1.6 Formal Charge to both nitrogen and hydrogen, and the third has a single bond to nitrogen and a tive charge. Nitrogen is positively charged. The positive and negative charges are alled formal charges, and the Lewis structure of nitric acid would be incomplete were they to be omitted le calculate formal charges by counting the number of electrons""owned"by each atom in a Lewis structure and comparing this electron count with that of a neutral atom. The r Figure 1.6 illustrates how electrons are counted for each atom in nitric acid. counting group element such electrons for the purpose of computing the formal charge differs from counting electrons nitrogen is equal to its group to see if the octet rule is satisfied. A second-row element has a filled valence shell if the gen this is 5 sum of all the electrons, shared and unshared, is 8. Electrons that connect two atoms by a covalent bond count toward filling the valence shell of both atoms. when calculating the formal charge, however, only half the number of electrons in covalent bonds can b considered to be"owned"by an atom To illustrate, let's start with the hydrogen of nitric acid. As shown in Figure 1.6, It will always be true that ydrogen is associated with only two electrons--those in its covalent bond to oxygen. covalently bonded hydrogen It shares those two electrons with oxygen, and so we say that the electron count of each has no formal charge(formal hydrogen is 2(2)=1. Since this is the same as the number of electrons in a neutral hydrogen atom, the hydrogen in nitric acid has no formal charge Moving now to nitrogen, we see that it has four covalent bonds(two single bonds t one double bond), and so its electron count is ) 4. A neutral nitrogen has five will always be true that a electrons in its valence shell. The electron count for nitrogen in nitric acid is I less than bonds has a formal charge of that of a neutral nitrogen atom, so its formal charge is +1 +1. (A nitrogen with four e Electrons in covalent bonds are counted as if they are shared equally by the atoms unshared pairs, because of they connect, but unshared electrons belong to a single atom. Thus, the oxygen which the octet rule is doubly bonded to nitrogen has an electron count of 6(four electrons as two unshared pairs+ two electrons from the double bond). Since this is the same as a neutral oxy- gen atom, its formal charge is 0. Similarly, the OH oxygen has two bonds plus two It will always be true that an unshared electron pairs, giving it an electron count of 6 and no formal charge The oxygen highlighted in yellow in Figure 1.6 owns three unshared pairs(six bonds and two unshared electrons)and shares two electrons with nitrogen to give it an electron count of 7. This is I more than the number of electrons in the valence shell of an oxygen atom, and so its formal charge is -l reasoning oxygen with one covalent The method described for calculating formal charge has been one of reasoning through a series of logical steps. It can be reduced to the following equation: pairs has a formal charge of Formal charge s group number in number of bonds -number of unshared electrons -Electron count(O)=1(4)+4=6 Electron count(H)=1(2)=1- Electron count(N)=5(8)=4 Electron count(O)=1(4)+4=6 Electron count(O)=1(2)+6=7 FIGURE 1.6 Counting electrons in nitric acid. The electron count of each atom is equal to half the number of electrons it r of electrons in its own unshared pairs. Back Forward Main MenuToc Study Guide ToC Student o MHHE Website
to both nitrogen and hydrogen, and the third has a single bond to nitrogen and a negative charge. Nitrogen is positively charged. The positive and negative charges are called formal charges, and the Lewis structure of nitric acid would be incomplete were they to be omitted. We calculate formal charges by counting the number of electrons “owned” by each atom in a Lewis structure and comparing this electron count with that of a neutral atom. Figure 1.6 illustrates how electrons are counted for each atom in nitric acid. Counting electrons for the purpose of computing the formal charge differs from counting electrons to see if the octet rule is satisfied. A second-row element has a filled valence shell if the sum of all the electrons, shared and unshared, is 8. Electrons that connect two atoms by a covalent bond count toward filling the valence shell of both atoms. When calculating the formal charge, however, only half the number of electrons in covalent bonds can be considered to be “owned” by an atom. To illustrate, let’s start with the hydrogen of nitric acid. As shown in Figure 1.6, hydrogen is associated with only two electrons—those in its covalent bond to oxygen. It shares those two electrons with oxygen, and so we say that the electron count of each hydrogen is 1 2 (2) 1. Since this is the same as the number of electrons in a neutral hydrogen atom, the hydrogen in nitric acid has no formal charge. Moving now to nitrogen, we see that it has four covalent bonds (two single bonds one double bond), and so its electron count is 1 2 (8) 4. A neutral nitrogen has five electrons in its valence shell. The electron count for nitrogen in nitric acid is 1 less than that of a neutral nitrogen atom, so its formal charge is 1. Electrons in covalent bonds are counted as if they are shared equally by the atoms they connect, but unshared electrons belong to a single atom. Thus, the oxygen which is doubly bonded to nitrogen has an electron count of 6 (four electrons as two unshared pairs two electrons from the double bond). Since this is the same as a neutral oxygen atom, its formal charge is 0. Similarly, the OH oxygen has two bonds plus two unshared electron pairs, giving it an electron count of 6 and no formal charge. The oxygen highlighted in yellow in Figure 1.6 owns three unshared pairs (six electrons) and shares two electrons with nitrogen to give it an electron count of 7. This is 1 more than the number of electrons in the valence shell of an oxygen atom, and so its formal charge is 1. The method described for calculating formal charge has been one of reasoning through a series of logical steps. It can be reduced to the following equation: Formal charge group number in number of bonds number of unshared electrons periodic table 1.6 Formal Charge 17 H±O±N O O œ ± Electron count (O) (4) 4 6 1 2 Electron count (N) (8) 4 1 2 Electron count (O) (2) 6 7 1 Electron count (O) (4) 4 6 2 1 2 Electron count (H) (2) 1 1 2 FIGURE 1.6 Counting electrons in nitric acid. The electron count of each atom is equal to half the number of electrons it shares in covalent bonds plus the number of electrons in its own unshared pairs. The number of valence electrons in an atom of a maingroup element such as nitrogen is equal to its group number. In the case of nitrogen this is 5. It will always be true that a covalently bonded hydrogen has no formal charge (formal charge 0). It will always be true that a nitrogen with four covalent bonds has a formal charge of 1. (A nitrogen with four covalent bonds cannot have unshared pairs, because of the octet rule.) It will always be true that an oxygen with two covalent bonds and two unshared pairs has no formal charge. It will always be true that an oxygen with one covalent bond and three unshared pairs has a formal charge of 1. Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CHAPTER ONE Chemical Bonding PROBLEM 1.9 Like nitric acid, each of the following inorganic compounds will be frequently encountered in this text. Calculate the formal charge on each of tructures gIv H--S—0-H SAMPLE SOLUTION (a)The formal charge is the difference between the num- ber of valence electrons in the neutral atom and the electron count in the lewis structure.(The number of valence electrons is the same as the group number in the periodic table for the main-group elements. Valence electrons of neutral atom Electron count Formal charge Sulfur: 667 4(6)+2=5 4(2)+6=7 chlorine 4(2)+6=7 The formal charges are shown in the Lewis structure of thionyl chloride Cl-S-CI So far we've only considered neutral molecules--those in which the sums of the positive and negative formal charges were equal. with ions, of course, these sums will not be equal. Ammonium cation and borohydride anion, for example, are ions with net harges of +I and -l, respectively. Nitrogen has a formal charge of +I in ammonium ion, and boron has a formal charge of -l in borohydride. None of the hydrogens in the Lewis structures shown for these ions bears a formal charge H HH—B-H H Ammonium ion Borohydride ion PROBLEM 1.10 Verify that the formal charges on nitrogen in ammonium ion and boron in borohydride ion are as shown Formal charges are based on Lewis structures in which electrons are considered to be shared equally between covalently bonded atoms. Actually, polarization of N-H bonds in ammonium ion and of B-H bonds in borohydride leads to some transfer of positive and negative charge, respectively, to the hydrogens PROBLEM 1.11 Use 8+ and 8- notation to show the dispersal of charge to the hydrogens in NHa and BHa Back Forward Main MenuToc Study Guide ToC Student o MHHE Website
PROBLEM 1.9 Like nitric acid, each of the following inorganic compounds will be frequently encountered in this text. Calculate the formal charge on each of the atoms in the Lewis structures given. SAMPLE SOLUTION (a) The formal charge is the difference between the number of valence electrons in the neutral atom and the electron count in the Lewis structure. (The number of valence electrons is the same as the group number in the periodic table for the main-group elements.) The formal charges are shown in the Lewis structure of thionyl chloride as Cl±S±Cl O W Sulfur: Oxygen: Chlorine: Valence electrons of neutral atom 6 6 7 Formal charge 1 1 0 Electron count 1 2(6) 2 5 1 2(2) 6 7 1 2(2) 6 7 Cl±S±Cl O W (a) Thionyl chloride H±O±S±O±H O W W O (b) Sulfuric acid H±O±NœO (c) Nitrous acid 18 CHAPTER ONE Chemical Bonding So far we’ve only considered neutral molecules—those in which the sums of the positive and negative formal charges were equal. With ions, of course, these sums will not be equal. Ammonium cation and borohydride anion, for example, are ions with net charges of 1 and 1, respectively. Nitrogen has a formal charge of 1 in ammonium ion, and boron has a formal charge of 1 in borohydride. None of the hydrogens in the Lewis structures shown for these ions bears a formal charge. PROBLEM 1.10 Verify that the formal charges on nitrogen in ammonium ion and boron in borohydride ion are as shown. Formal charges are based on Lewis structures in which electrons are considered to be shared equally between covalently bonded atoms. Actually, polarization of N±H bonds in ammonium ion and of B±H bonds in borohydride leads to some transfer of positive and negative charge, respectively, to the hydrogens. PROBLEM 1.11 Use and notation to show the dispersal of charge to the hydrogens in NH4 and BH4 . Ammonium ion H±N±H H W W H Borohydride ion H±B±H H W W H Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
1.7 Structural Formulas of Organic Molecules Determining formal charges on individual atoms of Lewis structures is an impor- tant element in good"electron bookkeeping. " So much of organic chemistry can be made more understandable by keeping track of electrons that it is worth taking some time at the beginning to become proficient at the seemingly simple task of counting electrons 1.7 STRUCTURAL FORMULAS OF ORGANIC MOLECULES Table 1. 4 outlines a systematic procedure for writing Lewis structures. Notice that the cess depends on knowing not only the molecular formula, but also the order in which the atoms are attached to one another. This order of attachment is called the constitu. tion, or connectivity, of the molecule and is determined by experiment. Only rarely is Organic chemists have devised a number of shortcuts to speed the writing of struc- tural formulas. Sometimes we leave out unshared electron pairs, but only when we are sure enough in our ability to count electrons to know when they are present and when theyre not. We've already mentioned representing covalent bonds by dashes In con densed structural formulas we leave out some, many, or all of the covalent bonds and use subscripts to indicate the number of identical groups attached to a particular atom These successive levels of simplification are illustrated as shown for isopropyl alcohol rubbing alcohol”) CHICHCH C-C-H written as or condensed even further to OH H :0: H (CH3)2CHOH PROBLEM 1.12 Expand the following condensed formulas so as to show all the bonds and unshared electron pair (a) HOCH2 CH2NH2 (d)CH3 CHCI2 (e)CH3 CH3 (c)ClCH2CH2CI ()(CH3)2CHCH=O SAMPLE SoLUTION (a) The molecule contains two carbon atoms, which are nded to each other. both carbons bear two hydrogens. One carbon bears the group HO- the other is attached to-NH2 When writing the constitution of a molecule, it is not necessary to self with the spatial orientation of the atoms. there are many othe to represent the constitution shown What is important is to show quence OCCN (or its equivalent NCCO)and to have the correct number of hydrogens present on each atom. n order to locate unshared electron pairs, first count the total number valence electrons brought to the molecule by its component atoms. Each hydro- gen contributes 1, each carbon 4, nitrogen 5, and oxygen 6, for a total of 26. There are ten bonds shown, accounting for 20 electrons; therefore 6 electrons must be contained in unshared pairs. Add pairs of electrons to oxygen and nitrogen so tha their octets are complete, two unshared pairs to oxygen and one to nitrogen Back Forward Main MenuToc Study Guide ToC Student o MHHE Website
Determining formal charges on individual atoms of Lewis structures is an important element in good “electron bookkeeping.” So much of organic chemistry can be made more understandable by keeping track of electrons that it is worth taking some time at the beginning to become proficient at the seemingly simple task of counting electrons. 1.7 STRUCTURAL FORMULAS OF ORGANIC MOLECULES Table 1.4 outlines a systematic procedure for writing Lewis structures. Notice that the process depends on knowing not only the molecular formula, but also the order in which the atoms are attached to one another. This order of attachment is called the constitution, or connectivity, of the molecule and is determined by experiment. Only rarely is it possible to deduce the constitution of a molecule from its molecular formula. Organic chemists have devised a number of shortcuts to speed the writing of structural formulas. Sometimes we leave out unshared electron pairs, but only when we are sure enough in our ability to count electrons to know when they are present and when they’re not. We’ve already mentioned representing covalent bonds by dashes. In condensed structural formulas we leave out some, many, or all of the covalent bonds and use subscripts to indicate the number of identical groups attached to a particular atom. These successive levels of simplification are illustrated as shown for isopropyl alcohol (“rubbing alcohol”). PROBLEM 1.12 Expand the following condensed formulas so as to show all the bonds and unshared electron pairs. (a) HOCH2CH2NH2 (d) CH3CHCl2 (b) (CH3)3CH (e) CH3NHCH2CH3 (c) ClCH2CH2Cl (f) (CH3)2CHCHœO SAMPLE SOLUTION (a) The molecule contains two carbon atoms, which are bonded to each other. Both carbons bear two hydrogens. One carbon bears the group HO±; the other is attached to ±NH2. When writing the constitution of a molecule, it is not necessary to concern yourself with the spatial orientation of the atoms. There are many other correct ways to represent the constitution shown. What is important is to show the sequence OCCN (or its equivalent NCCO) and to have the correct number of hydrogens present on each atom. In order to locate unshared electron pairs, first count the total number of valence electrons brought to the molecule by its component atoms. Each hydrogen contributes 1, each carbon 4, nitrogen 5, and oxygen 6, for a total of 26. There are ten bonds shown, accounting for 20 electrons; therefore 6 electrons must be contained in unshared pairs. Add pairs of electrons to oxygen and nitrogen so that their octets are complete, two unshared pairs to oxygen and one to nitrogen. H±O±C±C±N±H H W W H H W H W W H H±C±C±C±H written as or condensed even further to (CH3)2CHOH H W W H OH W CH3CHCH3 H W W H H W W O W H 1.7 Structural Formulas of Organic Molecules 19 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
TABLE 1.4 How to Write Lewis Structures Step Illustration 1. The molecular formula and the connectivity are Methyl nitrite has the molecular formula CH3No2. All determined experimentally and are included hydrogens are bonded to carbon, and the order of among the information given in the statement of atomic connections is CONO the problem 2. Count the number of valence electrons available Each hydrogen contributes 1 valence electron, car or a neutral molecule this is equal to the sum of on contributes 4, nitrogen contributes 5, and each the valence electrons of the constituent atoms oxygen contributes 6 for a total of 24 in CH3 NO 3. Connect bonded atoms by a shared electron pair For methyl nitrite we write the partial structure ond()represented by a dash(-) 4. Count the number of electrons in shared electron The partial structure in step 3 contains 6 bonds pair bonds (twice the number of bonds), and sub- equivalent to 12 electrons. Since CH3No, contains 24 tract this from the total number of electrons to electrons. 12 more electrons need to be added give the number of electrons to be added to com- plete the structure 5. Add electrons in pairs so that lany atoms as With 4 bonds, carbon already has 8 electrons. The re possible have 8 electrons(Hydrogen is limited to 2 maining 12 electrons are added as indicated. Both electrons. )When the number of electrons is insuffi- oxygens have 8 electrons, but nitrogen(less electro- cient to provide an octet for all atoms, assign elec- negative than oxygen) has only 6. trons to atoms in order of decreasing electronega- tivity. H-C-0—N-O H 6. If one or more atoms have fewer than 8 electrons, An electron pair on the terminal oxygen is shared se unshared pairs on an adjacent atom to form a with nitrogen to give a double bon double(or triple)bond d to complete the octet H|C|H The structure shown is the best (most stable) Lewis structure for methyl nitrite. All atoms except hydro- gen have 8 electrons(shared unshared) in their va lence shell 7. Calculate formal charges None of the atoms in the lewis structure shown in step 6 possesses a formal charge. An alternative Lewis structure for methyl nitrite Ithough it satisfies the octet rule, is less stable than the one shown in step 6 because it has a separation of positive charge from negative charge Back Forward Main MenuToc Study Guide ToC Student o MHHE Website
TABLE 1.4 How to Write Lewis Structures Step 1. The molecular formula and the connectivity are determined experimentally and are included among the information given in the statement of the problem. 2. Count the number of valence electrons available. For a neutral molecule this is equal to the sum of the valence electrons of the constituent atoms. 6. If one or more atoms have fewer than 8 electrons, use unshared pairs on an adjacent atom to form a double (or triple) bond to complete the octet. 7. Calculate formal charges. 4. Count the number of electrons in shared electron pair bonds (twice the number of bonds), and subtract this from the total number of electrons to give the number of electrons to be added to complete the structure. 5. Add electrons in pairs so that as many atoms as possible have 8 electrons. (Hydrogen is limited to 2 electrons.) When the number of electrons is insufficient to provide an octet for all atoms, assign electrons to atoms in order of decreasing electronegativity. Illustration Methyl nitrite has the molecular formula CH3NO2. All hydrogens are bonded to carbon, and the order of atomic connections is CONO. Each hydrogen contributes 1 valence electron, carbon contributes 4, nitrogen contributes 5, and each oxygen contributes 6 for a total of 24 in CH3NO2. An electron pair on the terminal oxygen is shared with nitrogen to give a double bond. For methyl nitrite we write the partial structure The structure shown is the best (most stable) Lewis structure for methyl nitrite. All atoms except hydrogen have 8 electrons (shared unshared) in their valence shell. None of the atoms in the Lewis structure shown in step 6 possesses a formal charge. An alternative Lewis structure for methyl nitrite, although it satisfies the octet rule, is less stable than the one shown in step 6 because it has a separation of positive charge from negative charge. The partial structure in step 3 contains 6 bonds equivalent to 12 electrons. Since CH3NO2 contains 24 electrons, 12 more electrons need to be added. With 4 bonds, carbon already has 8 electrons. The remaining 12 electrons are added as indicated. Both oxygens have 8 electrons, but nitrogen (less electronegative than oxygen) has only 6. H±C±O±N±O H W W H 3. Connect bonded atoms by a shared electron pair bond ( ) represented by a dash (±). H±C±O±N±O H W W H H±C±O±NœO H W W H H±C±OœN±O H W W H Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
1.7 Structural Formulas of Organic Molecules As you practice, you will begin to remember patterns of electron distribution. A neutral oxygen with two bonds has two unshared electron pairs. A neutral nitro- gen with three bonds has one unshared pa With practice, writing structural formulas for organic molecules soon becomes rou tine and can be simplified even more. For example, a chain of carbon atoms can be rep resented by drawing all of the C-C bonds while omitting individual carbons. The result ing structural drawings can be simplified still more by stripping away the hydrogens HHHH CH.,CH CH, becomes H H SImplified to In these simplified representations, called bond-line formulas or carbon skeleton dia- grams, the only atoms specifically written in are those that are neither carbon nor hydro- gen bound to carbon. Hydrogens bound to these heteroatoms are shown, however CH3CH2CH? CH,OH H, CH H PROBLEM 1.13 Expand the following bond-line representations to show all the atoms including carbon and hydrogen (a)∠ SAMPLE SoLUTIoN (a) There is a carbon ends of the chain. Each of the ten carbon of hydrogen substituents so that it has four H HHHHHHHH H HHHHHHH H Alternatively the structure could be written as CH3CH2 CH2CH2 CH2CH2 CH2CH2 CH2 CH3 or in condensed form as CH3(CH2)8 CH Back Forward Main MenuToc Study Guide ToC Student o MHHE Website
As you practice, you will begin to remember patterns of electron distribution. A neutral oxygen with two bonds has two unshared electron pairs. A neutral nitrogen with three bonds has one unshared pair. With practice, writing structural formulas for organic molecules soon becomes routine and can be simplified even more. For example, a chain of carbon atoms can be represented by drawing all of the C±C bonds while omitting individual carbons. The resulting structural drawings can be simplified still more by stripping away the hydrogens. In these simplified representations, called bond-line formulas or carbon skeleton diagrams, the only atoms specifically written in are those that are neither carbon nor hydrogen bound to carbon. Hydrogens bound to these heteroatoms are shown, however. PROBLEM 1.13 Expand the following bond-line representations to show all the atoms including carbon and hydrogen. (a) (c) (b) (d) SAMPLE SOLUTION (a) There is a carbon at each bend in the chain and at the ends of the chain. Each of the ten carbon atoms bears the appropriate number of hydrogen substituents so that it has four bonds. Alternatively, the structure could be written as CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 or in condensed form as CH3(CH2)8CH3. H±C±C±C±C±C±C±C±C±C±C±H H W W H H W W H H W W H H W W H H W W H H W W H H W W H H W W H H W W H H W W H HO CH3CH2CH2CH2OH becomes becomes OH Cl W C H2C W W H2C CH2 CH2 ± C H2 ± H ± ± ± Cl ± CH3CH2CH2CH3 becomes simplified to H H H H H H ±± H H ±± ± H H ± ± ± ±± H±O±C±C±N±H H W W H H W H W W H 1.7 Structural Formulas of Organic Molecules 21 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website