Section Activity #1: Fill out the following table for biology s most common elements assuming that each atom is neutrally charged.

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1 LS1a Fall 2014 Section Week #1 I. Valence Electrons and Bonding The number of valence (outer shell) electrons in an atom determines how many bonds it can form. Knowing the number of valence electrons present in hydrogen, carbon, nitrogen, oxygen, phosphorus, and sulfur in their neutral states allows us to predict the number of bonds each atom can make. Lone pair electrons are those that are not involved in a covalent bond (and therefore not shared) when the atom is covalently bonded to other atom(s). Section Activity #1: Fill out the following table for biology s most common elements assuming that each atom is neutrally charged. Lewis Dot Structure of Free Atom Hydrogen Carbon Nitrogen Oxygen Phosphorus* Sulfur* # of Valence Electrons # of Total Electrons # of Lone Pairs on Neutral Atom When Octet is Satisfied # of Bonds Neutral Atom Forms or 5 2, 4, or 6 * Phosphorus and sulfur each have an additional electron shell that carbon, nitrogen, and oxygen lack, and this additional electron shell allows phosphorus and sulfur to undergo octet expansion. II. Lewis dot structures and Formal charge Atoms acquire formal negative charges when they have additional valence electrons than are needed to be neutral; atoms acquire a formal positive charge when they have fewer valence electrons than are needed to be neutral. A quick way to assess formal charges is to compare the number of valence electrons present in the neutral, unbound atom to the number of valence electrons that it is assigned when bound to other atoms. An atom is formally assigned all of its lone pairs and half of the electrons in a covalent bond (because these electrons are being shared with the other atom to which it is bonded). Formal Charge = (# of valence electrons) (# of lone pairs electrons) ½ (# of electrons in bonds) Formal charge assumes that electrons in a covalent bond are shared equally between atoms, discounting the effect of the atoms relative electronegativities (discussed in the next section). 1

2 Section Activity #2: Draw Lewis dot structures for the following molecules including the lone pairs of electrons. Using your knowledge of how many valence electrons each atom has in its neutral state, assign appropriate formal charges to any relevant atoms within each molecule. For HNO 3, the atoms could be connected in other ways without violating the octet for each individual atom. All valid structures would be considered acceptable (even though nitric acid only has one molecular structure). Section Activity #3: a. Label any formal charges present in the following molecule. Assume that the molecule drawn below is a valid Lewis structure such that every atom has a filled octet and every carbon is making four bonds (i.e., not all of the hydrogen atoms are shown). b. Compare the structures of H-C N and H-N C by first drawing the Lewis diagrams of each structure. Label the formal charges on each structure. Which of the two Lewis structures that you drew would be the most stable and why? [Hint: Consider what it means to be relatively stable?] H-C N is more stable because in this bonding pattern all of the atoms are electrically neutral, whereas the H-N C bonding pattern produces a positive formal charge (+1) on the nitrogen and a negative formal charge (-1) on the carbon. Both molecules are overall electrically neutral, but since each atom of the H-C N structure is electrically neutral, it is lower in energy than the H-N C structure with the charged atoms. One reason H-N C is less stable is because this structure deprives nitrogen of an electron (making it positive) and gives it to carbon (making it negative). However, nitrogen is more electronegative than carbon, so nitrogen attracts electrons more strongly than carbon does. Depriving nitrogen of an electron so it can be provided to the less electronegative carbon is therefore unfavorable and makes the molecule less stable. 2

3 III. Electronegativity, Polar Bonds and Partial Charges Define electronegativity : The tendency of an atom to attract electrons. When there is a significant electronegativity difference ( ΔEN ) between two atoms in a bond, electrons are shared unevenly, causing there to be a separation of charge along the bond. This results in a polar covalent bond. Commonly observed polar covalent bonds in the molecules of life are N H, C O, P O and O H bonds. Relatively non-polar bonds generally have ΔEN values of 0.4 or less. Electronegativity values are located in the lecture notes. Polar bonds generate electrical dipoles. In contrast to monopoles, which features only polarity (+ or -), a dipole contains both positive and negative charge. Shown to the right is a hydroxyl group labeled with partial positive (δ + ) and partial negative (δ - ) charges. Section Activity #4: In the molecules shown below, determine which individual bonds are polar and label partial charges on all relevant atoms. IV. Hydrogen Bonding Hydrogen bonds are a special type of dipole-dipole interaction in which a hydrogen atom that is bonded to an electronegative atom (O, N, or F) is electrostatically attracted to the lone pair of another electronegative O, N, or F atom. Properties of Hydrogen Bonds: i. The donor must be a hydrogen atom with a partial positive charge. ii. The acceptor must have at least one lone pair and a partial negative charge. iii. Hydrogen bonds are (strongest/weakest) when they are linear. iv. The bond strength (increases/decreases) as bond distance increases. 3

4 Section Activity #5: a. Ammonia (NH 3 ) can hydrogen bond with ethanol (CH 3 CH 2 OH). Draw two ways in which this interaction can occur in the space below. For each hydrogen bond, circle the hydrogen bond donor and draw a square around the hydrogen bond acceptor. Be sure to use optimal bond angles in your drawing. Donors are circled in red, and acceptors are boxed in blue. The entire O-H group or the N-H group should be circled as the donors, and the electronegative atom (ether O or N) should be labeled as the acceptor. Notice that in this problem, both ethanol and ammonia can act as either H-bond acceptors or H-bond donors; therefore, the interaction can be drawn in two different ways. b. Could the same hydrogen bonds from part (a) occur if ammonium (NH 4 + ) were used instead of ammonia? If not, why? There can still be a hydrogen bond between a lone pair of electrons on the oxygen atom in ethanol and a properly oriented hydrogen atom on NH 4 +, similar to what occurs in NH 3. However, the ammonium ion no longer has a lone pair on the nitrogen atom and thus cannot act as a hydrogen bond acceptor. V. Intermolecular Interactions Molecules do not exist in isolation; they are involved in a range of intermolecular interactions with themselves and other molecules. All of the intermolecular interactions are electrostatic in nature, and the strength of these intermolecular interactions is typically proportional to the magnitudes of the charges of the molecules. [Note: hydrogen bonds are stronger than dipole-dipole interactions because they feature a vector/directional component.] We generally categorize the electrical charge of molecules and atoms as full charges, partial charges, or instantaneous charges. The interactions that we describe as intermolecular interactions (e.g., ionic bonds, dipole-dipole interactions, hydrogen bonds, van der Waals interactions, etc.) can also occur between distant parts of a single molecule. We will see this when start learning about protein structure. 4

5 Section Activity #6: Shown below are a series of intermolecular interactions. a. Identify the strongest type of interaction that is possible for each pair. Assume that all molecules lie in the same plane. Dipole-Induced Dipole Induced Dipole-Induced Dipole Ion-Dipole Ionic Hydrogen Bond (dipole-dipole) Dipole-Dipole b. Which pairs of molecules could form a hydrogen bond? Pairs D and E could form hydrogen bonds as long as the donors and acceptors are favorably oriented with respect to each other. c. The chart shown below represents a continuum of increasing interaction strength. Fill in the blanks on the chart with the letters from the examples above such that each interaction is ordered from weakest to strongest. 5

6 Section Activity #7: Challenge Question Shown below are two molecules: hexane and a related molecule, perfluorohexane, in which all of the hydrogen atoms are replaced with fluorine. Liquid perfluorohexane is known for its ability to dissolve relatively large volumes of gas (e.g., N 2, O 2, etc.). This property is due to the weak intermolecular interactions that form between molecules of perfluorohexane; these weak interactions result in open spaces between perfluorohexane molecules that can accommodate gas molecules. a. Label partial charges on the molecules shown above. a. Given the partial charges you drew in part (a), briefly explain why perfluorohexane does not form attractive dipole-dipole interactions with other perfluorohexane molecules in liquid phase. Perfluorohexane contains numerous polar C-F bonds, and as a result, the carbons in perfluorohexane are partially positively charged and the fluorine atoms are partially negatively charged. Although positively and negatively charged atoms are present, there are no significant dipole-dipole interactions between perfluorohexane molecules because only the negatively charged fluorine atoms are exposed to the surface of the molecule; the positively charged tetrahedral carbons are surrounded on all sides by negatively charged fluorine atoms, concealing the carbon atoms in the interior of the molecule. As a result, each molecule of perfluorohexane is only capable of interacting via its negatively charged fluorine atoms, which aren t attracted to one another due to their opposing charges. c. Between which pair of molecules, hexane-hexane or perfluorohexane-perfluorohexane, do stronger van der Waals interactions occur? Provide an atomic-level explanation to support your answer. The van der Waals interactions that form between hexane molecules are stronger than the ones that form between perfluorohexane molecules. The perfluorohexane interaction is weaker because the C-F bonds are quite polar because of the large electronegativity difference between carbon and fluorine. Because of this electronegativity difference, instantaneous changes in electron sharing are unlikely to occur and are especially unlikely to be sufficient in magnitude to reverse the polarity of the C-F bond, which would be necessary for significant van der Waals interactions to occur. 6

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