AP Chemistry Unit #7 Chemical Bonding & Molecular Shape Zumdahl Chapters 8 & 9 TYPES OF BONDING BONDING INTRA (Within (inside) compounds) STRONG INTER (Interactions between the molecules of a compound) WEAK Ionic (Metal + non-metal) Giant ionic lattice formed Covalent (Non-metals) Discrete molecules formed Hydrogen Bonding (H attached to N, O or F) Stronger permanaent dipoles van der Waals Forces (Attractions between dipoles) Permanent or induced Dative or Co-ordinate (Electron deficient species) Discrete molecules formed Dipole-Dipole (Polar molecules) Permanent dipoles London Dispersion Forces (Non-polar molecules) Induced dipoles
I. Ionic Bonding A. Lewis Structures & The Octet Rule 1. Lewis Symbols models used to show the valence electron configuration for elements, molecules and compounds Ex: Cl O 2 2. Octet Rule all atoms tend to gain, lose or share electrons in order to have eight electrons in their valence shell B. Energy of Ionic Bond Formation (link) 1. Lattice Energy the energy released when separated gaseous ions are packed together to form an ionic solid 2. Magnitude of Lattice Energy the energy released when ions form a solid is directly proportional to the charge on the ions and inversely related to the size of the ions; smaller ions with greater charge have greater lattice energy than larger ions with lesser charge. Where k is the Coulomb s law constant with a value of 8.99 x 10 9 J m/c 2 Q 1 and Q 2 are the magnitudes of the charge on the ions in coulombs, and d is the distance between ion centers in meters Sample Ex 8.1 Arrange the following compounds in order of increasing lattice energy: LiF, KBr, MgO
C. Electron Configurations of Ions (link) 1. Cations remove one or more electrons from the last number in the electron configuration of the neutral element 2. Anions add one or more electrons to the last number in the electron configuration of the neutral element D. Polyatomic Ion a charged particle that contains two or more covalently bonded atoms
E. Size of Ions 1. Isoelectronic Series a group of ions (and often a noble gas) that contain the same number of electrons Example: O -2, F -, Na +, Mg +2, Al +3, Ne 10 electrons each As nuclear charge increases, size Sample Problem Arrange the ions below in order of increasing size: Sr +2, As -3, Se -2, Rb +, Br II. Covalent Bonding A. Lewis Structures of Covalent Bonds Sample Problem Draw the Lewis structures for water, carbon dioxide, and ammonia. B. Multiple Bonds 1. Single Bond one pair, or 2 valence electrons shared between atoms 2. Double Bond two pair, or 4 valence electrons shared between atoms 3. Triple Bond three pair, or 6 valence electrons shared between atoms III. Bond Polarity and Electronegativity A. Bond Polarity 1. Nonpolar Covalent Bond a covalent bond in which the electrons are shared equally between atoms 2. Polar Covalent Bond a covalent bond in which the electrons are NOT shared equally between atoms because of a difference in electronegativity
3. Coordinate Covalent Bond a bond formed when one atom provides both electrons in a shared pair Ex: NH 4 + (draw the Lewis dot structure for this molecule) B. Electronegativity IV. Drawing Lewis Structures A. Procedure for drawing Lewis Structures 1. Count up the number of valence electrons 2. Write the atom symbols and connect the atoms with single bonds 3. Distribute electrons (in pairs) to complete octets of atoms 4. If there are not enough electrons, make multiple bonds to complete octets 5. If there are extra electrons, put them on the central atom
B. Sample Exercise 8.6 Draw the Lewis Structure for phosphorus trichloride. C. Sample Exercise 8.7 Draw the Lewis Structure for HCN V. Resonance Structures A. Resonance Structures/Resonance Forms resonance occurs when more than one valid Lewis structure can be written for a particular molecule; the actual structure of the molecule is an average of its resonance structures B. Formal Charge the difference between the number of valence electrons on the free atom and the number of valence electrons assigned to the atom in the molecule. C. Sample Exersise 8.9 Draw 2 resonance structures for the nitrite ion. D. Sample Problem Which has the shorter sulfur oxygen bonds, SO 3 or SO 3 2- and why?
VI. Exceptions to the Octet Rule A. Odd Number of Electrons B. Less than an Octet C. More than an Octet D. Sample exercise 8.10 Draw the Lewis Structure for ICl 4 -
VII. BOND ENERGIES (Look @ Table 8.4 and 8.5 on p. 372 1 st ) ΔH = sum of the ENERGIES REQUIRED to break old bonds plus the sum of the ENERGIES RELEASED in the forming of new bonds ΔH = ΣD (bonds broken) ΣD (bonds formed) energy required energy released Where D = bond energy per mole of bonds (always has a + sign) Ex: H 2 (g) + F 2 (g) 2HF(g) ΔH = D H-H + D F-F 2D H-F = 1 mol (432 kj/mol) + 1 mol (154 kj/mol) 2 mol (565 kj/mol) = -544 kj Sample Exercise 8.5: ΔH From Bond Energies Calculate ΔH for the rxn. of methane w/ chlorine and fluorine to give Freon- 12 (CF 2 Cl 2 ). Solution: Reactant bonds broken: Product bonds formed: CH 4 = 4 mol C-H = 4 mol (413 kj/mol) = 1652 kj 2Cl 2 = 2 mol Cl-Cl = 2 mol (239 kj/mol) = 478 kj 2F 2 = 2 mol F-F = 2 mol (154 kj/mol) = 308 kj Total energy required = 2438 kj CF 2 Cl 2 = 2 mol C-F = 2 mol (485 kj/mol) = 970 kj 2 mol C-Cl 2 mol (339 kj/mol) = 678 kj HF = 2 mol H-F = 2 mol (565 kj/mol) = 1130 kj HCl = 2 mol H-Cl = 2 mol (427 kj/mol) = 854 kj Total energy released = 3632 kj
ΔH = ΣD (bonds broken) ΣD (bonds formed) = 2438 kj 3632 kj = -1194 kj Therefore, 1194 kj of energy is released per mole of CF 2 Cl 2 formed. PROBLEMS: p.404 #47,49 A. Energy of Ionic Bond Formation 1. Lattice Energy 2. Magnitude of Lattice Energy E=k(Q 1 Q 2 /d) Sample Ex: Arrange the following compounds in order of increasing lattice energy: LiF, KBr, MgO.
IIX. POLAR vs. NONPOLAR MOLECULES: Molecule = A bonded substance; always 2 or more bonded together * compound = compound A.) POLAR Molecules: molecules sharing of Doesn t pass the mirror test Can t be folded to reflect itself 2 atoms different elements/electronegativities More than 2 atoms unbonded e - or pairs around the atom Ex: HCl Ex: H 2 O
B.) NONPOLAR Molecules: molecules sharing of or DOES pass the mirror test CAN be folded to reflect itself 2 Atoms same element/electronegativities More than 2 atoms unbonded e - or pairs around the atom Ex: Cl 2 Ex: CO 2 Ex: CCl 4 BEWARE! There are often POLAR BONDS inside NONPOLAR MOLECULES (look back at the previous 2 examples)
C. Intermolecular forces, A.K.A. IMF s ONLY IN COVALENT MOLECULES, NEVER IONIC COMPOUNDS! forces that act BETWEEN that hold molecules to EACH OTHER Only exist in & states Called WEAK forces because they are much weaker than CHEMICAL BONDS *REMEMBER: IMF s occur BETWEEN molecules, whereas BONDING occurs WITHIN molecules http://www.northland.cc.mn.us/biology/biology1111/animations/hydrogenbonds.html Just Remember IMF s ARE NOT BONDS!!!
Type of IMF London dispersion forces (LDF s) Description/Example(s) Weakest of all the IMF s Only important for NONPOLAR molecules Electron-electron repulsion creates BRIEF DIPOLES in atoms/molecules http://antoine.frostburg.edu/chem/senese/101/liquids/faq/h-bonding-vs-london-forces.shtml Dipole (dipole-dipole) Molecules such as HCl have both POSITIVE and a NEGATIVE ends, or POLES Two poles = Results from an UNEQUAL/ASYMMETRICAL sharing of electrons DIPOLE-DIPOLE = two molecules with permanent dipoles are attracted to one another DIPOLE MOMENT = measure of the of the dipole within a molecule (POLARITY) The GREATER the difference in ELECTRONEGATIVITY between atoms, the GREATER the POLARITY/DIPOLE MOMENT The HIGHER the dipole moment, the STRONGER the intermolecular forces (IMF s) The stronger the IMF s, the higher the m.p. and b.p. http://chemmovies.unl.edu/chemanime/dipoled/dipoled.html Hydrogen Bonds Specific type of interaction In a POLAR BOND, hydrogen is basically reduced to a BARE PROTON w/ almost no ATOMIC RADIUS of all IMF s by far Only occur in molecules containing Hydrogen AND,, or http://programs.northlandcollege.edu/biology/biology1111/animations/hydrogenbonds.html
IX. Lewis Structures & the Octet Rule Lewis Structures: Comments About the Octet Rule The second-row elements C, N, O, and F should always be assumed to obey the octet rule The second-row elements B and Be often have fewer than eight electrons around them in their compounds. These electron-deficient compounds are very reactive. The second-row elements never exceed the octet rule, since their valence orbitals (2s and 2p) can accommodate only eight electrons. Third-row and heavier elements often satisfy the octet rule but can exceed the octet rule by using their empty valence d orbitals. When writing the Lewis structure for a molecule, satisfy the octet rule for the atoms first. If electrons remain after the octet rule has been satisfied, then place them on the elements having available d orbitals (elements in Period 3 or beyond). Rules Governing Formal Charge To calculate the formal charge on an atom: 1. Take the sum of the lone pair electrons and ½ the shared electrons. This is the number of valence electrons assigned to the atom in the molecule. 2. Subtract the number of assigned electrons from the number of valence electrons on the free neutral atom to obtain the formal charge. The sum of the formal charges of all atoms in a given molecule or ion must equal the overall charge on that species. If nonequivalent Lewis structures exist for a species, those with formal charges closest to zero and with any negative formal charges on the most electronegative atoms are considered to best describe the bonding in the molecule or ion. *Delocalized bonding: bond pairs that appear to move between two or more different pairs of atoms; occurs in molecules with resonance structures.
X. The Localized Electron Model VSEPR Model (Valence Shell Electron Pair Repulsion) Used to predict shapes of molecules Steps to Apply the VSEPR Model 1. Draw the Lewis structure for the model. 2. Count the electron pairs and arrange them in the way that minimizes repulsion (that is, put the pairs as far apart as possible). 3. Determine the positions of the atoms from the way the electron pairs are shared. 4. Determine the name of the molecular structure from the positions of the atoms. Summary of the VSEPR Model The rules for using the VSEPR model to predict molecular structure: 1. Determine the Lewis structure(s) for the molecule. 2. For molecules with resonance structures, use any of the structures to predict the molecular structure. 3. Sum the electron pairs around the central atom. 4. In counting pairs, count each multiple bond as a single effective pair. 5. The arrangement of the pairs is determined by minimizing electronpair repulsions. These arrangements are shown in Table 8.6 (page 394) in your textbook. 6. Lone pairs require more space than bonding pairs do. Choose an arrangement that gives the lone pairs as much room as possible. Recognize that the lone pairs may produce a slight distortion of the structure at angles less than 120 degrees.
Chapter 9.1: Hybridization & the Localized Electron Model I. sp 3 Hybridization: Consider the molecule methane (CH 4 ) (Assume valence e- only in bonding) H = 1s C = 2s & 2p orbitals Problems: 1. CH 4 is supposed to have 4 identical C-H bonds 2. Three 2p orbitals want to be perpendicular, but CH 4 is tetrahedral Solution: The bonds in the 2s and 2p orbitals mix to form 4 equivalent orbitals HYBRIDIZATION Combine one s and three p orbitals four sp 3 hybrid orbitals *All 4 orbitals are identical in shape (1 lg. lobe, 1 sm. lobe) Principle: Whenever an atom is surrounded by 4 effective pairs, a tetrahedral set of sp 3 hybrid orbitals is required (the atom becomes sp 3 hybridized).
*The atom achieves minimum energy by hybridizing: Sample Exercise 9.1: The Localized Electron Model I Describe the bonding in the ammonia (NH 3 ) molecule using the localized electron model. Solution: 1. Draw the Lewis structure 2. Determine VSEPR shape 3. Determine hybridization 4 e - pairs = tetrahedral arrangement = sp 3 hybridization Three sp 3 orbitals are bonding One sp 3 orbital is lone pair PROBLEMS: p.441 #11
II. sp 2 Hybridization: Consider the molecule ethylene (C 2 H 4 ) 1. Draw the Lewis structure 2. Determine VSEPR shape 3. Determine hybridization *Double bond acts like 1 effective pair C 2 H 4 has 3 effective pairs around each C atom trigonal planar arrangements (120º) Principle: Whenever an atom is surrounded by 3 effective pairs, a trigonal planar set of sp 2 hybrid orbitals is required. 1. combine one s and two p orbitals sp 2 hybridization 2. remaining p orbital on each carbon exists in plane perpendicular to sp 2 orbitals 3. sp 2 orbitals form a sigma (σ) bond 4. one remaining 2p orbital (not hybridized) forms a pi (π) bond by sharing e- in space above and below the σ bond Solution: 3 e - pairs = trigonal planar arrangement = sp 2 hybridization Three sp 2 orbitals form sigma (σ) bonds One remaining p orbital forms pi a (π) bond
Click to view animation: http://www.youtube.com/watch?v=zsfcppl13so&feature=related&safe=active Click to view animation: http://www.youtube.com/watch?nr=1&v=c2w-ydpcpl4&feature=endscreen&safe=active
III. sp Hybridization: Consider the molecule carbon dioxide (CO 2 ) Principle: Whenever an atom is surrounded by 2 effective pairs, a linear set of sp hybrid orbitals is required. 1. Combine one s and one p orbital two sp hybrid orbitals 2. One s orbital & one p orbital hybridize, forming a set of two sp orbitals 3. Remaining two p orbitals will form π bonds Sample Exercise 9.2: The Localized Electron Model II Describe the bonding in the N 2 molecule. Solution: Each N is surrounded by 2 effective e- pairs (linear) 2 effective e- pairs = sp hybridization Each N has two sp orbitals, one bonding (σ), one lone (σ bond is buried in the diagram below) 2 remaining p orbitals form each form π bonds PROBLEMS: p.441 #13
IV. dsp 3 Hybridization: Consider PCl 5, which exceeds the octet rule Principle: Whenever an atom is surrounded by 5 effective pairs, a set of dsp 3 hybrid orbitals is required in a trigonal bipyramind arrangement. 4. One d orbital, one s orbital & three p orbitals hybridize, forming a set of five dsp 3 orbitals In this case, each Cl atom is surrounded by 4 electron pairs each Cl requires a tetrahedral arrangement tetrahedral arrangement requires four sp 3 orbitals each sp 3 orbital on Cl shares electrons with a dsp 3 orbital on P to form σ bonds the remaining sp 3 orbitals on Cl hold lone pairs Sample Exercise 9.3: The Localized Electron Model III Describe the bonding in the triiodide ion (I 3 - ).
Solution: The central iodine atom has five pairs of electrons A set of five electron pairs requires a trigonal bipyramidal arrangement, which requires a set of dsp 3 orbitals The outer atoms have four pairs of electrons, which class for a tetrahedral arrangement and sp 3 hybridization The central iodine is dsp 3 hybridized. Three of these hybrid orbitals hold lone pairs, and two of them overlap with sp 3 orbitals of the other two iodine atoms to form σ bonds PROBLEMS: p.442 #21 V. d 2 sp 3 Hybridization: Consider sulfur hexafluoride (SF 6 ), which has six pairs of electrons around its central atom Principle: Whenever an atom is surrounded by 6 effective pairs, they require a set of d 2 sp 3 hybrid orbitals arranged octahedrally. Each of the d 2 sp 3 orbitals on the sulfur atom is used to bond to a fluorine atom Each of the fluorine atoms has four electron pairs and they are assumed to be sp 3 hybridized Sample Exercise 9.4: The Localized Electron Model IV How is the xenon atom in XeF 4 hybridized?
Solution: The Xe atom has six pairs of electrons around it, arranged octahedrally to minimize repulsions An octahedral set of six electron pairs requires that Xe become d 2 sp 3 hybridized Four of the d 2 sp 3 orbitals form σ bonds with fluorine atoms The remaining two d 2 sp 3 orbitals hold lone pairs PROBLEMS: p.442 #20 XI. Molecular Orbitals A. Molecular Orbital Theory B. Bonding Molecular Orbital The Hydrogen Molecule C. Bond Order Sample EX:What is the bond order of the He 2 + ion? Would you expect this atom to be stable relative to the separated He atom and the He + ion? D. Second Period Diatomic Molecules 1. Paramagnetic 2. Diamagnetic