Chemical Bonds Ionic Bond Formation of Ions Electron Configurations of Ions Ionic Size and Charge density, Relative Strength of Ionic Bonds Lattice Energy Steps in the Formation of an Ionic Compound The Born-Haber Cycle

Chemical Bonds Covalent Bonds Electronegativity Polarity of Covalent Bonds Lewis Structures and the Octet Rule Exceptions to the Octet Rule Resonance Lewis Structures Bond Energies Calculating Enthalpy using Bond Energy Molecular Shape - The VSEPR Model

Review of Atomic Properties Effective nuclear charge & Atomic Size: 1.effective nuclear charge increases left to right and decreases down a group; 2.electronic shell gets smaller left-to-right across period and gets bigger down a group; 3.Atomic size decreases left to right across period and increases top to bottom down a group:

Review of Atomic Properties Atomic Size and Ionization Energy: 1.L-to-R: atomic size decreases; ionization energy increases; 2.Top-to-bottom: atomic size increases; ionization energy decreases; 3.Ionization energy increases across period (L-to- R), but decreases down a group;

Review of Atomic Properties Electron affinity increases left to right and decreases top to bottom: smaller atoms have stronger attraction of added electron than larger atoms Nonmetals have higher tendency to gain electrons than metals and become anion

Review of Atomic Properties Atomic Size and Electron Affinity: 1.L-to-R: atomic size decreases, electron affinity increases; 2.Top-to-bottom: atomic size increases, electron affinity decreases; 3.Electron affinity increases across period (L-to-R), but decreases down a group;

Ionic bonds Attractions between cations and anions; Bonds formed between metals and nonmetals

Formation of Cations Ions formed when metals react with nonmetals - metal atoms lose valence electrons to nonmetals; Atoms of representative metals lose valence electrons to acquire the noble gas electron configuration; Cations of representative group have noble gas electron configurations;

Formation of Cations From the alkali metals (1A): M  M + + e - From the alkaline Earth metals (2A): M  M e - From Group 3A metals: M  M e - ;

Formation of Ions The nonmetal atoms gain electrons to the noble gas electron configuration; Anions have noble gas electron configuration;

Formation of Anions From the halogen family (VIIA): X + e -  X - From the oxygen family (VIA): X + 2e -  X 2- From N and P (in Group VA): X + 3e -  X 3-

Common Ions of the Representative Elements Ions isoelectronic to He (1s 2 ): Li + & H - Ions isoelectronic to Ne (1s 2 2s 2 2p 6 ): Na +, Mg 2+, Al 3+, F -, O 2 -, and N 3 - Ions isoelectronic to Ar (1s 2 2s 2 2p 6 3s 2 3p 6 ): K +, Ca 2+, Sc 3+, Cl -, S 2 -, and P 3 -

Common Ions of the Representative Elements Ions isoelectronic to Kr (1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 ): Rb +, Sr 2+, Y 3+, Br -, and Se 2 - ; Ions isoelectronic to Xe (1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 ) Cs +, Ba 2+, La 3+, I -, and Te 2 - ;

Ionic Radii Relative size of isoelectronic ions: Al 3+ < Mg 2+ < Na + < Ne < F - < O 2- < N 3- ; Sc 3+ < Ca 2+ < K + < Ar < Cl - < S 2- < P 3- ; Trend of ionic radii within a group: Li + < Na + < K + < Rb + < Cs + ; F - < Cl - < Br - < I - ;

Cations From Transition Metals Transition metal atoms lose variable number of electrons; Cations have variable charges; Cations do not acquire noble gas electron configurations

Electron Configurations of Transition Metal Cations Examples: Cr: [Ar] 4s 1 3d 5 Cr  Cr e - ;Cr 2+ : [Ar] 3d 4 Cr  Cr e - ;Cr 3+ : [Ar] 3d 3 Fe: [Ar] 4s 2 3d 6 Fe  Fe e - ;Fe 2+ : [Ar] 3d 6 Fe  Fe e - ;Fe 3+ : [Ar] 3d 5

Charge Density and Strength Ionic Bond Charge density = charge/size of ion Greater charge but small ionic radius  High charge density  stronger ionic bond; Stronger ionic bond  High lattice energy; Stronger ionic bond  High melting point;

Lattice Energy (U L ) Lattice energy - energy released when gaseous ions combine to form solid ionic compound: M + (g) + X - (g)  MX (s) ; U L = Lattice energy Examples: Na + (g) + Cl - (g)  NaCl (s) ; U L = -787 kJ/mol Li + (g) + F - (g)  LiF (s) ; U L = kJ/mol

Lattice energy Lattice energy  k(q 1 q 2 /r 2 ) q 1 and q 2 = charge magnitude on ions; r = distance between nuclei, and k = proportionality constant. Lattice energy increases with charge magnitude but decreases with ionic size

Lattice Energies of Some Ionic Compounds Lattice Energy, U L (kJ/mol) The energy required to separate a mole of ionic solids into the gaseous/vapor ions; MX (s)  M + (g) + X - (g) M n+ /X n- F - Cl - Br - I - O 2 - Li Na K Mg Ca ______________________________________________________________________

The Born-Haber Cycle for NaCl Na + (g) + Cl (g) _______________ -349 kJ +496 kJ _______ Na + (g) + Cl - (g) Na (g) + Cl (g) ___________ +121 kJ Na (g) + ½Cl 2 (g) ________ ? kJ +108 kJ Na (s) + ½Cl 2 (g) ________ -411 kJ NaCl (s) _________________

Chemical Processes in the Formation of NaCl Na (s)  Na (g) ;  H s = +108 kJ ½Cl 2 (g)  Cl (g) ; ½BE = +121 kJ Na (g)  Na + (g) + e - ;IE = +496 kJ Cl (g) + e -  Cl - (g) ;EA = -349 kJ Na + ( g) + Cl - (g)  NaCl (s) ; U L = ? kJ Na (s) + ½Cl 2 (g)  NaCl (s) ;  H f = -411 kJ U L =  H f – (  H s + ½BE + IE + EA)  H s = Enthalpy of sublimation; IE = Ionization energy; BE = Bond energy; EA = Electtron affinity; U L = Lattice energy;  H f = Enthalpy of formation)

The Born-Haber Cycle for LiF Li + (g) + F (g) _______________ -328 kJ +520 kJ _______Li + (g) + F - (g) Li (g) + F (g) ___________ +77 kJ Li (g) + ½F 2 (g) ________ ? kJ +161 kJ Li (s) + ½F 2 (g) ________ -617 kJ LiF (s) _________________

Chemical Processes in the Formation of LiF Li (s)  Li (g) ;  H s = +161 kJ ½F 2 (g)  F (g) ; ½BE = +77 kJ Li (g)  Li + (g) + e - ;IE = +520 kJ F (g) + e -  F - (g) ;EA = -328 kJ Li + ( g) + F - (g)  LiF (s) ; U L = ? Li (s) + ½F 2 (g)  LiF (s) ;  H f = -617 kJ U L =  H f – (  H s + ½BE + IE + EA)  H s = Enthalpy of sublimation; IE = Ionization energy; BE = Bond energy; EA = Electtron affinity; U L = Lattice energy;  H f = Enthalpy of formation)

The Born-Haber Cycle for MgO Mg 2+ (g) + O 2- (g) _____________ +737 kJ Mg 2+ (g) + O (g) ________ kJ Mg (g) + O (g) _________ +247 kJ Mg (g) + ½O 2 (g) ________ ? kJ +150 kJ Mg (s) + ½O 2 (g) ________ -602 kJ MgO (s) _________________

Chemical Processes in the Formation of MgO Mg (s)  Mg (g) ;  H s = +150 kJ ½O 2 (g)  O (g) ; ½BE = +247 kJ Mg (g)  Mg 2+ (g) + 2 e - ;IE = kJ O (g) + 2 e -  O 2- (g) ;EA = +737 kJ Mg 2+ (g) + O 2- (g)  MgO (s) ; U L = ? kJ Mg (s) + ½O 2 (g)  MgO (s) ;  H f = -602 kJ U L =  H f – (  H s + ½BE + IE + EA)  H s = Enthalpy of sublimation; IE = Ionization energy; BE = Bond energy; EA = Electron affinity; U L = Lattice energy;  H f = Enthalpy of formation)

Covalent Bonds Bonds between two nonmetals or between a semimetal and a nonmetal atoms Bonds formed by sharing electron pairs; One, two or three pairs of electrons shared between two atoms; A pair of atoms may form single, double, or triple covalent bonds;

Potential energy of H-atoms during the formation of H 2 molecule

Polarity of Covalent Bonds 1.Covalent bonds - polar or nonpolar ; 2.Nonpolar covalent bonds - bonds between identical atoms or atoms having the same electronegativity. 3.Polar covalent bonds - bonds between atoms with different electronegativity;

Polar Covalent Bonds 1.Bonds have partial ionic character 2.Bond polarity depends on  EN;  EN = difference in electronegativity of bonded atoms

Electronegativity Electronegativity = relative ability of bonded atom to pull shared electrons. Electronegativity Trend: increases left-to-right across periods; decreases down the group.

Electronegativity Most electronegative element is at top right corner of Periodic Table Fluorine is most electronegative with EN = 4.0 Least electronegative element is at bottom left corner of Periodic Table Francium is least electronegative with EN = 0.7

General trends: Electronegativity increases from left to right across a period For the representative elements (s and p block) the electronegativity decreases as one goes down a group Electronegativity trend for transition metals is less predictable.

Electronegativity and Bond Polarity CompoundF2F2 HFLiF Electronegativity Difference = = = 3.0 Type of Bond Nonpolar covalent Polar covalent Ionic (non- covalent) In F 2 electrons are shared equally and bond is nonpolar In HF the fluorine is more electronegativity than hydrogen - electrons are drawn closer to fluorine. H ― F bond is very polar

Electronegativity and bond polarity The H-F bond can thus be represented as: The '  + ' and '  - ' symbols indicate partial positive and negative charges. The arrow indicates the "pull" of electrons off the hydrogen and towards the more electronegative atom. In lithium fluoride the much greater relative electronegativity of the fluorine atom completely strips the electron from the lithium and the result is an ionic bond (no sharing of the electron)

Predicting Bond Type From Electronegativity General rule of thumb for bonds   EN = 0-0.5, bond is non-polar covalent;  EN > 0.5, but < 2.0, bond is polar covalent  EN > 2.0, bond is considered ionic.

Potential Energy Diagram for Covalent Bond Formation

Bond Length... Bond length - distance between the nuclei of bonded atoms. The larger the atoms that are bonded, the greater the bond length. Bond length: single bonds > double bonds > triple bonds

Bond Energy Bond energy - the energy required to break the bonds between two atoms. The shorter the bond, the greater the bond energy. Bond energy: Triple bonds > double bonds > single bond

Bond Length and Bond Energies Bond length (pm) and bond energy (kJ/mol) Bond Length Energy _________________________________________________________________________________________________________ H─H H─C C─C H─N N─N H─O O─O H─F F─F H─Cl Cl─Cl H─Br Br─Br H─I I─I C─F C─S C─Cl C─C C─Br C─N C─I C─O C─C O─O C=C O=O C ≡ C C=O N=N ? 418 C=N N ≡ N

Bond Breaking and Bond Formation in the Reaction to form H 2 O

Using Bond Energy to Calculate Enthalpy Chemical reactions in the gaseous state only involve: the breaking of covalent bonds in reactants and the formation of covalent bonds in products. Bond breaking requires energy Bond formation releases energy  H reaction =  (Energy of bond breaking) +  (Energy of bond formation)

Calculating Enthalpy Reaction Using Bond Energy Example: use bond energy to calculate  H for the following reaction in gaseous state: CH 3 OH + 2 O 2  CO 2 + 2H 2 O;  H reaction =  {BE(in reactants)} -  {BE(in products)}

Using bond energy to calculate enthalpy  BE(in reactants) = 3 x BE(C─H) + BE(C─O) + BE(O─H) + 2 x BE(O═O) = (3 x 413) (2 x 495) = 3054 kJ  BE(in products) = 2 x BE(C═O)* + 4 x BE(O─H) = (2 x 799) + (4 x 495) = 3578 kJ  H reaction =  {BE(in reactants)} -  {BE(in products)}  H reaction = 3054  3578 = -524 kJ

Lewis Structures for Molecules or Polyatomic ions Step-1: Calculate number of valence electrons; For polyatomic ions, add one additional electron for each negative charge, or subtract one for each positive charge on the ion.

Lewis Structures for Molecules and Polyatomic ions Step-2: Choose a central atom (the least electronegative atom) (Hydrogen and Fluorine cannot become central atoms) Connect other atoms to the central atom with single bonds (a pair of electrons).

Lewis Structures for Molecules and Polyatomic ions Step-3: Complete the octet state of all terminal atoms, except hydrogen. Place remaining pairs of electrons (if present) on central atom as lone pairs.

Octet State of Central Atom Step-4: If central atom has not acquired octet state but no more electrons available, move lone-pair electrons from terminal atoms, one pair at a time, to form double or triple bonds to complete octet of the central atom.

Lewis Symbols and Formation of Covalent Molecules

Lewis Structures of CH 4, NH 3 and H 2 O

Lewis Structures of CO 2, HCN, and C 2 H 2

Resonance Lewis Dot Structures for CO 3 2-

Exception to Octet Rule 1.If central atom is from group 2A or 3A, octet state is not acquired - the central atom has incomplete octet. 2.Central atoms from periods 3, 4, 5, …may have more than 8 valence electrons (expanded octet) 3.Molecules with odd number of electrons will contain unpaired electrons.

Covalent Molecules with Central Atoms have Expanded Octet State

Evaluate Formal Charge Evaluate formal charges (fc) on each atom in the molecule to determine best correct or best Lewis structures. Formal charge is apparent charge on an atom in a Lewis formula; it is determined as follows: Formal charge = (Atom’s group #) – (# of lone-pair electrons on the atom) – (# of covalent bonds the atom forms)

Assigning Formal Charges

Choosing the correct or best Lewis structures based on formal charges If two or more Lewis dot structures that satisfy the octet rule can be drawn, the most stable one will be the structure in which: 1.The formal charges are as small as possible. 2.Any negative charges are located on the more electronegative atoms.

Which Lewis structures of CO 2 & N 2 O are correct?

The Shape of Water Molecules

Molecular Shapes of BeI 2, HCl, IF 2 -, ClF 3, and NO 3 -

Lewis Structures, Molecular Shapes & Polarity

The Shapes of Methane and Ammonia Molecules

The VSEPR Shapes

Linear and Trigonal Planar Electron-Pair Geometry

The Tetrahedral Electron-Pair Geometry

Trigonal Bipyramidal Electron-Pair Geometry

The Octahedral Electron-Pairs Geometry

Lewis Structures of HF, H 2 O, NH 3, & CH 4

Lewis Symbols for O, F, and Na

Lewis Model for the Formation of Covalent Bonds and Covalent Molecules

Covalent Bonds and Lewis Structures Some Molecules

Resonance Lewis Structures of PO 4 3-

Assigning Appropriate Formal Charges

Structures and Shapes of Formaldehyde and Ethylene