When more than one valid Lewis structure can be drawn for a particular molecule, formal charge is used to predict the most favorable structure.

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1 Homework 8 Chapter 8 Due: 11:59pm on Wednesday, November 9, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy Formal Charges and Resonance Formal charge ( FC ) keeps track of which electrons an atom "owns" in a chemical bond with the equation =( ) ( ) FC valence e in valence e in free atom bonded atom When more than one valid Lewis structure can be drawn for a particular molecule, formal charge is used to predict the most favorable structure. A bonded atom is considered to "own" all its nonbonding electrons but only half of the bonding electrons because these are shared with another atom. Therefore, the formal charge formula can be rewritten as follows: FC=( valence e in) 1( bonding e ) ( nonbonding e ) free atom 2 shown shown Part A HOFO For a molecule of fluorous acid, the atoms are arranged as. (Note: In this oxyacid, the placement of fluorine is an exception to the rule of putting the more electronegative atom in a terminal position.) What is the formal charge on each of the atoms? Enter the formal charges in the same order as the atoms are listed. Express your answers as charges separated by comma. For example, a positive one charge would be written as +1. Hint 1. Complete the Lewis electron dot structure HFO 2 Complete the structure for. Join the given atoms with the relevant number of bonds they form, making sure that each atom has an octet or duplet of electrons. Include all lone pairs of electrons. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Hint 1. How to draw a Lewis electron dot structure 1. Determine the total number of valence electrons. 2. Bond all of the atoms with a single bond and distribute the remaining electrons as lone pairs. 3. If any atom other than He, Be, or B lacks an octet, use multiple bonds to complete the octet around each atom. Remember to remove one lone pair for each extra bond added to retain the same total number of electrons. 4. An atom may have more than an octet if it meets two criteria: (1) It is the central atom and (2) it is in the third period of the periodic table or beyond. In HFO 2, none of the atoms will fit these criteria, so all will have at most an octet of electrons. For example, in the structure of the molecule, the total number of valence electrons will be eight: H 2 O 6 electrons on one oxygen atom + (2 1 electron on two hydrogen atoms) = 8 electrons Here the central atom is oxygen. When you draw a single bond between the oxygen atom and each hydrogen atom you will account for four electrons in the single bond and are left with four electrons that must be placed into lone pairs. These electrons are added as lone pairs on the oxygen atom to complete its octet. The hydrogen atoms completed their duplet by forming a single bond with oxygen. The structure looks like this: 1/17

2 Hint 2. Determine the number of valence electrons How many valence electrons are in a molecule of? Express your answer as an integer. 20 HFO 2 Hint 2. Determine the formal charge on Use the equation displayed above Part A to calculate the formal charge on. Enter the formal charge numerically. For example, a positive one charge would be written as +1. formal charge on H = 0 H H 2/17

3 Hint 3. Determine the formal charge on Use the equation displayed above Part A to calculate the formal charge on. Enter the formal charge numerically. For example, a positive one charge would be written as +1. formal charge on F = +1 F F Hint 4. Compare the oxygen atoms FC=( valence e in) 1( bonding e ) ( nonbonding e ) Compare the two oxygen atoms. In the equation which of the terms will be the same for the two oxygen atoms in calculating their formal charge? valence e in free atom bonding shown nonbonding e e shown free atom 2 shown shown Formal Charge for H, O, F, O = 0,0,+1, 1 Part B Two possible electron dot structures are shown for the cyanate ion, the structures are? NCO. What can you conclude about how favorable Hint 1. How to approach the problem Formal charge can provide a means of evaluating which of the possible structures is more favorable. Some rules for evaluation include the following: 1. Structures in which all atoms have zero (0) for their formal charges are lower in energy and therefore more favorable than those with nonzero formal charges. 3/17

4 2. Structures in which atoms have formal charges with a smaller magnitude are preferred to those with atoms having larger formal charges. For example, +1 and 1 would be preferred over +2 and Structures that place negative formal charges on the more electronegative atom are preferred to those that place them on the less electronegative atom. 4. Structures that place positive formal charges on the less electronegative atom are preferred to those that place them on the more electronegative atom. Hint 2. Determine formal charges in structure A What are the formal charges on,, and in structure A, respectively? Express your answers as charges separated by comma. For example, a positive one charge would be written as +1. N C Formal charges for N, C, O in A = 0,0, 1 O Hint 3. Determine formal charges in structure B What are the formal charges on,, and in structure B, respectively? Express your answers as charges separated by comma. For example, a positive one charge would be written as +1. N C Formal charges for N, C, O in B = 2,0,+1 O Structure A is more favored. Structure B is more favored. The structures are equally favored. Lewis Structures and the Octet Rule A Lewis structure shows the arrangement of atoms and valence electrons in a molecule. Most often, each bonding pair of electrons is shown as a line whereas nonbonding electrons are shown as dots as seen here. Notice that each atom is surrounded by a total of eight electrons, an octet. 4/17

5 Octet rule For elements in neutral molecules, the number of bonds needed to achieve an octet follows a trend by group, as shown in the following table: Group 4A 5A 6A 7A Number of bonds Number of lone pairs Part A Draw the Lewis structure for. To add the element Si SiCl 2 Br 2, either double click on any atom and type the element symbol, or access a periodic table of elements from the More button. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. You did not open hints for this part. 5/17

6 Exceptions to the octet rule Exceptions to the octet rule include hydrogen, which is stable with only two electrons (one bond), and boron, which is stable with only six electrons (three bonds). Additionally, elements in period 3 and beyond sometimes form more than the typical number of bonds, thus expanding their valence shells to more than eight electrons. Part B Consider the following four molecules. Which of these satisfy the octet rule and which do not? Drag the appropriate items to their respective bins. Hint 1. How to approach the problem To determine whether the octet rule is followed, the number of electrons around each atom in the formulas must be identified. This can best be done by drawing the Lewis structure for each atom. Here is a set of guidelines to follow in drawing electron dot structures. 1. Determine the total number of valence electrons using the number of valence electrons for each atom in the formula 2. If the species is a cation, subtract a number of electrons equal to the ionic charge. If the species is an anion, add a number of electrons equal to the ionic charge. 3. The central atom is usually the more electropositive element. This excludes H, which is never a central atom. 4. Bond all atoms in the appropriate skeletal arrangement so that all atoms (except H) have an octet (or their normal number of bonds). If there are not enough electrons to accomplish this, use double or triple bonds to achieve the octet. 5. If there are electrons left over, place the electrons as lone pairs on the central atom if the atom is in period 3 or higher. Hint 2. Determine the Lewis structure for PF 5 PF 5 Construct the Lewis structure for. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Hint 1. How to approach the problem 1. Determine the total number of valence electrons. 2. Identify the central atom (usually the least electronegative element, but never H). 3. Give all outer atoms their normal number of bonds and lone pairs. 4. If there are electrons left over, place them as lone pairs on the central atom. 6/17

7 Hint 3. Determine the Lewis structure for CS 2 CS 2 Construct the Lewis structure for. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Hint 1. How to approach the problem 1. Determine the total number of valence electrons. 2. Identify the central atom (usually the least electronegative element, but never H). 3. Give all outer atoms their normal number of bonds and lone pairs. 4. If there are electrons left over, place them as lone pairs on the central atom. 7/17

8 Hint 4. Determine the Lewis structure for BBr 3 BBr 3 Construct the Lewis structure for. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Hint 1. How to approach the problem 1. Determine the total number of valence electrons. 2. Identify the central atom (usually the least electronegative element, but never H). 3. Give all outer atoms their normal number of bonds and lone pairs. 4. If there are electrons left over, place them as lone pairs on the central atom. 8/17

9 Hint 5. Determine the Lewis structure for CO 3 2 Construct the Lewis structure for CO 2 3. You do not need to include brackets or charges. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons but do not include charges. Hint 1. How to approach the problem Drawing the structure of a polyatomic ion can be tricky. Use these steps as a guide. 1. Determine the total number of valence electrons. Don't forget to include the charge (a 1 charge means there is one extra electron). 2. Identify the central atom (usually the least electronegative element, but never H). 3. Start by giving all the outer atoms single bonds and then add lone pairs to the outer atoms until they all have an octet. 4. If there electrons left over, place them as lone pairs on the central atom. 5. If the central atom lacks an octet, turn one of the single bonds into a double bond and remove one lone pair from that outer atom. 9/17

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11 Sample Exercise 8.2 Practice Exercise 1 with feedback Part A Charges on Ions Which of these elements is most likely to form ions with a 2+ charge? Li Cl O P Ca Calcium is found in the second group of the periodic table and it is most likely to lose two electrons to form a 2+ charged cation, Ca 2+. Sample Exercise 8.3 Practice Exercise 1 with feedback 11/17

12 Part A Lewis Structure of a Compound Which of these molecules has the same number of shared and unshared electron pairs? CCl 2 F 2 HCl H 2 S Br 2 PF 3 H 2 S has two pairs of shared electrons and two pairs of unshared electrons. Sample Exercise 8.4 Practice Exercise 1 with feedback Part A Bond Polarity Which of the following bonds is the most polar? Ga Cl Se F N P H F H I H F is the most polar bond in the list because the difference between the electronegativity values for the fluorine atom and the hydrogen atom are the greatest, with the result that the electrons are shared unequally. The electronegativity difference for is 1.9. H F Sample Exercise 8.8 Practice Exercise 1 with feedback Part A Lewis Structure for a Polyatomic Ion How many nonbonding electron pairs are there in the Lewis structure of the peroxide ion, O 2 2? 12/17

13 The peroxide ion, O 2 2, contains 14 electrons in its Lewis structure of which there are 6 pairs on nonbonding electrons and 1 pair of bonding electrons. Sample Exercise 8.6 Practice Exercise 1 with feedback Part A Lewis Structure with a Multiple Bond Which of these molecules has a Lewis structure with a central atom having no nonbonding electron pairs? Check all that apply. SiF 4 S H 2 CO 2 PF 3 CO 2 SiF 4 Both and have central atoms with no nonbonding electrons. Sample Exercise 8.11 Practice Exercise 1 with feedback Part A Lewis Structure for an Ion with More than an Octet of Electrons In which of these molecules or ions is there only one lone pair of electrons on the central sulfur atom? SF 2 SOF 4 SO 4 2 SF 6 SF /17

14 The Lewis structure of atom. SF 4 has one lone pair of electrons and four bonding pairs of electrons on the central sulfur Sample Exercise 8.12 Practice Exercise 1 with feedback Part A Using Average Bond Enthalpies ΔH H 2 O(g) H (g) O 2 (g) Using the table below, estimate for the "water splitting reaction":. 417 kj 5 kj kj kj 242 kj 14/17

15 The enthalpy value can be calculated by subtracting the sum of the enthalpy associated with bonds forming from the sum of the enthalpy associated with bonds breaking: ΔH = = = [2D(O H)] [1D(H H)+ 1 2 D(O=O)] [2(463 kj)] [1(436 kj)+ 1 2 (495 kj)] 242 kj Problem 8.16 Part A Choose representation of the reaction that occurs between and atoms. Ca F Part B What is the chemical formula of the most likely product? Express your answer as a chemical formula. CaF 2 Part C 15/17

16 How many electrons are transferred? 2 electrons Part D Which atom loses electrons in the reaction? Express your answer as a chemical formula. Ca Problem 8.85 O P Te I Consider the collection of nonmetallic elements,,, and. B Part A Which two would form the most polar single bond? B O Te I I O P O Part B Which two would form the longest single bond? 16/17

17 P O Te I B O P I Part C Which two would be likely to form a compound of formula? XY 2 PI 2 BO 2 PO 2 TeI 2 Part D Which combinations of elements would likely yield a compound of empirical formula? X 2 Y 3 B 2 O 3 and P 2 O 3 B 2 I 3 and P 2 I 3 B 2 O 3 and P 2 I 3 B 2 I 3 and P 2 O 3 Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. 17/17