1() + 4() + 2 = 1(6) + 4(6) + 2 = 32e Lewis dot structure (LD) examples tep 3: build octets tep 4: check LD for #e 26e done! 32e done! 2() = 2(5) = 10e 14e too many e AWER 2nd attempt: 12e too many e 3rd attempt: 10e done! 3() = 3(6) = 18e 20e too many e 2nd attempt: 1() + 3() = 1(3) + 3(7) = 24e 2nd attempt: 3rd attempt: 1() + 4() = 1(6) + 4(7) = 34e 1(P) + 4() + 3() = 1(5) + 4(6) + 3(1) = 32e P (skeleton structure!) 8. (atom) 1() = 1(5) = 5e 9. a (ionic) 10. (odd #e = radical) 1(a) + 1() = 1(1) + 1(7) = 8e 1() + 1() = a change to a 6e P ionic compds have electrostatic bonds not covalent bonds; redraw with valence e 1(5) + 1(6) = 11e 2nd attempt: 3rd attempt: 4th attempt: 18e done! resonance place double bond in second location 26e too many e,, r, I can only have one bond unless they are the central atom back to original skeleton structure and e are elements where less than an octet are commonly seen in hem e too few e add e pairs to central atom 2 must include charge and brackets both structures are needed 6e 32e done! P 5e done! + a 8e done! ionic compound charged and needs brackets; both a and have octets; a 14e too many e 12e too many e 10e too few e back to double bond structure and remove 1e from and 1e from (1) (+1) has a [e] octet w/o e shown assign structure with 1e on is better; is more E than structure with 1e on is again better; both structures probably exist but structure with 1e on is better based on and E better LD: second LD:

4 RMAL ARGE (): The "charge" on an atom as determined from the LD with bonded e shared equally ormal harge = #valence e (from the Periodic Table) (#e around atom shown in LD) Example: Assign formal charges to each atom in (use this LD: ) Answer: = #e (valence) #e (LD) Replace bonds with e if needed to see #e around atom in LD; e in bonds are split between adjoining atoms; lone pairs of e stay with atom () = 5 6 = 1 : 5 valence e : 6 e in LD () = 4 4 = 0 : 4 valence e : 4 e in LD () = 6 6 = 0 : 6 valence e : 6 e in LD Σormal harges = Total harge on Molecule est Lewis dot structure based on ormal harge: 1. as the smallest (preferably zeros) formal charges. 2. as more zeros for formal charges. 3. If there is a choice between 2 structure with same formal charges and same number of zeros, the better structure has more negative formal charges on more electronegative atom. Example: Which 3 structure is best based on formal charges? E() = 2.7; E() = 3.4. I Answer: tep 1: Assign to all atoms in all structures. (1) II III IV (1) (+2) (1) (1) (+1) (1) (1) (1) I II III IV : 7 5 = +2 : 7 6 = +1 : 7 7 = 0 : 7 8 = 1 (single bond): 6 7 = 1 (single bond): 6 7 = 1 (single bond): 6 7 = 1 (single bond): 6 7 = 1 (double bond): 6 6 = 0 (double bond): 6 6 = 0 (double bond): 6 6 = 0 tep 2: Eliminate structures based on Rule 1, then Rule 2, then Rule 3. (1) (1) (+2) (1) I 1. Eliminate tructure I because of the larger on the (Rule 1: smaller are better). This is the only structure with a +2. (1) (+1) (1) II 2. Eliminate tructure II because it only has 1 zero while tructures III and IV each have 3 zeros (Rule 2: more zeros are better). (1) III est structure based on. Rule 1: smaller better. Rule 2: more zeros for are better. Rule 3: the more E atom gets the more negative. D ETALPIE (D): The energy needed to break or make a bond; break bonds: endothermic; form bonds: exothermic Δ r = ΣD bonds broken ΣD bonds formed Example: Using the values in Table 6.2, what is the enthalpy of reaction of 4 (g) combusting? 4 (g) (g) 2 (g) (g) Answer: tep 1: Draw LD of each molecule to determine whether bonds are single, double, etc. (1) IV 3. Eliminate tructure IV because the more E atom, (E() = 2.7; E() = 3.4), should have the more negative (Rule 3). tructure III: () = 0 and () = 1; tructure IV: () = 1 and () = tep 2: Determine bonds broken (reactants) and formed (products); keep in mind stoichiometric coefficients onds broken: 4 bonds; 2 = bonds (each 2 molecule has 1 = bond and there are 2 2 molecules so there are 2 = bonds) onds formed: 2 = bonds; 4 bonds (each 2 has 2 bonds and there are 2 2 molecules so there are 4 bonds) tep 3: alculate Δ r Δ r = [4(D ) + 2(D = )] [2(D = ) + 4(D )]; Δ r = [4(416) + 2(498)] [(2(803) + 4(467)] = 814kJ/mol

5 RGAI arbonbased structures that include and can commonly include and atoms. 1. Predict the order of increasing electronegativity in each of the following groups of elements: a.,, b., e, c. i, Ge, n d. Tl,, Ge 2. ased on the electronegativity differences, choose the molecule that is most likely to have ionic bonds? a. a 2 b. 2 c. 3 d. 3 e In each set of bonds, determine which has the most polar bond (i.e., largest bond dipole moment)? a.,,,, b. i, P, c. e, e, e r 4. Draw Lewis dot structures for each of the following: a. P 3 b. 3 c. 4 + d. 4 e. e 2 f. 4 2 g. 4 h. 2 i. j. P 5 k. r 3 l. 3 m. e 4 n. e 4 o. 3 p. q. 3 r. 2 4 s Draw Lewis dot structures for each of the following chemical species. (These types of chemical species are less commonly asked.) a. b. 2 2 c. 3 () 3 d. a e. Li 2 f. g. 6. Which of the following chemicals does not obey the octet rule? a. 4 b. i 4 c. e 4 d. 4 e. I Which molecule has the strongest bond? a. 3 2 b. c. 2 d. 2 ( is central atom) e. They are all the same strength. 8. What is the formal charge on the in 3 2? a. 2 b. 1 c. 0 d. 1 e What is the formal charge on the in 2? a. 2 b. 1 c. 0 d. 1 e Given the 3 drawings of 2, which would be the best structure based on formal charges? I. II. III. a. I b. II c. III d. I and III e. All equally important. 11. onsider the Lewis structures of the following molecules. Which molecules list below would include resonance in their Lewis structures? P a. P 5, 3 b. 3, 2 c. 2, 2 d. P 5, 2 e. 3, Which compound obeys the octet rule? a. 2 b. 3 c. d. 3 e. one obey the octet rule.

6 13. Which of the following statements is incorrect? a. Ionic bonding results from the complete transfer of electrons from one atom to another. b. Dipoles are the result of unequal electron distribution in a molecule. c. The electrons in a polar bond are found nearer to the more electronegative element. d. A molecule with very polar bonds can be nonpolar. e. Linear molecules cannot have a net dipole moment. 14. Using the bond enthalpies (D) below, calculate the change in enthalpy for the reaction shown. int: Draw Lewis dot structures for each chemical in the reaction; note that the atoms are bonded together in the product molecule. 3 = 2 + r 3 r = 598 r r If the change in enthalpy for the reaction below is 367kJ/mol, what is the bond enthalpy (D) of a carbonchlorine bond in 2 2? ond energies (kj/mol) are given below. int: Draw Lewis dot structures for each chemical in the reaction; note that the atoms are bonded together in the product molecule = AWER 1. a. < < b. e < < c. n < Ge < i d. Tl < Ge < 2. a 3. Greatest ΔE: a. b. i c. e 4. ee below. 5. ee below. 6. c 7. b 8. d {Draw Lewis dot structure; only 1 arrangement; : 6 5 = 1} 9. c {Draw Lewis dot structure; : 5 5 = 0; recall can't violate the octet rule} 10. c 11. b 12. e 13. e kJ/mol {Δ r = bonds broken bonds formed = [1(=) + 1( ) + 6( ) + 1( r)] [2( ) + 7( ) + 1( r)] = [1(598) + 1(356) + 6(416) + 1(366)] [2(356) + 7(416) + 1(285)] = [3816] [3909] = 93kJ/mol} + r r kJ/mol {Δ r = [2( ) + 1( ) + 2( )] [(2( ) + 4( ) + 1( )]; 367 = [2(416) + 1(813) + 2(242)] [(2(416) + 4(x) + 1(356)]; 367 = 2129 [ x]; 367 = 941 4x; 4x = 1308; x = 327kJ/mol} + 2

7 4. 5.