Presentation on theme: "1 Chapter 8: Ionic and Covalent Bonding RVCC Fall 2009 CHEM 103 – General Chemistry I Chemistry: The Molecular Science, 3 rd Ed. by Moore, Stanitski, and."— Presentation transcript:
1 Chapter 8: Ionic and Covalent Bonding RVCC Fall 2009 CHEM 103 – General Chemistry I Chemistry: The Molecular Science, 3 rd Ed. by Moore, Stanitski, and Jurs
2 Bonding – What holds atoms together? Octet rule: Octet rule: To form bonds, atoms gain, lose, or share e - to achieve a valence shell of 8 (or isoelectronic with a noble gas). Ionic bond – an electrostatic attraction between a cation and an anion that forms when electrons transfer from one atom to another. Covalent (Molecular) bond – the net attractive force that results from the sharing of electrons between atoms.
3 Ionic Bonds An ionic bond is formed by the transfer of electrons from one atom (metal with low EA) to another (nonmetal with high EA). The resultant ions are held together by electrostatic attraction. Na. : Cl :. : Na + Cl - [Ne]3s 1 [Ne]3s 2 3p 5 [Ne] [Ne]3s 2 3p 6 = [Ar] Each atom has satisfied the octet rule.
4 Ionic Bonds : : F. : : : F. : Mg.. F : : : : - 2+ F : : : : - MgF 2
5 Ionic Compounds - Properties crystalline high melting point high boiling point soluble in water electrolytes hard brittle Crystal Lattice
6 Covalent Bonding - G.N. Lewis (1916) share Some atoms share e - to form bonds. When two nonmetals bond, they often share electrons since they have similar attractions (EA) for them. This sharing of valence electrons is called the covalent bond. Attraction Stable bond Repulsion Number of bonds = Number shared e - pairs.
7 Covalent Bonds H.. : HHH + H2H2
9 Covalent Bonds - E potential “well”
10 single bond- one shared pair of e - H − H HH Lewis structures: show ALL valence electrons dot = 1 e - line = 1 pair of e- Single Covalent Bonds
11 # of e - shared Group # of to form an octetExample valence e - (8 - A group#) 4A44C in CH 4 5A53N in NF 3 6A62O in H 2 O 7A71F in HF H – F.. Single Covalent Bonds.. H H – C – H H H – O – H.. H – F.. F – N – F F.. # of e- shared to form an octet (8-A group#)
12 Lewis Structures Lewis electron-dot formulas or Lewis structures. An electron pair is either a bonding pair (shared between two atoms) or a lone pair (an electron pair that is not shared). bonding pair lone pairs :: HCl : : : H : : bonding pair lone pair :
13 Multiple Bonds In the molecules described so far, each of the bonds has been a single bond, that is, a covalent bond in which a single pair of electrons is shared. It is possible to share more than one pair. A double bond involves the sharing of two pairs between atoms. or C : C H H H H :: : : : C has octet. H OK with 2.
14 Triple bonds are covalent bonds in which three pairs of electrons are shared between atoms. Multiple Bonds CCor HH :: ::: Elements that form multiple bonds: C, O, N, S
15 The Procedure Using the molecular formula, count the total number of valence electrons available (bonding + lone pairs). Valence electrons for each atom corresponds to group # Adjust for charge (add electron for each minus, delete electron for each plus) 2(1) + 63(1) + 6 - 1
16 The Procedure Make a skeleton by connecting the atoms with single bonds only. When connecting atoms, remember… Put the least electronegative atom in the center. (Usually the first listed in the chemical formula.) Hydrogen is ALWAYS a terminal atom. More electronegative atoms are terminal (F, O…) Make the structure symmetric. 4 pairs 2 pairs left 1 pair left
17 The Procedure Put the left over electrons as lone pairs, preferably on the more electronegative atoms Is the octet rule satisfied? If YES, then you’re done…
18 The Procedure If you are electron-deficient (not enough electrons to complete an octet), then some atoms must share more than two electrons. “If you have a lone pair, make those two atoms share.” Ex. C 2 H 4
19 The Procedure If you have excess electrons, at least one atom must have an expanded valence Must be element from third period or lower Usually the central atom e.g. SF 4
20 The total number of valence electrons on an atom (from bonds & lone pairs) cannot exceed that atom’s maximum valence. First period: 2 electrons (s) Second period: 8 electrons (s,p) Third period & below: prefer to have 8, but can expand when necessary (s,p,d) General Rule
21 Writing Lewis Dot Formulas SCl 2 20 e - total or 10 pairs ClS 8 left : : ::: : : : 0 left
22 Cl C O Writing Lewis Dot Formulas COCl 2 24 e - total 12 pairs : : : : : : 0 left : : : Note that the carbon has only 6 electrons. 9 left
23 Cl C O Writing Lewis Dot Formulas COCl 2 12 pairs 9 e - left : : : : : : 0 e - left : : : To fulfill the octet rule… “If you have a lone pair, make those two atoms share!”
24 Writing Lewis Dot Formulas COCl 2 24 e - total Cl C 18 e - left : : : : : : 0 e - left O :: Note that the octet rule is now obeyed.
25 Writing Lewis Dot Formulas Practice N 2 SF 4 O 2 ClO 3 - HCNClO 2 -1 PO 4 -3 NO 3 -1
26 Question First evaluate the total valence electrons: 24 e - a.26 e -, wrong b.24 e-, looks OK c.24 e-, one F has too many d.24 e-, N not enough e.24 e-, but least electronegative has to be in the center f.24 e-, no bond between two N. no
27 Exceptions to the Octet Rule Although many molecules obey the octet rule, there are exceptions where the central atom has less or more than eight electrons. Incomplete octet – B, H BF 3 Boron has 3 valence electrons : F – B – F :.. :F::F:
28 Exceptions to the Octet Rule : F : : : F : : : : F : : P F : : : If a nonmetal is in the third period or greater it can accommodate as many as twelve electrons as the central atom. PF 5
29 Exceptions to the Octet Rule In sulfur tetrafluoride, SF 4, the sulfur atom must accommodate two extra lone pairs for a total of 5 electron pair (10 electrons) F : : : : F : : S F : : : : F : : :
30 Formal Charge and Lewis Structures In certain instances, more than one feasible Lewis structure can be illustrated for a molecule. For example, H CNCNH or :: The concept of “formal charge” can help us decide which structure is correct.
31 Formal Charge and Lewis Structures formal charge = valence e- before bonding– valence e- after bonding = valence e - - [1/2 bonding e - + lone pair e - ] H CNCNH or :: H: 1-½(2) = 0 C: 4 - ½(8) = 0 N: 5 – (½(8) + 2) = 0 H: 1-½(2) = 0 C: 4 – (½(6)+2) = -1 N: 5 – (½(8)) = +1
32 Formal Charge and Lewis Structures Smaller formal charges are more favorable More electronegative (or higher EA) atom should have negative formal charges Like charges should not be on adjacent atoms Net formal charge should be the overall charge on the molecule/ion. or H CN : 000 CNH : formal charges 0+1
33 Practice Determine the most stable structure for dinitrogen oxide. (All structures have 16 valence electrons.) N=N=ON-N≡ON ≡ N - O -1 +1 0 -2 +1 +1 0 +1 -1 formal charge= valence e - - [1/2 bonding e- + lone pair e-]
34 Practice - Formal Charge Which structure is correct? or O NCl 000 NO : 0+1
35 Delocalized Bonding: Resonance The structure of ozone, O 3, can be represented by two different Lewis electron-dot formulas. OO O : : : : : : OO O : : : : : : The bond lengths for the above structures are: O – O 132 pm O = O 112 pm However, experiments show that both bonds are identical.
36 Delocalized Bonding: Resonance According to theory, one pair of bonding electrons is spread (delocalized) over the region of all three atoms. In fact, the actual bond length is 127.8 pm (in between 132 and 112pm). The actual molecule is a hybrid or composite structure and not different structures that change back and forth… although, we often represent it that way. OO O
37 Delocalized Bonding: Resonance Lewis resonance structures, have the same atoms in the same positions. Only an electron pair position is different. O H O N O H O N O H O N O
38 Resonance Structures O=S O O S=O S=S O S O=S S=S O S O=S Which pair does NOT represent resonance structures?
39 Resonance Structures Draw resonance structure(s) for the following: O O C O -2 O O C O -2 O O C O -2
40 “All covalent bonds are created equal but some are more equal than others.” (We assumed equal sharing when we calculated formal charge.)
41 Electronegativity… …is a measure of the ability of an atom in a molecule to draw bonding electrons to itself when bonded. decreases down a group increases across a period Periodic Trend - Electronegativity
42 Electronegativity Notice, there are NO values for EN for the noble gases.
43 Types of Bonds Ionic: ΔEN >1.8 electron transfer Covalent: ΔEN <1.8 electron sharing Metallic: electron-sea model or band theory 0.9 3.0 Na + Cl - 2.5 2.1 C - H 1.6 Zn
47 Bond Polarity In HCl we have a partial negative charge on the chlorine (denoted -) and a partial positive charge on the hydrogen (denoted +) The bond is polar covalent. H : Cl : : :
48 Bond Polarity Arrange the following bonds from the most to the least polar: HH, HCl, HF, HI, HBr Compare the electronegativity of Cl, F, I and Br: Least Most Electronegative I Br Cl F Determine polarity: HH HI HBr HClHF non polarmost polar
49 Practice Which of the following bonds in each pair are more polar? C-S or C-O Cl-Cl or O=O N-H or C-H
50 Bond Length Bond length (or bond distance) is the distance between the nuclei of two bonded atoms (the sum of atomic radii).
51 The size of atoms that form the bond determine the length. C - N 147 pm C - C 154 pm C - P 187 pm(P, period 3) Bond Length
52 Bond Length – Multiple Bonds As the electron density between atoms increases the bond lengths decrease; the atoms are pulled together more strongly. decrease
53 Bond Enthalpy Bond Enthalpy – the enthalpy change that occurs when the bond between two bonded atoms in the gas phase is broken and the atoms are separated completely at constant pressure. As the electron density between two atoms increases, the bond gets shorter and stronger. Bond length Bond enthalpy C C C C CC 154 pm 134 pm 120 pm835 kJ/mol 602 kJ/mol 346 kJ/mol NaCl lattice energy -786 kJ/mol MgO lattice energy -3791 kJ/mol
54 Bond Enthalpy Hº = ∑[(moles of bonds) × D(bonds broken)] - ∑[(moles of bonds) × D(bonds formed)] Hº - standard enthalpy of reaction Bond Enthalpies can be used to calculate the standard enthalpies of reaction (gas phase, STP)
55 Bond Enthalpy Estimate the Hº for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + 2 :O = O: : O = C = O: + 2H – O - H : : ::: : 4 C – H bonds 1 O = O bond2 C = O bond 2 H – O bond per molecule per molecule
56 Hº = ∑[(moles of bonds) × D(bonds broken)] - ∑[(moles of bonds) × D(bonds formed)] + 2 :O = O: : O = C = O: + 2H – O - H : : ::: : 4 C – H bonds 1 O = O bond2 C = O bonds 2 H – O bonds per molecule per molecule = [4 × D (C-H) + 2 × D (O=O) ] – [2 × D (C=O) + 4 × D (H-O) ] = [4 × 416 + 2 × 498] – [2 × 803 + 4 × 467] = -814 kJ 4 C – H bonds 2 O = O bonds 2 C = O bonds 4 H – O bonds