Chemical Bonding I: Basic Concepts Chapter 9 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

9.1 Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons that particpate in chemical bonding. 1A 1ns 1 2A 2ns 2 3A 3ns 2 np 1 4A 4ns 2 np 2 5A 5ns 2 np 3 6A 6ns 2 np 4 7A 7ns 2 np 5 Group# of valence e - e - configuration

9.1 Lewis Dot Symbols for the Representative Elements & Noble Gases Lewis Dot Symbol: Chemical symbol + valence electrons represented by dots. Electrons can be placed on any of the 4 sides of the symbol They can be placed singly or in pairs

9.2 Li + F Li + F - The Ionic Bond Li Li + + e - e - + FF - F - Li + + Li + F - Lewis symbols of cations of 1A, 2A, 3A do not have any dots as the valence electrons have been lost. The common monoatomic anions of all the nonmetals except H have 8 dots in their Lewis symbol because they all have 8 valence electrons. The Lewis symbol of these anions is enclosed within square brackets & the charge is written outside bracket as a superscript. The ionic bond is the attraction between the cation and anion of the ionic compound.

Ionic Radii = Ionic Size Top to bottom along a group ionic sizes increase. Examples: Li + < Na + < K + < Rb + < Cs + ; F - < Cl - < Br - < I -. Among cations of a particular transition metal, the larger the charge, smaller the size. Fe +3 < Fe +2 Among isoelectronic ions, larger the Z, smaller the size. Example: Al +3 < Mg +2 < Na + < Ne < F - < O -2 < N -3. Z= If an ion is viewed as a sphere, the ionic radius is a measure of the ionic size. The distance between 2 ions, r, is the distance between the 2 corresponding spheres. Larger the ionic sizes, larger the distance r.

9.3 Lattice energy (E) increases as Q increases and/or as r decreases. cmpd lattice energy MgF 2 MgO LiF LiCl Q= +2,-1 Q= +2,-2 r F - < r Cl - Electrostatic (Lattice) Energy E = k Q+Q-Q+Q- r Q + is the charge on the cation Q - is the charge on the anion r is the distance between the ions Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.

9.3

FF + 7e - FF 8e - F F F F Lewis structure of F 2 lone pairs single covalent bond 9.4 Nonmetals share electrons to satisfy their octet. Sharing of electron pairs between 2 atoms is called a covalent bond. Lewis Structures of molecular compounds are obtained by combining Lewis symbols of constituent atoms. Shared electron pairs are called bond pairs. Unshared electron pairs are called lone pairs. Sharing of 1 electron pair = single bond; Sharing of 2 electron pairs = double bond; 3 electron pairs = triple bond Shared pairs can be replaced by lines. Single bond: 1 line; double bonds: 2 lines; triple bond: 3 lines.

H H O ++ O HH O HHor Lewis structure of water Double bond – two atoms share two pairs of electrons single covalent bonds O C O or O C O double bonds Triple bond – two atoms share three pairs of electrons N N N N triple bond or 9.4

Lengths of Covalent Bonds Bond Lengths Triple bond < Double Bond < Single Bond 9.4 Larger the atoms, longer the bond lengths

9.4

H F F H Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms electron rich region electron poor region e - riche - poor ++ -- 9.5

The Electronegativities of Common Elements Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond. Most electronegative: F; Least electronegative: Cs, Fr

Nonpolar Covalent share e - Polar Covalent partial transfer of e - Ionic transfer e - Increasing difference in electronegativity Classification of bonds by difference in electronegativity DifferenceBond Type 0Nonpolar Covalent  2 Ionic 0 < and <2 Polar Covalent 9.5

Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H 2 S; and the NN bond in H 2 NNH 2. Cs – 0.7Cl – – 0.7 = 2.3Ionic H – 2.1S – – 2.1 = 0.4Polar Covalent N – – 3.0 = 0Nonpolar Covalent 9.5

1.Draw skeletal structure of compound showing what atoms are bonded to each other. The central atom is generally the first atom in the formula or the least electronegative atom in the formula. H, F are never central atoms. C, S are generally central atoms. 2.Count total number of valence e -. Add 1 for each negative charge. Subtract 1 for each positive charge. 3.Connect end atoms to the central atom using SINGLE BONDS. 4.Complete an octet for all end atoms except hydrogen 5.Put any remaining electrons on the central atom even if doing so gives it an excess of 8 electrons 6.If the central atom does not still have an octet form multiple bonds between one or more end atoms & the central atom till octet is achieved. This is done by making lone pairs on the END ATOMS bond pairs between the end atom & the central atom. When more than 1 Lewis Structure can be drawn which differ only in the distributions of the double bonds, triple bonds they are called resonance structures Writing Lewis Structures 9.6

Write the Lewis structure of nitrogen trifluoride (NF 3 ). Step 1 – N is less electronegative than F, put N in center FNF F Step 2 – Count valence electrons N - 5 (2s 2 2p 3 ) and F - 7 (2s 2 2p 5 ) 5 + (3 x 7) = 26 valence electrons Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms. Step 4 - Check, are # of e - in structure equal to number of valence e - ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons 9.6

Write the Lewis structure of the carbonate ion (CO 3 2- ). Step 1 – C is less electronegative than O, put C in center OCO O Step 2 – Count valence electrons C - 4 (2s 2 2p 2 ) and O - 6 (2s 2 2p 4 ) -2 charge – 2e (3 x 6) + 2 = 24 valence electrons Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms. Step 4 - Check, are # of e - in structure equal to number of valence e - ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons 9.6 Step 5 - Too many electrons, form double bond and re-check # of e - 2 single bonds (2x2) = 4 1 double bond = 4 8 lone pairs (8x2) = 16 Total = 24

9.7 An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion. Or formal charge on an atom in a Lewis structure = total number of valence electrons in the free atom - total number of ‘dots’ on atom in structure - total number of bonds (lines) attached to atom in structure

Formal Charge and Lewis Structures For neutral molecules, a Lewis structure in which there are zero formal charges is preferable to one in which nonzero formal charges are present. 2.Lewis structures with large formal charges are less preferable to those with small formal charges. 3.Among Lewis structures having similar distributions of formal charges, the most preferred structure is the one in which negative formal charges are placed on the more electronegative atoms. Examples: 1. The best resonance structure for CO 2 is O=C=O not O-C≡O or O≡C-O. 2. The best resonance structure for the thiocyanate, SCN- is N=C=S not N≡C-S or N-C≡S (the least preferred)

A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. OOO + - OOO + - OCO O -- OCO O - - OCO O What are the resonance structures of the carbonate (CO 3 2 -) ion?

Exceptions to the Octet Rule The Incomplete Octet HHBe Be – 2e - 2H – 2x1e - 4e - BeH 2 BF 3 B – 3e - 3F – 3x7e - 24e - FBF F 3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total = B has an octet only in structures where it is bonded to 4 other atoms: BH 4 -

Exceptions to the Octet Rule Odd-Electron Molecules N – 5e - O – 6e - 11e - NO N O The Expanded Octet (central atom with principal quantum number n > 2) If the central atom is in period 3, 4, 5, 6, it can accommodate an excess of 8 electrons due to the participation of d orbitals. SF 6 S – 6e - 6F – 42e - 48e - S F F F F F F 6 single bonds (6x2) = lone pairs (18x2) = 36 Total =

The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. H 2 (g) H (g) +  H 0 = kJ Cl 2 (g) Cl (g) +  H 0 = kJ HCl (g) H (g) +Cl (g)  H 0 = kJ O 2 (g) O (g) +  H 0 = kJ OO N 2 (g) N (g) +  H 0 = kJ N N Bond Energy Bond Energies Single bond < Double bond < Triple bond 9.10

Average bond energy in polyatomic molecules H 2 O (g) H (g) +OH (g)  H 0 = 502 kJ OH (g) H (g) +O (g)  H 0 = 427 kJ Average OH bond energy = = 464 kJ 9.10

Bond Energies (BE) and Enthalpy changes in reactions  H 0 = total energy input – total energy released =  BE(reactants) –  BE(products) =Sum of the bond enthalpies of all bonds in reactants – sum of the bond enthalpies of all bonds in products Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. 9.10

H 2 (g) + Cl 2 (g) 2HCl (g)2H 2 (g) + O 2 (g) 2H 2 O (g)

Use bond energies to calculate the enthalpy change for: H 2 (g) + F 2 (g) 2HF (g)  H 0 =  BE(reactants) –  BE(products) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) HH FF Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) HF  H 0 = – 2 x = kJ 9.10