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101 CHEM CLASS NOTES & PROBLEMS Prof. Upali Siriwardane, Chemistry Program, Louisiana Tech University, Ruston, LA 71272.
Chapter 9. Molecular Structures. 9.1 Using Molecular Models 9.2 Predicting Molecular Shapes: VSEPR 9.3 Orbitals Consistent with Molecular Shapes: Hybridization 9.4 Molecular Polarity 9.5 Noncovalent Interactions and Forces Between Molecules 9.6 Biomolecules: DNA and the Importance of Molecular Structure 9.7 Chiral Molecules Objectives are as follows: Basic Skills Students should be able to: 9.1 Using Molecular Models 1. Recognize the various ways that the shapes of molecules are represented by models and on the printed page (Section 9. 1). 2. Recognize ball-and-stick, space-filling, perspective drawing of molecules and describe the three dimensional arrangement of atoms in the molecule. 9.2 Predicting Molecular Shapes: VSEPR 3. Predict shapes of molecules and polyatomic ions by using the VSEPR model (Section 9.2). 4. Recognize and describe different types of geometries created by molecules with 2,3,4,5 and 6 electron pairs. 5. Recognize non-bonding and bonding electrons in a molecule and predict the molecular shape. 6. Predict and recognize different shapes that a molecule can adopt from VSEPR theory. 9.3 Orbitals Consistent with Molecular Shapes: Hybridization 7. Predict and recognize different type of hybridization a central atom can adopt:sp, sp2, sp3, sp3d and sp3d2. 8. Determine the orbital hybridization of a central atom and the associated molecular geometry (Section 9 3). 9. Describe covalent bonding between two atoms in terms of sigma or pi bonds, or both (Section 9 3). 9.4 Molecular Polarity 10. Use molecular structure and electronegativities to predict the polarities of molecules (Section 9.4). 11. Predicting molecular polarity from bond polarities and molecular symmetry 9.5 Noncovalent Interactions and Forces Between Molecules 12. Describe different types of intermolecular forces between molecules. 13. Using noncovalent interactions and explain melting points and boiling points of molecular compounds (Section 9 5). 9.6 Biomolecules: DNA and the Importance of Molecular Structure 14. Identify the major components in the structure of DNA (Section 9.6). 9.7 Chiral Molecules 15. Define and describe the nature of chiral molecules and enantiomers (Section 9.7). 16. Describe the basis of infrared spectroscopy (Tools of Chemistry) and UTV visible spectroscopy (Tools of Cbemistry) and how they are used to determine molecular structures.
CHEM 101 CLASS NOTES & PROBLEMS Prof. Upali Siriwardane, Chemistry Program, Louisiana Tech University, Ruston, LA 71272.
CHAPTER 9 KEY CONCEPTS
Molecular Drawings & Models Molecular structure Molecular Structure of molecules with non-binding electron on the central atom Hybridization of atomic orbitals Predicting polarity from electeonegetivity and molecualr shapes Interpretation of b.p and m.p from intermolecular forces Ball-stick & Space Filling Models VSEPR theory Additional geometries to accommodate non bonding electrons Molecualr shapes resulting from: sp, sp2, sp3, sp3d and sp3d2. Molecualr polarity and dipole momentum Components of DNA Perspective drawings Basic molecular geometries Atomic orbitals and the molecualr shape Sigma and pi bonding Intermolecular forces between molecules Chiral Molecules and enantiomers
Molecular Models To properly picture chemistry at the molecular level, you need to have a good sense of what molecules look like. One way to develop the ability to visualize molecular geometries is through the use of models. Ball and Stick Models Ball and stick model gives a good idea of the spatial dimension of the structure of a molecule it still represents them as an assembly of little spheres (atoms) connected by tubes (bonds).
Space-filling Models Space-filling models give a much better idea of the compactness and three-dimensional shape of molecules.
Perspective Drawings It is often necessary to represent 3-dimensional molecular situations on a 2-dimensional surface (paper/blackboard/screen) and chemists use PERSPECTIVE DRAWINGS to focus attention on a particular atom of interest and to convey information to their audiences about the 3dimensional arrangement of atoms/groups/substituents about that atom, much as artists have since Renaissance times managed to insert 3-dimensional perspective into their 2-dimensional drawings and paintings
We often use perspective drawings in which the direction of a bond is specified by the line connecting the bonded atoms. In most cases the focus of configuration is a carbon atom so the lines specifying bond directions will originate there. As defined in the diagram on the right, a simple straight line represents a bond lying approximately in the surface plane. The two bonds to substituents A in the structure on the left are of this kind. A wedge shaped bond is directed in front of this plane (thick end toward the viewer), as shown by the bond to substituent B; and a hatched bond is directed in back of the plane (away from the viewer), as shown by the bond to substituent D. Some texts and other sources may use a dashed bond in the same manner as we have defined the hatched bond, but this can be confusing because the dashed bond is often used to represent a partial bond (i.e. a covalent bond that is partially formed or partially broken). The following examples make use of this notation, and also illustrate the importance of including nonbonding valence shell electron pairs (colored blue) when viewing such configurations
It is important to learn to draw perspective structures and to become proficient in understanding and manipulating them.
Carbon atoms are drawn in the perspective drawing as a wire frame. Where lines meet there is a carbon atom. If only tow line are there then there will be two hydrogen (not shown but assumed) attached to that carbon atoms. If the lines consists of a single and a double boon there will be a hydrogen atom (not shown but assumed) attached to that carbon atom.
Predicting Molecular Shapes: VSEPR The geometric shape of a given molecule depends on its skeleton structure and bond lengths, but also other factors. To determine the shape of a molecule, two things are important. The first is a correct Lewis Dot Structure; the second is the application of the Valence-Shell Electron-Pair Repulsion Theory ( also known as the VSEPR Theory). A simple method of predicting shapes of covalent molecules is the Valence shell electron pair repulsion theory (VSEPR) From chapter 8 you learned that are four rules to writing a good Lewis Dot Structure. 1. Add up all of the valence shell electrons in the molecule. (do not forget that ions have either gained or lost electrons 2. Pick the central atom. The central atom will be the atom with the largest bong capacity (or bond order).
3. Arrange the other atoms around the central atom. If the total number of electrons is even, the outer atoms will have a noble gas structure. Any extra electrons are put around the central atom. 4. Check the structure that you have drawn using Formal Charges Valence-Shell Electron-Pair Repulsion Theory (VSEPR Theory) In the theory, valence shell electron pairs are assumed to repel each other, assuming orientations to minimize repulsions and establish certain groups of molecular shapes. Basic Geometries created by molecules with 2,3,4,5 and 6 electron pairs.
As shown above, these shapes can be classified according to how many electron pairs surround the central atom. Two pairs - linear, three - trigonal planar, four-tetrahedral, five-trigonal bipyrimidal, six-octahedral. VSEPR notation: To help translate the Lewis Structure to a molecular structure, one can use the VSEPR notation. The VSEPR notation looks very much like the formula for a compound, with a few changes. First, the central atom is called "A." Secondly, all the outer atoms are designated with an "X." This is true even if the atoms are different (it only matters that these are atoms that surround the central atom. Also, molecules with double bonds are treated as if the double bond was merely a single bond, because there is still only one outer atom in the bond.) Finally, any lone pair electrons are designated with an "E." For instance, the water atom has 2 hydrogen atoms and 2 lone pair electrons surrounding the central atom. Therefore, its VSEPR notation is AX2E2.
This is the ideal tetrahedral structure. All the angles between the bonds are 109.5 degrees. Not all electron pair repulsions are equal. Lone pair electrons have a stronger repulsion than bonding pair electrons. The result of this is that lone pair electrons squeeze bonding pair electrons together. This causes the bond angle between atoms to become smaller than the ideal angle. Central atoms with expanded octet configurations: Hypervalent Compounds. A number of molecules have more electrons in the outer shell than in an inert gas configuration. This involves use of one or more d orbitals (so as to not violate the Pauli Exclusion Principle). A molecule with 5 pairs of electrons around its central atom can have the following possible molecular shapes: The PCl5 molecule has five bond pairs in the outer shell of P. The shape of this molecule is symmetrical trigonal bipyramidal. Note that there are two types of Cl atoms. The two Cl atoms which are on a straight line which passes through the P nucleus are said to occupy axial positions. The other three Cl are in equatorial positions. The axial Cl to P bond distance is slightly longer than the equatorial Cl to P bond distance. (Why?) The SF6 molecule is an example of a molecule with six bond pairs. The least repelling shape for non-bonding and bonding electrons in a molecule and predict the molecular shape this is octahedral, the F atoms being at the corners of a regular octahedron. Bonding and Non-bonding Electron on the Central Atom and the Molecular Structure Molecular Structures with Lone Pairs The molecular configuration is not exactly the same as the electron configuration. The reason for this is that lone pairs of electrons do not show up in the molecular configuration, although their effects are still seen (such as pushing bonding pair electrons closer together). This gives rise to 7 more configurations (in addition to the 5 basic electron configurations). 1) Angular or Bent: First case of angular structures has three electron pairs surrounding the central atom. The possible molecular configurations are either trigonal planar (if all of the electron pairs are
bonding pairs) or angular Angular configuration could be based on trigonal planar or triangular planar electron geometry or bent (If one is a lone pair) . The second case of angular structure has four electron pairs surrounding the central atom. The water molecule (H:O:H) has two lone pairs and two bond pairs (4 total pairs on O). The shape is angular or bent with a bond angle which is less than that in NH3. This is due to the greater repulsions with two lone pairs.
Triangular Pyramidal or Trigonal Bipyamid The molecule NH3 has a Lewis dot symbol which is much like BF3 but now with a lone pair in the outer shell of the central atom. The NH3 molecule has a shape which is called trigonal pyramidal (approximately tetrahedral minus one atom). Due to the greater repelling character of lone pairs, as also shown in the images above the H atoms are bent closer together than the regular tetrahedral angle of 109.5 . Other molecules with this one-lone, three- bond-pair configuration (:NCl3, :PCl3) have this same basic shape, but with slightly different bond angles from that of NH3 (but all less than 109.5 ).
Examples of molecules with 5 electron pairs, but one or more of which are lone pairs, are SF4 and ClF3, which form "See-saw", and "distorted "T" shapes respectively. Note that these are variations of the trigonal bipyramid. The lone pairs occupy equatorial positions so as to minimize repulsions. The greater repulsion of these lone pairs distorts the axial Cl atoms slightly to the opposite side of the molecule.
There is also possibility to make a T-shape from octahedral ideal electron pair geometry as show in the figure above.
6) Square pyramidal 7) Square Planar Square pyramidal Square Planar
Predict the molecular structure of the following molecules using VSEPR theory: H2O, NH3, CO2, SF6, PCl5, XeF4 a) Molecular Structure of H2O First get the Lewis structure of H2O. 2 bond pairs = 2 x 2 = 4 2 lone pairs = 2 x 2 = 4 Total 8 = 4 electron pairs (an Octet) Basic Structure: Arrangement of electron airs around the central atom is tetrahedral for H2O.
Molecular Structure: Arrangement of atom or groups around the central atom. Ignore the lone pairs and look at the basic structure. Molecular Structure: Angular
b) Molecular structure of NH3:
i) Lewis Structure:
ii)Basic Structure: tetrahedral pyramidal structure
iii) Molecule structure: Triangular
c) Molecular Structure of CO2 Lewis Structure of CO2
There are 4 electron pairs on the carbon, but the double bonds are counted as single electron pairs. Only consider one electron pair for double or triple bonds. This gives 2 pairs on C. 2 electron pairs gives linear basic structure. Since there are no lone pairs on C, basic structure is the same as the molecular structure.
Basic Structure: Linear d) Molecular Structure of SF6;
Molecular Structure: Linear
Lewis Structure i) Count electrons around central atom: 6 electron pairs ii) 6 electron pairs given octahedral basic structure iii) Since there are no lone pairs, basic structure is the same as the molecular structure. Molecular structure of SF6 is octahedral e) Molecular Structure of PCl5: 10
Lewis Structure: i) Count electron pairs on the central atom: 5 electron pairs. ii) Basic structure is trigonal bipyramidal iii) There are no lone pairs on central P atom. Therefore, basic structure is same as molecular structure. Molecular structure of PCl5 - Trigonal bipyramidal f) Molecular Structure of XeF4; First get the Lewis structure; i) Count the total valence electrons: 8 + 4 x 7 = 36 (18 pairs). ii) Xe is central atom and connect Xe to terminal Fs. iii) Give octets to four Fs. vi) Place extra electrons on Xe. iv) Count electron pairs. 6 pairs on Xe = 6 4 x 3 pairs on F = 12 18 electron pairs Lewis Structure To get the molecular structure i) Count electron pairs on central atom: 6 pairs ii) Basic structure is octahedral. The lone pairs occupy two opposite corners to avoid greater repulsion. First, get the stable basic structure by eliminating the unstable basic structures based on greater repulsion of lone pairs at smaller angles.
Stable Basic Structure (Octahedral) Unstable Basic Structure(Octahedral ) Molecular Structure(Square Planar)
More Examples: 5 electron pairs: examples (work Lewis dot structures on your own) PCl5 (trigonal bipyramidal) SF4 (see-saw: lone pair will be equatorial) ClF3 (T-shaped: two equatorial lone pairs) XeF2 (linear: three equatorial lone pairs) 6 electron pairs: examples (work Lewis dot structures on your own) SF6 (octahedral)
BrF5 (square pyramidal) XeF4 (square planar: lone pairs will be opposed) Predict shapes of molecules and polyatomic ions First draw the Lewis-Electron-Dot-Structure. The charge on the ion must be accounted for while computing the total number of valence electrons when writing Lewis structures of ions. For negative ions, one valence electron is added for each unit of negative charge and we subtract one valence electron for each unit of positive ion. a) Lewis Structure of CO32-: i) valence electrons:4 + 3 x 6 + 2(negative charge) = 24 (12 pairs) ii) central atom is carbon
iii) fill octet to C; fill octet to O
One oxygen does not have an octet. Share lone pair on C with the oxygens. The double bond could be on any other oxygen atoms.
Therefore, there are three resonance structures for CO32- ion To predict the Geometry of the CO32- ion: We look at one of the resonance structures and count number of electron pairs around the central atom. For counting electrons for the purpose of getting geometry we follow following set of rules: i) Single bond contribute one electron pair ii) Double, Triple and multiple bonds contribute only one electron pair CO32- ion has a central atom with 3 electron pairs therefore the geometry around the central atom is trigonal or triangular planar
Molecular structure of CO32- ion: Trigonal planar b) NO3-Lewis Structure: i) valence electrons: 5 + 3 x 6 + 1(negative charge) = 24 (12 pairs) iii) N is the central atom:
iii) Fill octet to N; Fill octet to Os
one oxygen has only 6 electrons share lone pair on N with the oxygen The double bond could be on any other oxygen atoms. Therefore, there are three resonance structures for the NO3- ion.
To predict the Geometry of the NO3- ion: We look at one of the resonance structures and count number of electron pairs around the central atom. For counting electrons for the purpose of getting geometry we follow following set of rules: i) Single bond contribute one electron pair ii)Double, Triple and multiple bonds contribute only one electron pair NO3- ion has a central nitrogen atom with 3 electron pairs therefore the geometry around the central atom is trigonal or triangular planar Molecular structure of NO3- ion: Trigonal planar c) NO2- Lewis Structure i) valence electrons: 5 + 2 x 6 + 1(negative charge) = 18 (9 pairs) ii) central atom is nitrogen. iii) Fill octet to N ; Fill octet to O 13
one oxygen has only six electrons share lone pair on N with oxygen. The double bond could be on any of the two oxygen atoms. Therefore, there are two resonance structures for NO2- ion.
To predict the Geometry of the NO2- ion: We look at one of the resonance structures and count number of electron pairs around the central atom. For counting electrons for the purpose of getting geometry we follow following set of rules: iii) Single bond contribute one electron pair iv) Double, Triple and multiple bonds contribute only one electron pair NO2- ion has a central atom with 3 electron pairs and one of them is non-bonding electron pair. Therefore the geometry around the central atom is angular or bent . Molecular structure of NO2- ion: Angular or Bent
Molecular Structure, Bonding and Valance Atomic Orbitals Valence Bond (VB) Theory VSEPR predicts only shapes or geometry of molecules, but does NOT explain bond formation using atomic orbitals from the quantum mechanical model. Valence Bond (VB) Theory provides an explanation for bonding VB Theory Valence Bond Theory attributes covalent bonding to overlapping atomic orbitals that share pairs of electrons. Overlap is when a portion of one orbital and a portion of a second occupy the same region of space. Bonding is due to the buildup of electron density between atoms caused by orbital overlap (using s, p, etc. atomic orbitals). They can be ss, sp, or pp overlap Ex: H2 (ss); HCl (sp); Cl2 (pp). The bonding is due to attraction between the nucleii and electrons in the shared (overlap) region overlap increases as nucleii get closer together, but if they get too close nucleus-nucleus repulsion outweighs it. Orbitals Consistent with Molecular Shapes: Hybridization Different type of hybridization a central atom can adopt: sp, sp2, sp3, sp3d and sp3d2 electron electron domain domains geometry possible molecular geometries unhybridize d orbitals
sp sp2 sp3
2 3 4
linear trigonal planar tetrahedral
linear trigonal planar; bent tetrahedral; trigonal pyramidal, bent
2p 1p 0p
Central atoms with d valence orbitals sp3d sp3d2 5 6 trigonal bypramid octahedral Seasaw, T-shape Square pyramid, square 0 p and 4 d 0 p and 3 d
Hybridization of Atomic Orbitals Experiments in quantum mechanics have shown that the orbitals in isolated atoms(electron configuration shown on the periodic table) change when two or more atoms come together to form a molecule. This is called hybridization, the mixing of atomic orbitals of an isolated atom to generate a new set of atomic orbitals called hybrid orbitals.
All hybrid orbitals are strongly "directed"--most of electron density in one direction directions of hybrid orbitals consistent with VSEPR theory. Nonbonding ("lone") electron pairs can also occupy hybrid orbitals (Ex: O in H2O is sp3 hybridized). Half-filled hybrid orbitals available to form bonds by orbital overlap Unhybridized p orbitals are involved with multiple bonding. Assigning Hybrid Orbitals to Central Atoms Determine Lewis structure Single bond contribute one electron pair Double, Triple and multiple bonds contribute only one electron pair Count number of electron pairs around the central atom Determine electron-pair geometry Linear: sp Trigonal: sp2 Tetrahedral: sp3 Trigonal bipyramid" sp3d Octahedral: sp3d2 6. Assign the set of hybridized orbitals that corresponds to this geometry 1. 2. 3. 4. 5.
BeCl2 Be: 1s22s2 has no unpaired electrons therefore it cannot form a bond since there can be no overlap with full orbitals (violates Pauli). However, Be "promotes" one 2s electron to one 2p orbital. Both the 2s and the one 2p orbital are now half-filled and thus available for overlap however, bonds formed with unhybridized 2s and 2p orbitals would not be identical (unlike reality). Now, mix or "hybridize" the two orbitals to give two equivalent "sp hybrid" orbitals equivalent hybrid orbitals give identical bonds and the correct bond angle.
Orbital hybridization of a central atom and the associated molecular geometry Predict the hybridization of the central atom(s) in the following molecules: H2O (H2S), CH4 (SiH4), CCl4 (CF4, CBr4), CO2, NH3 (PH3), PCl3, (PF3, NCl3), PCl5, SF6, XeF4 a) H2O: Valence electron structure: tetrahedral O: 2s2 2p4 has two unpaired electrons therefore it cannot form a bond could only overlap with two orbitals. However, C "promotes" one 2s2 electron to one 2p orbital. Thus four the 2s and the one 2p orbital are now half-filled and thus available for overlap with 1s atomic orbitals of four hydrogens. However, bonds formed with unhybridized 2s and three 2p orbitals would not be identical (unlike reality). Now, we mix or "hybridize" the two orbitals to give four equivalent "sp3 hybrid" orbitals. Equivalent hybrid orbitals give identical bonds and the correct bond angle. Number of electrons pairs around oxygen is four: four pairs arrange to minimize electron pair repulsion in a tetrahedron. Four electron pairs are in four sp3 hybridized orbitals in oxygen. The hybridization of the oxygen atom is sp3. H2S is similar to H2O Hybridization of S atom is sp3. b) CH4: C: 2s2 2p2 has two unpaired electrons therefore it cannot form a bond could only overlap with two orbitals. However, C "promotes" one 2s2 electron to one 2p orbital. Thus four the 2s and the one 2p orbital are now halffilled and thus available for overlap with 1s atomic orbitals of four 16
hydrogens. However, bonds formed with unhybridized 2s and three 2p orbitals would not be identical (unlike reality). Now, we mix or "hybridize" the two orbitals to give four equivalent "sp3 hybrid" orbitals. Equivalent hybrid orbitals give identical bonds and the correct bond angle.
Basic structure: tetrahedral number of electron pairs around the central atom is four. The hybridization of the C atom of CH4 is sp3. SiH4 - hybridization of Si atom is sp3. c) CCl4: basic structure: tetrahedral There are four electron pairs on C atom. The hybridization of the carbon atom is sp3. In CF4 and CBr4 the hybridization of C is sp3. d) CO2: basic structure: linear number of electron pairs 2 (what is really making directional bonds) Hybridization of the C atom is sp2 e) NH3: Four electron pairs around central N atom. sp3. PH3: sp3 hybridization of P f) PCl3: Four electron pairs around central P atom. Hybridization of P is sp3. PF3, NCl3: hybridization of central atom is sp3. g) PCl5: P: [Ne]3s23p34s03d0 17
4s orbital will not hybridize with orbitals in lower (n = 3) shell promote one 3s electron to 3d hybridize the one 3s, three 3p, and one 3d orbitals to give five sp3d hybrid orbitals trigonal pyramidal!
Five electron pairs around central P atom. Hybridization of P is sp3d. h) SF6 S: [Ne]3s23p44s03d0 4s orbital will not hybridize with orbitals in lower (n = 3) shell promote one 3s and one 3p electron to two 3d orbitals hybridize the one 3s, three 3p, and two 3d orbitals to give six sp3d2 hybrid orbitals octahedral!
Six electron pairs around central S atom. Hybridization of P is sp3d2. i) XeF4 Six electron pairs around central Xe atom. Hybridization of P is sp3d2. Types covalent bonding between two atoms in terms of sigma or pi bonds Sigma or Bonds single bonds in a Lewis structure. Occurs from the overlap of two s orbitals. The overlap of an s orbital and a p orbital, and the end-to-end overlap of two p orbitals. End to end configuration 18
is the strongest, any other configuration results in less overlap. Pi or Bonds a covalent bond that results from the side-by-side overlap of two p orbitals. A double bond in a Lewis structure will consist of one sigma bond, and one pi bond. A triple bond will consist of one sigma bond and two pi bonds. The overlap is less in a pi bond than in a sigma bond, therefore it is the weaker of the two bonds. The distinguishing feature of a sigma bond (or sigma bonding orbital) is that the overlap region lies directly between the two nuclei. First or single bond is sigma bond. Single is bond First bond in a double bond is a bond and second bond is bond First bond in a triple bond is a bond, second bond is bond and third bond is also a bond
The distinction between a sigma bond and a pi bond is diagrammed here. The sigma bond has orbital overlap directly between the two nuclei. The pi bond has orbital overlap off to the sides of the line joining the two nuclei. Acetylene C2H2 Lewis Structure
Three bonds and two bonds
In the sketch at the top here, the sp hybrid orbitals of one atom are shown as sticks so that you can concentrate on the unhybridized, or leftover, 2 p orbital. The bottom sketch shows you how the hybrid orbitals of two carbon atoms can come together, overlap, and form a regular sigma bond (shown as a stick). Notice that this also brings the leftover two 2 p orbitals so close that they can also overlap and form what we call two pi bond. Notice that the overlapping occurs in two places between two p orbitals , above and below the sigma bond. Two pi bond does not overlap in the region directly between the two carbon atoms where the sigma bond is formed. Ethylene C2H4 Lewis Structure: Five bonds and two bonds
The combination of a sigma and a pi bond between the same two carbon atoms is a double bond. A double bond consists of a sigma bond (using hybrid orbitals) and a pi bond (using p orbitals). sigma bond - a bond resulting from head-on overlap of atomic orbitals. Three sigma bonds in ethylene'
pi bond - a bond resulting from side-on overlap of atomic orbitals One Pi bonds in ethylene
The overlap of two p orbitals yields one pi bond. Another way of showing how two carbon atoms can form a double bond is indicated in the bottom (part c) of this diagram. Here we have two carbon atoms with sp2 hybridization, and each of those carbon atoms is bonded to two hydrogen atoms and also to the other carbon atom. Notice that this diagram shows how the carbon atoms and the hydrogen atoms are all in a flat plane. This is drawn in perspective. Notice that there is a sigma bond between the carbon atoms and each of the hydrogen atoms. Notice also a sigma bond between the two carbon atoms. The sigma bonds (or sigma orbitals) are shown as the dark shaded areas in this drawing. Each one of those sigma bonds uses one of the hybrid orbitals. Remember that with sp2 hybridization, there are three hybrid orbitals, and those are the ones used to form the sigma bonds between all the atoms. Remember also that we have a leftover p orbital for each carbon atom. Those p orbitals are the electron clouds or orbitals that are shown going up above and below each carbon atom. In this particular diagram, the shading between those p orbitals shows that the p orbitals overlap one another and allow the electrons in those p orbitals to be shared. This kind of bond is called a pi bond. The pi bond results when p orbitals overlap one another in this side-to-side fashion.
Multiple Bonds: double and triple bonds
all bonds discussed so far have lied along the internuclear axis we call these " (sigma) bonds" bonds can also form due to overlap between p orbitals oriented perpendicular to the internuclear axis. In these " (pi) bonds", the electron density is concentrated symmetrically about the internuclear axis single bonds are always sigma bonds multiple bonds have both sigma and pi bonds: double: one sigma bond, one pi bond triple: one sigma bond, two pi bonds
in order to form a pi bond, there must be at least one unhybridized p orbital on each of the two atoms p-orbitals must be half-filled to allow overlap and form a bond double bonds require sp2 hybridization (leaves one unhybridized p orbital) triple bonds require sp hybridization (leaves two unhybridized p orbitals) single bonds rotate (twist) freely--rotation doesn't alter orbital overlap multiple bonds do not rotate freely, because to do so you would have to break a pi bond
Molecular Polarity Molecules composed of covalently bonded atoms may also polar be or nonpolar. For the molecule to be polar, it must, of course, have polar bonds. But the key factor for determining the polarity of a molecule is its shape. If the polar bonds (dipoles) are symmetrical around the central atom, they offset each other and the resulting molecule is nonpolar. However, if the dipoles are not symmetrical around the central atom, the electrons will be pulled to one end of the molecule. The resulting molecule is polar. Ball and stick models are often used to demonstrate molecular shape. In this exercise you will build several covalent molecules and predict each molecule's polarity on the basis of its molecular shape. The geometry of the molecule affects its dipole moment or polarity - the asymmetric distribution of positive and negative charges. Polar Molecules will orient in a magnetic field When two elements with different electronegativities are bonded together, one obtains a polar bond. Example: HF
Depending on the molecular shape, polar bonds can give rise to a polar molecule, one with an overall uneven distribution of electron charge. We can see this charge imbalance when the molecules are placed in an electric field. Polarity is a key feature of a molecule because it can influence physical, chemical, and even biological properties.
For a diatomic molecule, a polar bond must lead to a polar molecule. Consider hydrogen fluoride, shown here as the Lewis structure changes to a ball-and-stick model enclosed within the space-filling shape. Note the polar arrows and the colors. If red indicates high electron density and blue indicates low, you can see that the F end of the molecule is much more negative than the H end, and thus HF is highly polar. Between two electric plates with the field off, the molecules lie every which way. With the field on, however, they become oriented with their negative ends facing the positive plate and their positive ends facing the negative plate. If a molecule has more than two atoms, its shape can affect the polarity in a crucial way. For example, in carbon dioxide, since oxygen is more electronegative than carbon, each bond is highly polar. But the linear molecular shape makes the bond polarities cancel each other, so the CO2 molecule is nonpolar. Notice that the orientation of the molecules is random, whether the field is off or on. The situation is very different for water. As in CO2, the highly electronegative oxygen pulls electron density toward itself, so each bond is polar. But the V-shape of the molecule allows the bond polarities to reinforce each other, so water is highly polar, as the large polar arrow shows. With the field off, the molecules are oriented randomly, but with it on, their poles become oriented toward the oppositely charged plates. The case of boron trifluoride is similar to that of CO2. Because the three highly polar bonds point to the corners of an equilateral triangle, the bond polarities cancel each other, and BF3 is nonpolar. The molecules are oriented randomly with the field off or on. Like BF3, ammonia has four atoms and three polar bonds, but the trigonal pyramidal shape means that the bond polarities reinforce each other. Thus, ammonia is highly polar for the same reason that water is. Sometimes two molecules have similar overall shapes, but their slightly different compositions lead to very different polarities. Carbon tetrachloride has four polar carbon-chlorine bonds that point to the corners of a tetrahedron, so they cancel each other and give a nonpolar molecule. Substituting a hydrogen for one of the chlorines gives chloroform, another tetrahedral molecule, but now the polar bonds reinforce each other. As a result, chloroform is highly polar. There are even cases when two molecules have similar shapes and identical compositions, but different polarities. Compounds such as cis- and trans-dichloroethylene behave this way. Both have the formula C2H2C2 and both have trigonal planar shapes around the two carbon atoms.
Yet the trans compound is nonpolar because the carbon-chlorine bond polarities cancel each other, while the cis compound is highly polar because they reinforce each other. Since greater polarity leads to
stronger attractions among the molecules, the cis compound boils over 18 higher than the trans compound.
Molecular polarity from bond polarities and molecular symmetry Why only H2O has a net dipole moment and SO3 has zero dipole moment? However, both the H-O and S-O bonds in these molecules have a polarity.
Molecular Structure of H2O is bent:
There is polarity alone O - H bonds: Polarity adds like other mechanical forces. If there are equal dipoles opposite (1800) to each other they cancel. In water two dipoles are added up to give water molecule a stronger net dipole. Therefore, water molecule is polar and have a dipole moment.
Molecular structure of SO3: First get the Lewis structure of SO3 i) Count the total valence lectrons: 6 + 3 x 6 = 24 (12 pairs) ii) S is central atom and connect S to terminal Os iii) Give octets to Os.
v) count electron pairs. 4 pairs on S 8 pairs on Os
= 4 = 8 12 electronpairs
Molecular structure of SO3: Trigonal planer There is a dipole along each S - O bond They are at 1200 to each other. Therefore, they cancel out. There is no net dipole moment in SO3 molecule.
Using VSEPR molecular structures predict the polarity (net dipole moment or zero 23
dipole moment) of the following molecules: H2O, NH3, CO2, SO3, SF6, PCl5, XeF4
a) H2O: Lewis Structure of H2O: Polarity adds like other forces. If there are equal dipoles opposite (1800) to each other they cancel. In water two dipoles are added up to give water molecule a stronger net dipole. Therefore, water molecule is polar and have a dipole moment. Two bonds are at about 1050 and the polarities adds up to a stronger dipole. H2O has a dipole moment. b) NH3: Lewis Structure of NH3: Molecular Structure: Pyramidal N - H bond is polar:
N - HS+
Bonds are not distributed symmetrically. Therefore, ammonia has a dipole moment. c) CO2: Lewis Structure of CO2:
Molecular Structure: Linear
C - O bond has a polarity: C - O However, bonds are aligned 1800 to each other canceling each other. There is no dipole moment in the molecule.
d) SO3: Lewis Structure of SO3: Molecular Structure of SO3: trigonal planer dipoles are arranged at 1200 to each other. Dipoles cancel and there is no dipole moment in the molecule.
e) SF6: Lewis Structure of SF6
Molecular Structure of SF6: octahedral S - F has a dipole. However, bonds are arranged symmetrically opposite to each other. There is no dipole moment in the molecule.
Lewis Structure of PCl5:
Molecular Structure: Trigonal bipyramid P - Cl: there is dipole moment in the bond However P - Cl bond are arranged symmetrically around P canceling each other PCl5 molecule do not have a dipole. e) XeF4: Lewis Structure of XeF4:
Molecular Structure of XeF4: Square planer Xe - F bond is polar. However bond are arranged symmetrically opposite to each other, cancelling the dipoles. There is no dipole moment in the molecule.
Noncovalent Interactions and Forces Between Molecules Intermolecular forces: dipole-dipole, hydrogen bond and London dispersion. Intermolecular forces are the attractions that exist among particles or molecules of matter in any of the three distinct phases of matter(gases, liquids and solids) found on earth. The strength of these forces increases as the matter changes phase from a gas to a liquid and then to a solid. Thus, gases have the weakest intermolecular forces compare to liquids and solids and solids have the strongest forces. The intermolecular forces in ionic solids are so strong that they exist only as solids even at very higher temperatures. Ionic solids have ionic interactions; the strong electrostatic attractions between oppositely charged ions form an ionic lattice as in sodium chloride, NaCl(s). Other forms of weaker intermolecular forces exist resulting from polar covalent bonds in molecules and the polarization of electrons on non-polar covalent molecules. These weaker intermolecular forces are classified according to their strength as shown below: i) Dipole-Dipole Intermolecular forces: This includes the attraction between all polar molecules through their dipoles (except F-H, O-H and N-H dipoles). Dipolar covalent bonds are formed by unequal sharing of electrons in bonds in a molecule. Because of symmetry of a molecular structure, canceling the dipoles in the molecule can make it non-polar even though it may have polar covalent bonds. Therefore, there are no dipole-dipole interactions in non-polar molecules. ii) Hydrogen bonding Three special cases of stronger dipole-dipole interactions resulting from F-H, O-H and N-H dipolar covalent bond are called hydrogen bonds. This is because F, O and N have the largest electronegativities among all other elements. Hydrogen bonding (between O-H and O-H dipoles) in water allows water to exist as a liquid at room temperature. N-H dipole in
proteins allows the formation of double helix structures in DNA and other complex structures found in living cells. iii) London Dispersion Forces: These forces are the weakest of all intermolecular attractions and occur in non-polar molecules without dipoles or dipolar covalent bonds. The London dispersion forces result from instantaneous shifts of electron clouds of non-polar molecules. These shifts in electron cloud create instantaneous dipoles with a very short lifespan. A weaker attractive force results because of the short lived dipole attractions between two molecules. Nonpolar molecules such as H2 and N2 can be cooled to liquids at very low temperature due to the existence of London Dispersion forces. As the number of electrons increases or the molecular weight of a substance increases, the London Dispersion forces tend to increase which makes these substances exist as solids. Even though London dispersion forces exist in ionic solids and polar covalent compounds, their effect is masked by stronger ionic and dipole-dipole interactions. Melting points and boiling points of molecular compounds The melting point of a compound is the temperature at which a compound turns from a solid to a liquid or a liquid to a solid. The boiling point of a compound is the temperature at which a compound turns from a liquid to a gas or a gas to a liquid. This temperature is a true measure of the forces of attractions between molecules as molecules separate from one another when they turn from a liquid to a gas. The stronger the attractions between particles (molecules or ions), the more difficult it will be to separate the particles. When substances melt, the particles are still close to one another but the forces of attraction that held the particles rigidly together in the solid state have been sufficiently overcome to allow the particles to move. When substances boil, the particles are completely separated from one another and the attractions between molecules are completely overcome. The energy required to cause substances to melt and to boil, and thus disrupt the forces of attraction, comes from the environment surrounding the material. If you place a piece of ice in your hand, the ice will melt more quickly than if it is placed on a cold counter top. The energy required to melt the ice comes from your hand, your hand gets colder and the ice gets warmer. Nitrogen Monoxide and Methanol Nitrogen monoxide (NO) and methanol (CH3OH) are similar in size and thus have similar London forces. Nitrogen monoxide and methanol are polar covalent molecules and thus have dipole-dipole forces. Since methanol has the higher melting point and boiling point, it must have the stronger intermolecular forces. The difference in these molecules is the presence of a certain extremely polar bond present in methanol that is not present in nitrogen monoxide. This is the oxygen - hydrogen bond. Oxygen is more electronegative than hydrogen and pulls the electron density in the oxygen hydrogen bond towards it. This leaves very little electron density around the hydrogen since hydrogen has no core electrons. The part of hydrogen directed away from the oxygen - hydrogen bond has very little electron density shielding the nucleus. Thus that part of the hydrogen nucleus which is exposed can interact with the non-bonding electrons on another methanol
molecule. This interaction of a non-bonding pair with a hydrogen attached to an electronegative element such as oxygen is called a hydrogen bond. For each of the following pairs of compounds, circle the compound that you would predict to have the higher boiling point. State a reason for your choice. a. CH4 b. PH3 c. HCl d. NaBr or or or or Ne AsH3 HF ClBr CH4 is more polarizable more elarger London forces
Hydrogen bonding dominates when present NaBr is ionic and ClBr is molecular
Biomolecules: DNA and the Importance of Molecular Structure Hydrogen-bonding pairs found in DNA. Note that the A-T pair has only two H-bonds because of the adenine molecule. (The lone pair of electrons on the oxygen atom in thymine molecule could act to form a third H-bond, but adenine does not have the required donor). The arrangement of donors and acceptors in thymine and guanine precludes formation of H-bonding pairs between these two compounds.
Major components in the structure of DNA Components of DNA DNA is a polymer. The monomer units of DNA are nucleotides, and the polymer is known as a "polynucleotide." Each nucleotide consists of a 5carbon sugar (deoxyribose), a nitrogen containing 27
base attached to the sugar, and a phosphate group. There are four different types of nucleotides found in DNA, differing only in the nitrogenous base. The four nucleotides are given one letter abbreviations as shorthand for the four bases. A is for adenine G is for guanine C is for cytosine T is for thymine
Pyrimidine Bases Cytosine and thymine are pyrimidines. The 6 stoms (4 carbon, 2 nitrogen) are numbered 1-6. Like purines, all pyrimidine ring atoms lie in the same plane. Purine Bases Adenine and guanine are purines. Purines are the larger of the two types of bases found in DNA. Structures are shown below: Ribose and Deoxyribose Sugars The deoxyribose sugar of the DNA backbone has 5 carbons and 4 oxygens. The carbon atoms are numbered 1', 2', 3', 4', and 5' to distinguish from the numbering of the atoms of the purine and pyrmidine rings. The hydroxyl groups on the 5'- and 3'- carbons link to the phosphate groups to form the DNA backbone. Deoxyribose lacks an hydroxyl group at the 2'-position when compared
Nucleosides A nucleoside is one of the four DNA bases covalently attached to the C1' position of a sugar. The sugar in deoxynucleosides is 2'-deoxyribose. The sugar in ribonucleosides is ribose. Nucleosides differ from nucleotides in that they lack phosphate groups. The four different nucleosides of DNA are deoxyadenosine (dA), deoxyguanosine (dG), deoxycytosine (dC), and (deoxy)thymidine (dT, or T) to ribose, the sugar component of RNA.
Nucleotides A nucleotide is a nucleoside with one or more phosphate groups covalently attached to the 3'and/or 5'-hydroxyl group(s).
DNA Backbone The DNA backbone is a polymer with an alternating sugar-phosphate sequence. The deoxyribose sugars are joined at both the 3'-hydroxyl and 5'-hydroxyl groups to phosphate groups in ester links, also known as "phosphodiester" bonds.
Double Helix Chiral Molecules Chirality: The word 'Chiral' comes from the ancient Greek 'Cheir' which means 'Hand'. The definition of Chirality is something which has a mirror image, but cannot be superimposed. Enantiomers: These two mirror images are known as 'Enantiomers'. They are, unsurprisingly, left and right 'Handed'. A mixture of these enantiomers in equal proportions is known as 'Racemic', but of course they can be separated. Examples of Chiral objects in everyday life are: Hands, shoes, ears. Examples of Achiral objects are things like: Balls, forks,hammers. If polarized light is passed through one enantiomer of a chiral substance, the plane of the light emerging from the substance is rotated from the original. A substance capable of causing this rotation is optically active. The degree to which the light is rotated by the enantiomer is determined by finding the angle another polarizer must be turned to in order to produce a dark field. (i.e. the amount it must be rotated from its original position perpendicular to the first polarizer.)
If polarized light is passed through one enantiomer of a chiral substance, the plane of the light emerging from the substance is rotated from the original. A substance capable of causing this rotation is optically active. The degree to which the light is rotated by the enantiomer is determined by finding the angle another polarizer must be turned to in order to produce a dark field. (i.e. the amount it must be rotated from its original position perpendicular to the first polarizer.) If one enantiomer rotates light to a certain degree one way, the other enantiomer will rotate it to the same degree but in the opposite direction. The name of a chiral compound will sometimes include a + or - to indicate the sign of its optical rotation. + indicates a clockwise rotation (in which case the molecule is dextrorotatory (D)), - indicates a counterclockwise rotation. (levorotatory(L)) The D-sugar and L-sugar are enantiomers. Enantiomers are mirror images of each other.
CHEM 101 HOMEWORK 4
HOMEWORK FOR CHAPTER 9
1. According to VSEPR theory, what is the molecular geometry of the PF5 molecule? a. Square pyramidal b. octahedral c. tetrahedral d. trigonal bipyramidal e. see-saw 2. According to VSEPR theory, what is the molecular geometry of the XeO3 molecule? a. trigonal planar c. tetrahedral e. see-saw b. T-shaped d. trigonal pyramidal 9. Nitrogen atoms in the nitrogen molecule (N2: lewis structure |NN|) has the following hybridization and geomerty: a. sp, linear. b. sp2, trigonal planar. c. sp3, tetrahedarl. d. sp3d, trigonal bipyramid. e. sp3d2, octahedral. 10. Hybrid orbitals resulting from linear combinations of one s and three p atomic orbitals would have the geometry of: a. 109.5o b. 180o c. 120o d. 90o e. 45o 11. Choose the correct hybridization for the central atom in SF2. a. sp b. sp2 c. sp3 d. sp3d e. sp3d2 12. The carbon-carbon bond in ethylene, CH2=CH2, results from the overlap of which of the following? a. sp hybrid orbitals b. sp2 hybrid orbitals c. sp3 hybrid orbitals d. s atomic orbitals e. p atomic orbitals 13. The number of "sigma" (s) bonds in the ammonia, NH3, molecule is; a. 0 b. 1 c. 2 d. 3 e. 4 14. How many carbon atoms in the structure shown below are chiral? a. 6 d. 3 b. 5 e. 2 *c. 4
3. Which molecular geometry is matched to CORRECT bond angles? a. octahedral - 90o and 180o b. trigonal planar - 90o and 120o c. tetrahedral - 120o d. trigonal bipyramidal- 109.5o e. linear - 120o 4. For which of the following molecules will the electron-pair geometry be different than the molecular geometry? a. CCl4 b. BH3 c. PCl3 d. Cl2 e. CO 5. Which of the following is a non-polar molecule? a. CCl4 b. NH3 c. H2S d. NF3 e. OF2
6. In the substance with this formula,
the number of s and p bonds is a. 3 and 19 bonds. b. 19 and 3 bonds. c. 19 and 19 bonds. d. 9 and 3 bonds. e. 16 and 3 bonds. 7. Which of the listed molecules possesses a square planar geometry? a. SiCl4 b. SF4 c. XeF4 d. CCl4 e. CH4 8. Which of the following molecules is no polar? a. NH3 b. H2S c. SF6 d.HCl e. CO
15. What is the major type of force that must be overcome to allow the processes below? I. the evaporation of propanol (CH3CH2CH2OH) II. the melting of solid Br2 III. the boiling of liquid BF3 IV. the boiling of liquid CH2Cl2 a. London forces, covalent bonding, dipole- dipole, dipole-dipole b. London forces, London forces, London forces, London forces c. hydrogen bonding, dipole-dipole, dipole-dipole, London forces d. hydrogen bonding, London forces, London forces, dipole-dipole e. dipole-dipole, dipole-dipole, London forces, London forces
CHEM 101 Sample Chapter 9, Test 2
SAMPLE TEST FOR CHAPTERS 9
Which of the following is matched with the CORRECT molecular geometry? a. H3O+1 - trigonal pyramidal b. CF2Cl2 - trigonal bipyramidal c. I3-1 - bent d. SO2 - linear e. HCl - octahedral 2. Which of the following statements concerning IO4-1 is FALSE? a. The Lewis structure shows iodine as the central atom with 4 oxygen atoms bonded to the iodine with single bonds. b. The Lewis structure has a total of 12 unshared pairs of electrons. c. The Lewis structure does not have unshared pairs on the central atom. d. The iodine is sp3 hybridized. e. The molecular shape is square planar. 3. In TeBr4 molecular geometry is a. linear. b. square planar. c. seesaw. d. tetrahedral. e. bent. 4. Which of the following is a non-polar molecule? a. I2 b. NaI c. IF d. CH2I2 e. NI3 5. All of the following molecules have polar bonds and are polar molecules except a. ICl3. b. BCl3. c. PCl3. d. H2CCl2. e. HCl. 6. Which response contains all the characteristics listed that should apply to phosphine, PH3? 1. trigonal planar 2. one unshared pair of electrons on P 3. sp2-hybridized phosphorus atom 4. polar molecule 5. polar bonds a. 1, 4, and 5 b. 2, 3, and 4 c. 1, 2, and 4 d. 2, 4, and 5 e. 1, 3, and 5 7. What is the total coordination number of the central atom in XeOF2? a. 2 b. 3 c. 4 d. 5 e. 6 8. When a carbon atom in a molecule has sp3 hybridization, it has a. four p bonds. b. three bonds and one bond. c. two bonds and two bonds. d. one bond and three bonds. e. four bonds. 9. The central atom in the iodite ion, IO2-1, is surrounded by a. two bonding pairs and two unshared pairs 1. of electrons. b. three bonding pairs and one unshared pair of electrons. c. one bonding pair and three unshared pairs of electrons. d. two double bonds and no unshared pairs of electrons. e. four bonding pairs and four lone pairs of electrons. 10. The total number of lone pairs in the Lewis structure for OF2 is a. 2. b. 4. c. 8. d. 9. e. 10. 11. Which one of the following has more than one equivalently good Lewis structure (resonance forms)? a. BrF5 b. CF4 c. NH3 d. CS2 e. IBr 12. Which bond would be expected to have the positive end of the bond dipole on the nitrogen atom? a. N-H b. N-N c. C-N d. Al-N e. N-O 13. The bonding in acetylene, C2H2, is best described as a. five s bonds. b. two s bonds and three p bonds. c. five p bonds. d. four s bonds and one p bond. e. three s bonds and two p bonds. 14. The carbon-carbon-carbon bond angle in H2CCCH2 is a. 90. b. 109.5. c. 120. d. 150 e. 180 15. The statement(s) which describe the bonding in the water molecule is(are) 1. nonpolar covalent. 2. polar covalent. 3. sigma bond. 4. sp3 hybridization. a. 1 only b. 2 only c. 1 and 4 only d. 2 and 3 only e. 2, 3, and 4 only 16. The H-C-C angle in acetylene, C2H2, is a. 180 b. 120 c. 90 d. 109.5 e. 90 and 180 17. The number of lone pairs of electrons about the central Cl atom in the chlorate ion, ClO3-1, is a. 0. b. 1. c. 2. d. 3. e. 4. 18. The number of electron pairs around the central nitrogen atom in the NH3 molecule is, a. 0 b. 1 c. 2 d. 3 e. 4 19. The molecular geometry of NH3 is, a. triangular pyramidal b. see-saw c. tetrahedral
d. square planar e. octahedral 20. Which of the following molecules is no polar? a. NH3 b. H2S c. SF6 d. HCl e. CO 21. Which of the listed ions and molecules possesses a square planar molecular geometry? a. SiCl4 b. SF4 c. XeF4d. CCl4 e. CH4 22. In the triple bond between two carbon atoms in acetylene (H-CC-H), a. three electrons are shared by the carbon atoms. b. three octets of electrons are shared by the atoms. c. triply charged ions, one +3 oc one carbon and the other -3 on other carbon are formed. d. two pairs of electrons are shared by the atoms. e. one (sigma) bond and a two (pi) bonds are formed between the carbon atoms. 23. In the substance with this formula,
the number of s and p bonds are a. 13 and 3 bonds. b. 13 and 4 bonds. c. 12 and 5 bonds. d. 5 and 12 bonds. e. 4 and 13 bonds. 24. In the substance with this formula,
the number of s and p bonds are a. 3 and 19 bonds. b. 19 and 3 bonds. c. 19 and 19 bonds. d. 9 and 3 bonds. e. 21 and 2 bonds. 25. The strongest intermolecular force that must be overcome in converting CH3CH2OH (ethanol) from a liquid to a gas is a. London Dispersion b. Metallic bonding c. Diople-dipole d. Hydrogen bonding e. Nework covalent bonding 26. The number of "sigma" (s) bonds in the methane, CH4, molecule is; a. 0 b. 1 c. 2 d. 3 e. 4
27. Which of the following molecules is polar? a. CH4 b. CO c. CF4 e. CO2 28. Given the substances: HF, H2O, and NH3. Arrange these compounds in order of increasing strength of intermolecular forces (lowest to greatest). a. HF<H2O<NH3 b. H2O<NH3<HF c. NH3<H2O<HF d. NH3<HF<H2O 29. The hybrid orbitals used by sulfur in SF6 are: a. sp b. sp2 c. sp3 d. sp3d e. sp3d2 30. For the molecule SO2 , the hybridization of sulfur's atomic orbitals and the molecular geometry can be described as, a. sp, linear (180) b. sp2, bent (120) c. sp3 , linear (180) d. sp3, bent (109.5) e. sp2d, bent (90) 31. Both Iodine, I2 , and bromine, Br2 are non polar moleculaes. But I2 has a higher boiling point Br2 because, a. of its weaker diploe forces. b. it has fewer electrons c. it has stronger dipole forces d. it has more electrons creating stronger dispersion forces 32. Which of the following central atoms would be expected to accommodate an octahedral or bipyramidal geometry? a. oxygen b. neon c. aluminum d. selenium e. silicon 33. Using VSEPR Theory, predict the X-A-X angles for the following compounds. CO2, GeF4, BCl3 a. 180, 109.5, 120 b. 180, 90, 120 c. 109.5, 109.5, 109.5 d. 109.5, 90, 90 e. 109.5, 90, 109.5 34. For which of the following molecules will the electron-pair geometry be the same as the molecular geometry? a NH3 b. TeBr4 c. O3 d. NO - e. SnCl4 2 35. Which of the following compounds does not have tetrahedral electron-pair geometry? a. H3O+ b.PCl3 c. H2O d.CCl4 e.NO236. Which of the following has a trigonal planar molecular geometry? I. nitrate ion
sulfite ion carbonate ion selenite ion a. I and II b. I and III c. II and III d. II and IV e. III and IV 37. Which of the following correctly list the relative strengths of electron-pair repulsions? a. lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair b. lone pair-lone pair > bonding pair-bonding pair > lone pair-bonding pair c. lone pair-bonding pair > lone pair-lone pair > bonding pair-bonding pair d. bonding pair-bonding pair > lone pair-lone pair > lone pair-bonding pair e. bonding pair-bonding pair > lone pair-bonding pair > lone pair-lone pair 38. Which of the following has the smallest bond angle? 2a. NO3- b. AsCl3 c. SiF4 d. SO 3 e. H2S 39. Label the hybridization at C#1, C#3, and C#4 in the molecule.
II. III. IV.
H2Te, H2S, H2O, HBr, HCl, HF, SiH4, CH4, HI, NH3, PH3, AsH3 a. H2S, H2O, HCl, HF b. AsH3, NH3, HF, H2S c. HBr, HCl, HF, H2O d. H2O, HF, NH3 e. CH4, H2O, HF, NH3 44. DNA is composed of many repeating units called nucleotides. Which of the following is not part of a nucleotide found in DNA? a. a nitrogen base b. deoxyribose c. an amino acid d. a sugar unit e. a phosphate unit 45. Which base pairs with thymine in DNA? a. adenine b. guanine c. cytosine d. a or c e. a or b 46. Which of the following is associated with "chiral" molecules? a. molecules that are superimposable on their mirror images b. ethylene c. bromochlorofluoromethane (BClFC) d. methane e. ethane 47. How many atoms in the structure shown below are chiral?
C#1 C#3 C#4 sp3 Option 1: sp sp2 3 sp3d Option 2: sp sp 2 3 Option 3: sp sp sp3 3 3 sp sp3 Option 4: sp 2 Option 5: sp sp sp2 a. Option 1 b. Option 2 c. Option 3 d. Option 4 e. Option 5 40. In what type of molecules do London forces exist? a. molecules with nonpolar bonds b. molecules with polar bonds c. molecules with metallic bonds d. molecules with hydrogen bonding e. all of the above 41. Which molecule below has London forces as its only intermolecular force? a. GeBr4 b. H2O c. HF d. PCl3 e. CH2Cl2 42. Which of the following has dipole-dipole interactions between its molecules? a. CO2 b. HF c. F2 d. CBr4 e.BH3 43. Of the molecules below, which ones undergo extensive hydrogen bonding?
1. > a. H3O+1 - trigonal pyramidal 2. > e. The molecular shape is square planar. 3. > c . seesaw. 4. > a I2 5. > b BCl3. 6. > d 2, 4, and 5 7. > d 5 8. > e 9. > a 10. > c 11. > d 12. > e 13. > e 14. > e
Answers to Questions
15. > e 16. > a 17. > b 18. > e 4 19. > a triangular pyramidal 20. > c SF6 21. > c XeF4 22. > e one (sigma) bond and a two (pi) bonds are formed between the carbon atoms. 23. > e 21 and 2 bonds. 24. > e. 21 and 2 bonds. 25. > d Hydrogen bonding 26. > e 4 27. > a CO 28. > c NH3<H2O<HF 29. > e sp3d2 30. > b sp2, bent (120) 31. > d it has more electrons creating stronger
dispersion forces 32. > d. selenium 33. > a. 180, 109.5, 120 34. > e. SnCl4 35. > e.NO236. > b. I and III 37. > a. lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair 38. > e. H2S 39. a. Option 1 40. > e. all of the above 41. > a. GeBr4 42. > b. HF 43. > d. H2O, HF, NH3 44. > c. an amino acid 45. > a. adenine 46. > c. bromochlorofluoromethane (BClFC) 47. > c. 4
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