UNIT IV: CHEMICAL BONDING Lewis Structures (Diagrams) of covalent compounds and polyatomic ions show the bonding arrangement of those compounds or ions. (Chang 13th edition, section 9.6 and 9.9 read pages 381-384 and 389-395) Practice Problems from the text: #9.43-9.47 pages 402-403 Lewis structures are based on the idea of electron sharing between atoms in order to complete an electron octet. They show a two-dimensional bonding picture of a molecule or an ion. The valence electrons are represented as follows: non-bonded electron pairs: (free or lone pairs) bonded electron pair(s): (shared between two atoms) – (the dash represents two bonded electrons) LEWIS STRUCTURES OF COVALENT COMPOUNDS AND POLYATOMIC IONS 1. A probable arrangement of the atoms can be deduced as follows: (note: this is for neutral molecules) The structure assumed must be consistent with the known valences of the elements: e.g. H forms one bond, F forms one bond, O forms two bonds, N forms three bonds, C forms four bonds, etc. ( In a molecule with the formula AXn the atom A is usually the central atom to which X are attached: e.g. CH4, BeH2, PCl5, SF6, etc. Atom A is also normally the least electronegative atom or the atom with the highest valence. ex. H2O, NH3, BF3, etc. 125 1. Draw a diagram of the molecule or polyatomic ion, showing the atoms connected by single bonds to the central atom 2. Add the number of valence electrons of each atom in the molecule to find the total number of valence electrons. Subtract the number of electrons needed to form the single bonds from the total number of electrons. Use the remainder to complete octets around each atom except hydrogen. Exceptions to the octet rule: compounds of Be and B have less than 8 electrons in their valence shell. Be ends up with 4 electrons and B has 6 electrons. The rule is that if there are insufficient electrons to complete all the octets, complete those of the more electronegative atoms first. 1) Incomplete octet: BeCl2, BF3, SnF2, and HgCl2 2) Molecules/ions with an odd number of electrons: NO, NO2 3) Expanded octet: PCl5, SF6, XeF4 etc… (Noble gases only have expanded octets) Examples: CH4 H2O NH3 126 Exceptions to the octet rule (Incomplete octet): BeCl2 BeH2 BF3 BH3 127 3. If the molecule is charged, and/or a polyatomic ion, add one electron for each negative charge or subtract one electron for each positive charge. Examples: H3O+ NH4+ BCl4– NH2– 128 4. Atoms from the 3rd or later period: If the central atom is from period 3 or a later period, the octet rule may not apply. These atoms can have more than 8 electrons in their valence shell: (Expanded octet) Examples: PCl5 SF6 If there are more than enough electrons to complete all octets, add the remaining electrons in pairs to the central atom (which must be from the 3rd or a later period). Examples: SCl4 ClF3 XeF2 BrF4+ SbCl52– 129 5. Double or triple bonds: If atoms of N, O, C, and S still have an incomplete octet, convert nonbonding electron pairs to bonding electron pairs. (Use nonbonding pairs to form double or triple bonds until each atom has an octet.) Examples: O2 C2H4 H2CO N2 CO CN – 130 6. Formal charges: (Chang 13th edition section 9.7, read pages 384-387) Practice problems #9.51 and 9.52 page 403 Formal charge can be assigned to any atom of a given Lewis structure. When calculating a formal charge, we are essentially comparing the number of electrons before and after bonding. Formal charge can be calculated using the following formula: Formal charge = group number of atom 1/2 number of shared electrons + number of unshared electrons Examples: H2O NH3 H3O+ SbCl5 2– 131 The following rules are helpful: For neutral molecules, the sum of the formal charges must add up to zero. For cations, the sum of the formal charges must equal the positive charge. For anions, the sum of formal charges must equal the negative charge. Formal charges are used to select the most appropriate Lewis structure when choice exists. The Lewis structure, which creates the least number of formal charges, is considered the most correct one. Examples: CO2 SCN – Rule: 1) Atoms in molecules, try to achieve formal charges as close to zero as possible. 2) Any negative formal charges are expected to reside on the most electronegative atoms. Homework Draw the Lewis structures for the following: H2S, AsF52–, BeF2, C2H2, TeF4, CS2, BrF3, ICl4–, N2H4, HCN, BrF5, XeF3+, XeF4, C2Cl4, BCl3, C2F2, N2H2. 132 LEWIS STRUCTURES: Isomers and Resonance Isomers are compounds with the same molecular formula but different structure. Isomerism can be observed when two or more Lewis structures can be written in which the atoms are arranged (connected) in different ways. To change one isomer to another, bonds between atoms have to be broken in order to rearrange the atoms. Examples: C2H2Cl2 C2H4O Homework Draw isomers of the following compounds: a) C2H4ClBr b) C3H5F 133 Resonance can be anticipated when two or more Lewis structures are equally plausible. In writing resonance structures, only electrons are shifted around, not atoms. Species having resonance are built up of one type of molecule whose structure is assumed to be in between (a hybrid of) those resonance structures. Species having resonance experience special stability. This is due to the delocalization of the electrons. (Chang 13th edition section 9.8 pages 387-389) Examples: C6H6: NO3- : O3: SO3: SeO2 134 LEWIS STRUCTURES: Bond Order and Bond Length Bond order (b.o.) can be pictured as the number of covalent bonds between bonded atoms. In resonance structures, fractions are possible. Bond length depends on the type of a bond: single bonds are the longest and triple bonds are the shortest. Examples: 1. C2H6 C2H4 C2H2 C6H6 C-C b.o.: CC bond length (nm) 2. S O b.o. _______ 0.154 _______ _______ _______ 0.134 0.120 0.139 SO42– SO3 _______ _______ Rank the above species according to their increasing S-O bond length: ______________________________________________________________________ shortest S-O bond longest S-O bond Conclusion: Bond length and bond order are inversely proportional to each other: When the bond order is high, bond length is short. When the bond order is low, bond length is long. 135 MOLECULAR GEOMETRY edition section 10.1 pages 411-420) Practice Problems from the text #10.7-10.14 page 453 (Chang 13th VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY: The shape of molecules depends on the number of electron pairs around the central atom. Electron pairs repel each other and will adopt a position in space to be so far apart as possible and thereby minimise the repulsion between them. The number of electron pairs around the central atom(s) is called the repulsion number or the number of regions of high electron density. Correct Lewis structures are important. Since the lone pairs exert a larger repulsive effect than do bonded pairs (they take up more space than a bonded pair) the bond angle will slightly decrease for the bonded pairs: Example: of CH4, NH3, and H2O Angles are specific for CH4, NH3 and H2O only figure 10.1, p. 416 Chang, R., Overby, J.(2019). Chemistry 13 th edition. New York: McGraw-Hill 136 Note: A dashed wedge represents the bond going into the paper (away from you) and a wedge represents the bond coming towards you. The other bon ds are in the plane of the paper. Molecules With Double and/or Triple Bonds: Electrons in a double or a triple bond occupy the same region of space and are counted as one region of high electron density. For the repulsion purposes, they count essentially as one pair. e.g. CO2 : O C O Two regions of high electron density, repulsion number is 2. Molecule of CO2 is linear. Table 10.1 in your textbook (13th edition) shows the geometry of some simple molecules (or ions) in which the central atom has no lone pairs and some molecules with one or more lone pairs. 137 Table 10.1, p. 413 Chang, R., Overby, J.(2019). Chemistry 13th edition. New York: McGraw-Hill 138 Table 10.1, 10.2, p. 413 418 Chang, R., Overby, J.(2019). Chemistry 13th edition. New York: McGraw-Hill 139 Study Sheet http://chemistry.ncssm.edu/labs/molgeom/vseprgeo.jpg 140 Polarity of Molecules: (Chang 13th edition section 10.2 read pages 421-426) Practice Problems from the text #10.19-10.22 page 454 Many molecules are polar, that is, one end of the molecule has a small net positive charge, while the other end has a small net negative charge. Refer to end of Unit III, covalent and ionic bonds and their electronegativity differences. Once the overall geometry is known, the polarity of a molecule or ion can be deduced. The dipole moment of a molecule or ion is the resultant of the individual bond dipoles. Each bond dipole can be considered a vector, represented by an arrow that points from the positive to the negative end. Do not include a vector towards the lone pairs, only towards atoms. The net dipole moment of the molecule is then obtained by the addition of all vectors. 141 Note: A polar bond has an electronegativity value of 0.5 or higher. Must have correct geometry before drawing in the vectors. Vectors are drawn between atoms only, no lone pairs. Examples: 142 HOMEWORK EXERCISE ON LEWIS STRUCTURES, GEOMETRY, BOND ANGLES AND POLARITY Complete the following table: Molecule or Species Lewis Structure Rep. # Arrangement of Electron Pairs # of Free el. pairs Geometry of Species (3D Sketch & Name) Bond Angle(s) Polarity (yes or no) Work On: OF2; CH4; BeCl2; CH2F2; BF3; NH3; PCl5; ClF3; SF6; IF5; BrF4–; XeF2; XeF4; ClF4+; SO2; H2CO; HCN; N2H4; CO2; NH2–; I3–; etc. 143 VALENCE BOND THEORY (Chang 13th edition section 10.3 pages 426-428) The covalent bond in H2 serves as a model for understanding all covalent bonds. When two atoms, each with an unpaired electron, come close enough together so that atomic orbitals begin to overlap, the two electrons are pulled into the region between the two nuclei. The greater the amount of overlap of the atomic orbitals in the region between the two nuclei, the stronger the bond between the two atoms will be. Maximum overlap has to be achieved. Overlap of Atomic Orbitals: a) b) H H c) H H HH (H2 molecule) Heitler-London diagram: Plot of potential energy vs distance between atoms Change in potential energy of two H atoms with their distance of separation. At the point of minimum potential energy, the H2 molecule is in its most stable state and the bond length is 74 pm. The spheres represent the 1s orbitals. figure 10.5, p. 427 Chang, R., Overby, J.(2019). Chemistry 13 th edition. New York: McGraw-Hill 144 Types of overlap: Orbital diagram: electron density diagram: bond: two 1s orbitals 1s 1s 1s and a p orbital two p orbitals (head to head) two hybrid orbitals (sp, sp2, sp3, sp3d, sp3d2) a hybrid orbital and 1s two p orbitals (side to side) bonds are symmetrical and free rotation about the axis that connects the two atoms exists: bonds are rigid and no free rotation is possible without twisting or breaking the bonds. 145 Hybrid Atomic Orbitals During the formation of H2 molecule, each hydrogen atom has one unpaired electron. Since unpaired electrons are required in the covalent bond formation, it could be assumed that the number of unpaired electrons in the atom’s valence shell is equal to the number of covalent bonds formed by that atom. But the above statement does not explain the number of covalent bonds in compounds of Be, B, C, and some other atoms, where the number of unpaired valence electrons does not equal the number of covalent bonds. The Concept of Hybridization: Chang 13th edition Section 10.4 pages 428-437 The hybridization theory considers that splitting electron pair(s) and promoting electrons to other orbitals fairly close in energy will provide us with the desired number of single electrons. These halffilled orbitals then form new orbitals through the process of hybridization. These new hybrid orbitals are used to explain the number of covalent bonds in many covalent compounds and/or, in some cases, the existing bond angles. Examples of such compounds are listed below. Atom: Box diagram: # of unpaired e-: compound(s): # of covalent bonds: Be [He 0 BeH2, BeCl2 2 B [He] 1 BH3, BCl3 3 C [He] 2 CH4, CCl4 4 P [Ne] 3 PCl5 5 146 Using the hybridized atoms on the following pages along with the following examples, orbital diagrams are constructed. All orbitals, bonds, bonds, electron spins, and lone pairs must be labeled. Compounds with Single () Bonds: Orbital Overlap Diagram H2 BeH2 BH3 CH4 NH3 NH3, CH4, C2H2, C2H4, H2, O2, X2, N2 & CO2, C3HX H2O HF, HCl PH3, PCl5 SF4, SF6 ClF3 XeF2, XeF4, KrF4 Compounds with One Double Bond (One Bond): C2H4 O2 N2H2 Compounds with One Triple Bond or Two Double Bonds (Two Bonds): C2H2 N2 CO2 Various Cations and Anions: SiF4 2ClF4+ AsF5 2BrF4IF6+ XeF3+ Practice Problems from the textbook #10.31-10.38 page 454 147 Examples: Molecule Central Atom BeH2 Be BH3 B NH3 N Box Diagram Geometry and bond angles 148 Molecule Central Atom H2O O HF F PCl5 P Box Diagram Geometry and bond angles 149 Molecule Central Atom SF4 S ClF3 Cl XeF2 Xe Box Diagram Geometry and bond angles 150 Molecule Central Atom KrF4 Kr Box Diagram Geometry and bond angles Ion SiF42- Si-2 ClF4+ Cl+ 151 Molecule Central Atom AsF52- As-2 BrF4- Br - IF6+ I+ Box Diagram Geometry and bond angles 152 Molecule Central Atom XeF3+ Xe+ CO2 C Box Diagram Geometry and bond angles Orbital Diagram 153 Molecule Central Atom O2 O CH4 C N2 N Box Diagram Orbital Diagram Geometry and bond angles 154 Molecule Central Atom C2H2 C C2H4 C Br2 Br Box Diagram Orbital Diagram Geometry and bond angles 155 Molecule Central Atom Box Diagram Orbital Diagram Geometry and bond angles H2 NH3 156 More than one central atom 1. Consider the following molecule whose Lewis structure is given below: H H H – O – C1– C2 ≡ C3 – C4 = O H a) Mow many sigma bonds are found in the above structure?_____________________ b) How many pi bonds?_________________________ c) Which carbon atom(s) is/are: (i) (ii) (iii) sp3 hybridized?_______ sp2 hybridized?_______ sp hybridized? _______ d) What is the C1– O – H theoretical bond angle? _______ Homework: 1. Give the 3D VSEPR sketch for the following examples. Give the hybridization of the central atom using the box diagram. CN-, CH2NH, CCl3NF2, CH3OH, CH3NH2, CCl4, C2H6, C3H8, CH3Cl, BeF2, BCl3 2. For each of the following structures, give the 3D VSEPR sketch. Give the hybridization of the central atom using the box diagram. H2S, AsF52–, BeF2, C2H2, TeF4, CS2, BrF3, ICl4–, N2H4, HCN, BrF5, XeF3+, XeF4, C2Cl4, BCl3, C2F2, N2H2 157
