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MOOREHW2 - Chapter 3 Covalent Bonding 3 If the elements A...

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Unformatted text preview: Chapter 3: Covalent Bonding 3. If the elements A and X have similar ability to take electrons, such as a nomnetal and a nonmetal, they will share electrons, and we'd find them bonded using a covalent bond. A polar covalent bond is formed between atom A and atom X when they are both nonmetals and have differing electronegativities. 4. Allianes have all single bonds between carbons. H H H H_l_l_l_H Allcenes have at least one double bond between two carbons. H H H H_.L l_l _H l Alliynes have at least one triple bond between two carbons. H H—cEc—clz—H l 5. (a) It shows several examples of molecules whose central atoms have more than eight electrons. None of them are from Period 2. (b) The maximum number of electrons around any central atom in the table is twelve, in six pairs. 3. Two resonance structures for N0; [O=l\'—O= ] _ H [RD—31:0 ]_ Table 3.1 has the bond lengths of single N—O bonds (126 pm) and double N=O bonds (115 pm). Since the two resonance structures are equivalent, and in one the bond is single and in the other the bond is double, we predict that the bond between N and 0 will be halfway between the length of single and double, or approximately 121 pm. 9. structures (a) and (b) are resonance structures, because the double bond is moved to alternate oxygen atoms. structure (c) is equivalent to structure (a). Placing the electrons below the symbol for C) does not change what atom they are associated with. 46. Ami-er: (a) N, C, Br, and O (b) 8—0 is most polar. Strateg= and Explain anon: (a) Look up electronegativity values (in Figure 8.6) for the atoms in these bonds. ENC = 2.5, ENN = 3.0, ENH = 2.1, ENE, =23, ENS = 25,1510 = 3.5 N is more electronegative in C—N, C is more electronegatit-‘e in C—H, Br is more electronegative in C—Br, and O is more electronegative in 3—0. (b) Bonds are more polar when the electronegativity difference is larger. Get AEN to find most polar. AENC_N = ENN — ENC = 3.0 — 2.5 = 0.5 AENGH = ENC — ENH = 2.4 — 2.1 = 0.4 AENC_B,. = ENE, — ENC = 2.8 — 2.5 = 0.3 AENS_O =EN0 — ENS = 3.5 — 2.5 = 1.0 mostpolar —1:0: 52. Anal-er: (a) P—S—p' (b) =NEC—CZN: (c) [OZN—O‘] 0 2+ —1 0 D 0 0 0 CI —1 Sfmteg; and Explanation: Given the formulas of molecules or ions write the correct Lewis structure and assign formal charges to each atom. Write the Lewis structures. Then determine the number of lone pair electrons and bonding electrons around each atom. Use the method described in Section 8.3 on page 355 to determine the formal charges on each atom. Formal charge = (armsber ofvafeme elections in an atom) — [(Prumber ofionepair electrons) + 1% number ofbonding efecfionsjj To get the (number efforts pair efecn'omj just count all the dots. To get (number ofboma’mg elections) just count two times the number lines representing covalent bonds. {a} Lewis structure for 303, molecule: 2.4 electrons total s:L_s The formal charges are calculated for each atom using the number of valence electrons, the number of lone pair electrons and the number of bonding electrons. Let's set up a chart for these value s: 3 =0 —0 Valence electrons s 6 :5 Lone pair electrons II} 4 s Bonding electrons E 4 2 Formalchare 6—D+4=-2 IS— 4+2= IS—IS-l =—l -1 : 0 : 9——S—{ci= El 2+ —1 There are other resonance structures that could be written for 303 with the double bond moved to each of the other two CI atoms and with more than one double bond to the sulfur atom. [A more involved description ofwriting andjudging the feasibility of resonance structures is found in the solutions to Questions 60-69.} {b} Lewis st1ucture for NECK molecule: 18 electrons total. = N_=C —C EN: The formal charges are calculated for each atom using the number of valence electrons, the number of lone pair electrons and the number of bonding electrons. Let's set up a chart for these value s: C IN" Valence electrons 4- 5 Lone pair electrons [II 2 Bonding electrons 8 t5 Formalchare 4—0+4='|I| 5— +3 =0 :NEC—CZN : {I Cl Cl' Cl' (C) Lewis st1ucture for N03‘ ion: 18 electrons total sis—s1 The formal charges are calculated for each atom using the number of valence electrons, the number of lone pair electrons and the number of bonding electrons. Let's set up a chart for these value s: 3' =0 —0 Valence electrons 5 6 6 Lone pair electrons 2 4 IS Bonding electrons t5 4 2 Fonnalchare 5— +3 =0 6— 4+2= 6— 6+l=— 78. 98. One other resonance structure could be written for Nfif with the double bond moved to the other 0 atom. [A more I'm-'otved description of writing resonance structures is found in the solutions to Questions EFU - 69.} 1" Reasonable Answer Check: The sum of the formal charges-is zero for the neutral molecules and the ionic charge for the charged ion. The atoms with more bonds have more positive formal diarges than those with fewer bonds and: more lone pairs. Anni-er: The bond in (r) Si—F, Si is farthest from F on the periodic table, so it is larger and has a lower electronegatirity. Strategv aaa’ Esptan arias: Use the periodic table and trends in electronegativities to determine which pair is farthest apart. That will mean that their electronegativities are most different and the bond will be most polar. Here, all the choices have P atom in connnon so f'md out which of the other elements is the farthest fi'om F. Si is farthest fi'om F on the periodic table; so it is larger and has a lower electronegativity. Therefore, (c) Si—F is more polar than these other choices (a) C—F. (b) S—R (d) D—F Ami-er: (31:31]; 8: 2.5; Br: 2.5; Se: 2.4; As: 2.1 Strategy and Explanation: Using the periodic trend for electronegativity (EN) smaller larger 4(— EN EN larger 5 Cl EN A5 Se Br smaller EN We certainly know that Cl*s EN is the largest; so we‘ll assign it the value of3.0. We certainly know that As‘s EN is the lowest so well assign it value of 2.1 The trend shows that EN’s of S and Br are both larger than that of Se, so well assign them both the value of 2.5. That leaves the EN value of 2.4 for Se. With the values of 2.5 and 2.4 so close together, the assignment of those values are uncertain. Chapter 9: Molecular Structures 13. Amt-er: See structures below; (a) linear (b) triangular planar (c) octahedral (d) triangular bipyramjdal Strateg= and Expianatton: Write Lewis structures for a list of formulas and identify their shape. Follow the systematic plan for Lewis structures given in the solution to Question 8.13, then determine the number of bonded atoms and lone paint on the central atom, determine the designated type (AXflEm) and use Table 9.1. Notice: It is not important to expand the octet of a centrall atom sotet'yfor the purposes oft'oweting itsformai charge, because theshape of the moiectde wih' not change. Hence, we with“ Itt'ite Lewis structures that foiiow the octet rate unless the atom needs more than eight eiectrons. (a) BeH2 (4 e‘) H—Be—H The type is AX2E0, so it is linear. (b) CHZCIE (20 e‘) The type is AX4E0, so it is tetrahedral. .. Cl :Cl= | .. l _ - — C :Cl—C—H ‘31 I \H .. l H H (e) BH3 (6 e‘) The type is AXSEU, so it is triangular planar. H H l H—B—H B H/ \H (d) SClfi (48 e‘) The type is AX6E0, so it is octahedral. :c1: Cl =c1 (:1: 4. \IS/ H C1~"-._iS.-"PC1 .g/| e1 Cl/l \CI :91: (:1 (e) P135 (40 e‘) The type is AXSEU, so it is triangular bipyramidal. .. = E: 1: :1: I F. ..\P—_E:1 '* _F 15. Amt-er: (a) electron—pair geometly and molecular geometry are tetrahedral (b) electron—pair geometlj is tetrahedral; molecular geometry is angular (109.50} (c) electron—pair geometry and molecular geometry are both triangular planar [d] electron—pair geometly and molecular geometry are both triangular planar Strategv and Explonofion: Adapt the method given in the solution to Question 13. (a) PH4+ (8 e‘) The type is AX4EU, so electron-pair geometry and the molecular geometry are both tetrahedral. T + H H —P—H _;|= | H' l \H H H (h) OIZZl2 (20 e‘) The type is AXzEE, so the electron—pair geometry is tetrahedral and the molecular geometry is angular (1095"). :Cl‘l : C‘l :0 —éi: o .. .. \\ C1 (c) 803 (24 e‘) The type is AxgEu, so electron—pair geometry and the molecular geometry are both triangular planar. 0 H H /5\ :9—5—9: 0 o (d) HECO (12 e‘] The type is AXSEU, so electron—pair geometry and the molecular geometry are both triangular planar. O =0 H t H—C—H / \ H H 48. Anni-er: Molecules (b) and (c) are polar; HBF2 has the F atoms on partial negative end and the H atom on the partial positive end; CHJCI has the Cl atom on partial negative end and the H atoms on the partial positive end. (see dipole moment diagrams below) Strategv and Explanation: We need the Lewis structines and molecular shapes of the molecules: (at (b) (c) .. (d) .. .. H i C1: :E): e=0=9 I | | "/B\-- H"'C\ --/S .. :F. ,F. = H/ H 9/ \Q: The molecules that are polar have asymmetrical atom arrangements; here, they are (b) 1-1231:2 and (c) CH3C1. The others= (a) CD2 and (d) 803, have symmetrical arrangements of identical atoms with the same partial charge. (Notice: Three equivalent resonance structures can be written for 803, making each S—O bond the same length and strength.) The bond polarities related to the difference in electronegativity. Use Figure 3.6 to get elecn'onegativity (EN) values: ENB = 2.0, ENH = 2.1= ENF = 4.0, ENC = 2.5, ENC1 = 3.0 AENEHI = ENH —:31\'B = 2.1 — 2.0 = 0.1 AENEH: = ENF — ENE = 4.0 — 2.0 = 2.0 AENGH = EN<3 —ENH = 2.5 — 2.1 = 0.4 AENGCI = ENC1 — ENC = 3.0 — 2.5 = 0.5 The bond pole arrows“ lengths are related to their AEN, and points toward the atom with the more negative EN. So, the B—F arrows are much longer than the I— 41 arrow, and a net dipole points toward the F atoms? side of the molecule, making the F atoms‘ side of the molecule the partial negative end and the H atom‘s side of the molecule the partial positive end. The C—Cl arrow points toward Cl, and the C—H points toward the C, so all the arrows point toward the C. (Left right, and back forward cancel due to the syimnetry of the triangular orientation of the H atoms). That means a net dipole points toward the Cl atom’s side of the molecule, making the Cl atom‘s side of the molecule the partial negative end and the H atoms? side of the molecule the partial positive end. £3111 J; H“)! \H H | B {/10po 58. Anni-er: Molecules (c), (d) and (e) will folm hydrogen bonds. Strategv and Exptanon'on: Hydrogen bonds form between very electronegative atoms in one molecule to H atoms bonded to a very electronegative atom (EN 2 3.0) in another molecule. If H atoms are present in a molecule but they are bonded to lower electronegativity atoms, such as C atoms, the molecule cannot use those H atoms for hydrogen bonding. (a) The H atoms are bonded to C atoms and the highest electronegativity atom in the molecule is Br (EN = 2.8), so this molecule cannot form hydrogen bonds. (1)) The H atoms are bonded to C atoms, so this molecule cannot form hydrogen bonds with other molecules of the same compound. It does have a high electronegativity O atom (EN = 3.5), so this molecule could interact with other molecules, such as H20, which can provide the H atoms for hydrogen bonding to the O atom on this molecule. H H I | HH/C'\~.Q/C\H*H (c) Three H atoms are bonded to high electronegativity atoms (ENCI = 3.5 and ENN = 30). Those H atoms (circled in the structure below) can form hydrogen bonds with the N and O atoms in neighboring molecules. Dene/9'69 | ®"/C§"‘ N db (d) Two H atoms are bonded to high electronegativity O atoms (EN = 3.5). Those H atoms (circled in the structure below) can form hydrogen bonds with the O atoms in neighboring molecules. TQ (”5,63 (e) One H atom is bonded to a high electronegativity O atom (EN = 3.5). That H atom (circled in the structure below) can form hydrogen bonds with the O atoms in neighboring molecules. H \/\/@ H‘H/ 62. Anni-er: (a) London forces (b) London forces (c) Covalent bonds ((1) Dipole—dipole forces (e) Intermolecular hydrogen bonding forces Sirotegv and Eaplon anon: (a) (b) (C) (d) Hydrocarbons have no capability of forming hydrogen bonds and the bonds are close to non-polar. The molecules interact primarily with London forces. The London forces between the molecules must be overcome to sublime C1oHs- As described in (a), propane molecules, a hydrocarbon, would need to overcome London forces to melt. Decomposing molecules of nitrogen and oxygen require breaking covalent bonds, so the forces that must be overcome are the intramolecular (covalent) forces. To evaporate polar PC13 requires that the molecules overcome dipole-dipole forces. ,5 :it5_1"'l\c'1_ (-3.1 .. l (e) The double strands of DNA are shown in Figure 9.25. The interaction between the two strands are C P9 .A. CP9.B. hydrogen bonds. These hydrogen bonding forces must be overcome to “unzip“ the DNA double helix. VSEPR theory helps predict the particular geometric shape of a combination of covalently bonded atoms. Lewis structures only reliably predict the atom connectivity and the type of bonding (single, double, triple bonds} in a covalent structure. Combining the VSEPR model with Lewis structure refines a prediction of the molecules shape. Several answers exist for this problem, depending on how the students interpret the word “evidence.” Experiments can show that molecules like HCN, C02, and HZBe are all linear, even though they have a variety of single, double and triple bonds. The experimental evidence is usually spectroscopic. You will need to decide to what extent you wish to go beyond the scope of this course to describe specific types of evidence identifying a molecule as linear in shape. ...
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