CH 121 Worksheet 4 - Part 1 Notes PDF

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Chemistry 121 Oregon State University Worksheet 4 - Part 1 Notes 1. What is the ground state electron configuration for each of the following ions? a. F- the ground state configuration of the F atom is: [He] 2s2 2p5 . When it ionizes, it gains one electron to be: [He] 2s2 2p6 which is the same as [Ne} b. Mg2+ the ground state configuration of the Mg atom is [Ne] 3s2 . When it ionizes, it loses two electrons to be: [Ne] c. V2+ the ground state configuration of the V atom is [Ar] 4s2 3d3. It is a transition metal, so when it ionizes it loses the ns electrons first (in this case, the 4s), so the configuration will be [Ar] 3d3 d. V3+ as before, the ground state configuration will be [Ar] 4s2 3d3. The first two electrons lost will be from the ns; once that is empty we remove from the (n-1)d (in this case the 3d). That means the configuration will be [Ar] 3d2 e. Cr3+ the ground state electron configuration of Cr is [Ar] 4s2 3d4. The first electrons removed are from the ns, and then we remove the remaining electrons from the 3d. That means the configuration is [Ar] 3d3 2. Predict the formula of the ionic compound that will form between each of the following pairs of elements: a. Li and N Li forms the Li+ ion, and N forms the N3- ion. That means the formula will be Li3N b. K and F K forms the K+ ion, and F forms the F- ion. The formula will be KF c. Ca and S Ca forms the Ca2+ ion and S forms the S2- ion. The formula will be CaS d. Ca and Br Ca forms Ca2+ and Br forms Br- . The formula will be CaBr2

e. Mg and N Mg forms the Mg2+ ion and N forms the N3- ion. The formula will be Mg3N2 3. For each of the following pairs of atoms, use the electronegativity difference to determine whether they would form a covalent, polar covalent, or ionic bond: To solve these, we look up the electronegativity value of each, and find the absolute value of the difference between the two. Then we compare them to the chart in Fig. 5 that classifies bond types based on electronegativity difference to determine the bond type. a. C and H 2.5 (for C) - 2.1 (for H) = 0.4 This would be classified as a covalent bond. (Note: in some texts, C-H will be classed as a polar bond. This is all a matter of degree, but it is generally not considered to be a polar bond). a. N and P 3.0 (for N) - 2.1 (for P) = 0.9 This would be classified as a polar covalent bond. b. O and O 3.5 (for O) - 3.5 (for O) = 0.0 This is a covalent bond. We have the same atom with the same electronegativity, so the electrons will be evenly shared. c. Li and C 2.5 (for C) - 1.0 (for Li) = 1.5 This is a polar covalent bond, which is not what we might expect for a bond between a metal and a non-metal. This helps illustrate the importance of using electronegativity values (and not just general trends) in determining the nature of bonds.

d. B and H 2.1 (for H) - 2.0 (for B) = 0.1 This is a covalent bond. The electrons are not shared completely equally, but the difference is not substantial. e. C and F 4.0 (for F) - 2.5 (for C) = 1.5 This is a polar covalent bond. f. C and O 3.5 (for O) - 2.5 (for C) = 1.0 This is a polar covalent bond, although the electrons are not as unequally shared as in the C-F bond in f. g. Na and F 4.0 (for F) - 0.9 (for Li) = 3.1 This is an ionic bond.

4. Fill in the blanks on the following table of ionic compounds: There are three important things to remember here when getting a name from a formula: 1. List the cation first 2. Do not use prefixes to denote how many of each ion are in an ionic compound (it’s implied by the formula) 3. Change the suffix on the anion to -ide (unless it’s a polyatomic ion - don’t change anything about the polyatomic ion) When getting a formula from a name, remember that you’ll need to determine how many of each ion will give a neutral formula. Also remember that you only include the polyatomic ion in parentheses if there are more than one of that ion in the formula.

Compound Name

Compound Formula

magnesium chloride

MgCl2

lithium sulfate

Li2SO4

sodium hydroxide

NaOH

calcium sulfide

CaS

magnesium nitride

Mg3N2

potassium fluoride

KF

sodium oxide

Na2O

sodium carbonate

Na2CO3

5. Fill in the blanks on the following table of ionic compounds containing a transition metal cation: These are very similar to question 4, but we need more information for metal cations that don’t have a single possible charge. For that reason, we need to specify in the name what the charge on the metal cation is. When you have the name, use that information to find the correct formula.

Compound Name

Compound Formula

iron(III) bromide

FeBr3

cobalt(II) nitride

Co3N2

chromium(III) chloride

CrCl3

vanadium(IV) oxide

VO2

vanadium(V) oxide

V2O5

manganese(II) sulfate

MnSO4

copper(I) oxide

Cu2O

titanium(IV) flouride

TiF4

6. Give the correct name for each compound from their formula: With covalent (molecular) compounds, the elements can combine in different ratios in the formula. For this reason, we include prefixes to designate how many of each type of atom is present. The exception to this is that we don’t need to include “mono” on the first element if there is only one. a. CO2 - carbon dioxide b. CO

- carbon monoxide (note that we still need mono on the second element

c. PF3 - phosphorus trifluoride d. SF6 - sulfur hexafluoride e. N2O4 - dinitrogen tetraoxide f. I2O5 - diiodine pentoxide

7. Give the correct formula for each compound based on the name For these compounds, the formula is dictated by the elements present and the prefixes used to denote how many of each atom are there. Remember that if the first element listed does not have a prefix, there is only one in the formula. Also remember that we don’t use “1”s as subscripts in the formula - if there is no subscripted number, that means there is one.

a. chlorine dioxide - ClO2 b. silicon tetrafluoride - SiF4 c. iodine heptafluoride - IF7 d. diiodine pentaoxide - I2O5 e. disulfur decafluoride - S2F10 8. Give the Lewis dot structures for H and He

Helium has paired electrons despite only having two, because the only electrons are in the 1s subshell. 9.

Give the Lewis dot structures for Li and Li+

10.

Give the Lewis dot structures for: a. N

b. O

c. F

d. F-

e. Ne

f. Na+

g. Mg2+

h. Al3+

11.

Give the Lewis dot structures for the following ionic compounds: a. LiCl The Lewis dot structure for an ionic compound will depict the ions with the valence electrons present after the electron transfer.

b. NaCl Sodium has more electrons than lithium, but in this case it’s only the valence electrons that matter, so the Lewis dot structure differs from that of LiCl only in the atomic symbol for the cation.

c. MgCl2 With MgCl2, there are two chloride ions. This is depicted in the Lewis dot structure of the compound simply by adding a coefficient in front of the anion’s dot symbol.

d. Na2O In a similar fashion to c., the multiple sodium cations are depicted using a coefficient ahead of the cation’s dot symbol:

e. MgO

12. Draw the Lewis Structures for the following molecules: a. NH3 The basic steps in drawing a Lewis structure for a molecule are: - Calculate the total number of valence electrons in the molecule - Place the atoms in the expected arrangement, with the least electronegative atom in the center (H is never the central atom) - Place pairs of electrons to make single bonds between the atoms - Assign lone pairs of electrons to complete the octets (or duet, in the case of H) of the outer atoms - Assign remaining lone pairs to the central atom until valence electrons have all been assigned - Check for complete octets. If necessary, move lone pairs from outer atoms to form double or triple bonds.

For this molecule: - Total valence electrons: Atom N H

How many? 1 3

Valence e5 1 Total

Total valence e5 3 8

- Expected arrangement of atoms: H can’t be in the center, so we arrange the H atoms around the N:

- Make single bonds to connect the atoms. This uses up 6 of the valence electrons, leaving us with 2 to place.

- No more electrons can go on the H atoms, so we next place remaining electrons on the central atom. This completes the octet of N, and uses up our remaining electrons, and completing our Lewis structure.

b. CH3CH3 - Our total valence electrons are 14 (4 each from 2 C atoms, and 1 each from 6 H atoms). The C atoms go in the center. Only single bonds are necessary to complete all octets (and duets), and no lone pairs need to be assigned.

c. CH2CH2 This is the same approach as in part b, but the two fewer H atoms (and two fewer electrons) requires us to use a double bond between the carbons to complete the carbons’ octets.

d. CHCH Again, we have two fewer hydrogen atoms (and two fewer electrons) than in the previous structure, so we now need a triple bond between the carbon atoms for the octets to be complete.

e. H2

f. O2

g. N2

13.

Draw the Lewis Structure for the following polyatomic ions: a. OHWith polyatomic ions, we approach the Lewis structure in a similar fashion. The first significant difference is including the charge on the ion in the count of valence electrons:

Atom O H

Valence eHow many? 1 6 1 1 Overall charge: -1 Total

Total valence e6 1 1 8

When the resulting Lewis structure is drawn, the charge is depicted as being on the polyatomic ion as a whole, rather than assigning it to one specific atom.

b. CN-

c. NH4+ When the polyatomic ion is positively charged, we take the same approach, but remember that a positive charge means the ion has an electron deficit, so we’ll be subtracting valence electrons from the total.

Atom N H

Valence eHow many? 1 5 4 1 Overall charge: +1 Total

Total valence e5 4 -1 8

The resulting structure is this:

d. H3O+

e. O22This ion differs from the uncharged O2 molecule in 3f by two electrons. These additional two electrons allow the ion to have complete octets for each atom without a double bond between the two.

14. Draw the Lewis Structure for methane. Calculate the formal charge on each atom in the molecule.

All of the H atoms are equivalent, so we can just calculate the formal charge once and it will apply to each H. Valence eAssigned eFormal Charge

C 4 4 0

H 1 1 0

T The sum of the formal charges equals 0, which matches the charge on the methane molecule.

15.

Calculate the formal charge on each atom for each of the Lewis Structures shown below: a.

Valence eAssigned eFormal Charge

C 4 5 -1

O 6 5 +1

Again, the sum of the formal charges equals 0, which matches the charge on the carbon monoxide molecule.

b.

-

Valence e Assigned eFormal Charge

Cl 7 6 +1

O 6 7 -1

Both oxygen atoms have a formal charge of -1, so the sum of all of the charges is -1. This matches the overall charge on the ion. c.

Valence eAssigned eFormal Charge

S 6 4 +2

O (sides) 6 7 -1

O (bottom) 6 6 0

Two of the three oxygen atoms have a formal charge of -1, so the sum of all of the charges is 0, which is the same as the overall charge on the molecule.

16.

Draw the possible resonance structures for sulfur dioxide. Calculate the formal charge on each atom for each. Will one structure be favored over another? We can draw two equivalent structures that each have one single and one double bond

S Valence eAssigned eFormal Charge

6 5 +1

O

O

(double bond)

(single bond)

6 6 0

6 7 -1

Both of these structures have the same development of formal charge - the only difference is which O has the single bond (and hence the -1 formal charge). Since the formal charge development is the same, neither structure is favored. We can also draw a structure with two double bonds by expanding the octet of S:

Calculating the formal charge gives:

S

O (both equivalent)

Valence eAssigned eFormal Charge

6 6 0

6 6 0

In theory, as there is no formal charge on any of the atoms in this structure, it would be predicted to contribute most strongly to the overall hybrid. Interestingly, experimental data suggest that reality is actually more of a hybrid of the first two structures.

17.

Draw the possible resonance structures for the thiocyanate ( SCN- ) ion (use C as the central atom). Calculate the formal charge on each atom for each. Rank the resonance forms in order of likely importance. Potential structure I (these are not listed in any particular order):

Valence eAssigned eFormal Charge

S 6 7 -1

C 4 4 0

N 5 5 0

S 6 5 +1

C 4 4 0

N 5 7 -2

S 6 6 0

C 4 4 0

N 5 6 -1

Potential structure II:

Valence eAssigned eFormal Charge Potential structure III:

Valence eAssigned eFormal Charge

Based on the development of formal charge, structure II is the least important resonance form. Structures I and III have the same development of formal

charge, so when evaluating the resonance structures, the more important one to the overall structure will be the structure with a negative charge on the most electronegative atom. This means that structure III has the greatest contribution to the overall structure of the molecule, and from most to least important, the order is: III > I > II.

18.

Draw the Lewis Structure for each of the following species: a. BCl3

The central boron atom will have an incomplete octet.

b. SeF4

For this molecule to have enough electrons the central Se atom needs to have an expanded octet (the extra lone pair is highlighted in red). This is allowed because Se is in the third period and will have 3d orbitals that can be used in bonding.

c. IF5

For this molecule to have enough bonds, and the correct number of electrons, the central I atom needs to have an expanded octet. In this case, it doesn’t just have an additional lone pair (highlighted in red), but an additional I-F covalent bond....