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Making the Transition to University Chemistry

Making the Transition to University Chemistry

Michael Clugston, Malcolm Stewart, and Fabrice Birembaut
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date: 17 June 2024

p. 106 Chapter 11 Trends Across the Periodic Tablefree

p. 106 Chapter 11 Trends Across the Periodic Tablefree

  • Michael Clugston, Michael Clugstonformerly Tonbridge School
  • Malcolm StewartMalcolm StewartUniversity of Oxford
  •  and Fabrice BirembautFabrice BirembautCaen, France

Abstract

This chapter explores trends across the periodic table. The physical trends across the periodic table include metal, non-metal, metalloids, and electrical conductivity. The chemical trends across the periodic table involve the oxidation state as elements shift drastically due to their reaction to oxygen. Phosphorus, sulfur, and chloride for a variety of oxides. The chapter also tackles the melting points, ionization energies, and atomic radii across periods 2 and 3 of the periodic table. The melting point of a metallic element is dependent on the strength of the metallic bonding and the attraction between metal ions and a delocalized sea of electrons.

11.1 Physical trends across the periodic table

The nature of the elements changes significantly across the periodic table.

The most commonly studied periods are Period 2 (Li to Ne) and especially Period 3 (Na to Ar).

On the left of the table the elements are metals, whereas on the right of the table the elements are non-metals. The stepped line drawn rather like a staircase in Figure 11.1 roughly marks the division between the metals and non-metals. Taking the Next Step 11.1 explains that a more careful approach would acknowledge the existence of the metalloids.

In Period 3 the electrical conductivity is high for the metallic elements sodium, magnesium, and aluminium. Silicon is a semiconductor. The other elements are all insulators, having very low conductivities. Across any period, the elements change from conductors to insulators.

FIGURE 11.1 The standard modern form of the periodic table. The deeper green colour identifies the metalloids. The s block is in yellow, the d block in blue, the p block in green, and the f block in red.

Taking the Next Step 11.1 p. 107Which elements are metalloids?

Five of the chemical elements are intermediate in nature between metals and non-metals. Silicon (Si) and germanium (Ge), for example, are semiconductors rather than conductors or insulators. Together with arsenic (As), antimony (Sb), and tellurium (Te), they are called metalloids. Any element to the left or below the metalloids is a metal. Any element to the right or above the metalloids is a non-metal.

11.2 Melting points across Periods 3 and 2

The melting point of a metallic element depends on the strength of the metallic bonding (Section 4.4) it experiences, and hence the attraction between the metal ions and the delocalized sea of electrons.

Aluminium has the highest melting point because its ion Al3+ has the highest charge density (Section 2.7) and aluminium contributes the largest number of electrons (3) into the delocalized sea.

The melting point gradually rises from Na to Mg to Al (although not smoothly), as shown in Figure 11.2.

FIGURE 11.2 Trends in the melting and boiling points of the Period 3 elements.

The next element after aluminium is silicon. Silicon is a giant covalent solid (Section 4.4) and as such its very strong covalent bonds need to be weakened to allow melting.

Si therefore has the highest melting point in Period 3, as seen in Figure 11.2.

Figure 11.2 shows that the next four elements (phosphorus, sulfur, chlorine, and argon) all have very low melting points, because the forces of attraction between the molecules are only dispersion forces.

p. 108We can explain the variation in melting point across these four elements by knowing the molecular formula of the element: P4, S8, Cl2, and Ar.

As dispersion forces increase as the number of electrons in a molecule increases (Section 4.2), sulfur (S8) has the highest melting point, followed by phosphorus (P4) then chlorine (Cl2) then argon (Ar).

Figure 11.3 shows that the trend in Period 2 is very similar to the trend in Period 3, except that the melting points for nitrogen, oxygen, and fluorine are very similar as they are all diatomic molecules.

FIGURE 11.3 Trends in the melting and boiling points of the Period 2 elements are very similar to the Period 3 trends, except that the values for nitrogen, oxygen, and fluorine are almost identical as they are all diatomic (N2, O2, F2).

11.3 Atomic radii across Period 3

Going across a period, the atoms get smaller, as shown in Figure 11.4.

FIGURE 11.4 Trends in atomic radius for Periods 2 and 3.

Each successive element has one more proton and one more electron. The extra electron does not shield the other electrons in the same shell effectively. Taking the Next Step 11.2 explains the concept of shielding in more detail.

Taking the Next Step 11.2 p. 109How does shielding change across Period 3?

As we move across a period, each successive element has one more proton and one more electron. The extra electron does not shield the other electrons in the same shell effectively. To understand why this is the case, we need to understand the concept of effective nuclear charge.

We define the effective nuclear charge Z eff as the actual nuclear charge minus the shielding constant (which takes into account the shielding effect of all the other electrons). Going across a period the effective nuclear charge increases, hence there is a stronger attraction to the nucleus felt by the outermost electrons. As a result, the atom becomes smaller.

John Slater introduced a simple model for approximating the value of the effective nuclear charge. Slater’s rules suggest that an electron in the same shell contributes only about 0.35 to the shielding constant, electrons in the penultimate shell contribute about 0.85, and electrons in shells even closer to the nucleus contribute 1 (hence shielding effectively perfectly). Slater’s rules applied to Na would give a Z eff value of 11 − (8 × 0.85) − (2 × 1) = 2.2. The rules would give a Z eff value for Si of 14 − (3 × 0.35) − (8 × 0.85) − (2 × 1) = 4.15 and a Z eff value for Ar of 18 − (7 × 0.35) − (8 × 0.85) − (2 × 1) = 6.75. Notice how these approximate values are close to the more exact calculated values shown in Table 11.1.

Table 11.1 Calculated effective nuclear charges across Period 3.

Atom

Electron ionized

Z eff

Na

3s

2.51

Mg

3s

3.31

Al

3p

4.07

Si

3p

4.29

P

3p

4.89

S

3p

5.48

Cl

3p

6.12

Ar

3p

6.76

11.4 Ionization energies across Period 3

Figure 11.5 shows how sodium has a low ionization energy (498 kJ mol−1) compared with the preceding element (neon, 2081 kJ mol−1), because the electron being ionized is in a new shell (shell 3 rather than shell 2), further away from the nucleus.

FIGURE 11.5 Plot of first ionization energy against atomic number for Z = 11 (Na) to Z = 18 (Ar). The plot shows that there is a periodic variation in ionization energy across Period 3.

p. 110 Figure 11.5 shows that there is another, less significant, drop in ionization energy for aluminium compared with magnesium, because the electron being ionized is in a new subshell (3p rather than 3s), which is higher in energy.

The rise from sodium to magnesium, the rise from aluminium to phosphorus, and the rise from sulfur to argon, Figure 11.5, all occur because the number of protons increases by one for each successive element and the extra electron added as well is poor at shielding (see Taking the Next Step 11.2).

The drop in ionization energy from phosphorus to sulfur is much more difficult to explain.

You will probably have come across the explanation that this drop occurs because there is additional electron–electron repulsion between the two electrons in one orbital for sulfur. Such an explanation, while sounding convincing, is not in agreement with detailed calculations.

11.5 Chemical trends across the periodic table

The oxidation state (Section 8.2) of the elements changes significantly across the periodic table, as judged especially by their reaction with oxygen.

In Period 3, the oxidation number of the highest oxide gradually increases by one from Ox(Na) = +1 in Na2O to Ox(Cl) = +7 in Cl2O7.

It was this smooth trend which convinced Mendeleyev that his periodic table would be very valuable.

Phosphorus, sulfur, and chlorine form a variety of oxides. See Taking the Next Step 11.3.

Phosphorus forms P4O6 and P4O10.

Sulfur forms SO2 and SO3.

Chlorine forms several oxides including Cl2O, ClO2, and Cl2O7.

Taking the Next Step 11.3 Naming the oxides

p. 111The way of naming the oxides depends on the nature of the element. For the metals (sodium, magnesium, and aluminium), the names are sodium oxide, magnesium oxide, and aluminium oxide. We do not need to include an oxidation state, as we need to do in the case of iron(III) oxide for example, because these elements only have one oxidation state.

For metalloids and non-metals, the name usually reflects the formula of the compound—either the molecular formula for a gaseous oxide or the empirical formula for a solid oxide. So silicon dioxide is the name for SiO2, sulfur dioxide for SO2, sulfur trioxide for SO3, dichlorine oxide for Cl2O, chlorine dioxide for ClO2, and dichlorine heptoxide for Cl2O7.

An exception to this rule is that P4O6 and P4O10, which would consistently be called tetraphosphorus hexoxide and tetraphosphorus decoxide, have recommended names of phosphorus(III) oxide and phosphorus(V) oxide.

Phosphorus(V) oxide, an important dehydrating agent, was traditionally called phosphorus pentoxide as the solid has an empirical formula (Section 3.1) of P2O5.

11.6 Reaction with oxygen across Period 3

All the elements from sodium to sulfur burn in oxygen or air, forming an oxide.

Sodium burns with a yellow flame (Section 1.3 discusses the strong yellow line in sodium’s atomic emission spectrum).

Most of them (magnesium, aluminium, silicon, and phosphorus) generate white light from the strongly exothermic reaction and produce a white smoke containing the element’s oxide.

Sulfur burns with a blue flame and produces a pungent colourless gas.

4Na(s)+O2(g)2Na2O(s)(seeA Deeper Look 11.1)2Mg(s)+O2(g)2MgO(s)4Al(s)+3O2(g)2Al2O3(s)Si(s)+O2(g)SiO2(s)P4(s)+5O2(g)P4O10(s)S(s)+O2(g)SO2(g)

This is the usual equation for the combustion of sulfur. (In the manufacture of sulfuric acid, the temperature is such that the sulfur is liquid and consists of individual atoms.) Given that the solid contains S8 molecules (Section 11.2), it would also be correct to write S8(s) + 8O2(g) → 8SO2(g).

Taking the Next Step 11.4 discusses the nature of the oxides across Period 3.

A Deeper Look 11.1 What else can Group 1 elements form with oxygen?

In addition to the oxide, sodium also forms some sodium peroxide, Na2O2, which contains the peroxide ion O2 2−. In addition to the oxide and peroxide, potassium forms some potassium superoxide KO2, which contains the superoxide ion O2 (Section 2.1).

p. 112Potassium superoxide is used for purifying the air in submarines by reacting with carbon dioxide to form oxygen:

4KO2(s)+2CO2(g)2K2CO3(s)+3O2(g)

Taking the Next Step 11.4 How do the oxides change across Period 3?

Going across the table, we find significant changes in the nature of the bonding in the oxides and hence in the state in which we find the oxide, as seen in Figure 11.6.

FIGURE 11.6 Plot of melting point for the oxides of the elements of Period 3. Note that, where an element has more than one oxide, we choose the oxide corresponding to the highest oxidation number.

The metals sodium, magnesium, and aluminium form oxides (Na2O, MgO, and Al2O3) which are predominantly ionic and so exist as solids with high melting points.

The metalloid silicon forms silicon dioxide (SiO2) which is predominantly covalent and adopts a giant covalent structure, shown in Figure 11.7, which is therefore also a solid with a high melting point.

FIGURE 11.7 The silicon atoms in one form of silica adopt the diamond structure (Figure 4.19); there are oxygen atoms between each pair of silicon atoms.

The oxides of the non-metals phosphorus and sulfur also have covalent bonding, but these oxides are simple molecular: P4O10 is a low melting- point solid (see A Deeper Look 11.2). Sulfur dioxide SO2 is a gas.

Taking the Next Step 11.5 discusses the changes in pH of solutions containing the oxides.

Silicon dioxide, also called silica, is a major constituent of sand.

A Deeper Look 11.2 What are the structures of P4O6, P4O10, and SO3?

The structure of P4O6 in the gas phase, shown in Figure 11.8(a), is based on a tetrahedron of phosphorus atoms with each phosphorus atom linked to three neighbours by oxygen atom bridges. (Note how the bottom part of the structure resembles the shape of cyclohexane.)

The structure of P4O10 in the gas phase, shown in Figure 11.8(b), is based on that of P4O6 but with each phosphorus atom having an additional double bond to an oxygen atom.

p. 113Sulfur trioxide molecules join together in the solid state to form chains and rings (as shown in Figure 11.9). The ring structure explains why the melting point of SO3 (see Figure 11.6) falls between those of P4O10 and Cl2O7 as the solid is effectively S3O9.

FIGURE 11.8 The structures of (a) P4O6 and (b) P4O10.

FIGURE 11.9 (a) The structure of an SO2 molecule. (b) SO3 molecules join together to make chains and rings.

Taking the Next Step 11.5 How does the pH of the oxide’s aqueous solution change?

There is a significant change in the pH of aqueous solutions containing the oxides from alkaline for Na2O to acidic for SO3 and the chlorine oxides. A Deeper Look 11.3 explains why this happens.

Na2O reacts with water, as illustrated in Figure 11.10, to form the strong alkali (Section 7.2) sodium hydroxide. The pH of the solution is close to 14:

Na2O(s)+H2O(l)2NaOH(aq)

FIGURE 11.10 Solid sodium oxide Na2O(s) reacts with water exothermically. The product is aqueous sodium hydroxide NaOH(aq), which turns the indicator blue.

MgO also reacts very slowly to form the weak base (Section 7.3) magnesium hydroxide, hence the pH of the solution is only 10:

MgO(s)+H2O(l)Mg(OH)2(aq)

Neither Al2O3 nor SiO2 dissolves in water, so water remains at pH 7.

p. 114 P4O10 reacts with water to form the weak acid (Section 7.3) phosphoric acid, and the pH of the solution is about 1:

P4O10(s)+6H2O(l)4H3PO4(aq)

You may have been told that phosphoric acid is a strong acid. In fact its pK a1 value is 2.15. The equilibrium measured by pK a1 (that is, the pK a value for its first ionization) is H3PO4(aq) + H2O(l) ⇌ H3O+(aq) + H2PO4 (aq).

SO2 reacts with water to form the weak acid sulfurous acid, and the pH of the solution is about 1:

SO2(g)+H2O(l)H2SO3(aq)

High levels of sulfur dioxide in the atmosphere contribute to acid rain.

You may have been told that sulfur dioxide is ‘moderately soluble’. It is in fact the third most soluble gas.

SO3 reacts with water to form the strong acid (Section 7.2) sulfuric acid, so the pH of the solution is close to 0:

SO3(s)+H2O(l)H2SO4(aq)

There are a number of chlorine oxides, which all dissolve in water to form acidic solutions. For example, dichlorine oxide Cl2O forms HOCl (Section 13.6).

A Deeper Look 11.3 How can we explain the trend from alkaline to acidic oxides?

Think of a part of a compound involving an element E and a hydroxide (E–O–H).

For an element of low electronegativity such as sodium, the Na–O bond is highly polarized towards the oxygen, and so it is reasonable to imagine this producing an Na+ ion and a hydroxide ion OH. The hydroxide ion will make the solution alkaline.

The O–H bond is highly polarized towards the oxygen. An element of high electronegativity such as chlorine enhances this polarization by also withdrawing electron density towards itself. This therefore allows the loss of a proton H+ from the O–H bond. The proton will make the solution acidic.

© Michael Clugston, Malcolm Stewart, and Fabrice Birembaut 2021