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CHAPTER TWENTY
THE REPRESENTATIVE ELEMENTS:
GROUPS 5A THROUGH 8A
For Review
- Group 5A: ns^2 np^3 ; 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.
NH 3 , 5 + 3(1) = 8 e−^ AsCl 5 , 5 + 5(7) = 40 e−
As Cl
Cl
Cl Cl
Cl
H
N H H
trigonal pyramid; sp^3 trigonal bipyramid; dsp^3
PF 6 −, 5 + 6(7) + 1 = 48 e−
P F
F
F
F
F
F
Octahedral; d^2 sp^3
Nitrogen does not have low energy d orbitals it can use to expand its octet. Both NF 5 and NCl 6 −^ would require nitrogen to have more than 8 valence electrons around it; this never happens.
- N: 1s^2 2s^2 2p^5 ; 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 1s^2 2s^2 2p^6. 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 N−O 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 electronegative atom; this is oxygen in N−O 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.
NH 3 : fertilizers, weak base properties, can form hydrogen bonds; N 2 H 4 : rocket propellant, blowing agent in manufacture of plastics, can form hydrogen bonds; NH 2 OH: weak base properties, can form hydrogen bonds; N 2 : makes up 78% of air, very stable compound with a very strong triple bond, is inert chemically; N 2 O: 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; N 2 O 3 , least common of nitrogen oxides, a blue liquid that readily dissociates into NO(g) and NO 2 (g); NO 2 : another odd electron species, dimerizes to form N 2 O 4 , plays a role in smog production; HNO 3 : important industrial chemical, used to form nitrogen-based explosives, strong acid and a very strong oxidizing agent.
- Hydrazine also can hydrogen bond because it has covalent N−H 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.
N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (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 NH 3 (g) is removed, the reaction shifts right to produce more NH 3 (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 has fewer 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.
- White phosphorus consists of discrete tetrahedral P 4 molecules. The bond angles in the P 4 tetrahedrons are only 60°, which makes P 4 very reactive, especially towards oxygen. Red and black phosphorus are covalent network solids. In red phosphorus, the P 4 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 PH 3 are only 94Ε, well below the predicted tetrahedral bond angles of 109.5Ε. PH 3 is an unusual exception to the VSEPR model.
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 π electrons are delocalized over the entire surface of the molecule in SO 2 and SO 3 , all of the S−O 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 (C 12 H 22 O 11 ) it removes the hydrogen and oxygen in a 2:1 ratio even though there are no H 2 O molecules in sugar. H 2 SO 4 is indeed a powerful dehydrating agent.
- Group 7A: ns^2 np^5 ; The diatomic halogens (X 2 ) 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 F 2 to I 2 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 F 2 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 ∆S° term is a positive ∆G° 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 X 2 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), HClO 2 (+3), HClO 3 (+5), and HClO 4 (+7). Note that these are all acids. HCl is a strong acid, and of the oxyacids, only HClO 4 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 < HClO 2 < HClO 3 < HClO 4.
- 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: 120Ε, sp^2 , e.g., BX 3 V-shape: < 109.5Ε, sp^3 , e.g., OF 2 , OCl 2 , OBr 2 , SF 2 , SCl 2 , SeCl 2 trigonal pyramid: < 109.5Ε, sp^3 , e.g., NX 3 , PX 3 , AsF 3 , SbF 3
tetrahedral: 109.5Ε, sp^3 , e.g., BF 4 −, CX 4 , SiF 4 , SiCl 4 , GeF 4 , GeCl 4 T-shape: 90Ε, dsp^3 , e.g., ClF 3 , BrF 3 , ICl 3 , IF 3 see-saw: 90Ε and ~120Ε, dsp^3 , e.g., SF 4 , SCl 4 , SeF 4 , SeCl 4 , SeBr 4 , TeBr 4 , TeCl 4 , TeBr 4 , TeI 4 trigonal bipyramid: 90Ε and 120Ε, dsp^3 , e.g., PF 5 , PCl 5 , PBr 5 , AsF 5 , SbF 5 square pyramid: 90Ε, d^2 sp^3 , e.g., ClF 5 , BrF 5 , IF 5 octahedral: 90Ε, d^2 sp^3 , e.g., SiF 62 −, GeF 62 −, SF 6 , SeF 6 , TeF 6
ICl, IBr, BrF, BrCl, and ClF have no molecular structure or bond angles. The predicted hybridization for each halogen is sp^3. N 2 F 4 is trigonal pyramid about both nitrogens, with < 109.5Ε bond angles and sp^3 hybridization. O 2 F 2 , S 2 Cl 2 , S 2 F 2 , and S 2 Cl 2 is V-shape about both central oxygens or sulfurs with < 109.5Ε bond angles and sp^3 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 obey the octet rule.
- The noble gases have filled s and p valence orbitals (ns^2 np^6 = valence electron configura- tion). They don’t need to react like other representative elements in order to achieve the stable ns^2 np^6 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.
- XeF 2 : 180 Ε, dsp^3 ; XeO 2 F 2 : ~90Ε and ~120Ε, dsp^3 ; XeO 3 : < 109.5Ε, sp^3 ; XeO 4 : 109.5Ε, sp^3 ; XeF 4 : 90Ε, d^2 sp^3 ; XeO 3 F 2 : 90Ε and 120Ε, dsp^3 ; XeO 2 F 4 , 90Ε, d^2 sp^3
