(CHAPTER 8) BASIC CONCEPTS OF CHEMICAL BONDING Prof. Dr. Nizam M. El-Ashgar CHEMICAL BONDS, LEWIS SYMBOLS, AND THE OCTET RULE A chemical Bond: A Chemical Bond is a link between atoms or attaching two atoms or ions strongly to each other. Why do atoms bond? Atoms gain stability when they share, gain or lose electrons which lead to reach noble-gas electron configurations. A bond will form if the energy of the aggregate is lower than that of the separated atoms. ١
TYPES OF CHEMICAL BONDS Ionic : Electrons are transferred from a metal to a nonmetal Covalent: Electrons are shared between 2 nonmetals Metallic: Formed between metal atoms The electrons involved in bonding are called valence electrons. Valence electrons: are found in the incomplete, outermost shell of an atom. IONIC BOND An ionic bond is the electrostatic forces that exist between ions of opposite charge. Ionic bonds normally occur between a metal and a nonmetal. ٢
COVALENT BOND A covalent bond results from the sharing of electrons between two atoms. Covalent bonds normally occur between two or more nonmetals. METALLIC BOND Metallic bonds are the attractions between the positive nuclei of metal atoms and the sea of electrons (from the outer shell of atom). ٣
LEWIS SYMBOLS The valence electrons: The electrons involved in chemical bonding which present in the outermost occupied shell of an atom. The Lewis symbol for an element: Consists of the chemical symbol for the element plus a dot ( ), circle (ο) or cross(x) for each valence electron. Example: [Ne]3s 2 3p 4 (sulfur) (6 valence electrons) The number of valence electrons: Is the same for elements in the same group. LEWIS DOT SYMBOL 1A 2A 3A 4A 5A 6A 7A 8A H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar For representative (main group) elements: group number = number of valence electrons ٤
OCTET RULE Most elements follow the octet rule. Octet Rule: Atoms tend to gain, lose, or share electrons until they are surrounded by eight valence electrons. An octet of electrons consists of full s and p subshells in an atom. In doing so, they achieve the configuration of a noble gas and achieve stability. Exceptions: Elements that do not follow the octet rule include hydrogen and helium. They follow the duet rule. IONIC BONDING An ionic bond involves the transfer of electrons between acation and ananion. Metals will lose electrons to achieve a stable outer valence level, and nonmetals will take those electrons to also achieve a stable outer valence. The loss of electrons is always an endothermic process. The gaining of electrons is generally an exothermic process. When ions come together, energy is released, so ionic compounds are stable. Na + Cl Na + + Cl ٥
ENERGETICS OF IONIC BONDS The formation of NaCl is extremely exothermic. Why? Na(s) +½ Cl 2 (g) NaCl(s) H f = 410.9 kj/mol Loss of e - is always endothermic takes energy to remove an electron from an orbital. Na(s) Na + (g)+ e H f =+495kJ/mol Whenane - isgained,theprocess isexothermic. 1/2Cl(g) + e Cl (g) H f =-349kJ/mol The overall process would be an endothermic process that requires. 495-349 = 146 kj/mol. These numbers don t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic. It could be explained by the Lattice Energy ٦
LATTICE ENERGY Lattice energy: Is the energy required to completely separate a mole of a solid ionic compound into its gaseous ions. The energy associated with electrostatic interactions is governed by Coulomb s law: E el = κ Q 1Q 2 d Lattice energy, increases: -with the charge on the ions. - with decreasing size of ions. ٧
The stronger the ionic bond the higher melting point the SrF 2 +2, -2 66, 133 66, 140 113, 133 1261 o C 2852 o C 1477 o C r 1 r 2 SAMPLE EXERCISE 8.1 Arrange the following ionic compounds in order of increasing lattice energy: NaF, CsI, and CaO. NaF consistsof Na + and F ions, CsI consistsof Cs + and I ions, CaO consistsof Ca 2+ and O 2 ions. CsI< NaF < CaO. ٨
PRACTICE EXERCISE Which substance would you expect to have the greatest lattice energy, MgF 2, CaF 2, or ZrO 2? Answer: ZrO 2 IONIC BONDS AND ELECTRONS Most atoms form ionic bonds in order to achieve noble gas configuration. e.g.li=[he]2s 1, e.g.f =[Ne]2s 2 3p 5, Na + =[He] F =[Ne] ٩
SAMPLE EXERCISE 8.2 Predict the ion generally formed by a. Sr b. S c. Al (a) Strontium is a metal in group 2A and will therefore form a cation. Sr = [Kr]2s 2 2e + Sr 2+ = [Kr] (b) Sulfur is a nonmetal in group 6A and will thus tend to befoundasananion. S= [Ne]3s 2 3p 4 + 2e S 2 = [Ne] (c) Aluminum is a metal in group 3A. We therefore expect itto form Al 3+ ions. Al = [Ne]3s 2 3p 3 3e + Al 3+ = [Ne] PRACTICE EXERCISE Predict the charges on the ions formed when magnesium reactswithnitrogen. Answer: Mg 2+ and N 3 ١٠
TRANSITION METAL IONS Ionization energies increase rapidly for each successive electron removed. The lattice energies of ionic compounds are generally large enough to compensate for the loss of up to 3 electrons. This means that charges above 3+ are uncommon. TRANSITION METAL IONS Since transition metals have d sublevel electrons: they do not generally form ions that have a noble gas configuration. When forming ions, transition metals lose the valence-shell s electrons first, then as many d electrons as are required to reach the charge of the ion. e.g.: Fe = [Ar]3d 6 4s 2 Fe 2+ = [Ar]3d 6 Fe 3+ = [Ar]3d 5 ١١
Characteristic of Ionic Compound: Ionic compounds tend to be brittle, have high melting points, are crystalline, and can be cleaved (break with a smooth surface). COVALENT BONDING A covalent bond: A chemical bond formed by sharing a pair of electrons.. There are several electrostatic interactions in covalent bond: Attractions between electrons and nuclei. Repulsions between. Repulsions between nuclei. The attractive forces of the nuclei and electrons must overcome the repulsion between electrons and nuclei ١٢
LEWIS STRUCTURES The formation of covalent bonds can be represented using Lewis symbols. The unpaired electrons are drawn as dots. The bonding electrons are drawn as a dash. Single Covalent Bonds Multiple Covalent Bonds ١٣
SAMPLE EXERCISE 8.3 Predict the formula of the stable binary compound formed when nitrogen reacts with fluorine, and draw its Lewis structure. Practice Exercise: Compare the Lewis symbol for neon with the Lewis structure for methane, CH 4. In what important ways are the electron arrangements about neon and carbon alike and different? Answer: Both atoms have an octet of electrons about them. However, the electrons about neon are unshared Ne electron pairs, whereas those about carbon are shared with four hydrogen atoms. SINGLE AND MULTIPLE BONDS A single bond: consists of 1 pair of electrons. A double bond: consists of 2 pairs of electrons. A triple bond: consists of 3 pairs of electrons. ١٤
BOND LENGTH Bond Length: The distance between the nuclei oftheatomsinvolveinabond. As a general rule, the distance between bonded atoms decreases as the number of shared electrons increases. BOND POLARITY AND ELECTRONEGATIVITY The concept of bond polarity helps describe the sharing of electrons between atoms. When two identical atoms bond, the electron pairs mustbeshared equally.(e.g.cl 2 andh 2 ) A nonpolar covalent bond: is one in which the electrons are shared equally (diatomic molecules of same electronegativity). ١٥
POLAR BOND In a polar covalent bond: one of the atoms exerts a greater attraction for the bonding electrons than the other. If the difference in relative ability to attract electrons is large enough, an ionic bond is formed. The fluorine end of the molecule has more electron density than the hydrogen end: Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does. ELECTRONEGATIVITY Electronegativity: is the ability of an atom in a molecule to attract electrons(of bond) to itself. ١٦
ELECTRONEGATIVITY TRENDS IN P.T. On the periodic chart, electronegativity increases as you go:- Fromlefttorightacrossarow. Fromthebottomtothetopofacolumn. BOND POLARITY There is no sharp dividing line between ionic and covalent bonding: The greater the difference in electronegativity between two atoms, the more polar their bond. Bond Electronegativity Difference nonpolar covalent < 0.5 polar covalent 0.5 2.0 ionic >2 δ+ δ- ١٧
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The greater the difference in electronegativity, the more polar is the bond. Bond length increases according to size EXERCISE! Arrange the following bonds from most to least polar: a) N F O F C F C F > N F > O F b) C F N O Si F Si F > C F > N O c) Cl Cl B Cl S Cl B Cl > S Cl > Cl Cl Copyright Cengage Learning. All rights reserved 38 ١٩
SAMPLE EXERCISE 8.4 In each case, which bond is more polar? Indicate in each case which atom has the partial negative charge. a) B Cl or C Cl Referring to the periodic table: The chlorine atom is common to both bonds which is of highest electronegativity. B < Cin electronegativity So,theB ClbondismorepolarthanB Cl Check: B-Cl: 3.0 2.0 = 1.0 and C-Cl: 3.0 2.5 = 0.5. Hence the B Clbond is more polar (b) P Cl or P Cl The phosphorus is common to both bonds. F > Cl in electronegativity. So,P FismorepolarthanP Cl Check: P-Cl: 3.0 2.1 = 0.9 and P-F: 4.0 2.1 = 1.9. Hence the bond is more polar PRACTICE EXERCISE Which of the following bonds is most polar: S-Cl, S-Br, Se-Cl, or Se-Br? Answer: Se Cl ٢٠
DIPOLE MOMENT Whenever a distance separates two electrical charges of equal magnitude but opposite sign, a dipole is established. Dipole moment (µ): The quantitative measure of the magnitude of a dipole. µ= Q r µ = dipole moment (debyes (D) = 3.34 x 10-30 coulomb. meters) Q = charge (unitsof1.60x10-19 C) r = distance(m) The dipole moment increases: as the magnitude of charge that is separated increases. as the distance between the charges increases. ٢١
SAMPLE EXERCISE 8.5 Thebondlengthinthe HClmoleculeis1.27Å. a. Calculate the dipole moment, in debyes, that would result if the charges on the H and Cl atoms were 1+ and 1-, respectively. Solution: Q = 1 which is e = 1.60 10 19 C. The separation r =1.27 Å. b.the experimentallymeasureddipole moment of HCl (g) is 1.08D.Whatmagnitude of charge, in units of e, on the H andcl atomswouldleadtothisdipolemoment? µ= 1.08 D, r = 1.27 Å. So we calculate Q then convert to units of e: Polar Covalent ٢٢
PRACTICE EXERCISE The dipole moment of chlorine monofluoride, ClF (g), is 0.88D.Thebondlengthofthemoleculeis1.63Å. a. Which atom is expected to have the partial negative charge? Answer: F b.whatisthe chargeon thatatom,inunitsofe? Answer: 0.11 Moleculear compounds: MOLECULAR COMPOUNDS have low melting and boiling points. exhibit behavior. nonelectrolyte Ionic compounds: are brittle solids. have a crystal lattice structure, have high melting points exhibit strong electrolyte behavior. ٢٣
A simple approach : PREDICTING IONIC OR COVALENT? Thebondbetweenametalandanonmetalisionic. The bond between two nonmetals is covalent. Many exceptions. Example: SnCl 4 =metalandnonmetal but = covalent -Exists as a colorless liquid at room temperature. - Itfreezes at-33 o Candboilsat114 o C. - Clearly this substance does not have the characteristics of an ionic substance. Electroneg. Diff. of Sn Cl bond = 1.2(Polar Covalent) TRANSITION METALS There is no difference in the electronegativity values for the transition metals that have multiple charges. The higher the charge (+4 or higher) on a transition metal, the more covalent the bond. Examples: in case of Mn compounds. MnO: Ox. No = +2, Diff in Electroneg. = 2 So: normally ionic; Melting point = 1842 o C. Mn 2 O 7 Ox. No. = +7 green liquid m.p= 5.9 o C The following are polyatomic ions MnO 4 - Ox. No = +7 CrO 4 2- Ox. No = +6 ٢٤
DRAWING LEWIS STRUCTURES Rules for drawing Lewis dot structures: Example:PCl 3 1) Find the sum of valence electrons of all atoms in the polyatomic ion or molecule. If it is an anion, add one electron for each negative charge. If it is a cation, subtract one electron for each positive charge. TV e s = 5 + 3(7) = 26 2. Identify the central atom the least electronegative element that isn t hydrogen. P In many molecules or polyatomic ions central atoms writtenfirst CCl 4, CO 3-, SO 4 2-3- Connect the outer atoms to the central atom by singlebonds(eachbond ( )=2e s). 4- Obtain the difference: between Tve s and Bonding e s. Diff. = 26 6 = 20 ٢٥
3. Complete octets around all atoms bonded to the central atom. (Exception: Hydrogen only needs 1 bond so no extra electrons should ever be added.) 4. Place any leftover electrons on the central atom. Excess = 20-18 = 2 5.Iftherearenotenoughelectronstogivethecentral atom an octet, try multiple bonds. WRITING LEWIS STRUCTURES: HCN TVe s= 1 + 4 + 5=8 B e s= 2 x 2 = 4 Diff.= 8-4 = 4 H C N But the central atom not completed by 8 e s so form multiple bond ٢٦
PRACTICE EXERCISE How many valence electrons should appear in the Lewis structure for CH 2 Cl 2? Answer: 20 Draw the Lewis structure. EXAMPLE: Draw the Lewis structure for CCl 4 TV e s=4 + (7 x4) = 32 Cl Cl C Cl B e s= 4 x 2 = 8 Cl Diff. = 32 8 = 24 X X X X Cl X X X X X X Cl C X X X X Cl X X X X X X Cl X X X X ٢٧
PRACTICE EXERCISE Draw the Lewis structure for a. NO + ion b. C 2 H 4 SAMPLE EXERCISE 8.8 Draw the Lewis structure for the BrO 3 ion. For oxyanions: BrO 3, SO 4 2, NO 3, CO 3 2, the oxygen atoms surround the central nonmetal atoms. For BrO 3 TV e s= 7 + (6 x 3) +1 = 26 B e s= 26 6 = 20 O Br O O Excess e s= 2 located on the Central atom ٢٨
PRACTICE EXERCISE Draw the Lewis structure for: a. ClO 2- ion b. PO 4 3- ion FORMAL CHARGE The formal charge of an atom in a molecule: is the charge the atom would have if all the atoms in the molecule had the same electronegativity. زع ات وي ا رات (**Remember that formal charges are not real charges!) Formal charge of any atom = valence electrons (Group Number) number of bonds around the atom number of unshared e s electrons on the atom 6 2 4 = 0 ٢٩
FORMAL CHARGE When more than one Lewis structure can be drawn for a molecule, Formal charges help us choose which Lewis Structure is most likely, from a group of Lewis Structures that all fulfill the octet rule. 1. We generally choose the Lewis structure in which the atoms bear formal charges closest to zero. 2. Any negative charges reside on the more electronegative atoms. Best Structure Bonds 2 4 2 1 4 3 unshared e s 4 0 4 6 0 2 Formal charge = 0 0 0-1 0 +1 SAMPLE EXERCISE 8.9 a) The following are three possible Lewis structures for the thiocyanate ion, NCS : a. Determine the formal charges of the atoms in each structure. b. Which is the preferred structure? N is more electronegative than C or S. Therefore, we expect that any negative formal charge will reside on the N atom. The middle structure is the preferred Lewis structure ofthencs ion. ٣٠
PRACTICE EXERCISE The cyanate ion (NCO ) has three possible Lewis structures. Draw all three Lewis structures, label the formal charges on the atoms, and indicate the preferred structure. Structure (iii), which places a negative charge on oxygen, the most electronegative of the three elements, is the preferred Lewis structure. RESONANCE STRUCTURES Resonance structures in which the position of the atoms are the same but the placement of the electrons is different. Neither structure is correct on its own and the structure does not oscillate between the two. The true structure is a hybrid of the two structures because experimental evidences shows that the O-O bonds in O 3 are of equal length. Resonance Structures Resonance Hybrid ٣١
BOND ORDER Divide the number of bonds between individual atoms by the total number of bonds. Bond Order = 4/3 = 1.33 The bond length is between the double and single bond OZONE Observed structure of ozone, in which both O-O bonds are the same length. both outer oxygens have a charge of -1/2. ٣٢
In writing resonance structures, the only difference is in the arrangement of the electrons. The resonance structures are separated by a doubleheaded arrow.() SAMPLE EXERCISE 8.10 Which is predicted to have the shorter sulfur-oxygen bonds,so 3 orso 3 2-? ForSO 3 ForSO 3 2- The experimentally measured S O bond lengths are 1.42 Å ( BO = 1) in SO 3 and 1.51 Å ( B O = 1.33) in SO 3 2. ٣٣
PRACTICE EXERCISE Draw two equivalent resonance structures for the formateion,hco 2-. RESONANCE OF BENZENE Benzene is an aromatic (ring) organic molecule with the formula C 6 H 6. The resonance of benzene is represented in many ways. ٣٤
EXCEPTIONS TO THE OCTET RULE There are 3 main categories of exceptions to the octet rule: 1. Odd number of electrons 2. Less than an octet 3. More than an octet ODD NUMBER OF ELECTRONS When an odd number of electrons occurs, the central atom usually gets the unpaired electron. If there is no central atom, formal charges can be used to pick the correct structure. ٣٥
LESS THAN AN OCTET Molecules with less than an octet normally occur with boron and beryllium ( The other bonded atom is X (F, Cl, Br, or I). MORE THAN AN OCTET Molecules with more than an octet can only occur for atoms in the third period and beyond because of the d orbitals. Common elements that have more than an octet are S,P,As, some halogens,andsome noblegases. ٣٦
The better structure puts a double bond between the phosphorus and one of the oxygens (Formal charge = zero) SAMPLE EXERCISE 8.11 Draw the Lewis structure for ICl 4-. Total V e s = 7 + (7 x 4) + 1 = 36 B e s= 4 x 2 = 8 Diff. = 36 8 = 28 ٣٧
PRACTICE EXERCISE Which of the following atoms is never found with more than an octet of valence electrons around it: S, C, P, or Br? Draw the Lewis structure for XeF 2. STRENGTHS OF COVALENT BONDS Thestabilityofamoleculeisrelatedtothestrengthsof the covalent bonds it contains. The strength of a covalent bond between two atoms: is determined by the energy required to break that bond. The bond enthalpy: is the enthalpy change, H, for the breaking of a particular bond in one mole of a gaseous substance. ٣٨
BOND ENTHALPY We use the designation D(bond type) to represent the bond enthalpies. The bond enthalpy for a Cl-Cl bond, D(Cl-Cl), is measured to be 242 kj/mol. When molecules contain multiple identical bonds, then we use the average bond enthalpy. BOND ENTHALPY The bond enthalpy is always a positive quantity; energy is always required to break chemical bonds. The greater the bond enthalpy is, the stronger the bond. A molecule with strong chemical bonds generally has less tendency to undergo chemical change than does one with weak bonds. ٣٩
BOND ENTHALPY AND REACTIONS When calculating the H of a reaction using bond enthalpies you use the following equation: H rxn = Σ(bonds broken) Σ(bonds formed) Reactants Products CH 4 (g) +Cl 2 (g) CH 3 Cl (g)+hcl (g) In this example, one C-H bond and one Cl-Cl bondarebroken;onec-clandoneh-clbondare formed. So, H = [D(C-H) + D(Cl-Cl)] - [D(C-Cl) + D(H-Cl)] =[(413 kj) +(242kJ)] - [(328kJ) +(431 kj)] =(655 kj) -(759 kj) =-104 kj ٤٠
SAMPLE EXERCISE 8.12 Using Table 8.4, estimate H for the following reaction: PRACTICE EXERCISE Using Table 8.4, estimate H for the following reaction: Answer: 86 kj ٤١
BOND ENTHALPY AND BOND LENGTH We can also measure an average bond length for different bond types. As the number of bonds between two atoms increases, the bond length decreases. SAMPLE INTEGRATIVE EXERCISE Phosgene has the following elemental composition: 12.41% C, 16.17% O, and 71.69% Cl by mass. Its molar mass is 98.9 g/mol. a. Determine the molecular formula of this compound. The empirical formula, COCl 2. ٤٢
b. Draw three Lewis structures for the molecule that satisfy the octet rule for each atom. c) Using formal charges, determine which Lewis structure is the most important one. The first structure is expected to be the most important one. Writing the chemical equation in terms of the Lewis structures of the molecules, we have: e. Using average bond enthalpies, estimate H for the formationofgaseousphosgene from CO (g) andcl 2(g). ٤٣
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