chemistryatkanehs.weebly.com · Web viewBoth NF5 and NCl6( would require nitrogen to have more than 8 valence electrons around it; this never happens. 2. N: 1s22s22p5; The extremes

CHAPTER TWENTYTHE REPRESENTATIVE ELEMENTS: GROUPS 5A THROUGH 8A

For Review

1. Group 5A: ns2np3; As with groups IIIA and IVA, metallic character increases going down a group. Nitrogen is strictly a nonmetal in properties, while bismuth, the heaviest Group 5A element, has mostly metallic physical properties. The trend of increasing metallic character going down the group is due in part to the decrease in electronegativity. Nitrogen, with its high electronegativity, forms covalent compounds as nonmetals do. Bismuth and antimony, with much lower electronegativities, exhibit ionic character in most of their compounds. Bismuth and antimony exist as +3 metal cations in these ionic compounds.

NH3, 5 + 3(1) = 8 e AsCl5, 5 + 5(7) = 40 e

trigonal pyramid; sp3 trigonal bipyramid; dsp3

PF6, 5 + 6(7) + 1 = 48 e

Octahedral; d2sp3

Nitrogen does not have low energy d orbitals it can use to expand its octet. Both NF 5 and NCl6

would require nitrogen to have more than 8 valence electrons around it; this never happens.

2. N: 1s22s22p5; The extremes of the oxidation states for N can be rationalized by examining the electron configuration of N. Nitrogen is three electrons short of the stable Ne electron configuration of 1s22s22p6. Having an oxidation state of 3 makes sense. The +5 oxidation state corresponds to N “losing” its 5 valence electrons. In compounds with oxygen, the NO bonds are polar covalent, with N having the partial positive end of the bond dipole. In the world of oxidation states, electrons in polar covalent bonds are assigned to the more

751

752 CHAPTER 20 THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A

electronegative atom; this is oxygen in NO bonds. N can form enough bonds to oxygen to give it a +5 oxidation state. This loosely corresponds to losing all of the valence electrons.

NH3: fertilizers, weak base properties, can form hydrogen bonds; N2H4: rocket propellant, blowing agent in manufacture of plastics, can form hydrogen bonds; NH2OH: weak base properties, can form hydrogen bonds; N2: makes up 78% of air, very stable compound with a very strong triple bond, is inert chemically; N2O: laughing gas, propellant in aerosol cans, effect on earth’s temperature being studied; NO: toxic when inhaled, may play a role in regulating blood pressure and blood clotting, one of the few odd electron species that forms; N2O3, least common of nitrogen oxides, a blue liquid that readily dissociates into NO(g) and NO2(g); NO2: another odd electron species, dimerizes to form N2O4, plays a role in smog production; HNO3: important industrial chemical, used to form nitrogen-based explosives, strong acid and a very strong oxidizing agent.

3. Hydrazine also can hydrogen bond because it has covalent NH bonds as well as having a lone pair of electrons on each N. The high boiling point for hydrazine’s relatively small size supports this.

N2(g) + 3 H2(g) ⇌ 2 NH3(g) + heat

a. This reaction is exothermic, so an increase in temperature will decrease the value of K (see Table 13.3 of text.) This has the effect of lowering the amount of NH 3(g) produced at equilibrium. The temperature increase, therefore, must be for kinetics reasons. When the temperature increases, the reaction reaches equilibrium much faster. At low temperatures, this reaction is very slow, too slow to be of any use.

b. As NH3(g) is removed, the reaction shifts right to produce more NH3(g).

c. A catalyst has no effect on the equilibrium position. The purpose of a catalyst is to speed up a reaction so it reaches equilibrium quicker.

d. When the pressure of reactants and products is high, the reaction shifts to the side that hasfewer gas molecules. Since the product side contains 2 molecules of gas compared to 4 molecules of gas on the reactant side, the reaction shifts right to products at high pressures of reactants and products. Also, a high pressure indicates that reactants are present in large quanties. The more reactants present, the further right the reaction shifts.

The pollution provides nitrogen and phosphorous nutrients so the algae can grow. The algae consume oxygen, causing fish to die.

4. White phosphorus consists of discrete tetrahedral P4 molecules. The bond angles in the P4

tetrahedrons are only 60°, which makes P4 very reactive, especially towards oxygen. Red and black phosphorus are covalent network solids. In red phosphorus, the P4 tetrahedra are bonded to each other in chains, making them less reactive than white phosphorus. They need a source of energy to react with oxygen, such as when one strikes a match. Black phosphorus is crystalline, with the P atoms tightly bonded to each other in the crystal, and is fairly unreactive towards oxygen.Even though phosphine and ammonia have identical Lewis structures, the bond angles of PH3

are only 94, well below the predicted tetrahedral bond angles of 109.5. PH3 is an unusual exception to the VSEPR model.

CHAPTER 20 THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A 753

The acidic hydrogens in the oxyacids of phosporus all are bonded to oxygen. The hydrogens bonded directly to phosphorus are not acidic. H3PO4 has three oxygen bonded hydrogens, and it is a triprotic acid. H3PO3 has only two of the hydrogens bonded to oxygen and it is a diprotic acid. The third oxyacid of phosphorus, H3PO2, has only one of the hydrogens bonded to an oxygen; it is a monoprotic acid.

5. Group 6A: ns2np4; As expected from the trend in other groups, oxygen has properties which are purely nonmetal. Polonium, on the other hand, has some metallic properties. The most significant property differences are radioactivity and toxicity. Polonium is only composed of radioactive isotopes, unlike oxygen, and polonium is highly toxic, unlike oxygen.

The two allotropic forms of oxygen are O2 and O3.

O2, 2(6) = 12 e O3, 3(6) = 18 e

The MO electron configuration of O2 has two unpaired electrons in the degenerate pi antibonding ( ) orbitals. A substance with unpaired electrons is paramagnetic (see Figure 9.40). Ozone has a V-shape molecular structure with bond angles of 117, slightly less than the predicted 120 trigonal planar bond angle.

In the upper atmosphere, O3 acts as a filter for UV radiation:

O3 is also a powerful oxidizing agent. It irritates the lungs and eyes, and, at high concentra-tion, it is toxic. The smell of a "fresh spring day" is O3 formed during lightning discharges. Toxic materials don't necessarily smell bad. For example, HCN smells like almonds.

6. Both rhombic and monoclinic sulfur exist in S8 rings. The difference between the two is that the S8 rings are stacked together differently giving different solid structures.

Oxygen forms strong bonds and, because of this, exists in nature as O2 molecules. Sulfur forms much stronger sigma bonds than bonds. Therefore, elemental sulfur is found in nature singly bonded to other sulfur atoms. We assume SO doesn’t form because of the difference in ability of oxygen and sulfur to form bonds. Sulfur forms relatively weak bonds as compared to oxygen.

SO2(aq) + H2O(l) → H2SO3(aq); SO3(aq) + H2O(l) → H2SO4(aq); SO2 and SO3 dissolve in water to form the acids H2SO3 and H2SO4, respectively. Figures 20.18 and 20.19 show the Lewis structures for SO2 and SO3. The molecular structure of SO2 is bent with a 119 bond angle (close to the predicted 120 trigonal planar geometry). The molecular structure of SO3

is trigonal planar with 120 bond angles. Both SO2 and SO3 exhibit resonance. Both sulfurs in SO2 and SO3 are sp2 hybridized. To explain the equal bond lengths that occur in SO2 and SO3, the molecular orbital model assumes that the electrons are delocalized over the entire surface of the molecule. The orbitals that form the delocalized bonding system are


unhybridized p atomic orbitals from the sulfurs and oxygens in each molecule. When all of the p atomic orbitals overlap together, there is a cloud of electron density above and below the entire surface of the molecule. Because the electrons are delocalized over the entire surface of the molecule in SO2 and SO3, all of the SO bonds in each molecule are equivalent.

A dehydrating agent is one that has a high affinity for water. Sulfuric acid grabs water whenever it can. When it reacts with sugar (C12H22O11) it removes the hydrogen and oxygen in a 2:1 ratio even though there are no H2O molecules in sugar. H2SO4 is indeed a powerful dehydrating agent.

7. Group 7A: ns2np5; The diatomic halogens (X2) are nonpolar, so they only exibit London dispersion intermolecular forces. The strength of LD forces increases with size. The boiling points and melting points steadily increase from F2 to I2 because the strength of the intermolecular forces are increasing.

Fluorine is the most reactive of the halogens because it is the most electronegative atom and the bond in the F2 molecule is very weak.

One reason is that the H ‒ F bond is stronger than the other hydrohalides, making it more difficult to form H+ and F. The main reason HF is a weak acid is entropy. When F (aq) forms from the dissociation of HF, there is a high degree of ordering that takes place as water molecules hydrate this small ion. Entropy is considerably more unfavorable for the formation of hydrated F than for the formation of the other hydrated halides. The result of the more unfavorable ΔS° term is a positive ΔG° value, which leads to a Ka value less than one.

HF exhibits the relatively strong hydrogen bonding intermolecular forces, unlike the other hydrogen halides. HF has a high boiling point due to its ability to form these hydrogen bonding interactions.

The halide ion is the 1 charged ion that halogens form when in ionic compounds. As can be seen from the positive standard reduction potentials in Table 20.6, the halogens energetically favor the X form over the X2 form. Because the reduction potentials are so large, this give an indication of the relative ease to which halogens will grab electrons to form the halide ion. In general, the halogens are highly reactive; that is why halogens exist as cations in various minerals and in seawater as opposed to free elements in nature.

Some compounds of chlorine exhibiting the 1 to +7 oxidation state are: HCl(1), HOCl (+1), HClO2 (+3), HClO3 (+5), and HClO4 (+7). Note that these are all acids. HCl is a strong acid, and of the oxyacids, only HClO4 is a strong acid. The oxyacid strength increases as the number of oxygens in the formula increase. Therefore, the order of the oxyacids from weakest to strongest acid is HOCl < HClO2 < HClO3 < HClO4.

8. Most of the compounds in Table 20.11 have the following molecular structures, bond angles, and hybridization. Examples found in Table 20.11 are listed for each.trigonal planar: 120, sp2, e.g., BX3

V-shape: < 109.5, sp3, e.g., OF2, OCl2, OBr2, SF2, SCl2, SeCl2

trigonal pyramid: < 109.5, sp3, e.g., NX3, PX3, AsF3, SbF3

tetrahedral: 109.5, sp3, e.g., BF4, CX4, SiF4, SiCl4, GeF4, GeCl4

T-shape: 90, dsp3, e.g., ClF3, BrF3, ICl3, IF3


see-saw: 90 and ~120, dsp3, e.g., SF4, SCl4, SeF4, SeCl4, SeBr4, TeBr4, TeCl4, TeBr4, TeI4

trigonal bipyramid: 90 and 120, dsp3, e.g., PF5, PCl5, PBr5, AsF5, SbF5

square pyramid: 90, d2sp3, e.g., ClF5, BrF5, IF5

octahedral: 90, d2sp3, e.g., SiF62, GeF6

2, SF6, SeF6, TeF6

ICl, IBr, BrF, BrCl, and ClF have no molecular structure or bond angles. The predicted hybridization for each halogen is sp3. N2F4 is trigonal pyramid about both nitrogens, with < 109.5 bond angles and sp3 hybridization. O2F2, S2Cl2, S2F2, and S2Cl2 is V-shape about both central oxygens or sulfurs with < 109.5 bond angles and sp3 hybridization.

Some of the compounds in Table 20.11 are exceptions to the octet rule, like ICl 3. The row three halogens (Cl) and heavier (Br and I) have low lying empty d-orbitals available to expand their octet when they have to. Fluorine, with its valence electrons in the n = 2 level, does not have low energy d-orbitals available to expand its octet. When F is the central atom, its compounds always obeys the octet rule.

9. The noble gases have filled s and p valence orbitals (ns2np6 = valence electron configura-tion). They don’t need to react like other representative elements in order to achieve the stable ns2np6 configuration. Noble gases are unreactive because they do not want to lose their stable valence electron configuration.

Noble gases exist as free atoms in nature. They only exhibit London dispersion forces in the condensed phases. Because LD forces increase with size, as the noble gas gets bigger, the strength of the intermolecular forces get stronger leading to higher melting and boiling points.

Helium is unreactive and doesn't combine with any other elements. It is a very light gas and would easily escape the earth's gravitational pull as the planet was formed.

In Mendeleev's time, none of the noble gases were known. Since an entire family was missing, no gaps seemed to appear in the periodic arrangement. Mendeleev had no evidence to predict the existence of such a family. The heavier members of the noble gases are not really inert. Xe and Kr have been shown to react and form compounds with other elements.

10. XeF2: 180, dsp3; XeO2F2: ~90 and ~120, dsp3; XeO3: < 109.5, sp3; XeO4: 109.5, sp3; XeF4: 90, d2sp3; XeO3F2: 90 and 120, dsp3; XeO2F4, 90, d2sp3

Questions

1. This is due to nitrogen’s ability to form strong bonds whereas heavier group 5A elements do not form strong bonds. Therefore, P2, As2, and Sb2 do not form since two bonds are required to form these diatomic substances.

2. Nitrogen fixation is the process of transforming N2 to other nitrogen-containing compounds. Some examples are:

N2(g) + 3 H2(g) → 2 NH3(g)

N2(g) + O2(g) → 2 NO(g)


N2(g) + 2 O2(g) → 2 NO2(g)

3. There are medical studies that have shown an inverse relationship between the incidence of cancer and the selenium levels in soil. The foods grown in these soils and eventually digested are assumed to somehow furnish protection from cancer. Selenium is also involved in the activity of vitamin E and certain enzymes in the human body. In addition, selenium de-ficiency has been shown to be connected to the occurrence of congestive heart failure.

4. Chlorine is a good oxidizing agent. Similarly, ozone is a good oxidizing agent. After chlorine reacts, residues of chloro compounds are left behind. Long term exposure to some chloro compounds may cause cancer. Ozone would not break down and form harmful substances. The major problem with ozone is that because virtually no ozone is left behind after initial treatment, the water supply is not protected against recontamination. In contrast, for chlorination, significant residual chlorine remains after treatment, thus reducing (eliminating) the risk of recontamination.

5. +6 oxidation state: SO42, SO3, SF6

+4 oxidation state: SO32, SO2, SF4

+2 oxidation state: SCl2

0 oxidation state: S8 and all other elemental forms of sulfur2 oxidation state: H2S, Na2S

6. sp3 hybridization: HBr, 1 + 7 = 8 e IBr, 7 + 7 = 14 e

Diatomic molecules don’t have a molecular structure because they have no bond angles.

dsp3 hybridization: BrF3, 7 + 3(7) = 28 e

T-shaped

d2sp3 hybridization: BrF5, 7 + 5(7) = 42 e

square pyramid

7. a. H2(g) + Cl2(g) → 2 HCl(g); this reaction produces a lot of energy which can be used ina cannon apparatus to send a stopper across the room. To initiate this extremely slow reaction, light of specific wavelengths is needed. This is the purpose of lighting the magnesium strip. When magnesium is oxidized to MgO, an intense white light is pro-duced. Some of the wavelengths of this light can break ClCl bonds and get the reaction started.


b. Br2 is brown. The disappearance of the brown color indicates that all of the Br2 has reacted with the alkene (no free Br2 is remaining).

c. 2 Al(s) + 3 I2(s) → 2 AlI3(s); This is a highly exothermic reaction, hence the sparks that accompany this reaction. The purple smoke is excess I2(s) being vaporized [the purple smoke is I2(g)].

8. One would expect RnF2 and RnF4 to form in fashion similar to XeF2 and XeF4. The chemistry of radon is difficult to study because radon isotopes are all radioactive. The hazards of dealing with radioactive materials are immense.

Exercises

Group 5A Elements

9. NO43

Both NO43 and PO4

3 have 32 valence electrons, so both have similar Lewis structures. From the Lewis structure for NO4

3, the central N atom has a tetrahedral arrangement of electron pairs. N is small. There is probably not enough room for all 4 oxygen atoms around N. P is larger, thus, PO4

3 is stable.

PO3

PO3 and NO3

each have 24 valence electrons so both have similar Lewis structures. From the Lewis structure for PO3

, PO3 has a trigonal planar arrangement of

electron pairs about the central P atom (two single bonds and one double bond). P=O bonds are not particularly stable, while N=O bonds are stable. Thus, NO3

is stable.

10. a. PF5; N is too small and doesn't have low energy d-orbitals to expand its octet to form NF5.

b. AsF5; I is too large to fit 5 atoms of I around As.

c. NF3; N is too small for three large bromine atoms to fit around it.

11. a. NO: %N = × 100 = 46.68% N

b. NO2: %N = × 100 = 30.45% N

3-O

N

OOO

P

O

OO


c. N2O4: %N = × 100 = 30.45% N

d. N2O: %N = × 100 = 63.65% N

The order from lowest to highest mass percentage of nitrogen is: NO2 = N2O4 < NO < N2O.

12. 1.0 × 106 kg HNO3 × = 1.6 × 107 mol HNO3

We need to get the relationship between moles of HNO3 and moles of NH3. We have to use all 3 equations.

Thus, we can produce 16 mol HNO3 for every 24 mol NH3 we begin with:

1.6 × 107 mol HNO3 × = 4.1 × 108 g or 4.1 × 105 kg

This is an oversimplified answer. In practice, the NO produced in the third step is recycled back continuously into the process in the second step. If this is taken into consideration, then the conversion factor between mol NH3 and mol HNO3 turns out to be 1:1, i.e., 1 mol of NH3

produces 1 mol of HNO3. Taking into consideration that NO is recycled back gives an answer of 2.7 × 105 kg NH3 reacted.

13. a. NH4NO3(s) N2O(g) + 2 H2O(g)

b. 2 N2O5(g) → 4 NO2(g) + O2(g)

c. 2 K3P(s) + 6 H2O(l) → 2 PH3(g) + 6 KOH(aq)

d. PBr3(l) + 3 H2O(l) → H3PO3(aq) + 3 HBr(aq)

e. 2 NH3(aq) + NaOCl(aq) → N2H4(aq) + NaCl(aq) + H2O(l)

14. 4 As(s) + 3 O2(g) → As4O6(s); 4 As(s) + 5 O2(g) → As4O10(s)

As4O6(s) + 6 H2O(l) → 4 H3AsO3(aq); As4O10(s) + 6 H2O(l) → 4 H3AsO4(aq)

15. Unbalanced equation:

CaF23Ca3(PO4)2(s) + H2SO4(aq) → H3PO4(aq) + HF(aq) + CaSO42H2O(s)


Balancing Ca2+, F-, and PO43:

CaF23Ca3(PO4)2(s) + H2SO4(aq) → 6 H3PO4(aq) + 2 HF(aq) + 10 CaSO42H2O(s)

On the right hand side, there are 20 extra hydrogen atoms, 10 extra sulfates, and 20 extra water molecules. We can balance the hydrogen and sulfate with 10 sulfuric acid molecules. The extra waters came from the water in the sulfuric acid solution. The balanced equation is:

CaF23Ca3(PO4)2(s) + 10 H2SO4(aq) + 20 H2O(l) → 6 H3PO4(aq) + 2 HF(aq) + 10 CaSO42H2O(s)

16. a. NO2, 5 + 2(6) = 17 e- N2O4, 2(5) + 4(6) = 34 e-

plus other resonance structures plus other resonance structures

b. BH3, 3 + 3(1) = 6 e NH3, 5 + 3(1) = 8 e

BF3NH3, 6 + 8 = 14 e

In reaction a, NO2 has an odd number of electrons, so it is impossible to satisfy the octet rule. By dimerizing to form N2O4, the odd electron on two NO2 molecules can pair up, giving a species whose Lewis structure can satisfy the octet rule. In general, odd electron species are very reactive. In reaction b, BH3 is electron deficient. Boron has only six electrons around it. By forming BH3NH3, the boron atom satisfies the octet rule by accepting a lone pair of electrons from NH3 to form a fourth bond.

17. 2 NaN3(s) → 2 Na(s) + 3 N2(g)

= 3.12 mol N2 needed to fill air bag.

NOO

N NOO

OO

N

H HH


mol NaN3 reacted = 3.12 mol N2 × = 2.08 mol NaN3

18. For ammonia (in one minute):

= 1.1 × 103 mol NH3

NH3 flows into the reactor at a rate of 1.1 × 103 mol/min.

For CO2 (in one minute):

= 6.6 × 102 mol CO2

CO2 flows into the reactor at 6.6 × 102 mol/min.

To react completely with 1.1 × 103 mol NH3/min, we need:

= 5.5 × 102 mol CO2/min

Because 660 mol CO2/min are present, ammonia is the limiting reagent.

= 3.3 × 104 g urea/min

19.

Bonds broken: Bonds formed:

1 N‒N (160. kJ/mol) 1 N≡ N (941 kJ/mol)4 N‒H (391 kJ/mol) 2 × 2 O‒H (467 kJ/mol)1 O=O (495 kJ/mol)

ΔH = 160. + 4(391) + 495 [941 + 4(467)] = 2219 kJ 2809 kJ = 590. kJ

20. 5 N2O4(l) + 4 N2H3CH3(l) → 12 H2O(g) + 9 N2(g) + 4 CO2(g)

H O H (g) + 2O O+N NH

H

H

HN N (g)(l) (g)


ΔH° =

= 4594 kJ

Using bond energies, ΔH = 5.0 × 103 kJ (from Sample Exercise 20.2). When using bond energies to calculate ΔH, the enthalpy change is assumed to be due only to the difference in bond strength between reactants and products. Bond energies generally give a very good estimate for ΔH for gas phase reactions. However, when solids and liquids are present, ΔH estimates from bond energy differences are not as good. This is because the difference in the strength of the intermolecular forces between reactants and products is not considered when using bond energies. Here, the reactants are in the liquid phase. The loss in strength of the intermolecular forces as the liquid reactants are converted to the gaseous products was not considered when using bond energies in Sample Exercise 20.2; hence, the large difference between the two calculated ΔH values.

21. 1/2 N2(g) + 1/2 O2(g) → NO(g) ΔG° = = 87 kJ/mol; By definition, for a compound equals the free energy change that would accompany the formation of 1 mol of that compound from its elements in their standard states. NO (and some other oxides of nitrogen) have weaker bonds as compared to the triple bond of N2 and the double bond of O2. Because of this, NO (and some other oxides of nitrogen) have higher (positive) standard free energies of formation as compared to the relatively stable N2 and O2 molecules.

22. ΔH° = 2(90. kJ) [0 + 0] = 180. kJ; ΔS° = 2(211 J/K) [192 + 205] = 25 J/K

ΔG° = 2(87 kJ) [0] = 174 kJ

At the high temperatures in automobile engines, the reaction N2 + O2 → 2 NO becomes spontaneous since the favorable ΔS° term will become dominate. In the atmosphere, even though 2 NO → N2 + O2 is spontaneous at the cooler temperatures of the atmosphere, it doesn't occur because the rate is slow. Therefore, higher concentrations of NO are present in the atmosphere as compared to what is predicted by thermodynamics.

23. MO model:

NO+: (σ2s)2(σ2s*)2(π2p)4(σ2p)2, Bond order = (8 2)/2 = 3, 0 unpaired e (diamagnetic)

NO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1, B.O. = 2.5, 1 unpaired e (paramagnetic)

NO-: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2, B.O. = 2, 2 unpaired e (paramagnetic)

N O N ON O


Lewis structures: NO+:

NO:

NO:

The two models give the same results only for NO+ (a triple bond with no unpaired electrons). Lewis structures are not adequate for NO and NO . The MO model gives a better representation for all three species. For NO, Lewis structures are poor for odd electron species. For NO, both models predict a double bond, but only the MO model correctly predicts that NO is paramagnetic.

24. For NCl3 → NCl2 + Cl, only the N‒Cl bond is broken. For O=N‒Cl → NO + Cl, the NO bond gets stronger (bond order increases from 2.0 to 2.5) when the N‒Cl bond is broken. This makes ΔH for the reaction smaller than just the energy necessary to break the N‒Cl bond.

25. a. H3PO4 > H3PO3; The strongest acid has the most oxygen atoms.

b. H3PO4 > H2PO4 > HPO4

2; Acid strength decreases as protons are removed.

26. TSP = Na3PO4; PO43 is the conjugate base of the weak acid HPO4

2 (Ka = 4.8 × ). All conjugate bases of weak acids are effective bases (Kb = Kw/Ka = 1.0 × /4.8 × = 2.1 × ). The weak base reaction of PO4

3 with H2O is: PO43 + H2O ⇌ HPO4

2 + OH Kb = 2.1 × .

27. The acidic protons are attached to oxygen.

H4P2O6 (50 valence e): H4P2O5 (44 valence e):

N O +

N O

H P O P O H

O O

O O

H H

H O P O P O H

O O

H H


28. a. SbF5, 40 valence e HSO3F, 32 valence e- H2SO3F+, 32 valence e

dsp3 sp3 sp3

F5SbOSO2FH, 72 valence e F5SbOSO2F-, 72 valence e

Sb: d2sp3; S: sp3 Sb: d2sp3; S: sp3

b. The active protonating species is H2SO3F+, the species with two OH bonds.

Group 6A Elements

29. O=O‒O → O=O + O

Break O‒O bond: ΔH = = 2.42 × kJ = 2.42 × J

A photon of light must contain at least 2.42 × J to break one O‒O bond.

Ephoton = = 8.21 × m = 821 nm

30. From Figure 7.2 in the text, light from violet to green will work.

31. a. 2 SO2(g) + O2(g) → 2 SO3(g) b. SO3(g) + H2O(l) → H2SO4(aq)

c. 2 Na2S2O3(aq) + I2(aq) → Na2S4O6(aq) + 2 NaI(aq)

d. Cu(s) + 2 H2SO4(aq) → CuSO4(aq) + 2 H2O(l) + SO2(aq)

32. H2SeO4(aq) + 3 SO2(g) → Se(s) + 3 SO3(g) + H2O(l)

33. a. SO32, 6 + 3(6) + 2 = 26 e b. O3, 3(6) = 18 e

-

Sb

F

F

F O S F

O

O

F

FSb

F

F

F O S F

O

O H

F

F

S

O

O

F O H

S

O

O

F O H

H+

Sb F

F

FF

F


trigonal pyramid; ≈ 109.5°; sp3 V-shaped; ≈ 120°; sp2

c. SCl2, 6 + 2(7) = 20 e d. SeBr4, 6 + 4(7) = 34 e

V-shaped; ≈ 109.5°; sp3 see-saw; a ≈ 120°, b ≈ 90°; dsp3

e. TeF6, 6 + 6(7) = 48 e

octahedral; 90°; d2sp3

34. S2N2 has 2(6) + 2(5) = 22 valence electrons.

35. 1.50 g BaO2 × = 8.86 × mol BaO2

25.0 mL × = 1.87 × mol HCl

The required mole ratio from the balanced reaction is 2 mol HCl to 1 mol BaO2. The actual ratio is:

= 2.11

Because the actual mole ratio is larger than the required mole ratio, the denominator (BaO2) is the limiting reagent.

S

OO O

2-O

O O

O

O O

a

b

b

Se

Br

BrBr

BrSCl Cl

TeF

F F

F

F

F

S NN S

S NN S

S NN S

S NN S

Formal Charge

Oxid. Number

F O O F

0 0 0 0-1 +1+1 -1


8.86 × mol BaO2 × = 0.301 g H2O2

The amount of HCl reacted is:

8.86 × mol BaO2 × = 1.77 × mol HCl

excess mol HCl = 1.87 × mol 1.77 × mol = 1.0 × mol HCl

mass of excess HCl = 1.0 × mol HCl × = 3.6 × g HCl

36. 1.00 L = 18.8 g

AgBr

Group 7A Elements

37. O2F2 has 2(6) + 2(7) = 26 valence e; From the following Lewis structure, each oxygen atom has a tetrahedral arrangement of electron pairs. Therefore, bond angles ≈ 109.5° and each O is sp3 hybridized.

Oxidation numbers are more useful. We are forced to assign +1 as the oxidation number for oxygen. Oxygen is very electronegative, and +1 is not a stable oxidation state for this element.

38. a. CCl2F2, 4 + 2(7) + 2(7) = 32 e b. HClO4, 1 + 7 + 4(6) = 32 e

tetrahedral; 109.5°; sp3 About Cl: tetrahedral; 109.5°; sp3

About O: V-shaped; ≈ 109.5°; sp3

c. ICl3, 7 + 3(7) = 28 e d. BrF5, 7 + 5(7) = 42 e

C

F

ClF Cl

Cl

OO O

O H


T-shaped; ≈ 90°; dsp3 square pyramid; ≈ 90°; d2sp3

39. a. BaCl2(s) + H2SO4(aq) → BaSO4(s) + 2 HCl(g)

b. BrF(s) + H2O(l) → HF(aq) + HOBr(aq)

c. SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l)

40. a. F2 + H2O → HOF + HF; 2 HOF → 2 HF + O2; 3 HOF + H2O → 3 HF + H2O2 + O2; HOF + H2O → HF + H2O2 (dilute acid)

In dilute base, HOF exists as OF and HF exists as F. The balanced reaction is:

(2e + H2O + OF → F + 2 OH) × 2 4 OH → O2 + 2 H2O + 4e

________________________________ 2 OF → O2 + 2 F

b. HOF: Assign +1 to H and 1 to F. The oxidation state of oxygen is then zero. Oxygen is very electronegative. A zero oxidation state is not very stable since oxygen is a very good oxidizing agent.

41. ClO + H2O + 2 e → 2 OH + Cl E° = 0.90 V 2 NH3 + 2 OH → N2H4 + 2 H2O + 2 e E° = 0.10 V

_________________________________________________________ ClO(aq) + 2 NH3(aq) → Cl(aq) + N2H4(aq) + H2O(l) = 1.00 V

Because is positive for this reaction, ClO , at standard conditions, can spontaneously oxidize NH3 to the somewhat toxic N2H4.

42. A disproportion reaction is an oxidation-reduction reaction in which one species will act as both the oxidizing agent and reducing agent. The species reacts with itself , forming products with higher and lower oxidation states. For example, 2 Cu+ → Cu + Cu2+ is a disproportion reaction.

HClO2 will disproportionate at standard conditions because > 0:

HClO2 + 2 H+ + 2 e → HClO + H2O E° = 1.65 V HClO2 + H2O → ClO3

+ 3 H+ + 2 e E° = 1.21 V _________________________________________________________________

2 HClO2(aq) → HClO(aq) + ClO3(aq) + H+(aq) = 0.44 V

Group 8A Elements

I

Cl

Cl

Cl BrF

F F

F

F


43. Xe has one more valence electron than I. Thus, the isoelectric species will have I plus one extra electron substituted for Xe, giving a species with a net minus one charge.

a. IO4b. IO3

c. IF2 d. IF4

e. IF6

44. a. KrF2, 8 + 2(7) = 22 e b. KrF4, 8 + 4(7) = 36 e

linear; 180°; dsp3 square planar; 90°; d2sp3

c. XeO2F2, 8 + 2(6) + 2(7) = 34 e

All are: see-saw; ≈ 90° and ≈ 120°; dsp3

d. XeO2F4, 8 + 2(6) + 4(7) = 48 e

All are: octahedral; 90°; d2sp3

45. XeF2 can react with oxygen and water to produce explosive xenon oxides and oxyfluorides, and react with water to form HF.

46. 10.0 m × 10.0 m × 10.0 m = 1.00 × 103 m3; From Table 20.12, volume % Ar = 0.9%.

1.00 × 103 m3 × = 9 × 103 L of Ar in the room

PV = nRT, n = = 4 × 102 mol Ar

KrF

F

F

F

F Kr F

Xe

F

F

O

OXe

O

O

F

FXe

O

F

F

Oor or

orXe

O

O

F

F

F

FXe

F

F

F

F

O

O


4 × 102 mol Ar × = 2 × 104 g Ar in the room

4 × 102 mol Ar × = 2 × 1026 atoms Ar in the room

A 2 L breath contains: 2 L air × = 2 × L Ar

n = = 8 × mol Ar

8 × mol Ar × = 5 × 1020 atoms of Ar in a 2 L breath

Because Ar and Rn are both noble gases, both species will be relatively unreactive. However,all nuclei of Rn are radioactive, unlike most nuclei of Ar. It is the radioactive decay products of Rn that can cause biological damage when inhaled.

47. Release of Sr is probably more harmful. Xe is chemically unreactive. Strontium is in the same family as calcium and could be absorbed and concentrated in the body in a fashion similar to Ca. This puts the radioactive Sr in the bones, and red blood cells are produced in bone marrow. Xe would not be readily incorporated into the body.

The chemical properties determine where a radioactive material may concentrate in the body or how easily it may be excreted. The length of time of exposure and what is exposed to radiation significantly affects the health hazard.

48. a. U → Rn + ? He + ? ; To account for the mass number change, 4 alpha particles are needed. To balance the number of protons, 2 beta particles are needed.

Rn → He + Po; Polonium-218 is produced when 222Rn decays.

b. Alpha particles cause significant ionization damage when inside a living organism. Because the half-life of 222Rn is relatively short, a significant number of alpha particles will be produced when 222Rn is present (even for a short period of time) in the lungs.

Additional Exercises

49. As the halogen atoms get larger, it becomes more difficult to fit three halogen atoms around the small nitrogen atom, and the NX3 molecule becomes less stable.

50. a. The Lewis structures for NNO and NON are:


The NNO structure is correct. From the Lewis structures, we would predict both NNO and NON to be linear. However, we would predict NNO to be polar and NON to be nonpolar. Since experiments show N2O to be polar, NNO is the correct structure.

b. Formal charge = number of valence electrons of atoms [(number of lone pair electrons) + 1/2 (number of shared electrons)].

The formal charges for the atoms in the various resonance structures are below each atom. The central N is sp hybridized in all of the resonance structures. We can probably ignore the third resonance structure on the basis of the relatively large formal charges compared to the first two resonance structures.

c. The sp hybrid orbitals on the center N overlap with atomic orbitals (or hybrid orbitals) on the other two atoms to form the two sigma bonds. The remaining two unhybridized p orbitals on the center N overlap with two p orbitals on the peripheral N to form the two π bonds.

51. OCN has 6 + 4 + 5 + 1 = 16 valence electrons.

Only the first two resonance structures should be important. The third places a positive formal charge on the most electronegative atom in the ion and a -2 formal charge on N.

CNO:

N N O N N O N N O

N O N N O N N O N

N N O N N O N N O+1+1-2-1+100+1-1

z axis

x2p

y2p

spsp

O C N O C N O C N

Formalcharge 00 -1 -1 0 0+10 -2


All of the resonance structures for fulminate (CNO) involve greater formal charges than in cyanate (OCN), making fulminate more reactive (less stable).

52. a. (2e + NaBiO3 → BiO33 + Na+) × 5

(4 H2O + Mn2+ → MnO4 + 8 H+ + 5e) × 2

____________________________________________________________________8 H2O(l) +2Mn2+(aq) + 5 NaBiO3(s) → 2 MnO4

(aq) + 16 H+(aq) + 5 BiO33(aq)

+ 5 Na+(aq) 5 Na+(aq)

b. Bismuthate exists as a covalent network solid: (BiO3)x.

53. 1.0 × 104 kg waste

= 3.4 × 104 g tissue if all NH4+

converted

Since only 95% of the NH4+ ions react:

mass of tissue = (0.95) (3.4 × 104 g) = 3.2 × 104 g or 32 kg bacterial tissue

54. This element is in the oxygen family as all oxygen family members have ns2np4 valence electron configurations.

a. As with all elements of the oxygen family, this element has 6 valence electrons.

b. The nonmetals in the oxygen family are O, S, Se and Te, which are all possible identitiesfor the element.

c. Ions in the oxygen family are 2 charged in ionic compounds. Li2X would be the for-mula between Li+ and X2 ions.

d. In general, radii increase from right to left across the periodic table and increase going down a family. From this trend, the radius of the unknown element must be smaller than the Ba radius.

e. The ioniation energy trend is the opposite of the radii trend indicated in the previous answer. From this trend, the unknown element will have a smaller ionization energy than fluorine.

C N O C N O C N O

Formalcharge +1-2 0 -1 -1 +1-3+1 +1


55. TeF5 has 6 + 5(7) + 1 = 42 valence electrons.

The lone pair of electrons around Te exerts a stronger repulsion than the bonding pairs, pushing the four square planar F's away from the lone pair and thus reducing the bond angles between the axial F atom and the square planar F atoms.

56. a. AgCl(s) Ag(s) + Cl; The reactive chlorine atom is trapped in the crystal. When light is removed, Cl reacts with silver atoms to reform AgCl, i.e., the reverse reaction occurs. In pure AgCl, the Cl atoms escape, making the reverse reaction impossible.

b. Over time, chlorine is lost and the dark silver metal is permanent.

57. As temperature increases, the value of K decreases. This is consistent with an exothermic reaction. In an exothermic reaction, heat is a product and an increase in temperature shifts the equilibrium to the reactant side (as well as lowering the value of K).

58. 7.15 = log (6.2 × ) + log , 7.15 = 7.21 + log

= = 0.9, = 1.1 ≈ 1

A best buffer has approximately equal concentrations of weak acid and conjugate base so that pH ≈ pKa for a best buffer. The pKa value for a H3PO4/H2PO4

buffer is log (7.5 × ) = 2.12. A pH of 7.1 is too high for a H3PO4/H2PO4

buffer to be effective. At this high a pH, there would be so little H3PO4 present that we could hardly consider it a buffer. This solution would not be effective in resisting pH changes, especially when a strong base is added.

59. MgSO4(s) → Mg2+(aq) + SO42(aq); NH4NO3(s) → NH4

+(aq) + NO3(aq)

Note that the dissolution of MgSO4 used in hot packs is an exothermic process, and the dissolution of NH4NO3 in cold packs is an endothermic process.

60. Strong acids have a Ka >> 1 and weak acids have Ka < 1. Table 14.2 in the text lists some Ka

values for weak acids. Ka values for strong acids are hard to determine so they are not listed in the text. However, there are only a few common strong acids, so if you memorize the strong acids, then all other acids will be weak acids. The strong acids to memorize are HCl, HBr, HI, HNO3, HClO4 and H2SO4.

TeF

F F

F

F -


a. HClO4 is a strong acid.b. HOCl is a weak acid (Ka = 3.5 × ).c. H2SO4 is a strong acid.d. H2SO3 is a weak diprotic acid with Ka1 and Ka2 values less than one.

61. EO3 is the formula of the ion. The Lewis structure has 26 valence electrons. Let x = number

of valence electrons of element E.

26 = x + 3(6) + 1, x = 7 valence electronsElement E is a halogen because halogens have 7 valence electrons. Some possible identities are F, Cl, Br and I. The EO3

ion has a trigonal pyramid molecular structure with bond angles 109.5°.

62. The formula is EF2O2- and the Lewis structure has 28 valence electrons.

28 = x + 2(7) + 6 + 2, x = 6 valence electrons for element E

Element E must belong to the group 6A elements since E has 6 valence electrons. E must also be a row 3 or heavier element since this ion has more than 8 electrons around the central E atom (row 2 elements never have more than 8 electrons around them). Some possible identities for E are S, Se and Te. The ion has a T-shaped molecular structure with bond angles of 90°.

63. 8 corners × + 1 Xe inside cell = 2 Xe; 8 edges × + 2 F inside cell = 4 F

Empirical formula is XeF2. This is also the molecular formula.

Challenge Problems

64. In order to form a π bond, the d and p orbitals must overlap “side to side” instead of “head to head” as in sigma bonds. A representation of the “side to side” overlap follows. For a bonding orbital to form, the phases of the lobes must match (positive to positive and negative to negative).

65. For the reaction:

+ NO NO2N N O

O

O


the activation energy must in some way involve breaking a nitrogen-nitrogen single bond.For the reaction:

at some point nitrogen-oxygen bonds must be broken. N‒N single bonds (160. kJ/mol) are weaker than N‒O single bonds (201 kJ/mol). In addition, resonance structures indicate that there is more double bond character in the N‒O bonds than in the N‒N bond. Thus, NO2 and NO are preferred by kinetics because of the lower activation energy.

66. Mg2+ + P3O105 ⇌ MgP3O10

3 K = 4.0 × 108

[Mg2+]o = = 2.1 × M

[P3O105]o = = 0.11 M

Assume the reaction goes to completion because K is large. Then solve the back equilibrium problem to determine the small amount of Mg2+ present.

Mg2+ + P3O105 ⇌ MgP3O10

3

Before 2.1 × M 0.11 M 0Change 2.1 × 2.1 × → +2.1 × React completelyAfter 0 0.11 2.1 × New initial condition

x mol/L MgP3O103 dissociates to reach equilibrium

Change +x +x ← xEquil. x 0.11 + x 2.1 × x

K = 4.0 × 108 = (assume x << 2.1 × )

4.0 × 108 ≈ x = [Mg2+] = 4.8 × M; Assumptions good.

67. a. NO is the catalyst. NO is present in the first step of the mechanism on the reactant side, but it is not a reactant because it is regenerated in the second step and does not appear in the overall balanced equation.

N N OO

OO2 N2O+


b. NO2 is an intermediate. Intermediates also never appear in the overall balanced equation. In a mechanism, intermediates always appear first on the product side while catalysts always appear first on the reactant side.

c. k = A exp (Ea/RT);

= e0.85 = 2.3

The catalyzed reaction is approximately 2.3 times faster than the uncatalyzed reaction at 25°C.

d. The mechanism for the chlorine-catalyzed destruction of ozone is:

O3(g) + Cl(g) → O2(g) + ClO(g) slow ClO(g)+ O(g) → O2(g) + Cl(g) fast __________________________________ O3(g) + O(g) → 2 O2(g)

e. Because the chlorine atom-catalyzed reaction has a lower activation energy, the Cl cat-alyzed rate is faster. Hence, Cl is a more effective catalyst. Using the activation energy, we can estimate the efficiency that Cl atoms destroy ozone as compared to NO mole-cules.

At 25°C:

= e3.96 = 52

At 25°C, the Clcatalyzed reaction is roughly 52 times faster (more efficient) than the NOcatalyzed reaction, assuming the frequency factor A is the same for each reaction and assuming similar rate laws.

68. 3 O2(g) ⇌ 2 O3(g); ΔH° = 2(143 kJ) = 286 kJ; ΔG° = 2(163 kJ) = 326 kJ

ln K = = 131.573, K = = 7.22 ×

Note: We carried extra significant figures for the K calculation.

We need the value of K at 230. K. From Section 16.8 of the text: ln K =

For two sets of K and T:

ln K1 = ln K2 =


Subtracting the first expression from the second:

ln K2 ln K1 =

Let K2 = 7.22 × , T2 = 298; K1 = K230, T1 = 230. K; ΔH° = 286 × 103 J

ln = 34.13 (Carrying extra sig. figs.)

= e34.13 = 6.6 × 1014, K230 = 1.1 ×

K230 = 1.1 × = = 3.3 × atm

The volume occupied by one molecule of ozone is:

V = , V =

9.5 × 1017 L

Equilibrium is probably not maintained under these conditions. When only two ozone molecules are in a volume of 9.5 × 1017 L, the reaction is not at equilibrium. Under these conditions, Q > K and the reaction shifts left. But with only 2 ozone molecules in this huge volume, it is extremely unlikely that they will collide with each other. In these conditions, the concentration of ozone is not large enough to maintain equilibrium.

69. NH3 + NH3 ⇌ NH4+ + NH2

K = [NH4+] [NH2

] = 1.8 ×

NH3 is the solvent, so it is not included in the K expression. In a neutral solution of ammonia:

[NH4+] = [NH2

]; 1.8 × = [NH4+]2, [NH4

+] = 1.3 × M = [NH2]

We could abbreviate this autoionization as: NH3 ⇌ H+ + NH2, where [H+] = [NH4

+].This abbreviation is synonomous to the abbreviation of the autoionization of water (H2O ⇌H+ + OH). So: pH = pNH4

+ = log(1.3 × ) = 5.89.

70. Let’s consider a reaction between 3.00 x mol N2 and 3.00 x mol H2 (equimolar).

N2(g) + 3 H2(g) → 2 NH3(g)

Before 3.00 x mol 3.00 x mol 0Change 1.00 x mol 3.00 x mol +2.00 x molEquil. 2.00 x mol 0 2.00 x mol


When an equimolar mixture is reacted, the number of moles of gas present decreases from 6.00 x moles initially to 4.00 x moles after completion.

a. The total pressure in the piston apparatus is a constant 1.00 atm. After the reaction, we have 2.00 x moesl N2 and 2.00 x moles NH3. One-half of the moles of gas present are NH3 molecules, so one-half of the total pressure is due to the NH3 molecules. = 0.500 atm.

b. = 0.500

c. At constant P and T, volume is directly proportional to n. Because n decreased from 6.00 x moles to 4.00 x moles, the volume will decrease by the same factor.

Vfinal = 15.0 L (4/6) = 10.0 L

71. Let = initial mol SO2 present. The reaction is summarized in the following table (O2 is in excess).

2 SO2 + O2(g) → 2 SO3(g)

Initial 2.00 mol 0

Change /2 +

Final 0 2.00 - /2

d = mass/volume; Let di = initial density of gas mixture and df = final density of gas mixture after reaction. Because mass is conserved in a chemical reaction, mass i = massf.

At constant P and T, V n, so: ; Setting up an equation:

= 1.059, 1.059 = = =

Solving: = 0.25 mol; so, 0.25 moles of SO3 formed


0.25 mol SO3 × = 20. g SO3

72. Ca(IO3)2(s) ⇌ Ca2+(aq) + 2 IO3(aq) Ksp = [Ca2+][ IO3

]2

Initial s = solubility (mol/L) 0 0Equil. s 2s Ksp = s(2s)2 = 4s3

mol IO3 present = 0.0149 L ×

= 2.48 × mol IO3

[IO3] = = 2.48 × M = 2s, s = 1.24 × M

Ksp = 4s3 = 4(1.24× )3 = 7.63 ×

Integrative Problems

73. a. 307 kJ = [1136 + x] – [(-254 kJ) + 3(96 kJ)], x = = 287 kJ/mol

b. IF2+, 7 + 2(7) – 1 = 20 e BF4

, 3 + 4(7) + 1 = 32 e

V-shaped; sp3 tetrahedral; sp3

74. a. Because the hydroxide ion has a 1 charge, Te has a +6 oxidation state.

b. mol Te = (0.545 cm)3 = 7.92 × mol Te

mol F2 = n = = 0.101 mol F2

= 12.8


The balanced reaction requires a 3:1 mol ratio of F2 to Te. Because actual > theoretical, the denominator (Te) is limiting. Assuming 115 mL of solution:

[TeF6]o = = 6.89 × M

Because , the amount of protons produced by the reaction will be insignificant.

Te(OH)6 ⇌ Te(OH)5O + H+ = = 2.1 ×

Initial 0.0689 M 0 ~0Equil. 0.0689 x x x

= 2.1 × = , x = [H+] = 3.8 × M

pH = log(3.8 × ) = 4.42; Assumptions good.

Marathon Problem

75. The answers to the clues are:

(1) HI has the second highest boiling point; (2) HF is the weak hydrogen halide acid; (3) He was first discovered from the sun’s emission spectrum; (4) Both Bi and Sb form MOCl precipitates. For a message that makes sense, Bi is the correct choice; (5) Te is a semicon-ductor; (6) S has both rhombic and monoclinic solid forms; (7) Cl2 is a yellow-green gas and Cl- forms the indicated precipitates; (8) O is the most abundant element in and near the earth’s crust; (9) Se appears to furnish some form of protection against cancer; (10) Kr forms compounds. The symbol in reverse order is rk; (11) As forms As4 molecules; (12) N2 is a major inert component of air and N is often found in fertilizers and explosives.

Filling in the blank spaces with the answers to the clues, the message is “If he bites, close ranks.”

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chemistryatkanehs.weebly.com · Web viewBoth NF5 and NCl6( would require nitrogen to have more than 8 valence electrons around it; this never happens. 2. N: 1s22s22p5; The extremes