AP Chemistry Chapter 8: Bonding Study Guide - Enlow
Types of Chemical Bonds: bonds result from the tendency of a system to seek its lowest possible energy Ionic Bonds:
– e- transferred in order to attain noble gas e- configuration
Solid state at room temperature due to the large forces of attraction between oppositely charged ions, and the sheer density of the ions in a given space forms a solid
Occur between M and NM, when ions are paired they have lower energy than the separated ions, thus greater stability {don’t forget that all compounds prefer to be lower energy due to this stability and this is also favored because there is an increase in free energy in the universe}
o Gaseous ions are much less stable than their ionic solid counterparts
Driving forces for this process are the strong mutual attraction of the oppositely charged ions and the EA attraction that Cl has for the additional e-.
This energy of attraction = Lattice Energy = calculated from Coulomb’s Law o E = energy in joules
o Q1 and Q2 are numerical ion charges o distance between ion center in nanometers o negative sign indicates an attractive force
o if comparing the same charged ions, then positive, which represents repulsive force
Any compound that conducts an electric current when melted is an ionic compound.
Electrically neutral compounds overall because equal e- exchanged between the atoms
Ions generally adopt noble gas electron configuration (gain or lose e-) Covalent Bonds:
What conditions favor bond formation? If there is a decrease in system energy, then the bond will form despite the p+ p+ repulsion and e- e- repulsion
o The distance where energy is minimal is called the bond length between the atoms
o Very close = higher PE due to repulsion;
Lower PE when at appropriate bond length
Ex: The H2 molecule is more stable with less energy than 2 separated H atoms by a certain
quantity of energy {THE BOND ENERGY!!}; it is equally attracted to each nucleus with minimal repulsion
Energy is given off (bond energy) when two atoms achieve greater stability together than apart
o Nonpolar Covalent Bonds – There are e- that are shared between two atom nuclei; mutually attracting shared e-
o Polar Covalent bonds – unequal sharing and atoms end up with disproportionate charge distribution
o Fraction charge written as lowercase delta d+ or d-
Electronegativity – the ability of an atom in a molecule to attract shred e- to itself; Pauling scale is what is used most commonly, shows polar nature of bonds.
o The attraction between the partially charged atoms will lead to a greater bond strength; therefore, if the larger the EN different = stronger bond
o This EN difference ranges from ionic (large diff) to polar covalent (med diff) to nonpolar covalent (0 difference)
These EN differences will tell the extent of ionic character vs covalent character
0. 0 NPC 0.4 PC 0.9 1.4 1.7 Ionic 1.9
Dipole Moment – arrow points towards the negative center and hash mark at the positive center o May result in a polar molecule if the atoms are not arranged so that the dipole moments cancel
Diatomic Molecules with a dipole moment are ALWAYS polar!
Polar Molecule = H2O (bent) ; Nonpolar Molecule (has dipole moment though) = CO2 (linear, tetrahedral)
Ionic Compounds and More!
o Ion sizes
o These sizes are determined based on assumption and cannot be 100% determined, but we can talk in trends
o Cations are smaller than parent ions, anions are larger than parent ions
o Isoelectric ions – ions containing the same number of e-, however, the difference in proton number allows for assumption that the greater the # of protons, the smaller the atomic radius
o Binary Cmpds
o Lattice energy is the change in energy that takes place when separated gaseous ions are packed together to form an ionic solid M+ (g) + X- (g) à MX (s)
o From the system’s perspective, it is losing energy and is a negative value
o The weird thing about the formation of ionic solids from their gaseous ions is that the energy needed to form the gaseous ions is exothermic (unfavorable), then the formation is exothermic (favorable). This is the overall driving force
o The lattice formed by alkali metals and halogens is cubic and each ion is surrounded by 6 of the opposite ion, exception cesium salts
o Lattice energy calculation is modified from Coulomb’s Law LE = k (q1)(q2) / r
r = shortest distance between centers of the cations and the anions
Opposite charges yield negative value due to its exothermic nature
Same charges yield positive value due to endothermic nature o How are charges of ions determined? It all depends on the payoff of
decreased ionic energy after the lattice energy is released once the solid is formed.
Partial Ionic Character
o No individual bonds are completely ionic
o Ionic cmpds generally have greater than 50% ionic character, due to this being very EN (greater than 1.6)
o
o Polyatomic ions are composed of covalent bonds, but can assemble “ionic cmpds”
Models: Scientists use models to understand nature; they are man-made and not always correct!!!!!!!
Covalent Bond Model: The Energy of stabilization is equal to the amount of energy that the molecule lost when it was formed from its ions divided by the number of bonds formed. This allows us to associate quantities of energy with the formation of bonds between elements and allows us to draw structural relationships. These bond energies are expected to be the same in any environment; this has been supported in many situations, but does not explain all phenomena
o Ex: the idea that bonds have the same average energy in any environment is INCORRECT, but they still use this assumption to aid in qualitative observations.
Bonds
o Single – share a pair of e-, longest and weakest o Double - share 2 pairs of e-, medium length
o Triple – share 3 pairs of e-, shortest and strongest
Bond Enthalpies
o Bonds broken – endothermic because energy must be put into the system o Bonds formed – exothermic because this is an energy lowering process o Enthalpy of reaction = bonds broken – bonds formed
o H D(Bonds broken) D(Bonds formed) The Localized e- Bonding Model:
o Assumes a molecule is composed of atoms that are bound together by sharing pairs of e- using the atomic orbitals of the bound atoms
o e- pairs are assumed to be localized on a particular atom of in the space between two atoms (lone pair- on an atom ; bonding pairs- between atoms)
o This model uses: valence e- Lewis structures, prediction of molecule geometry using the VSEPR theory, and descriptions of the atomic orbitals that the e- are held in
Lewis Structures:
o These structures come from the assumption that the most stable cmpd comes from atoms that achieve noble gas config
o Rules to the Theory
o only valence e- are included
o H follows the duet rule; as does He, Be, Li, and B…. because this is the closest noble gas e- config o Octet rule is followed by all other elements because this is the noble gas e- config
o Rules of Drawing Lewis Structures
o Add up the TOTAL number of valence electrons from all atoms
o Use a pair of electrons to form a bond between each pair of bound atoms. Lines instead of dots are used to indicate each pair of bonding electrons
o Arrange the remaining atoms to satisfy the duet rule for hydrogen and the octet rule for the second row elements
o If ions are present, the add e- to anions and minus e- from cations in total #ve
** the best way to look at a molecule is to regard it as a new entity that uses all of available
ve- of atoms to achieve the lowest possible energy
o Exceptions
o Boron –
It is common for B to form molecules in which B atom is e- deficient
BF3 = B has 6ve- around it, but can also double bond to attain octet
This acts as a lewis acid (e- pair acceptor) o Lower Period 3 Elements
SF6 =S exceeds the octet rule because there are different d orbitals available for additional ve-
o More About the Octet Rule
o 1. Second row elements C, N, O and F should always obey the octet rule
Another version of Lewis Structures e-:
To determine where all the electrons are to be placed, apply the N - A = S rule where: N = sum of valence electrons needed for each atom. The two allowed values are two for hydrogen and eight for all other elements. A = sum of all available valence electrons S = # of electrons shared and S/2 = # bonds
o 2. B and Be (second row) often have fewer then eight electrons around them, and form electron deficient, highly reactive molecules
o 3. Second row elements never exceed the octet rule
o 4. Third row and heavier elements often satisfy (or exceed) the octet rule
o 5. Satisfy the octet rule first. If extra electrons remain, place them on elements having available d orbitals
a. When necessary to exceed the octet rule for one of several third row elements, assume that the extra electrons be placed on the central atom
o Resonance
o Sometimes more than one lewis structure is valid, so we write them as resonance structures with a double-headed arrow between each possible structure
o The correct description of the structure is not given by any of the structures but IS a superposition of all possible structures = it is an average of all possible structures (DO NOT SAY THAT THESE STRUCTURES FLIP BACK AND FORTH BETWEEN VERSIONS !!)
o It has been observed that where one thought there would be a single bond or a double bond, there is actually a bond that an intermediate between single bonds and double bonds in strength and length.
o Odd e- Molecules
o Results in the formation of free radicals most of the time, but this model does not support odd e- o Formal Charge
o For an atom in a molecule = (number of ve- on free atom) – (number of ve- assigned to the atom in the molecule)
Where the “number of ve- assigned to the atom in the molecule “ is equal to the number of nonbonding electrons + 1/2 number of bonding electrons
o If it has more ve- in the molecule than its normal ve-, then (-) charge o If it has less ve- in the molecule that its normal ve-, then (+) charge
o A formal charge of zero for each atom in a molecule is a very common result for a favorable Lewis structure. In other cases, a favorable Lewis structure will follow these rules:
1. Small numbers, preferably 0.
2. No like charges are adjacent to each other, but unlike charges are close together.
3. The more electronegative element(s), the lower the formal charge(s) will be.
4. The total of the formal charges equals the charge on the ion.
** Formal charges are estimated charges and are not to be taken as atomic charges
** Experimental evidence can conclusively determine the correct bonding situation Molecular Structure – The VSEPR Model
o Valence-shell- electron- pair-repulsion theory = useful in predicting geometries of molecules formed from nonmetals
o the structure around a given atom is determined principally by minimizing e- pair repulsion o Bonding and nonbonding pairs around a molecule will be positioned as far apart as possible o Applying VSEPR
o Draw lewis structure
o Count e- pairs and arrange them in a way that minimizes repulsion o Determine the positions of the atoms from the way the e- are shared
o Determine the name of the molecular structure from the position of the atoms
This means that the e- play a role, but do not count as a position for the name o Geometries
o Linear – flat - bond angle 180º
I3- has linear geometry because the 3 lone pair must be as far away from each other as possible forcing the bonding atoms to arrange in a linear fashion
o Bent – V shaped due to nonbonding pairs – bond angle 104.5º o Trigonal planar – flat – bond angles 120º
o Square planar – flat - 90º
o **Tetrahedral gives angles that are farther apart and more desirable – 109.5º
When there are 4 pairs of e- present around an atom, they should always be tetrahedral o Trigonal Planar
3 bonding pairs and 1 nonbonding pair is like tetrahedral without a 4th atom
Bond angle - 107º o Trigonal Bipyramidal
5 pairs of e- o Octehedral
6 pairs of e-
XeF4 – arranged so that the lone pair of e- are opposite of one another to limit repulsive forces. This causes the molecule to be square planar in shape, allowing the polar bonds to cancel out and leads to an overall nonpolar molecule
o The effect of unshared lone pairs causes distortion of the geometry because the lone pairs are thought to take up more space (due to greater repulsive forces), which tends to compare the angles between bonding pairs of e-.
o Lone pairs do NOT cause distortion when the bond angles are 120º and greater
o VSEPR Model and Multiple Bonds
o Each double or triple bond between two atoms is thought of as an “effective pair” because they are located between the same two atoms and “act as one”. This allows us to determine geometry for multiple bond molecules as well!
o When a molecule exhibits resonance, ANY of the resonance structures can be used to predict the molecular structure using the VSEPR model
o Molecules Containing No Singe Central Atom
o Apply the principal of distancing shared and unshared electron pairs o Look at real 3-dimensional, rotatable models to develop predictive skills
Chapter 9 Study Guide – Covalent Bonding in Orbitals- Enlow Hybridization and the Localized Electron Model
o VSEPR theory is only one way in which the molecular geometry of molecules may be determined. Another way involves the valence bond theory.
o The valence bond theory describes covalent bonding as the mixing of atomic orbitals to form a new kind of orbital, a hybrid orbital.
o Hybrid orbitals are atomic orbitals formed as a result of mixing the atomic orbitals of similar energies of the atoms involved in the covalent bond.
o The number of hybrid orbitals formed is the same as the number of atomic orbitals mixed, and the type of hybrid orbital formed depends on the types of atomic orbitals mixed.
o Evidence for hybridization SUMMARY OF VSEPR:
Two geometries can be determined; the electron-group geometry, in which all electron pairs surrounding a nucleus are considered, and molecular geometry, in which the nonbonding electrons become “invisible” and only the geometry of the atomic nuclei are considered. For the purposes of geometry, double and triple bonds count the same as single bonds. To determine the geometry:
1. Write the Lewis electron-dot formula of the compound.
2. Determine the number of electron-pair groups surrounding the central atom(s). Remember that double and triple bonds are treated as a single group.
3. Determine the geometric shape that maximizes the distance between the electron groups. This is the geometry of the electron groups.
4. Mentally allow the nonbonding electrons to become invisible. They are still there and are still repelling the other electron pairs, but we don’t “see” them. The molecular geometry is determined by the remaining arrangement of atoms (as determined by the bonding electron groups) around the central atom.
o Ex: CH4 is known to have 4 identical bonds, however, according to its valence e-, it should have 2 variations of bonds. One between the 2p orbital of C and 1s orbital of H and the other between the 2s orbital of C and 1s orbital of H
It is assumed that C hybridizes its 2s and 2p orbitals to create 4 sp3 hybrid orbitals to equally bond to H atoms
o We assume that that C’s atomic orbitals are rearranged to accommodate the best e- configuration for the MOLECULE as a whole. This total # of ve- arranged around the entire molecule is what is important for the energetics of the molecule
o ** A molecule is not just simply the sum of its parts ---individual atoms respond as needed to achieve the minimum energy for the molecule***
o
The different types of Hybridized orbitals:
o sp hybridization results from the overlap of an s orbital with one p orbital. Two sp hybrid orbitals are formed with a bond angle of 180°. This is a linear orientation. Two effective pair around an atom will always require sp hybridization!
o sp2 hybridization results from the overlap of an s orbital with two p orbitals. Three sp2 hybrid orbitals are formed with a trigonal planar orientation and a bond angle of 120°. One place this type of bonding occurs is in the formation of the C to C double bond. The orientation of p orbitals used determines the orientation of the hybridized orbitals! 3 effective pair around an atom will always require sp2 hybridization!
o Note that there is one unhybridized p orbital that is still available for bonding
o sp3 hybridization results from the mixing of one s orbital and three p orbitals, giving four sp3 hybrid orbitals with a tetrahedral geometric orientation. This sp3 hybridization is found in carbon when it forms four single bonds.
o sp3d hybridization results from the blending of an s orbital, three p orbitals, and one d orbital. The result is five sp3d orbitals with a trigonal bipyramidal orientation. This type of bonding occurs in compounds like PCl5. Note that this hybridization is an exception to the octet rule. A set of 5 effective pairs around a given atom always requires sp3d hybridization!
o sp3d2 hybridization occurs when one s, three p, and two d orbitals are mixed, giving an octahedral arrangement. SF6 is an example. Again, this hybridization is an exception to the octet rule. If one starts with this structure and one of the bonding pairs becomes a lone pair, then a square pyramidal shape results, while two lone pairs gives a square planar shape. A set of 6 effective pairs around a given atom always requires sp3d2 hybridization!
Other bonding stuff to know:
o Sigma bonds (bond)
o Bond in which the electron pair is shared in an area centered on a line running between the atoms
o Lobes of bonding orbital point toward each other
Ex: All bonds in methane are sigma bonds o Pi bonds (bonds)
o e- pair above and below the bond
Ex: Created by overlapping of nonhybridized 2p orbitals on each carbon o Multiple bonds
o Double bonds always consist of one bond and one bond
A C to C triple bond results from the overlap of an sp hybrid orbital and two p orbitals on one carbon, with the same on the other carbon. In this situation there will be one bond (overlap of the sp hybrid orbitals) and two bonds (overlap of two sets of p orbitals).
Molecular Orbital Theory: ANOTHER model to represent the bonding that takes place in covalent compounds is the molecular orbital theory.
o In the molecular orbital (MO) theory of covalent bonding, atomic orbitals (AOs) on the individual atoms combine to form orbitals that encompass the entire molecule. These are called molecular orbitals (MOs).
o These molecular orbitals have definite shapes and energies associated with them. They can hold 2 e- with opposite spins (relate to e- probabaility) o When two atomic orbitals are added, 2 molecular orbitals are formed, one
bonding and one antibonding. The bonding MO is of lower energy than the antibonding MO.
o In the MO model the AOs are added together to form the MOs. Then the e- are added to the MOs, following the rules used previously when filling orbitals:
lowest-energy orbitals get filled first, maximum of 2 e- per orbital, and half fill orbitals of equal energy before pairing e-
o When s AOs are added, one sigma bonding (σ) and one sigma antibonding (σ*) molecular orbital are formed.
o The MO bond order is the number of e-s in bonding MOs minus the number of electrons in antibonding MOs, divided by 2. For H2 , the bond order would be (2
− 0)/2 = 1. A stable bonding situation exists between 2 atoms when the bond order is greater than zero. The larger the bond order, the stronger the bond, the shorter the bond strength, the greater the bond energy!
o Ex: The length and strength of a covalent bond is related to its bond order. The greater the bond order, the shorter and
stronger the bond. Diatomic nitrogen, for example, has a short, extremely strong bond due to its nitrogen-to-nitrogen triple bond.
Paramagnetism and Dimagnetism:
o One of the advantages of the molecular orbital model is that it can predict some of the magnetic properties of molecules.
o If molecules are placed in a strong magnetic field, they exhibit one of two magnetic behaviors—attraction or repulsion.
o Paramagnetism, the attraction to a magnetic field, is due to the presence of unpaired electrons o Diamagnetism, the slight repulsion from a magnetic field, is due to the presence of only paired
electrons.
From 5 Steps to a 5: Common Mistakes to Avoid:
1. Remember that metals + nonmetals form ionic bonds, while the reaction of two non-metals forms a covalent bond.
2. The octet rule does not always work, but for the representative elements it works the majority of the time.
3. Atoms that lose electrons form cations; atoms that gain electrons form anions.
4. In writing the formulas of ionic compounds, make sure the subscripts are in the lowest ratio of whole numbers.
5. When using the crisscross rule be sure the subscripts are reduced to the lowest whole-number ratio.
6. When using the N - A = S rule in writing Lewis structures, be sure you add electrons to the A term for a polyatomic anion, and subtract electrons for a polyatomic cation.
7. In the N - A = S rule, only the valence electrons are counted.
8. In using the VSEPR theory, when going from the electron-group geometry to the molecular geometry, start with the electron-group geometry; make the nonbonding electrons mentally invisible; and then describe what is left.
9. When adding electrons to the molecular orbitals, remember: lowest energy first. On orbitals with equal energies, half fill and then pair up.
10. When writing Lewis structures of polyatomic ions, don’t forget to show the charge.
11. When you draw resonance structures, you can move only electrons (bonds). Never move the atoms.
12. When answering questions, the stability of the noble-gas configurations is a result, not an explanation. Your answers will require an explanation, i.e., lower energy state.
From 5 Steps to a 5: Rapid Review
• Compounds are pure substances that have a fixed proportion of elements.
• Metals react with nonmetals to form ionic bonds, and nonmetals react with other non-metals to form covalent bonds.
• The Lewis electron-dot structure is a way of representing an element and its valence electrons.
• Atoms tend to lose, gain, or share electrons to achieve the same electronic configuration (become isoelectronic to) as the nearest noble gas.
• Atoms are generally most stable when they have a complete octet (eight electrons).
• Ionic bonds result when a metal loses electrons to form cations and a nonmetal gains those electrons to form an anion.
• Ionic bonds can also result from the interaction of polyatomic ions.
• The attraction of the opposite charges (anions and cations) forms the ionic bond.
• The crisscross rule can help determine the formula of an ionic compound.
• In covalent bonding two atoms share one or more electron pairs.
• If the electrons are shared equally, the bond is a nonpolar covalent bond, but unequal sharing results in a polar covalent bond.
• The element that will have the greatest attraction for a bonding pair of electrons is related to its electronegativity.
• Electronegativity values increase from left to right on the periodic table and decrease from top to bottom.
• The N − A = S rule can be used to help draw the Lewis structure of a molecule. • Molecular geometry, the arrangement of atoms in three-dimensional space, can be predicted using the VSEPR theory. This theory says the electron pairs around a central atom will try to get as far as possible from each other to minimize the repulsive forces.
• In using the VSEPR theory, first determine the electron-group geometry, then the molecular geometry.
• The valence bond theory describes covalent bonding as the overlap of atomic orbitals to form a new kind of orbital, a hybrid orbital.
• The number of hybrid orbitals is the same as the number of atomic orbitals that were mixed together.
• There are a number of different types of hybrid orbital, such as sp, sp2, and sp3.
• In the valence bond theory, sigma bonds overlap on a line drawn between the two nuclei, while pi bonds result from the overlap of atomic orbitals above and below a line connecting the two atomic nuclei.
• A double or triple bond is always composed of one sigma bond and the rest pi bonds.
• In the molecular orbital (MO) theory of covalent bonding, atomic orbitals form molecular orbitals that encompass the entire molecule.
• The MO theory uses bonding and antibonding molecular orbitals.
• The bond order is (# electrons in bonding MOs – # electrons in anti-bonding MOs)/2.
• Resonance occurs when more than one Lewis structure can be written for a molecule. The actual structure of the molecule is an average of the Lewis resonance structures.
• The higher the bond order, the shorter and stronger the bond.
• Paramagnetism, the attraction of a molecule to a magnetic field, is due to the presence of unpaired electrons.
Diamagnetism, the repulsion of a molecule from a magnetic field, is due to the presence of paired electrons.
From 5 Steps to a 5: Multiple Choice Questions:
You have 20 minutes, and you may not use a calculator.
1. VSEPR predicts an SbF5 molecule will be which of the following shapes?
(A) tetrahedral (B) trigonal bipyramidal (C) square pyramid (D) trigonal planar (E) square planar 2. The shortest bond would be present in which of the following substances?
(A) I2 (B) CO (C) CCl4 (E) SCl2
3. Which of the following does not have one or more π bonds? (A) H2O (B) HNO3 (C) O2 (D) N2 (E) NO2-
4. Which of the following is polar? (A) SF4 (B) XeF4 (C) CF4 (D) SbF5 (E) BF3
5. Resonance structures are needed to describe the bonding in which of the following?
(A) H2O (B) ClF3 (C) HNO3 (D) CH4 (E) NH3
For questions 6 and 7, pick the best choice from the following:
(A) ionic bonds (B) hybrid orbitals (C) resonance structures (D) hydrogen bonding (E) van der Waals attractions 6. An explanation of the equivalent bond lengths of the nitrite ion is:
7. Most organic substances have low melting points. This may be because, in most cases, the intermolecular forces are:
8. Which of the following has more than one unshared pair of valence electrons on the central atom?
(A) BrF5 (B) NF3 (C) IF7 (D) ClF3 (E) CF4
9. What is the expected hybridization of the central atom in a molecule of TiCl4? This molecule is tetrahedral.
(A) sp3d2 (B) sp3d (C) sp (D) sp2 (E) sp3
10. The species in the following set do not include which of the following geometries?
(A) square planar (B) tetrahedral (C) octahedral (D) trigonal pyramidal (E) linear 11. The only substance listed below that contains ionic, σ, and π bonds is:
(A) Na2CO3 (B) HClO2 (C) H2O (D) CO2 (E) NaCl
For problems, 12–14 choose a molecule from the following list: (A) C2 (B) F2 (C) B2 (D) O2 (E) Ne2 12. The paramagnetic molecule with a bond order of two.
13. The diamagnetic molecule with no antibonding electrons.
14. The paramagnetic molecule with antibonding electrons.
15. The electron pairs point toward the corners of which geometrical shape for a molecule with sp2 hybrid orbitals?
(A) trigonal planar (B) octahedron (C) trigonal bipyramid (D) trigonal pyramid (E) tetrahedron 16. Regular tetrahedral molecules or ions include which of the following?
I. CH4 II. SF4 III. CO2
(A) I, II, and III (B) I and III only (C) I only (D) I and II only (E) II only 17. Which molecule or ion in the following list has the greatest number of unshared electrons around the central atom? (A) CF4 (B) ClF3 (C) BF3 (E) IF5
18. Which of the following molecules is the least polar? (A) PH3 (B) CH4 (C) H2O (D) NO2 (E) HCl 19. What types of hybridization of carbon are in the compound 1,4-butadiene, CH2CHCHCH2?
I. sp3 II. sp2 III. sp
(A) I and II (B) I, II, and III (C) I and III (D) I only (E) II only 20. Which of the following molecules is the most polar? (A) C2H2 (B) N2 (C) CH3I (D) BF3 (E) NH3
21. Which of the following processes involves breaking an ionic bond?
(A) H2O(g) + Cl2(g) 2 HCl(g)
(B) I2(g) 2 I(g) (C) Na(s) Na(g)
(D) 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g) (E) 2 KBr(s) 2 K(g) + Br2(g)
From 5 Steps to a 5: Free-response Questions First Free-Response Question
Answer the following questions. You have 15 minutes, and you may not use a calculator. You may use the tables at the back of the book. Answer each of the following with respect to chemical bonding and structure.
a. The nitrite ion, NO2- , and the nitrate ion, NO3- , both play a role in nitrogen chemistry.
i. Draw the Lewis (electron-dot) structure for the nitrite ion and the nitrate ion.
ii. Predict which ion will have the shorter bond length and justify your prediction.
b. Using Lewis (electron-dot) structures, explain why the ClF3 molecule is polar and the BF3 molecule is not polar.
c. Consider the following substances and their melting points:
Explain the relative values of the melting points of these substances.
SrS = MP greater than 2000ºC KCl = MP 770 ºC
H2O = MP 0 ºC H2S = MP -85.5 ºC CH4 = MP -182 ºC
Second Free-Response Question
Answer the following questions. You have 15 minutes, and you may not use a calculator. You may use the tables in the back of the book. Answer the following questions about structure and bonding.
a. Which of the following tetrafluorides is nonpolar? Use Lewis electron-dot structures to explain your conclusions.
SiF4 SF4 XeF4
b. Rank the following compounds in order of increasing melting point. Explain your answer. Lewis electron-dot structures may aid you. SnF2 SeF2 KrF2
c. Use Lewis electron-dot structures to show why the carbon–oxygen bonds in the oxalate ion are all equal.
d. When PCl5 is dissolved in a polar solvent, the solution conducts electricity. Explain why. Use an appropriate chemical equation to illustrate your answer.
Answers and Explanations:
1. B—The Lewis (electron-dot) structure has five bonding pairs around the central Sb and no lone pairs. VSEPR predicts this number of pairs to give a trigonal bipyramidal structure.
2. B—All the bonds except in CO are single bonds. The CO bond is a triple bond. Triple bonds are shorter than double bonds, which are shorter than single bonds. Drawing Lewis structures might help you answer this question.
3. A—Answers B—E contain molecules or ions with double or triple bonds. Double and triple bonds contain π bonds.
Water has only single (σ) bonds. If any are not obvious, draw a Lewis structure.
4. A—The VSEPR model predicts all the other molecules to be nonpolar.
5. C—All the other answers involve species containing only single bonds. Substances without double or triple bonds seldom need resonance structures.
6. C—Resonance causes bonds to have the same average length.
7. E—Many organic molecules are nonpolar. Nonpolar substances are held together by weak London dispersion forces.
8. D—Lewis structures are required. You do not need to draw all of them. A and B have one unshared pair, while C and E have no unshared pairs. D has two unshared pairs of electrons.
9. E—Tetrahedral molecules are normally sp3 hybridized.
10. D—SiCl4, is tetrahedral. is square planar. C2H2 is linear. TeF6 is octahedral. is trigonal planar. If you are uncertain about any of these, Lewis structures and VSEPR are needed.
11. A—Only A and E are ionic. The chloride ion has no internal bonds, so σ and π bonds are not possible. 12–14—
Sketch a molecular orbital energy-level diagram. Use the same diagram to save time, unless it becomes too messy.
12. D 13. A 14. D
15. A—This hybridization requires a geometrical shape with three corners.
16. B—One or more Lewis structures may help you. I and III are tetrahedral, and II is an irregular tetrahedron (see-saw).
17. B—A has 0. B has 2. C and D have 0. E has 1. You may need to draw one or more Lewis structures.
18. B—All the molecules are polar except B.
19. E—The structure is: All the carbon atoms have one double and two single bonds. This combination is sp2.
20. E—Drawing one or more Lewis structures may help you. Only C and E are polar. Only the ammonia has hydrogen bonding, which is very, very polar.
21. E—C is metallic bonding. All the others involve covalently bonded molecules.
First Free-Response Question
a. Give yourself 1 point for each structure that is correct. The double bonds could be between the nitrogen and any of the oxygens, not just the ones shown. Only one double bond per structure is allowed.
b. ii. If you predicted the nitrite ion has the shorter bond length, you have earned 1 point. The explanation must invoke resonance. You do not need to show all the resonance structures. You need to mention that the double bond “moves” from one oxygen to another. In the nitrite ion, each N–O bond is a double bond half the time and a single bond the other half. This gives an average of 1.5 bonds between the nitrogen and each of the oxygens. Similarly, for the nitrate ion, each N–O bond spends one-third of the time as a double bond, and two-thirds of the time as a single bond. The average N–O bond is 1.33. The larger the average number of bonds, the shorter the bond is. This explanation will get you 1 point. b. Give yourself 1 point for each correct Lewis structure. The BF3, with three bonding pairs and zero nonbonding pairs on the central atom, is not polar. The ClF3, with five pairs about the central atom, is polar because of the two lone pairs.
Give yourself 1 point for this explanation. c. You get 1 point for saying that the two compounds with the highest melting points are ionic and the other compounds are molecular. You get 1 point if you say that SrS is higher than KCl because the charges on the ions in SrS are higher. You get 1 point if you say H2O is higher than the lowest two because of hydrogen bonding. You get 1 point if you say H2S is higher than CH4 because H2S is polar and CH4 is nonpolar. There are a maximum of 11 points.
Second Free-Response Question
a. Silicon tetrafluoride is the only one of the three compounds that is not polar. You get 1 point if you correctly predict only SiF4 to be nonpolar. You get 1 additional point for each correct Lewis structure.
b. The order is: KrF2 < SeF2 < SnF2. The Lewis structure indicates that KrF2 is nonpolar. Thus, it only has very weak London dispersion forces between the molecules. SeF2 is polar and the molecules are attracted by dipole–
dipole attractions, which are stronger than London forces. SnF2 has the highest melting point, because of the presence of strong ionic bonds. You get 1 point for the order, and 1 point for the discussion.
c. The following resonance structures may be drawn for the oxalate ion. The presence of resonance equalizes the bonds. You get 1 point for any correct Lewis structure for , and 1 point for showing or discussing resonance. d. PCl5 must ionize. There are two acceptable equations. You get 1 point for the explanation, and you get 1 point for either of the equations. Total your points for the different parts. There are 10 points possible.
