Ch 10 Lewis Structures PDF

Ch. 10 – Shapes of Molecules: 1. Lewis Structures Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution p...

Ch. 10 – Shapes of Molecules: 1. Lewis Structures Chemistry The Molecular Nature of Matter and Change Ninth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Lewis Molecular Structures A Lewis structure of a molecule is a simplified, symbolic representation of the type of bonds (single, double, etc.) and lone electrons (pair or individual) in molecules. A Lewis structure consists of: – Periodic Table symbols of all elements present in a molecule according to the Molecular Formula, representing the nuclei and the inner electrons of the atoms; – The element symbols are arrayed according to the experimental evidence of the molecular structure in: linear or branched chain, rings, or a combination. In limited cases, the Lewis theory can predict the array, especially for small molecules. – Connectors (lines, sticks) between element symbols representing covalent bonds: 1 connector for each pair of shared electrons; multiple sticks between 2 atoms represent a multiple bond; – Dots around element symbols, representing un-shared valence electrons, i.e., not involved in covalent bonding. Most common are the lone pairs of electrons. ©McGraw-Hill Education. Steps In Converting a Molecular Formula Into a Lewis Structure Lewis theory is good (but not perfect) at predicting the type of covalent bonds in molecules starting from the molecular formula and the number of valence electrons of each atom. The steps for creating a Lewis structure (shown below) represent a unified system of counting and distributing the valence electrons of all atoms. ©McGraw-Hill Education. Writing Lewis Structures of Molecules 1. Write the correct skeletal structure for the molecule following these rules: – Hydrogen is always terminal atom because they form duets. – More electronegative atoms are placed in terminal positions. – Fluorine is always a terminal atom as the most electronegative of all elements. Fluorine is not known to form multiple bonds. 2. Calculate the total number of electrons for the Lewis structure by summing the valence electrons of each atom in the molecule. 3. Distribute the electrons among the atoms, giving octets (or duets in the case of hydrogen) to as many atoms as possible. C, N, O and F (located in period 2: 2s2 2px electron configuration) strictly follow the octet rule, i.e., never more than 8 electrons. The next available empty atomic orbitals, 3s and 3p are too high in energy to allow expanded octets. 4. If any atoms lack an octet, form double or triple bonds from lone pairs as necessary to give them octets. © 2017 Pearson Education, Inc. Example: NF3 Step 1: Nitrogen has a lower EN than F, so N is placed in the center. The array does not have to be symmetric for the purpose of writing a Lewis structure; however, this is the real geometry of this molecule determined experimentally. Step 2: N has five valence electrons, and each F has seven: [1 × N(5e−)] + [3 × F(7e−)] = 5e− + 21e− = 26 valence e(or 13 pairs) Step 3: Draw a single bond from each surrounding atom to the central atom: 3 pairs of electrons are used up. Step 4: Distribute the remaining 10 electron pairs so that each atom ends up with an octet. ©McGraw-Hill Education. Writing Lewis Structures for Species with Single Bonds and One Central Atom Sample Problem 10.1 PROBLEM: Write a Lewis structure for the following: a) CCl2F2 b) PCl2- ©McGraw-Hill Education. Sample Problem 10.1 - Solution (CCl2F2) SOLUTION (a): CCl2F2 Step 1: Place the atoms relative to each other. Carbon has the lowest EN and is the central atom. The other atoms are placed around it. Step 2: [1 x C(4e–)] + [2 x F(7e–)] + [2 x Cl(7e–)] = 32 valence e– Steps 3-4: Add single bonds, then give each atom a full octet. ©McGraw-Hill Education. Sample Problem 10.1 - Solution (PCl2-) SOLUTION (b): PCl2 Step 1: Place atoms relative to each other. Phosphorus has the lower EN and is the central atom. The other atoms are placed around it. Step 2: [1 x P(5e–)] + [2 x Cl(7e–)] + [1 negative charge (1e–)] = 20 valence e– Steps 3-4: Add single bonds, then give each atom a full octet. Step 5: Since PCl2– is an ion, draw square brackets as shown and add the negative sign at the top (outside the brackets). ©McGraw-Hill Education. Practice Problems 10.1A and B PROBLEM: Write a Lewis structure for the following: − H2S, AlCl− , BH 4 4 OF2, CH2Br2, IBr2+ ©McGraw-Hill Education. Writing Lewis Structures for Molecules with Single Bonds and More Than One Central Atom APPROACH: Identify the inner (“central”) and terminal atoms based on the duet rule for Hydrogen, and the octet rule for Carbon and Oxygen. Hydrogen is a terminal atom for all Organic Chemistry molecules. (There are 2-3 exceptions in Inorganic Chemistry.) In Organic Chemistry, chemists prefer condensed structural formulas for molecular formulas. Example: H3COH, rather than CH4O. The condensed structure is chosen to identify and emphasize the central atoms. Many compounds may have elements obeying the octet rule in terminal positions. The latter situation requires experimental data to identify the correct layout of the atoms in the molecular structure. ©McGraw-Hill Education. Sample Problem 10.2 - Problem PROBLEM: Write the Lewis structure for methanol (molecular formula CH4O), an important industrial alcohol that can be used as a gasoline alternative in cars. ©McGraw-Hill Education. Sample Problem 10.2 - Solution SOLUTION: Step 1: Place the atoms relative to each other. H can only form one bond, so C and O must be central and adjacent to each other. Step 2: [1 x C(4e–)] + [1 x O(6e–)] + [4 x H(1e–)] = 14 valence e– (7 pairs) Step 3-4: Add single bonds, then give each atom (other than H) a full octet. Follow-up practice: NH2OH, (CH3)2O, N2H4, CH3NH2 ©McGraw-Hill Education. Multiple Bonds If there are not enough electrons for the central atom to attain an octet, one or more multiple bonds are present. Step 5: If the central atom does not end up with a full octet, convert a lone pair on a surrounding atom to another bonding pair to the central atom to make a multiple bond. ©McGraw-Hill Education. Sample Problem 10.3 Writing Lewis Structures for Molecules with Multiple Bonds PROBLEM: Write Lewis structures for the following: a) Ethylene (C2H4), the most important reactant in the manufacture of polymers b) Nitrogen (N2), the most abundant atmospheric gas ©McGraw-Hill Education. Sample Problem 10.3 - Solution (C2H4) SOLUTION (a): – Step 1: Place the atoms relative to each other. In C2H4,the two C atoms must be bonded to each other since H atoms can have only one bond. – Step 2: Determine the total number of valence electrons (C is in Group 4A; H is in Group 1A): [2 × C(4e−)] + [4 × H(1e−) = 12 valence e− – Step 3: Add single bonds and subtract 2e− for each bond: 5 bonds × 2e− = 10e− so 12e− − 10e−= 2e− remaining – Step 4: Distribute the remaining valence electrons to attain octets. – Step 5: Change a lone pair to a bonding pair. The right C has an octet, but the left C has only 6e−, so we change the lone pair on the right C to another bonding pair between the two C atoms. ©McGraw-Hill Education. Sample Problem 10.3 - Solution (N2) SOLUTION (b): – Step 1: Place the atoms relative to each other. – Step 2: Determine the total number of valence electrons (N is in Group 5A): [2 × N(5e−)] = 10 valence e− – Step 3: Add single bonds and subtract 2e− for each bond: 1 bond × 2e− = 2e− so 10e− − 2e− = 8e− remaining – Step 4: Distribute the remaining valence electrons to attain octets. – Step 5: Neither N ends up with an octet, so we change a lone pair to a bonding pair. In this case, moving one lone pair to make a double bond still does not give both N atoms an octet, so we also move another lone pair from the other N to make a triple bond. ©McGraw-Hill Education. Writing Lewis Structures of Molecules and Ions Practice exercise. Write the Lewis structure for: CO, H2CO, CO2, NO+, N2H2, ClO-. © 2017 Pearson Education, Inc. Resonance Structures: Lewis Structures O3 can be drawn in 2 ways: These are two different reasonable Lewis structures for the same molecule. Neither structure depicts O3 accurately, because in reality the O-O bonds are equal in both length and energy. Experimental data show that the bond length in ozone is intermediate between the O=O and O-O bond lengths. ©McGraw-Hill Education. Resonance Structures: Arrows The structure of O3 is shown more correctly using both Lewis structures, called resonance structures. – A two-headed resonance arrow is placed between them. Resonance structures have the same relative placement of atoms but different locations of bonding and lone electron pairs. ©McGraw-Hill Education. Resonance Hybrid, Electron Delocalization A species like O3, which can be depicted by more than one valid Lewis structure, is best represented as a resonance hybrid. Resonance forms are not real bonding depictions: O3 does not change back and forth between its two resonance forms. The real structure of the resonance hybrid for O3 is an average of its contributing resonance forms. Lewis structures depict electrons as localized either on an individual atom (lone pairs) or in a bond between two atoms (shared pair). In a resonance hybrid, electrons are delocalized: their density is “spread” over a few adjacent atoms. Dotted lines are used to show delocalized electrons. Molecular Orbital theory (ch. 11) calculations show that the delocalization is real. ©McGraw-Hill Education. Hybrid A mule is a genetic mix, a hybrid, of a horse and a donkey. It is not a horse one instant and a donkey the next. Likewise, a resonance hybrid has a single structure although it retains characteristics of its resonance forms. ©McGraw-Hill Education. Fractional Bond Orders Resonance structures often have fractional bond orders due to partial bonding. # of shared electron pairs Bond Order = # of bonded − atom pairs For O3 = © McGraw Hill 3 shared electron pairs = 1 12 2 bonded − atom pairs Electron Delocalization In a resonance hybrid, electrons are delocalized: their density is “spread” over a few adjacent atoms. Benzene, C6H6 Carbonate, CO2− 3 9 6 Bond order in C6H6 = = 1.5 4 3 Bond order in CO2− 3 = = 1.33 ©McGraw-Hill Education. Resonance Structures: Example Practice exercise: Write the Lewis structure for the 𝑁𝑂2− nitrite anion. © 2017 Pearson Education, Inc. Formal Charge In some Lewis structures it is not clear which atom should be the central atom. In other Lewis structures it is not clear whether double or triple bonds are appropriate. In some resonance structures, one Lewis structure is closer to the hybrid structure than all other structures, so it is a more accurate description. What distinguishes various possible competing Lewis structures is a concept called the formal charge. © 2017 Pearson Education, Inc. Formal Charge Formal charge is a fictitious charge assigned to each atom in a Lewis structure that helps us to distinguish among competing Lewis structures. In a Lewis structure, we calculate an atom’s formal charge as the difference between the number of the atom’s valence electrons and the number of electrons it still “owns” in a Lewis structure: Formal Charge(atom) = # valence e− − [nonbonding e− + ½ bonding e−] Sum of all the formal charges in a neutral molecule must equal 0. In an ion, the sum of all formal charges equals the charge of the ion. © 2017 Pearson Education, Inc. Evaluating Resonance Structures Follow this hierarchy: 1. Better structures have fewer formal charges. Best Lewis structures do not have formal charges. 2. If formal charges are unavoidable, better structures have smaller formal charges. 3. Better structures have the negative formal charge on the more electronegative atom. © 2017 Pearson Education, Inc. Formal Charge: All Resonance Forms Formal charges must sum to the actual charge on the species for all resonance forms. Resonance Structure I Resonance Structure II OA [6 – 4 – ½(4)] = 0 OA [6 – 6 – ½(2)] = –1 OB [6 – 2 – ½(6)] = +1 OB [6 – 2 – ½(6)] = +1 OC [6 – 6 – ½(2)] = –1 OC [6 – 4 – ½(4)] = 0 For both these resonance forms the formal charges sum to zero, since O3 is a neutral molecule. ©McGraw-Hill Education. Preferred Resonance Form of NCO- Resonance forms with smaller formal charges are preferred. Resonance form I is therefore not an important contributor. A negative formal charge should be placed on a more electronegative atom, so resonance form III is preferred to resonance form II. The overall structure of the NCO– ion is a weighted average of all three forms; resonance forms II and III contribute more than I, and III contributes more than II. ©McGraw-Hill Education. Formal Charge Versus Oxidation Number For a formal charge, bonding electrons are shared equally by the atoms. – The formal charge of an atom may change between resonance forms. For an oxidation number, bonding electrons are considered as transferred to the more electronegative atom, irrespective of the actual Lewis structure. The oxidation number of an atom is the same in all resonance forms. ©McGraw-Hill Education. Writing Resonance Structures and Assigning Formal Charges Sample Problem 10.4 PROBLEM: Write resonance structures for the nitrate ion, NO3-​ and assign formal charges to the atoms in each resonance structure. PLAN: We write a Lewis structure, remembering to add 1e− to the total number of valence electrons because of the 1− ionic charge. Then we move lone and bonding pairs to write other resonance structures and connect them with resonance arrows. We assign formal charges. ©McGraw-Hill Education. Sample Problem 10.4 - Solution SOLUTION: Steps 1 and 2. Nitrogen has the lower group number and is placed in the center. There are [1 × N(5e−)] + [3 × O(6e−)] + [charge (1e−)] = 24 valence e−. Steps 3 and 4 give us: Step 5: Since N does not have a full octet, we change a lone pair from O to a bonding pair to form a double bond. Follow-up exercises: CH3NO2, SCN- ©McGraw-Hill Education. Exceptions to the Octet Rule The octet rule does not always apply. Notable exceptions: 1. Molecules with Electron-Deficient Atoms: B and Be are commonly electron-deficient. These elements can only share their valence electrons. The resulting Lewis structure gives less than an octet around these elements. 2. Odd-Electron Species: a molecule with an odd number of electrons is called a free radical. It happens when the sum of the valence electrons of all atoms plus charges in an ion is an ODD number. 3. Expanded Valence Shells: more covalent bonds are formed than allowed by the octet rule. Typical for period 3 and higher non-metals (P, S, Cl, Br, I, etc.) in combination with Oxygen. ©McGraw-Hill Education. Electron-Deficient and Odd-Electron Species Molecules with Electron-Deficient Atoms – B, Al, and Be are commonly electron-deficient. BeCl2 is a molecule, not an ionic compound (salt). Not likely! – Experimental evidence of electron deficiency in BF3 (formation of an adduct): ©McGraw-Hill Education. Odd-Electron Species Odd-Electron Species – NO2 has an odd number of valence electrons: 6 + 6 + 5 = 17. One electron cannot be paired up. There are 8 electron pairs and 1 odd electron. – A molecule with an odd number of electrons is called a free radical. 0 +1 -1 0 – Which resonance structure is more likely? – Experimental evidence shows that the dimerization of this radical always occurs at the N atom, in apparent contradiction to the formal charge theory: ©McGraw-Hill Education. Expanded Valence Shells An expanded valence shell is only possible for nonmetals from Period 3 or higher because these elements have available empty 3d (or nd) orbitals. ©McGraw-Hill Education. Expanded Valence Shells: ClF3 Determine the Lewis structure for ClF3: – #VE = 7(Cl) + 3 x 7(F) = 28e- (14 pairs) – 14 pairs: 3 single bonds(Cl-F) + 11 lone pairs – Distribution of 11 lone pairs: 9 lone pairs on 3 F atoms to generate octets 2 lone pairs on Cl, but this yields more than an octet around Cl Practice problems: XeF6, IF5 ©McGraw-Hill Education. Expanded Valence Shells: oxoacids, oxoanions Practice problems: HClO4, ClO3©McGraw-Hill Education. Limitations of Lewis Structures and Formal Charge Cannot choose the correct location for the odd electron in NO2. Octet rule and non-zero formal charges in H2SO4, SO42- seem to be the preferred structures, according to quantum-mechanical calculations. Shorter bonds arise not from double bonds but from stronger attraction between partially charged S and O atoms. ©McGraw-Hill Education. Sample Problem 10.5 - Problem and Plan Writing Lewis Structures for Octet-Rule Exceptions PROBLEM: Write a Lewis structure and identify the octet-rule exception for a) XeF4 b) H3PO4 c) BFCl2 PLAN: We write each Lewis structure and examine it for exceptions to the octet rule. a) The central atoms are in Periods 5 and 3, respectively, so they can have more than an octet. b) The central atoms are in Periods 5 and 3, respectively, so they can have more than an octet. c) ©McGraw-Hill Education. The central atom is B, which can have fewer than an octet of electrons. Sample Problem 10.5 - Solution (XeF4) SOLUTION (a): a) XeF4 has a total of 36 valence electrons: 8 form the bonds and 24 complete the octets of the F atoms. The remaining 4 valence electrons are placed on the central atom, resulting in an expanded valence shell: ©McGraw-Hill Education. Sample Problem 10.5 - Solution (H3PO4) SOLUTION (b): b) H3PO4 has 32 valence electrons: 14 form bonds and 18 complete the octets of the O and P atoms (structure I, shown with formal charges). Another resonance structure (structure II) can be drawn in which a lone pair from the O atom with nonzero formal charge is changed to a bonding pair. Structure I obeys the octet rule but has nonzero formal charges. Structure II has an expanded valence shell with zero formal charges. According to formal charge rules, structure II is the more important structure. ©McGraw-Hill Education. Sample Problem 10.5 - Solution (BCl2F) SOLUTION (c): c) BFCl2 has 24 valence electrons: 6 form bonds and 18 complete the octets of the F and Cl atoms. This molecule has an electron-deficient atom (incomplete octet); B has only six electrons surrounding it: Follow-up exercises: POCl3, ClO2, IBr-4, BeH2, I-3, XeO3 ©McGraw-Hill Education.