1. Lecture 9: Covalent Bonding: Part 1 Lewis Structures (Ch. 7)

2. Recap Ionic compounds are formed by the transfer of electrons from a metal to a nonmetal The result is a cation-anion pair. Each ion has reached its nearest noble gas configuration Na ([Ne] 3s 1 ) + Cl ([Ne] 3s 2 3p 5 ) ---> Na + Cl - [Ne] [Ar] The Coulombic attraction between the oppositely charged ions is what holds the molecule together. This is why ionic compounds are solids at room temperature. Cl - Na +

3. Covalent Bonding Covalent bonding is electron sharing (not transfer) between nonmetals Consider Cl 2 (g) Each Cl has 7 valence electrons Chlorine is 1 electron short of a full octet ([Ar] configuration) Since both atoms are the same, they have the same electron affinities and ionization energies, so 1 chlorine will not donate an electron to the other. Cl [Ne]3s 2 3p 5

4. Covalent Bonding For both atoms to achieve an [Ar] configuration (full octet), they share a pair of electrons between them. Elements in a covalent bond will react so that at least eight electrons occupy the valence shell. This is the OCTET RULE. To indicate the covalent bond, we use a solid line. This is a single bond (2 electrons). Bonding represents an overlap of the valence orbitals. The electrons not involved in bonding are called lone pairs. Covalent bond Cl [Ar] Cl

5. Notes About the Octet Rule Nonmetals with valence shells having a principle quantum number (n) of 2 must obey the octet rule. They will bond to obtain exactly 8 electrons in their valence shell. NO MORE, NO LESS. (Boron is a metalloid and is exempt). n=2, L = 0, 1 (ex. C, N, O, F) These elements possess both valence s and valence p orbitals s orbitals hold a max of 2 electrons, p orbitals can hold 6 (8 total) Elements beyond the 2 nd row may violate the octet rule if need be. This will be discussed later.

6. Covalent Bonding Ex. OF 2 Oxygen has 6 valence electrons (needs 2) Fluorine has 7 (each F needs 1) To achieve octets, each F will share its electron with O F F O F O F F O F [Ne] [He]2s 2 2p 5 [He]2s 2 2p 4

7. Electronegativity Before we learn about how to draw a Lewis structure, it is very important to consider what makes ionic and covalent bonds so different: electronegativity Electronegativity is the ability of an atom to attract electrons to itself. A more electronegative atom is in greater need of electrons, and will attract them more strongly than a less electronegative atom (think “tug-of-war”) Thus, when there is a difference in electronegativities between atoms in a molecule, the electrons are NOT equally shared.

8. Electronegativity of atoms increases up and to the right. Table of Electronegativities

9. Although we have discussed ionic and covalent bonds, most chemical bonds are neither purely ionic or purely covalent Most compounds are an intermediate between the two. Electronegativity Bond TypeCovalent Polar Covalent (Partially polar) Ionic (Totally Polar) Difference In Electronegativity < 0.40.4 – 2.0> 2.0

10. Electronegativity Let’s consider NaCl. The difference in electronegativity between Na and Cl is: Because the difference in electronegativity is so big, the Na electron is completely pulled away by the Cl atom. So, the molecule is totally polar (ionic) This is why there is no actual bond in ionic compounds, only coulombic attraction. 3.16 – 0.93 = 2.23 ionic Cl - Na +

11. Partial negative character Partial positive character Electronegativity Let’s look at another molecule, like HCl: The HCl molecule is not purely ionic or covalent, but BOTH. The electron density is unevenly distributed, such that more of the electron cloud is on the Cl than on the H. Therefore, the H and Cl have partial charges. The arrow depicts the direction of electron “pull”, or dipole. Any molecule with a net dipole is polar. We will discuss dipoles later. H Cl δ - δ + 3.16 – 2.1 = 1.06 polar covalent

12. General Rules of Drawing Covalent Lewis Structures 1. Arrange atoms together. Put the least electronegative atom in the center. Hydrogen is an exception to this rule. Hydrogen atoms are always terminal (at the ends of molecules), as hydrogen only possesses a 1s orbital and can therefore accommodate a maximum of two electrons (one bond) 2. Compute the total number of valence electrons. Account for charges. 3. Represent bonds with solid lines. Ensure octets around all atoms, except hydrogen. Hydrogens can only accommodate 2 electrons. 4. Add in remaining lone pairs 5. If there are not enough electrons to complete an octet, consider multiple bonds or formal charges.

13. Other Rules 1. Fluorine is always a terminal atom. 2. Terminal halogens will only make single bonds. 3. Carbon atoms should always have four bonds around them (no lone pair). A few exceptions exist, like CO, CN-, and carbon ions (you will learn about these in organic chemistry) 4. Never draw a Lewis structure possessing an unpaired electron, unless told explicitly that the molecule is a radical!

14. Example Draw the Lewis structure of CCl 4 Carbon is the least electronegative atom (also, there is only one C), so C is the central atom Compute the valence electrons C: [He] 2s 2 2p 2 Cl: [Ne] 3s 2 3p 5 C Cl Each Cl needs 1, the C needs 4. Therefore, the C will share each of its 4 valence electrons with Cl Cl C C

15. Hydrogen Hydrogen CAN ONLY ACCOMMODATE 2 ELECTRONS Ex. H 2 (g) HX where X = F, Cl, Br, I (halogens) In chemical structures, hydrogens are always terminal atoms, meaning that they are at the ends of a molecule H H X H H H X H

16. Organic Molecules (Carbon-based) Ex. Methanol, CH 3 OH O C H H H H Hydrogens are terminal, which means there is a C—O bond in the center C O Now, C needs 3 more electrons around it to achieve an octet. O needs 1. H H H H C O H H H H C O

17. Double and Triple Bonds There are instances where single bonds (2 e - bonds) are not enough to satisfy the octet rule. Ex. C 2 H 4 (ethene) C H H H H C As you can see, each C only possesses 7 total electrons, including a lone electron. Never draw a Lewis structure with unpaired electrons. Lone electrons migrate into the C-C bond to form a 4e - double bond. CC H H H H C C H H H H C C H H HH C C H H HH

18. 6e - triple bonds are also possible, as in the case of N 2 (g) NNN Two unpaired electrons each N NN 6 electron Triple Bond The N atoms are sharing 6 electrons, so each N now has a full octet. Double and Triple Bonds N N

19. Group Examples Draw the following Lewis Structures. Include all lone pairs. CH 3 Cl NH 3 H 2 O CO 2 C 2 H 2 H 2 CO NOF

20. Formal Charges So far, all of the atoms we’ve seen in covalent compounds have had a zero charge. However, nonmetals can assume positive or negative charges in order to facilitate the formation of a covalent bond Thus, to satisfy the octet rule in certain cases, formal charges have to be applied. This is most commonly observed in polyatomic ions.

21. Formal Charges Consider ammonium, NH 4 + -Lets start by looking at the valence electrons of the neutral atoms N HH HH - As is, N can only make 3 bonds. N H H H H X - N loses an electron to accommodate the 4 th H, attains a charge of 1 +. N H H H H + N H H H H + N +

22. Formal Charge Allows Us to Rule Out Structures Formal charges can help us determine if we’ve arranged a molecule incorrectly The preferred arrangement of atoms in a molecule is always the arrangement that requires the least amount of formal charge If formal charges must exist, the more electronegative elements prefer negative formal charges, and less electronegative elements prefer positive ones.

23. Ex. Hydrogen cyanide, HCN We know that C is the central atom in HCN. But how can we rule out HNC? Lets consider both possibilities Ambiguity in Atomic Arrangement As we determined earlier, we have a CN triple bond and an HC bond. All octets are satisfied. There are no formal charges required C N H 8 total 8 total 2 total

24. If we rearrange N and C, we have the following structure. As shown, C is 1 electron short of an octet. N can not make anymore bonds because it can not exceed an octet. N has a 1 + formal charge because it must lose a valence electron. N C H C must take on a 1 - formal charge to reach an octet by accepting the N electron. Now, all the atoms are satisfied, and the molecule is neutral. 7 total 8 total 2 total + - N C H + 8 valence 8 total 2 valence Ambiguity in Atomic Arrangement

25. Now, our two possibilities are: Both structures satisfy the octet rule, and both are charge neutral, but the HNC structure has formal charges on both N and C, while the HCN structure has zero formal charges. Thus, the HCN structure is preferred. - N C H + 8 total 8 total 2 total C N H 8 total 8 total 2 total Ambiguity in Atomic Arrangement

26. Some Hints For Drawing Polyatomic Ions For anionic molecules containing oxygens, the negative charge will most likely be on the oxygens. A negatively charged O will only require a single bond. Positive charges are most likely found on the central atom, which is the least electronegative (not including hydrogen) Remember, terminal halogens only require one electron. They will not make multiple bonds or take on formal charges.

27. Group Examples Ex. Nitrite (NO 2 - ) Draw the Lewis structures of carbonate The correct structure of the NOCl is a central N with a double bond to O, a single bond to Cl, and a lone pair on the N. Use formal charges to show why O is not the central atom. Single bonded oxygen requires an additional electron (blue) to reach octet. Now has 7 valence electrons instead of 6, giving it a 1 - charge. 6 valence 8 total 0 charge 5 valence 8 total 0 charge 7 valence 8 total charge = 1-

28. Resonance When a covalent molecule has an overall charge, the charge is said to be delocalized, meaning that the charge is spread over the whole molecule, as opposed to being localized on a single atom (wave-particle duality) Ex. Nitrite NO 2 You can’t distinguish between one O and the other. So, you could write either of the following: N O O N O O - -

29. Resonance These structure are called resonance structures. We can account for both structures by writing the resonance form, as shown below. All resonance structures possess a multiple bond. ONOONO - Since the double bond exists simultaneously at both locations, we can imagine that there are 1 ½ bonds at each site. Therefore, we would say that the BOND ORDER (average number of bonds per site) is 1 ½.

30. Expansion into the d-orbital Nonmetals with available d-orbitals (n> 3) can use these orbitals to hold more than 8 electrons. Unless Carbon is present, assume that these atoms are the central atoms. When drawing Lewis structures of molecules with expanded octets, put octets around the outer atoms first, then account for any remaining electrons on the central atom. Ex. SF 4 S F F F F 10 electrons around Sulfur After completing octets on the F atoms, there is a lone pair remaining on S

31. Group Examples Draw the Lewis structures for the following “expanded octet” molecules PCl 5 SF 6 IF 5 Always be mindful when dealing with atoms of n>3 of the possibility of surpassing 8 valence electrons. Draw the following polyatomic ions and report the bond orders: sulfite phosphate perchlorate

32. Revisiting Dipoles: Polar vs. Nonpolar Molecules Dipole moments describe the coulombic attraction that is formed between the partial positive and negative charges on nonmetals when electrons are unevenly shared. Dipole moments are vectors, quantities with both magnitude and direction. Imagine pulling on a rope. The force is in the direction of the pull. If two people pull on opposing ends of a rope with equal force, the net force is zero. Polar molecules have an overall dipole moment, nonpolar molecules don’t.

33. Electronegativity of atoms increases up and to the right. Table of Electronegativities

34. COO δ + Equal force of attraction in opposite directions: Non polar Here, electronegativity is labeled as β Δβ = 0.89 COS δ + Δβ = 0.03 Unequal dipole moments. Thus, there is an overal dipole: Polar + = 0 + = δ - δ - δ - δ -