CHAPTER 1 CHEMICAL BONDING SOLUTIONS TO TEXT PROBLEMS 1. 1 The element carbon has atomic number 6. and so it has a total of six electrons Two of these elec- trons are in the ls level. The four electrons in the 2s and 2p levels(the valence shell) are the valence electrons. Carbon has four valence electrons 1. 2 Electron configurations of elements are derived by applying the following principles (a) The number of electrons in a neutral atom is equal to its atomic number Z. (b) The maximum number of electrons in any orbital is 2 (c) Electrons are added to orbitals in order of increasing energy, filling the ls orbital before any electrons occupy the 2s level. The 2s orbital is filled before any of the 2p orbit 3s orbital is filled before any of the 3p orbitals (d) All the 2p orbitals(2p, 2py, 2p )are of equal energy, and each is singly occupied before any is doubly occupied. The same holds for the 3p orbitals With this as background, the electron configuration of the third-row elements is derived (Z=11)1s22p3s5 ls22p53s23 (Z=15)152s2p323y3p,l3p S(z=16)1s22p3323y23p3p )122D32323923p Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
1 CHAPTER 1 CHEMICAL BONDING SOLUTIONS TO TEXT PROBLEMS 1.1 The element carbon has atomic number 6, and so it has a total of six electrons. Two of these electrons are in the 1s level. The four electrons in the 2s and 2p levels (the valence shell) are the valence electrons. Carbon has four valence electrons. 1.2 Electron configurations of elements are derived by applying the following principles: (a) The number of electrons in a neutral atom is equal to its atomic number Z. (b) The maximum number of electrons in any orbital is 2. (c) Electrons are added to orbitals in order of increasing energy, filling the 1s orbital before any electrons occupy the 2s level. The 2s orbital is filled before any of the 2p orbitals, and the 3s orbital is filled before any of the 3p orbitals. (d) All the 2p orbitals (2px, 2py, 2pz) are of equal energy, and each is singly occupied before any is doubly occupied. The same holds for the 3p orbitals. With this as background, the electron configuration of the third-row elements is derived as follows [2p6 2px 2 2py 2 2pz 2 ]: Na (Z 11) 1s 2 2s 2 2p6 3s 1 Mg (Z 12) 1s 2 2s 2 2p6 3s 2 Al (Z 13) 1s 2 2s 2 2p6 3s 2 3px 1 Si (Z 14) 1s 2 2s 2 2p6 3s 2 3px 1 3py 1 P (Z 15) 1s 2 2s 2 2p6 3s 2 3px 1 3py 1 3pz 1 S (Z 16) 1s 2 2s 2 2p6 3s 2 3px 2 3py 1 3pz 1 Cl (Z 17) 1s 2 2s 2 2p6 3s 2 3px 2 3py 2 3pz 1 Ar (Z 18) 1s 2 2s 2 2p6 3s 2 3px 2 3py 2 3pz 2 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CHEMICAL BONDING 1.3 The electron configurations of the designated ions are Number of Electrons Electron Configuration in lon of lo (c) H (d) O 522s-2pr 2px 2p, 9 ()c 1522p°3s5p Those with a noble gas configuration are H, F, and Ca2+ 1. 4 A positively charged ion is formed when an electron is removed from a neutral atom. The equation representing the ionization of carbon and the electron configurations of the neutral atom and the ion A negatively charged carbon is formed when an electron is added to a carbon atom. The addi- tional electron enters the 2p, orbital. C s232p242p 152322p p, 2p Neither Ct nor C has a noble gas elect 1.5 Hydrogen has one valence electron, and fluorine has seven. The covalent bond in hydrogen fluoride arises by sharing the single electron of hydrogen with the unpaired electron of fluorine Combine H and F: to give the Lewis structure for hydrogen fluoride H: F: 1.6 We are told that Ch has a carbon-carbon bond to write the HH Thus, we combine two . C. and six H Lewis structure H: C: C: H of ethane There are a total of 14 valence electrons distributed as shown. Each carbon is surrounded by eight electrons 1.7(b) Each carbon contributes four valence electrons, and each fluorine contributes seven. Thus, CFA has 36 valence electrons. The octet rule is satisfied for carbon only if the two carbons are at- tached by a double bond and there are two fluorines on each carbon. The pattern of connections shown(below left) accounts for 12 electrons. The remaining 24 electrons are divided equally (six each) among the four fluorines. The complete Lewis structure is shown at right below (c) Since the problem states that the atoms in C3 H3N are connected in the order CCCn and all hy drogens are bonded to carbon, the order of attachments can only be as shown(below left)so as to have four bonds to each carbon. Three carbons contribute 12 valence electrons, three hy drogens contribute 3, and nitrogen contributes 5, for a total of 20 valence electrons. The nine Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
2 CHEMICAL BONDING 1.3 The electron configurations of the designated ions are: Number of Electrons Electron Configuration Ion Z in Ion of Ion (b) He 2 11s 1 (c) H 1 21s 2 (d) O 8 91s 2 2s 2 2px 2 2py 2 2pz 1 (e) F 9 10 1s 2 2s 2 2p6 ( f ) Ca2 20 18 1s 2 2s 2 2p6 3s 2 3p6 Those with a noble gas configuration are H, F, and Ca2. 1.4 A positively charged ion is formed when an electron is removed from a neutral atom. The equation representing the ionization of carbon and the electron configurations of the neutral atom and the ion is: A negatively charged carbon is formed when an electron is added to a carbon atom. The additional electron enters the 2pz orbital. Neither C nor C has a noble gas electron configuration. 1.5 Hydrogen has one valence electron, and fluorine has seven. The covalent bond in hydrogen fluoride arises by sharing the single electron of hydrogen with the unpaired electron of fluorine. 1.6 We are told that C2H6 has a carbon–carbon bond. There are a total of 14 valence electrons distributed as shown. Each carbon is surrounded by eight electrons. 1.7 (b) Each carbon contributes four valence electrons, and each fluorine contributes seven. Thus, C2F4 has 36 valence electrons. The octet rule is satisfied for carbon only if the two carbons are attached by a double bond and there are two fluorines on each carbon. The pattern of connections shown (below left) accounts for 12 electrons. The remaining 24 electrons are divided equally (six each) among the four fluorines. The complete Lewis structure is shown at right below. (c) Since the problem states that the atoms in C3H3N are connected in the order CCCN and all hydrogens are bonded to carbon, the order of attachments can only be as shown (below left) so as to have four bonds to each carbon. Three carbons contribute 12 valence electrons, three hydrogens contribute 3, and nitrogen contributes 5, for a total of 20 valence electrons. The nine C C F F F F C F F C F F H H C H H H H Thus, we combine two C H C to write the Lewis structure of ethane and six Combine H and F to give the Lewis structure for hydrogen fluoride H F C 1s2 2s2 2px 1 py 1 2pz 1 C 1s2 2s2 2px 1 2py 1 e C 1s2 2s2 2px 1 2py 1 C 1s2 2s2 2px 1 e Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CHEMICAL BONDING bonds indicated in the partial structure account for 18 electrons. Since the octet rule is satis- fied for carbon, add the remaining two electrons as an unshared pair on nitrogen(below right) H C≡N 1. 8 The degree of positive or negative character at carbon depends on the difference in electronegative ity between the carbon and the atoms to which it is attached. From Table 1. 2, we find the elec tronegativity values for the atoms contained in the molecules given in the problem are 1.0 H 2.1 C2.5 C13.0 Thus, carbon is more electronegative than hydrogen and lithium, but less electronegative than chlorine. When bonded to carbon, hydrogen and lithium bear a partial positive charge, and carbon bears a partial negative charge. Conversely, when chlorine is bonded to carbon, it bears a partial neg- ative charge, and carbon becomes partially positive. In this group of compounds, lithium is the least electronegative element, chlorine the most electronegative H 十 H H Chloroma most negative most posIt haracter at carbon character at carbon 1.9(b) The formal charges in sulfuric acid are calculated as follows Valence electrons Neutral Atom Electron Count Formal Charge Hydrogen: 1 Oxygen(of Oh) 1 4)+4=6 Oxygen 3(2)+6=7 Sulfur: 5(8)+0=4 H—Q (c) The formal charges in nitrous acid are calculated as follows Valence electrons in Neutral Atom Electron Count Formal Charge 2(2)=1 Oxygen(of Oh) 5(4)+4=6 (6)+2=5 H-(-N= Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
CHEMICAL BONDING 3 bonds indicated in the partial structure account for 18 electrons. Since the octet rule is satis- fied for carbon, add the remaining two electrons as an unshared pair on nitrogen (below right). 1.8 The degree of positive or negative character at carbon depends on the difference in electronegativity between the carbon and the atoms to which it is attached. From Table 1.2, we find the electronegativity values for the atoms contained in the molecules given in the problem are: Li 1.0 H 2.1 C 2.5 Cl 3.0 Thus, carbon is more electronegative than hydrogen and lithium, but less electronegative than chlorine. When bonded to carbon, hydrogen and lithium bear a partial positive charge, and carbon bears a partial negative charge. Conversely, when chlorine is bonded to carbon, it bears a partial negative charge, and carbon becomes partially positive. In this group of compounds, lithium is the least electronegative element, chlorine the most electronegative. 1.9 (b) The formal charges in sulfuric acid are calculated as follows: Valence Electrons in Neutral Atom Electron Count Formal Charge Hydrogen: 1 1 2 (2) 1 0 Oxygen (of OH): 6 1 2 (4) 4 6 0 Oxygen: 6 1 2 (2) 6 7 1 Sulfur: 6 1 2 (8) 0 4 2 (c) The formal charges in nitrous acid are calculated as follows: Valence Electrons in Neutral Atom Electron Count Formal Charge Hydrogen: 1 1 2 (2) 1 0 Oxygen (of OH): 6 1 2 (4) 4 6 0 Oxygen: 6 1 2 (4) 4 6 0 Nitrogen: 5 1 2 (6) 2 5 0 H O N O S 2 O O H O O H H H H C Li H H H C H H H H C Cl Methyllithium; most negative character at carbon Chloromethane; most positive character at carbon N C H H C H C N C C H H H C Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CHEMICAL BONDING 1.10 The electron counts of nitrogen in ammonium ion and boron in borohydride ion are both 4(one half f& electrons in covalent bonds ). H N-H H-B-H H Borohydride Since a neutral nitrogen has 5 electrons in its valence shell, an electron count of 4 gives it a formal harge of +l. A neutral boron has 3 valence electrons, and so an electron count of 4 in borohydride ponds to a form of-1. 1.11 As shown in the text in Table 1. 2, nitrogen is more electronegative than hydrogen and will draw the electrons in N-H bonds toward itself. Nitrogen with a formal charge of +I is even more elec- tronegative than a neutral nitrogen H H H-N一HH—N-H Boron(electronegativity=2.0)is, on the other hand, slightly less electronegative than hydrogen (electronegativity=2. 1). Boron with a formal charge of -l is less electronegative than a neutral boron. The electron density in the B--H bonds of BHa is therefore drawn toward hydrogen and away from boron 1.12(b) The compound( CH,) Ch has a central carbon to which are attached three CH, groups and a hydrogen H-C—H H H H Four carbons and 10 hydrogens contribute 26 valence electrons. The structure shown has 13 covalent bonds, and so all the valence electrons are accounted for. The molecule has no unshared electron pairs (c) The number of valence electrons in CICH, CH, CI is 26(2CI= 14: 4H= 4; 2C=8). The constitution at the left below shows seven covalent bonds accounting for 14 electrons. The re- maining 12 electrons are divided equally between the two chlorines as unshared electron pairs. The octet rule is satisfied for both carbon and chlorine in the structure at the right below HH -Cl Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
1.10 The electron counts of nitrogen in ammonium ion and boron in borohydride ion are both 4 (one half of 8 electrons in covalent bonds). Since a neutral nitrogen has 5 electrons in its valence shell, an electron count of 4 gives it a formal charge of 1. A neutral boron has 3 valence electrons, and so an electron count of 4 in borohydride ion corresponds to a formal charge of 1. 1.11 As shown in the text in Table 1.2, nitrogen is more electronegative than hydrogen and will draw the electrons in N@H bonds toward itself. Nitrogen with a formal charge of 1 is even more electronegative than a neutral nitrogen. Boron (electronegativity 2.0) is, on the other hand, slightly less electronegative than hydrogen (electronegativity 2.1). Boron with a formal charge of 1 is less electronegative than a neutral boron. The electron density in the B@H bonds of BH4 is therefore drawn toward hydrogen and away from boron. 1.12 (b) The compound (CH3)3CH has a central carbon to which are attached three CH3 groups and a hydrogen. Four carbons and 10 hydrogens contribute 26 valence electrons. The structure shown has 13 covalent bonds, and so all the valence electrons are accounted for. The molecule has no unshared electron pairs. (c) The number of valence electrons in ClCH2CH2Cl is 26 (2Cl 14; 4H 4; 2C 8). The constitution at the left below shows seven covalent bonds accounting for 14 electrons. The remaining 12 electrons are divided equally between the two chlorines as unshared electron pairs. The octet rule is satisfied for both carbon and chlorine in the structure at the right below. H H C H H Cl C Cl H H C H H Cl C Cl H H H C H H C H H H C H H C H H H B H H H B H H H H H N H H H N H H H H N H H H H B H H Ammonium ion Borohydride ion 4 CHEMICAL BONDING Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
structure. It, too, has 26 valence electrons, and again only chlorine has unshared CHEMICAL BONDING (d) This compound has the same molecular formula as the compound in part (c), but a different H (e) The constitution of CHaNHCH, CH3 is shown(below left). There are 26 valence electrons, and 24 of them are accounted for by the covalent bonds in the structural formula. The remaining two electrons complete the octet of nitrogen as an unshared pair(below right) HH (Oxygen has two unshared pairs in(CH3),CHCH=O. H H H C=0 1. 13 (b) This compound has a four-carbon chain to which are appended two other carbons. is equivalent to CH- which may be rewritten as (CH,)2CHCH( CH (c) The carbon skeleton is the same as that of the compound in part(b), but one of the terminal carbons bears an OH group in place of one of its hydrogens HO—C-H which may be CHOH is equivalent to written as 一C CH CHCH(CH3) (d) The compound is a six-membered ring that bears a-C(CH3)3 substituent. HH H-C-C H is equivalent to which may be CH rewritten as C(CH3)3 CH3 HH CH3 problem specifies that nitrogen and both oxygens of carbamic acid are bonded to carbon and of the carbon-oxygen bonds is a double bond. Since a neutral carbon is associated with four Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
CHEMICAL BONDING 5 (d) This compound has the same molecular formula as the compound in part (c), but a different structure. It, too, has 26 valence electrons, and again only chlorine has unshared pairs. (e) The constitution of CH3NHCH2CH3 is shown (below left). There are 26 valence electrons, and 24 of them are accounted for by the covalent bonds in the structural formula. The remaining two electrons complete the octet of nitrogen as an unshared pair (below right). ( f ) Oxygen has two unshared pairs in (CH3)2CHCH?O. 1.13 (b) This compound has a four-carbon chain to which are appended two other carbons. (c) The carbon skeleton is the same as that of the compound in part (b), but one of the terminal carbons bears an OH group in place of one of its hydrogens. (d) The compound is a six-membered ring that bears a @C(CH3)3 substituent. 1.14 The problem specifies that nitrogen and both oxygens of carbamic acid are bonded to carbon and one of the carbon–oxygen bonds is a double bond. Since a neutral carbon is associated with four which may be rewritten as H C C C C C C H H C H H H H H H H H CH3 CH3 CH3 is equivalent to C(CH3)3 CH2OH CH3CHCH(CH3)2 H H CH3 CH3 C HO H C CH3 H C is equivalent to which may be rewritten as HO CH3 CH3 H H CH3 C C CH3 (CH3)2CHCH(CH3)2 is equivalent to which may be rewritten as H H H C H H C H H C H C O H H H C H H H C H H H N C H H H C H H H C H H N C H H H C H H C Cl Cl Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website