Bonding – General Concepts

The ability of an atom in a molecule to attract shared electrons to itself. Electronegativity: The ability of an atom in a molecule to attract shared electrons to itself.

Chemical Bonds bonds form when the potential energy of the system will be lowered chemical bond – mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together

Ionic Bonds  Electrons are transferred  Electronegativity differences are generally greater than 1.7  The formation of ionic bonds is always exothermic!

Determination of Ionic Character Compounds are ionic if they conduct electricity in their molten state Electronegativity difference is not the final determination of ionic character

Coulomb’s Law “The energy of interaction between a pair of ions is proportional to the product of their charges, divided by the distance between their centers”

Table of Ion Sizes

Sodium Chloride Crystal Lattice Ionic compounds form solids at ordinary temperatures. Ionic compounds organize in a characteristic crystal lattice of alternating positive and negative ions.

Estimate  H f for Sodium Chloride Na(s) + ½ Cl 2 (g)  NaCl(s) Lattice Energy-786 kJ/mol Ionization Energy for Na495 kJ/mol Electron Affinity for Cl-349 kJ/mol Bond energy of Cl kJ/mol Enthalpy of sublimation for Na109 kJ/mol Na(s)  Na(g) kJ Na(g)  Na + (g) + e kJ ½ Cl 2 (g)  Cl(g) + ½(239 kJ) Cl(g) + e -  Cl - (g) kJ Na + (g) + Cl - (g)  NaCl(s) -786 kJ Na(s) + ½ Cl 2 (g)  NaCl(s) -412 kJ/mol

Bond Length Diagram

Covalent Bonding Forces  Electron – electron repulsive forces repulsive forces Proton – proton  Proton – proton repulsive forces repulsive forces Electron – proton  Electron – proton attractive forces attractive forces

Polar-Covalent bonds Nonpolar-Covalent bonds Covalent Bonds  Electrons are unequally shared  Electronegativity difference between.3 and 1.7  Electrons are equally shared  Electronegativity difference of 0 to 0.3

Polarity uneven distribution of charge Dipole Moment a molecule with a imbalance in electrons has a dipole moment polyatomic molecules have dipole moments if their polar bonds don’t cancel out

Practice Determine whether each of the following bonds will be: ionic, polar covalent, OR nonpolar covalent S and H – =0.4 –polar covalent S and Cs – =1.8 –ionic Cl and O – =0.5 –polar covalent Cl and Br – –nonpolar covalent

Practice For each of the following molecules, indicate which ones have a dipole moment. HClH: 2.1 < Cl: 3.0 dipole moment Cl 2 Cl: 3.0 = Cl: 3.0 no dipole moment ++++ ----

Practice CH 4 C: 2.5 > H: 2.1 no dipole moment H 2 S H: 2.1 < S: 2.5 dipole moment 4-4-4-4- ++++ ++++ ++++ ++++ 2-2-2-2- ++++ ++++

Ions: Electron Configurations and Sizes –nonmetals sharing to make covalent bonds taking electrons from metals to form anions –metals giving electrons to nonmetals to form cations Sizes of Ions trends are most important cation < parent atom –more protons attracting electrons in anion > parent atom –more electrons without additional protons

Sizes of Ions isoelectronic ions: ions with the same number of electrons Ex: O 2- F - Na + Mg 2+ Al 3+ electrons have greater attraction to nucleus as # of protons increases so the size decreases as nuclear charge (Z) increases

Example Give three ions that are isoelectronic with neon. Place them in order of increasing size. N 3-, O 2-, F -, Na +, Mg 2+, Al 3+ 7 > 8 > 9 > 11 > 12 > 13

Example for each of the following groups, place atoms in order of decreasing size –Cu, Cu +, Cu 2+ all have 29 p, lower #e- mean smaller ion Cu > Cu + > Cu 2+ –Ni 2+, Pd 2+, Pt 2+ different amounts of e- and p-energy levels Pt 2+ > Pd 2+ > Ni 2+ –O, O -, O 2- same #p, more electrons means larger ion O 2- > O - > O

Bond Length and Energy BondBond type Bond length (pm) Bond Energy (kJ/mol) C - CSingle C = CDouble C  CTriple C - OSingle C = ODouble C - NSingle C = NDouble C  NTriple Bonds between elements become shorter and stronger as multiplicity increases.

1. The compound most likely to be ionic is: A.KF B.CCl 4 C.CO 2 D.ICl E.CS 2

2. Ranking the ions S 2-, Ca 2+, K +, Cl - from smallest to largest gives the order as: A.S 2-, Cl -, K +, Ca 2+ B.Ca 2+, K +, Cl -, S 2- C.K +, Ca 2+, Cl -, S 2- D.Cl -, S 2-, K +, Ca 2+ E.Ca 2+, K +, Cl -, S 2-

3. The type of bonding within a water molecule is: A.ionic bonding B.polar covalent bonding C.nonpolar covalent bonding D.metallic bonding E.hydrogen bonding

4. Which of the following would represent a non-polar molecule containing polar bonds: A.I 2 B.CO 2 C.PF 3 D.SO 2 E.H 2 O

5. What type of bond would you expect in CsI?: A.Ionic B.Covalent C.Hydrogen D.Metallic E.van der walls

The compound most likely to be ionic is: A.KF B.CCl 4 C.CO 2 D.ICl E.CS 2 Answer: A Bonds formed from elements with greater differences in electronegativity are most likely ionic in nature. (remember trends!)

Ranking the ions S 2-, Ca 2+, K +, Cl - from smallest to largest gives the order as: A.S 2-, Cl -, K +, Ca 2+ B.Ca 2+, K +, Cl -, S 2- C.K +, Ca 2+, Cl -, S 2- D.Cl -, S 2-, K +, Ca 2+ E.Ca 2+, K +, Cl -, S 2- Answer A Note that all ions in this isoelectronic series have an Ar electron configuration; therefore the nuclear charge determines the size.

The type of bonding within a water molecule is: A.ionic bonding B.polar covalent bonding C.nonpolar covalent bonding D.metallic bonding E.hydrogen bonding Answer B Consider the nature of the bonding between hydrogen and oxygen in water.

Which of the following would represent a non-polar molecule containing polar bonds: A.I 2 B.CO 2 C.PF 3 D.SO 2 E.H 2 O Answer B You tell me why???

What type of bond would you expect in CsI?: A.Ionic B.Covalent C.Hydrogen D.Metallic E.van der walls Answer A You tell me why???

Bond Energy and Enthalpy D = Bond energy per mole of bonds Energy requiredEnergy released Breaking bonds always requires energy Breaking = endothermic (+  H) Forming bonds always releases energy Forming = exothermic (-  H)

Example Calculate the change in energy for the following reaction: H 2 (g) + F 2 (g)  2HF(g) ∆H = ∑ D (bonds broken) – ∑ D (bonds formed) ∆H = (D H-H + D F-F ) – 2D H-F ∆H = (432 kJ kJ) – 2 (565 kJ) ∆H = -544 kJ

Example Using bond energies, calculate ∆H for the reaction: CH 4 (g) + 2Cl 2 (g) + 2F 2 (g)  CF 2 Cl 2 (g) + 2HF(g) + 2HCl(g) ∆H = ∑ D (bonds broken) – ∑ D (bonds formed) ∑ D (bonds broken) = 4 (413 kJ) + 2 (239 kJ) + 2 (154 kJ) ∑ D (bonds formed) = 2 (485 kJ) + 2 (339 kJ) + 2 (565 kJ) + 2 (427 kJ) ∆H = 2438 kJ – 3632 kJ = kJ

The Octet Rule Combinations of elements tend to form so that each atom, by gaining, losing, or sharing electrons, has an octet of electrons in its highest occupied energy level. Diatomic Fluorine

Formation of Water by the Octet Rule

Comments About the Octet Rule  2nd row elements C, N, O, F observe the octet rule (HONC rule as well).  2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive.  3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals.  When writing Lewis structures, satisfy octets first, then place electrons around elements having available d orbitals.

Shows how valence electrons are arranged among atoms in a molecule. Reflects central idea that stability of a compound relates to noble gas electron configuration. Lewis Structures

C H H H Cl.. Completing a Lewis Structure - CH 3 Cl Add up available valence electrons: C = 4, H = (3)(1), Cl = 7 Total = 14 Join peripheral atoms to the central atom with electron pairs. Complete octets on atoms other than hydrogen with remaining electrons Make carbon the central atom..

Multiple Covalent Bonds: Double bonds Two pairs of shared electrons Ethene

Multiple Covalent Bonds: Triple bonds Three pairs of shared electrons Ethyne

Example CH 3 Cl methyl chloride C: 1 x 4e - = 4e - H: 3 x 1e - = 3e - Cl: 1 x 7e - = 7e - 14 e - Carbon is central atom H H C Cl H duet octet

Example CH 2 O Formaldehyde C: 1 x 4e - = 4 e - H: 2 x 1e - = 2 e - O: 1 x 6e - = 6 e - 12 e - Carbon is central

Practice Time!

Resonance Resonance is invoked when more than one valid Lewis structure can be written for a particular molecule. The actual structure is an average of the resonance structures. Benzene, C 6 H 6 The bond lengths in the ring are identical, and between those of single and double bonds.

Resonance Bond Length and Bond Energy Resonance bonds are shorter and stronger than single bonds. Resonance bonds are longer and weaker than double bonds.

Resonance in Ozone, O 3 Neither structure is correct. Oxygen bond lengths are identical, and intermediate to single and double bonds

Resonance in a carbonate ion: Resonance in an acetate ion: Resonance in Polyatomic Ions

How about the nitrate ion(NO 3 - ) ?: How about the nitrite ion (NO 2 - ) ?: Resonance in Polyatomic Ions Lewis Structure Nitrate ion Lewis Structure Nitrite ion

Resonance The lewis structure of which molecule requires resonance structures? A.MgCl 2 B.PCl 5 C.SiO 2 D.SO 2 E.OCl 2 Answer: D In this case there is a single bond between sulfur and one of the oxygens, and a double bond between sulfur and the other oxygen.

Localized Electron Model Lewis structures are an application of the “Localized Electron Model” L.E.M. says: Electron pairs can be thought of as “belonging” to pairs of atoms when bonding Resonance points out a weakness in the Localized Electron Model.

Formal Charge Helps you to compare various Lewis Structures and choose the best or most likely structure FC = (# valence e - ) – (#e - assigned to it) Assigned: –all of the lone pairs belong to that atom –plus half of the electrons involved in bonding Valence: –from periodic table

Formal Charge Not REAL: but provides less extreme charges than oxidation numbers sum of the FC on molecule must equal overall charge on molecule atoms try to achieve a FC as close to zero as possible any negative FC must be on most electronegative atom

Example SO 4 2- sulfate ion 6e - + 4(6e - )= 30 e = 32 electrons

Example NO 2 nitrogen dioxide 17 electrons

Formal Charge (FC) Dinitrogen oxide, N 2 O has two double bonds. The general structure is N=N=O. The formal charge on the oxygen atom in this molecule is? A.zero B.Positive 1 (+1) C.Positive 2 (+2) D.Negative one (-1) E.Negative two (-2) Answer: A Use the formula! If necessary, draw the lewis structure.

VSEPR – Valence Shell Electron Pair Repulsion Steric # Overall Structure Forms 2 LinearAX 2 3 Trigonal PlanarAX 3, AX 2 E 4 TetrahedralAX 4, AX 3 E, AX 2 E 2 5 Trigonal bipyramidalAX 5, AX 4 E, AX 3 E 2, AX 2 E 3 6 OctahedralAX 6, AX 5 E, AX 4 E 2 Steric # = X+E X = atoms bonded to A E = nonbonding electron pairs on A A = central atom

Molecular Shapes VSEPR Theory (Valence Shell Electron Pair Repulsion) pairs of electrons are arranged as far apart from each other as possible. 1.Linear: 180 o between bonds Ex: O 2 O 2. Trigonal Planar: “flat triangle” – one central atom surrounded by three atoms with NO LONE PAIRS. Ex: BCl 3 B Cl

3. Tetrahedral: four atoms bonded to a central atom. C H H H H 4. Trigonal pyramidal: three atoms bonded to a central atom with one lone pair of electrons about the central atom. N H H H

5. Bent: a central atom bonded to two other atoms AND one or two LONE PAIRS. O HH

VSEPR: Linear AX 2 CO 2 Shape: Linear

VSEPR: Trigonal Planar AX 3 AX 2 E BF 3 SnCl 2 Shape: Trigonal Planar Shape: Bent

VSEPR: Tetrahedral AX 4 AX 3 E AX 2 E 2 CCl 4 PCl 3 Cl 2 O Shape: Tetrahedral Shape: Trigonal Pyramidal Shape: Bent

VSEPR: Trigonal Bi-pyramidal AX 5 AX 4 E AX 3 E 2 AX 2 E 3 PCl 5 SF 4 ClF 3 I3-I3-I3-I3- Shape: Trigonal Bipyramidal Shape: Seesaw Shape: T-shaped Shape: Linear

VSEPR: Octahedral AX 6 AX 5 E AX 4 E 2 SF 6 ICl 4 - BrF 5 Shape: Octahedral Shape: Square Pyramidal Shape: Square planar

VSEPR The predicted geometry (shape) of PH 3 according to the VSEPR theory is? A.linear B.bent C.Trigonal pyramidal D.tetrahedral E.Trigonal planar Answer: C Easy one!