CHAPTER 1
Elements of the p Block
edited by Charlie Harding, Rob Janes and David Johnson
Based on 'The Chemistry of Hydrogen, the Halogens and the Noble Gases' by David Johnson (1994) and 'The Chemistry of Groups III-VI' by David Johnson and Lesley Smart (1994)
INTRODUCTION 1
Chemistry is an immensely diverse subject. Nowhere is this more evident than in the properties of the p-Block elements, those in which the outer electronic configuration is s2px. In this Book, we will study the typical elements, focusing on the p-Block elements (Figure 1.1).
Much of the chemistry of the metallic s-Block elements is consistent with an ionic model. With the p-Block elements, this is no longer the case: the proportion of obviously covalent substances is much larger. Before we begin our study of these elements, it is necessary to introduce some essential principles, mainly concerning bonding and thermochemistry, which apply particularly to covalent substances. Following this, we consider the chemistry of the Groups of p-Block elements, beginning with the halogens and noble gases, as these Groups most clearly display covalent properties. These are the two Groups that contain no metallic elements.
1.1 Group numbers and the Periodic Table
Because this Book concentrates upon the typical elements, we shall make particular use of the mini-Periodic Table of Figure 1.1 that contains the typical elements alone. In Figure 1.1, the Groups are numbered in two different ways. In the Mendeléev numbering scheme, which is given in Roman numerals, the Group numbers increase smoothly across each row from I to VIII. These numbers are useful because, for each element, they are almost always equal to the number of outer electrons in the atom, and to the highest oxidation state (see Section 2.1) that is reached in the compounds that the element forms. Use is made of these relationships in this Book. Most modern textbooks and chemistry research journals, however, use a numbering system recommended by IUPAC (International Union of Pure and Applied Chemistry). In this system, the Groups are numbered in Arabic numerals as Groups 1–18 (Figure 1.2). The p-Block elements of Groups III–VIII are, therefore, numbered as Groups 13–18. Because both numbering schemes are important, we include both alternatives in the headings of the Sections and the figures describing the Groups in this Book. Thus the Group of elements headed by boron is designated Group III/13. Notice that for the typical elements of Figure 1.1, the second digit of a Group number in the IUPAC scheme is equal to the Mendeléev number. This provides an easy way of shifting from one system to another.
OXIDATION AND REDUCTION 2
Oxidation may be defined as a loss of electrons, and reduction as a gain of electrons. In a reaction such as
Mg(s) + Cl2(g) = MgCl2(s) (2.1)
the ionic model implies that magnesium dichloride is composed of the ions Mg2+ and Cl-. Consequently, Reaction 2.1 is a redox reaction: magnesium has been oxidized because a magnesium atom loses two electrons; chlorine has been reduced because a chlorine atom gains an electron. However, this argument assumes that MgC12 conforms to an ionic model. While this may be justified for compounds such as MgC12, a difficulty arises when we turn to covalent compounds. Consider the reaction between sulfur and chlorine:
S(s) + Cl2(g) = SCl2(l) (2.2)
Reactions 2.1 and 2.2 appear similar, but at room temperature, SCl2 is a red liquid containing discrete SCl2 molecules, whose atoms are held together by covalent bonds (Structure 2.1); there cannot be complete transfer of two electrons per sulfur atom from sulfur to chlorine. Consequently, our definitions of oxidation and reduction prevent the classification of both Reactions 2.1 and 2.2 as redox reactions. To obtain a broader definition of oxidation and reduction, we must abandon the implication that complete electron transfer takes place.
2.1 Oxidation numbers or oxidation states
The solution that has been devised for this problem is to imagine that the shared electron pairs in covalent bonds between different elements are completely transferred to one of the two bound atoms. The transfer is assumed to occur to the more electronegative of the two.
* Consider Structure 2.1. Which is the more electronegative type of atom?
* Chlorine is more electronegative than sulfur; it lies to the right of sulfur in Period 3 (Figure 1.1).
If the shared electron pair in each bond in Structure 2.1 were transferred to chlorine, each chlorine would have eight outer electrons, one more than in the free atom. It would therefore become Cl-, and carry a charge of -1.
* What number of outer electrons would the sulfur be left with, and what charge would it carry?
* Sulfur would be left with four outer electrons, two less than in the free atom, and would have a charge of +2.
These imaginary charges are called oxidation numbers or oxidation states. The oxidation number of an atom in its elemental form, e.g. graphite, aluminium metal and chlorine gas, is taken to be zero.
Oxidation is now defined as an increase in oxidation number, and reduction as a decrease in oxidation number.
* Why is Reaction 2.2 a redox reaction? What has been oxidized and reduced in it?
* Sulfur has been oxidized because its oxidation number has increased from zero to +2; chlorine has been reduced because its oxidation number has decreased from zero to -1.
To deduce the oxidation numbers of the elements in a compound or ion, it is unnecessary to write down a Lewis structure, as described above. Instead a set of rules allows the rapid assignment of oxidation numbers in most of the substances that you will normally encounter (Box 2.1).
Remember that only in ionic substances does the oxidation number of an element equal the charge that the element carries. In NaCl, which is regarded as an ionic compound, we do indeed take the charge carried by sodium to be equal to its oxidation n umber of +1.
But in the covalent compound SC12, you have seen that the sulfur oxidation number of +2 was obtained by an imaginary distortion of the distribution of electrons, so that the shared electron pairs in the covalent bonds are transferred completely to the more electronegative chlorine. The charge carried by sulfur in the real SCl2 molecule must therefore be less than the oxidation number of +2. To emphasize this, oxidation numbers are often written as Roman numerals in parentheses after the element. Thus, SCl2 is written as sulfur(II) chloride, and SO2 as sulfur(IV) oxide. Oxidation numbers can be used to balance redox equations, which we consider next.
2.2 Balancing redox equations
Much of the chemistry of the p-Block elements involves change in oxidation number. For example, in both of the reactions given above (Reactions 2.1 and 2.2), one element, Mg or S, is oxidized and another, Cl, is reduced. Balancing equations of this kind is often a trivial exercise which can be done by inspection. Consider the reaction of antimony with iodine, in which antimony is oxidized and iodine is reduced. The equation (2.3) is balanced simply by ensuring that the ratio Sb : I is 1 : 3 while avoiding the inconvenience of including fractions:
2Sb + 3I2 = 2SbI3 (2.3)
For more complicated reactions, we can use oxidation numbers to obtain a properly balanced equation, as shown below.
A convenient method of preparing a small amount of chlorine gas in the laboratory is the reaction of potassium permanganate, KMnO4, with concentrated hydrochloric acid. As the chloride ion, Cl-(aq), is oxidized to chlorine gas, the permanganate ion, MnO4-(aq), is reduced to give Mn2+(aq) ions. A preliminary equation for this reaction, in which only the element being oxidized and the element being reduced are balanced, is
[FORMULA NOT REPRODUCIBLE IN ASCII] (2.4)
The first step is to use the rules in Box 2.1 to calculate the oxidation numbers of the elements which are oxidized or reduced, in this case manganese and chlorine:
Rule 1 tells us that in Mn2+(aq) and Cl-(aq), the oxidation numbers are +2 and -1, respectively.
Rule 2 tells us that in Cl2(g), the oxidation number is zero.
* What is the oxidation number of manganese in MnO4-(aq)?
* +7. By rule 4, we assign an oxidation number of -2 to each of four oxygen atoms. The oxidation number of manganese must be +7, because when 7 is added to -8, it gives -1, which is the charge on MnO4-(aq) (rule 7).
We can now write Equation 2.4 with the oxidation numbers in place:
[FORMULA NOT REPRODUCIBLE IN ASCII] (2.4)
Notice that as the equation stands, the oxidation number of manganese falls from +7 to +2, and the oxidation number of chlorine rises from -1 to zero. Notice too, that the total change in oxidation number is -4, because the value for manganese falls by five and that for chlorine rises by one. The following rule now applies:
In a balanced equation , the total change in oxidation number i zero.
So, in the second step we adjust the ratio MnO4-:Cl- to make this so.
* What number of Cl-(aq) ions must be written to make this true?
* Five. If each of five chlorine ions increases its oxidation state by one, giving +5 in all, this balances the change of -5 for manganese.
The equation now becomes
[FORMULA NOT REPRODUCIBLE IN ASCII] (2.5)
In the third step, we balance the equation in the elements that are not oxidized or reduced. Since the reaction takes place in aqueous acid solution, we do this by including the necessary H+(aq) or H2O. There are four oxygen atoms on the left side (in MnO4-), and so four molecules of H2O must be added to the right-hand side.
[FORMULA NOT REPRODUCIBLE IN ASCII] (2.6)
We now add the appropriate number of hydrogen ions, H+(aq), to complete the balancing.
* How many H+(aq) are required?
* 8H+(aq) must be added to the left-hand side to balance the hydrogen atoms in four water molecules on the right.
[FORMULA NOT REPRODUCIBLE IN ASCII] (2.7)
For convenience, we double this equation to eliminate fractions:
[FORMULA NOT REPRODUCIBLE IN ASCII] (2.8)
The equation is now balanced with respect to all elements and oxidation numbers. If we have done this correctly, it should also be balanced with respect to charge. This is the fourth and final step.
* Find the total charge on each side of the equation. Is the equation balanced with respect to charge?
* On the left the charge is: -2 - 10 + 16 = +4. This balances the charge of +4 on the right.
2.3 Summary of Section 2
1 Oxidation states or oxidation numbers may be regarded as charges, usually fictitious, which are assigned to elements in compounds by assuming that shared electron pairs in covalent bonds are completely transferred to one of the two bound atoms. Transfer is assumed to occur to the more electronegative of the two.
2 Oxidation is an increase in oxidation number; reduction is a decrease.
3 Redox reactions may be represented by balanced equations. Oxidation numbers are assigned to those substances involved in the oxidation/reduction. The total change in oxidation number is set to zero by adjusting the numbers of reactants and products. In aqueous solution, H2O, H+(aq) or OH-(aq) ions are included as necessary to balance the number of hydrogen and oxygen atoms.
QUESTION 2.1
The ten substances listed below contain sulfur. Work out the oxidation number of sulfur in each one. Then present your results in the form of a table, showing the substances in order of increasing sulfur oxidation number.
(i) solid sulfur, S(s)
(ii) sulfide ion, S2-(aq)
(iii) sulfur dioxide, SO2 (g)
(iv) sulfur trioxide, SO3(s)
(v) hydrogen sulfide, H2S(g)
(vi) sulfur dichloride, SCl2(l)
(vii) sulfuric acid, H2SO4(l)
(viii) sulfite ion, SO32-(aq)
(ix) sulfur hexafluoride, SF6(g)
(x) hydrogen sulfate ion, HSO4-(aq)
QUESTION 2.2
By working out the relevant oxidation numbers, specify which of the following are redox reactions and state which element is oxidized, and which is reduced.
(i) C(s) + 2Cl2 (g) = CCl4(l)
(ii) N2 (g) + 3H2(g) = 2NH3(g)
(iii) H+(aq) + OH-(aq) = H2O(l)
(iv) 2H2S(g) + SO2(g) = 3S(s) + 2H2O(l)
QUESTION 2.3
During the manufacture of bromine, bromine gas is concentrated by mixing it with sulfur dioxide, and passing the mixture into water, when a concentrated solution of bromide is produced. This can later be re-oxidized with chlorine. An unbalanced equation for the concentrating process is:
[FORMULA NOT REPRODUCIBLE IN ASCII]
By assigning oxidation numbers, identify the elements that are oxidized and reduced. Then balance the equation by the addition of H2O(l) and H+(aq).
QUESTION 2.4
Bromine can also be made by adding acid to a mixture of solutions of sodium bromide (NaBr) and sodium bromate (NaBrO3). Starting with the unbalanced equation,
BrO3-(aq) + Br-(aq) = Br2(aq)
give a balanced equation by using oxidation numbers, and adding H2O(l) and H+(aq).
DEFINING ACIDS AND BASES 3
The simplest definition of acids and bases is that of Arrhenius: acids dissociate to give H+ (aq) in an aqueous solution; bases dissociate to give OH-(aq). In this section we consider the strengths of acids and bases, and two examples of attempts to broaden the definitions of an acid and a base.
3.1 Strengths of acids and bases
When the gas HCl dissolves in water, its solution is called hydrochloric acid. According to the Arrhenius theory, it is an acid because the HCl molecules break down to give aqueous H+ ions:
HCl(aq) = H+(aq) + Cl-(aq) (3.1)
In this reaction the equilibrium lies far to the right, and so hydrochloric acid is called a strong acid. Acid strength may be expressed by the equilibrium constant which describes an equation such as 3.1. For acids, the equilibrium constant, which is also referred to as the dissociation constant, is denoted in general by Ka. So for Reaction 3.1,
Ka = [H+(aq)][Cl-(aq)]/[HCl(aq)] (3.2)
For HCl, Ka has a value of about 107 mol litre-1, and as [H+(aq] = [Cl-(aq)], an aqueous solution of HCl (1 mol litre-1) contains more than 103 as many H+ ions as HCl molecules. By contrast, when acetic (ethanoic) acid, CH3COOH, dissolves in water, the equilibrium lies far to the left:
CH3COOH(aq) = CH3COO-(aq) + H+(aq) (3.3)
So acetic acid is only slightly dissociated, and is described as a weak acid (Ka = 1.8 × 10-5 mol litre-1).
Similarly we recognize strong and weak bases by the extent to which they dissociate. Thus NaOH dissolves in water with virtually total dissociation to give Na+(aq) and OH-(aq) ions, and so is referred to as a strong base.
3.2 Brønsted–Lowry theory
By the Arrhenius definition, an acid such as HCl dissociates to give H+ ions:
HCl(aq) = H+(aq) + Cl-(aq) (3.1)
This definition does not give much prominence to the solvent: water only appears in the equation as the symbol (aq). Yet the solvent is important; HCl also dissolves in benzene, for example, yet the solution is non-conducting and fails to give characteristic acid tests. This is because it contains solvated HCl molecules rather than ions. By contrast, in aqueous solution, the hydrogen ion is present as the hydrated proton, which we represent as H3O+. The realization that the aqueous proton is bound to a solvent water molecule enables us to put this neglect of the solvent to rights.
* Rewrite Equation 3.1 so H3O+(aq) rather than H+(aq) appears on the right-hand side.
* HCl(aq) = H2O(l) = H3O+(aq) + Cl-(aq) (3.4)
This equation recognizes the solvent by including a solvent molecule on the left-hand side. What happens is that a proton, H+, is transferred from HCl to H2O (Figure 3.1). This transfer of protons is the basis of the Brønsted–Lowry theory:
