Ionic Bonds and Structure | Chemistry | PDF

Structure 2.1 The ionic model What determines the ionic nature and properties of a compound? Ionic compounds are characterized by the presence the strong electrostat...

Structure 2.1 The ionic model What determines the ionic nature and properties of a compound? Ionic compounds are characterized by the presence the strong electrostatic attractions between oppositely of positive and negative ions, which attract each other charged ions. Once liquid, however, ionic compounds are electrostatically. In solid ionic compounds, these ions electrical conductors due to the presence of mobile ions. are arranged in rigid crystalline lattices. Melting these Due to their charge, ions interact strongly with polar water solids requires a large amount of thermal energy due to molecules, so ionic compounds are oen water-soluble. Understandings Structure 2.1.1 — When metal atoms lose electrons, they Structure 2.1.3 — Ionic compounds exist as three- form positive ions c alled c ations. When non-metal atoms dimensional lattice structures, represented by empiric al gain electrons, they form negative ions c alled anions. formulas. Structure 2.1.2 — The ionic bond is formed by electrostatic attractions between oppositely charged ions. Binary ionic compounds are named with the c ation rst, followed by the anion. The anion adopts the sux “ide”. Introduction to bonds and structure Atoms rarely exist in isolation. They are connected together in several dierent ways. Atoms c an be bonded to atoms of the same type, or to atoms of dierent elements. The varying arrangements of atoms and features of the bonds between them give rise to certain dierent properties. For example, 78% of the air around us is nitrogen, N. However, in agriculture, nitrogen fertilizers are added to soils 2 to help crops grow. This is bec ause the structure and bonding of nitrogen in air are dierent to that of the nitrogenous compounds found in fertilizers. Atoms are held together by chemical bonds. This chapter discusses three dierent bonding models: ionic, covalent and metallic. These lead to four types of structure: ionic, molecular covalent, covalent network and metallic. You may be wondering why there are four types of structure, given that there are only three types of bonds. This is because covalent substances can be found in two arrangements: a continuous 3D network, or discrete groups of atoms known as molecules. types of ionic covalent metallic bond types of molecular covalent ionic metallic structure covalent network metal ion deloc alised electron tFigure 1 There are three types of bonds and four types of structure 95  Structure 2 Models of bonding and structure Chemic al bonds Models Chemic al bonds are strong forces of attraction that hold atoms or ions together Structure 2.1 , 2.2 and 2.3 d i s c u ss in a substance. All chemic al bonds occur due to electrostatic attractions between models of bonding and s t r u c t u r e. positively charged species and negatively charged species. The type of bonding Scientific models simplify and depends on which species are involved (table 1). represent c o m p l ex phenomena. Sometimes models help us to visualize things that we c annot The electrostatic attraction between… observe d i r e c t l y. This is one of Type of bonding Positively charged Negatively charged the re asons bonding models are species species so useful. ionic c ations anions All models have limitations. covalent atomic nuclei shared pair of electrons This does not necessarily make metallic c ations deloc alized electrons the models inadequate, but it is important to understand the Table 1 All bonding types involve a positively charged species and a negatively charged weaknesses of a model. As you species that are electrostatic ally attracted to each other work through these sections, identify some of the strengths and limitations of the various bonding models. Ions (Structure 2.1.1) Sodium chloride and copper(II) sulfate are examples of ionic compounds. They are crystalline and brittle, which are properties characteristic of ionic compounds. Ionic compounds are also poor electric al conductors when solid, but good electric al conductors when molten or dissolved. The reactions and properties of these ionic compounds are very dierent to those of their constituent elements. For instance, sodium chloride, the main ingredient in table salt, is water-soluble. However, elemental sodium is a so metal that reacts violently with water, and chlorine is a poisonous gas. Figure 3 Sodium chloride crystals on a tree branch and copper(II) sulfate crystals. Sodium chloride and copper(II) sulfate are ionic compounds Figure 2 Oshore oil platform in C alifornia, USA. What examples of structure Before discussing ionic bonds and the characteristics of ionic structures, we will and bonding are present in the photo? rst look into what ions are. 96  Structure 2.1 The ionic model a b C ations and anions + Sodium chloride contains sodium ions, not sodium atoms. Sodium atoms and sodium ions have dierent numbers of electrons, and therefore behave dierently. Na Na + You will notice three dierences between Na and Na : 1. number of electrons Figure 4 (a) sodium atom (b) sodium ion 2. electron arrangement 3. charge. Sodium atoms are neutral. Sodium ions have a 1+ charge, indic ated by a + superscript + sign next to the symbol: Na. Worked example 1 Determine the number of subatomic particles to show that a. sodium atoms are neutral b. sodium ions have a 1+ charge. Solution In Structure 1.2, you learned a. In a sodium atom there are: that protons have a 1+ charge 11 protons (charge = 11+) and electrons have a 1– charge. 11 electrons (charge = 11 ) You c an ignore neutrons in ionic O verall charge is 11 11 = 0 charge c alculations as these are b. In a sodium ion there are: uncharged. 11 protons (charge = 11+) 10 electrons (charge = 10 ) O verall charge is 11 10 = 1+ Worked example 2 Deduce the electron conguration of a sodium atom and a sodium ion. a b 2– Solution 2 2 6 1 Na: 1s 2s 2p 3s S S + 2 2 6 Na : 1s 2s 2p C ations are ions with more protons than electrons. This means that c ations are Figure 5 (a) sulfur atom (b) sulde ion positively charged, as the combined positive charge of protons is greater than the combined negative charge of electrons. As sodium ions have 11 protons and 10 electrons, the overall charge is 1+ Activity Anions are negatively charged ions. They contain a greater number of electrons than protons. Figure 5 shows a sulfur atom and a sulde ion. The sulde ion has a Show that the sulfur atom is neutral 2 2 charge, denoted by the superscript in the symbol S. Note that anions adopt and the sulde ion has a charge a slightly dierent name: the rst part corresponds to the name of their parent of 2– by counting their subatomic atom. This is followed by the sux ide. particles. Determine the electron conguration of the sulfur atom For now, we will only consider monatomic ions. You will look at charged groups and sulde ion. of atoms (c alled polyatomic ions) in a later section. 97  Structure 2 Models of bonding and structure Predicting the charge of an ion The main group elements are in periodic table groups 1, 2, 13, 14, 15, 16, 17 and 18. The electron congurations for some main group element atoms and their corresponding c ations are shown below: 1e + Na Na 2 2 6 1 2 2 6 1s 2s 2p 3s 1s 2s 2p + Na has the same electron conguration as neon, Ne. Two dierent species with + same electron conguration are c alled isoelectronic. Therefore, Na and Ne are isoelectronic. 1e 2e + 2+ Li Li Ca Ca 2 1 2 2 2 6 2 6 2 2 2 6 2 6 1s 2s 1s 1s 2s 2p 3s 3p 4s 1s 2s 2p 3s 3p 2+ + Li is isoelectronic with helium, He. Ca is isoelectronic with argon, Ar. The resulting c ations all have noble gas congurations. Noble gases have full (or “closed”) sublevels. When main group elements form ions, they oen achieve this noble gas electron conguration. The atoms above have all done so by losing their outermost valence electrons. As they have lost electrons, and electrons are negatively charged, the resulting c ations are positively charged. C ation formation is an example of oxidation bec ause it involves the loss of electrons. Anions are formed when atoms gain electrons. Look at the examples below where the parent atoms gain electrons in order to achieve a noble gas electron conguration: +1e +2e 2 Cl Cl O O 2 2 6 2 5 2 2 6 2 6 2 2 4 2 2 6 1s 2s 2p 3s 3p 1s 2s 2p 3s 3p 1s 2s 2p 1s 2s 2p – 2– Cl is isoelectronic with argon, Ar. O is isoelectronic with neon, Ne. Atoms that gain electrons become anions. As reduction is the gain of electrons, To obtain a noble gas conguration, the formation of anions is a reduction process. a chlorine atom gains an electron. The formation of an ionic compound from its elements is a redox reaction. Chlorine would also have a Consider the formation of sodium chloride from its elements: noble gas conguration if it lost the seven outermost electrons. 2Na(s) + Cl (g) 2NaCl(s) 2 However, the removal of so many + electrons from the attractive pull Sodium chloride, NaCl, is made up of sodium c ations, Na , and chloride of the positively charged nucleus anions, Cl. The half equations are shown below. The rst is an oxidation and the requires a large amount of energy other is a reduction and therefore the formation of NaCl from its elements is a while the addition of a single redoxreaction. electron releases energy. This is + 2Na 2Na + 2e Electron loss = oxidation why chlorine instead will gain an Cl + 2e 2Cl Electron gain = reduction electron to become a chloride ion. 2 The energetics of these processes, Once you have learned about oxidation states (Structure 3.1), you should also be called ionization energy and able to see that the sodium is undergoing oxidation bec ause its oxidation state electron anity, are discussed in increases (from 0 to +1) and the chlorine is reduced bec ause its oxidation state Structure 3.1 and relevant in the decreases (from 0 to 1). construction of Born–Haber cycles (Reactivity 1.2). 98  Structure 2.1 The ionic model Atoms tend to achieve a noble gas electron conguration through gaining, TOK losing, or, as we will see in Structure 2.2, sharing electrons. This is oen referred to as the octet rule. It is c alled the octet rule bec ause most noble gases have General rules in chemistry (such eight electrons in their outer shell. as the octet rule) oen have There exists a relationship between the charge of the ion formed by a main group exceptions. How many exceptions element and its periodic table group. In general: have to exist for a rule to cease to be useful? Elements in groups 1, 2 and 13 form 1+, 2+ and 3+ ions, respectively Elements in groups 15, 16 and 17 form 3–, 2– and 1– ions, respectively The electron conguration of 2 2 2 Elements in group 18 (noble gases) do not form ions c arbon, 1s 2s 2p , suggests that c arbon atoms could lose or gain The relationship between periodic table group and ionic charge is illustrated in four electrons in order to achieve gure 6. a noble gas conguration. This would result in the formation 1 18 4+ 4– of C or C ions, respectively. Although this is possible, c arbon 1 2 13 14 15 16 17 more commonly forms compounds through a process c alled covalent + Li 3 2 2 N O F bonding, which does not involve ion formation. Covalent bonding is + 2+ 3+ Na Mg Al 3 2 discussed in Structure 2.2 3 P S Cl + 2+ K Ca 2 4 Se Br + 2+ Rb Sr 2 5 Te I + 2+ 6 Cs Ba Figure 6 The charges of some common ions Hydrogen atoms have only one electron in the 1s sublevel. They form ions by either losing that electron or gaining one. Electron loss leads to the formation of + H , which is simply a hydrogen nucleus: a proton with no electrons surrounding + it. The charge density of a H ion is therefore very high, so these c ations readily combine with other species. One such example is the formation of acidic + hydronium ions, H O , formed when hydrogen c ations bond with water. 3 Hydrogen atoms c an also gain an electron to achieve a noble gas conguration, Hydride anions are very strong – thus forming hydride anions, H. bases. You will learn more about bases in Reactivity 3.1 hydrogen + – + e ion, H –1 + + 1 e hydride hydrogen + ion, H atom + Figure 7 The formation of H and H 99  Structure 2 Models of bonding and structure Practice questions 1. Determine the charge of the ion formed by each of the following elements. a. lithium, Li b. magnesium, Mg c. aluminium, Al d. uorine, F e. nitrogen, N f. selenium, Se g. barium, Ba 2. State the name of ions d, e and f above. 3. Complete the table: Number Number Electron Name Symbol of of Charge conguration protons electrons beryllium 0 + K 8 2 15 18 + H 4. For each electron conguration given, identify three isoelectronic species: 2 2 6 2 6 a. 1s 2s 2p 3s 3p 2 b. 1s 5. Explain why noble gases do not form ions. A transition element is an element with a partially lled d sublevel. In contrast to main group elements, a transition element c an form multiple ions with 2+ 3+ dierent charges. For example, iron commonly forms Fe and Fe ions (gure 8). + 2 Fe : [Ar] – e 2 – 3d 4s Fe: [Ar] iron(II) ion 3d 4s – 3 e – iron atom + 3 Fe : [Ar] 3d 4s iron(III) ion Figure 8 Iron atoms c an form ions with a 2+ charge and ions with a 3+ charge 100  Structure 2.1 The ionic model Consider the electron congurations of the rst-row transition elements. As seen in chapter Structure 1.3, 2+ the 4s sublevel lls up before the Most of them contain two 4s electrons, which are lost when the M ions 3d sublevel. areformed. This helps to explain why most of these elements commonly form 2+c ations. tTable 2 Electron congurations of the Electron conguration rst-row transition elements Symbol Element Atom 2+ ion 2 1 1 Sc sc andium [Ar] 4s 3d [Ar] 3d 2 2 2 Ti titanium [Ar] 4s 3d [Ar] 3d 2 3 3 V vanadium [Ar] 4s 3d [Ar] 3d 1 5 4 Cr chromium [Ar] 4s 3d [Ar] 3d 2 5 5 Mn manganese [Ar] 4s 3d [Ar] 3d 2 6 6 Fe iron [Ar] 4s 3d [Ar] 3d 2 7 7 Co cobalt [Ar] 4s 3d [Ar] 3d 2 8 8 Ni nickel [Ar] 4s 3d [Ar] 3d 1 10 9 Cu copper [Ar] 4s 3d [Ar] 3d When the rst row transition elements are ionized, the 4s electrons are lost before Practice questions the 3d electrons. Further successive ionizations occur in many of these elements 6. Deduce the abbreviated electron bec ause the 3d sublevel is similar in energy to the 4s sublevel. conguration of each of the Transition elements have variable oxidation states (Structure 3.1). This following: characteristic c an be explored by examining successive ionization energy data. 2+ a. Mn Let’s focus on the transition elements in period 4. They have variable oxidation 3+ b. V states bec ause the 4s and 3d sublevels are close together in energy, as shown + c. Cu by successive ionization energy data (gure 9). It is important to realize that 2+ d. Cu ionization rarely happens in isolation. Ionization absorbs energy, but this energy 7. Zinc only forms 2+ ions. investment is usually oset by other processes that release energy, such as lattice formation. If a certain ionization only requires a small amount of additional energy a. Deduce the full electron 2+ compared to the previous ionization, then it could be energetic ally favourable if it conguration of Zn. leads to a subsequent exothermic process. b. Explain why zinc is not a transition element. 8. The ion of a transition metal 3 40 has mass number 55, electron lom Jk 5 conguration [Ar] 3d and a charge of 2+. 3 30 01 / ygrene a. Write its symbol using nuclear notation. 20 b. Identify a 1+ ion that has the same electron conguration n o it a z i n o i 10 as the above. 0 0 2 4 6 8 10 12 number of electrons removed Figure 9 D ata for the rst 12 ionization energies of iron. As you c an see, the 4s and 3d electrons are very close together in energy. The large jump between the 8th and 9th electrons occurs bec ause the 9th electron is removed from the 3p energy level, which is closer to the nucleus 101  Structure 2 Models of bonding and structure ATL Communic ation skills You may have noticed that we c an refer to charge using dierent formats depending on context. When using chemical symbols, charges appear as 3+ a superscript number followed by + or –, for example, Fe. In speech or writing, we say “the ion has a 3+ charge”. Charge is related to oxidation state (Structure 3.1), where the + or – sign is given rst followed by the magnitude. For example, “oxygen has an oxidation state of –2”. Roman numerals are also used to indic ate oxidation states in the names of compounds (Structure 3.1 and Reactivity 3.2). For example, the symbol for a 2+ copper(II) ion is Cu , its charge is 2+, and its oxidation state is +2. Write your own example to help you remember these distinct ways of referring to ionic charge. Linking questions How does the position of an element in the periodic table relate to the charge of its ion(s)? (Structure 3.1) How does the trend in successive ionization energies of transition elements explain their variable oxidation states? (Structure 1.3) Ionic bonds (Structure 2.1.2) C ations and anions are electrostatic ally attracted to each other bec ause of their opposite charges. This attraction results in the formation of ionic bonds. Therefore, if a given element forms c ations, and another forms anions, they c an bond ionic ally to form an ionic compound. ecnereffid ionic 3.0 Electronegativity (χ) greater bonding ionic One way to estimate whether a bond between two given elements is ionic is to character look at the dierence in electronegativity between the two. Electronegativity ytivitagenortcele 2.0 ( χ) is a measure of the ability of an atom to attract a pair of covalently bonded electrons. Within the periodic table, electronegativity increases across the periods and up the groups. This means that uorine is the most electronegative 1.0 element, so it has a high tendency to attract pairs of covalently bonded electrons. One of the electronegativity sc ales used by chemists is c alled the Pauling sc ale. 0 Values in the Pauling sc ale are dimensionless and range from 0.8 to 4.0. Fluorine has an electronegativity value of 4.0. The electronegativity of c aesium, one of the Figure 10 If two elements have an least electronegative elements, is 0.8. Noble gases are generally not assigned electronegativity dierence greater than 1.8, electronegativity values. the bonding between them will have a high ionic character The larger the dierence in electronegativity between two elements in a compound, the greater the ionic character of the bond between them. Ionic bonding is assumed to occur when the dierence in electronegativity is greater Electronegativity and other than 1.8 (gure 10). In reality, bonding occurs across a continuum, so above 1.8 periodic trends are discussed in the main type of bonding in the compound is ionic, but there may be other types greater detail in Structure 3.1 of bonding present. 102  Structure 2.1 The ionic model Data-based question Predict which of the compounds in table 3 will have ionic structure. χ Compound Dierence in electronegativity (∆ ) χ χ sodium uoride, (Na) = 0.9 and ∆ (F) = 4.0 χ NaF ∆ = 3.1 χ χ sodium chloride, (Na) = 0.9 and (Cl) = 3.2 χ NaCl ∆ = 2.3 χ χ aluminium (Al) = 1.6 and (Cl) = 3.2 χ chloride, AlCl ∆ = 1.6 3 Table 3 Electronegativity dierences for selected metal chlorides Periodic table position You c an qualitatively approximate how ionic a compound will be by looking at the positions of its constituent elements in the periodic table. Elements with large dierences in electronegativity are generally found at a greater horizontal distance from each other. Worked example 3 Compare the ionic character of bonding in the following pairs of compounds: a. c aesium uoride, C sF, and c aesium iodide, C sI b. magnesium oxide, MgO, and c arbon monoxide, CO Solution a. Qualitative comparison: b. Qualitative comparison: Cs and F are a gre ater distance from e ach other In the periodic table, Mg and O are further away than Cs and I are in the periodic table. Therefore, from e ach other than C and O are. Therefore, the the dierence in electronegativity is larger between electronegativity dierence between Mg and O Cs and F, me aning the bond between them is more must be larger than that between C and O, and the ionic. bond between Mg and O must be more ionic. Q uantitative comparison: Q uantitative comparison: χ(C s) = 0.8 and χ(F) = 4.0 χ(Mg) = 1.3 and χ(O) = 3.4 ∆χ(C sF) = 3.2 ∆χ (MgO) = 2.1 χ(C s) = 0.8 and χ(I) = 2.7 χ(C) = 2.6 and χ(O) = 3.4 ∆χ(C sI) = 1.9 ∆χ(CO) = 0.8 Both ∆χ v alues are gre ater than 1.8, so the bonds Mg and O bond ionic ally bec ause ∆χ is gre ater in both compounds are ionic. However, C sF has a than 1.8 for this compound. C and O do not bond higher percentage ionic character than C sI. ionic ally bec ause ∆χ is lower than 1.8. 103  Structure 2 Models of bonding and structure Activity Determine whether the following pairs of elements are likely to bond ionic ally using the following two methods: i. look at their positions in the periodic table ii. refer to their electronegativity values in the data booklet. a. Li and F d. As and S b. Rb and Ga e. P and Cl c. C a and I f. Ag and Br It is oen incorrectly said that only ionic bonds form when a metallic element and a non-metallic element bond together. There are substances, such as aluminium chloride, AlCl , that do not t this description. Aluminium is a metal and chlorine 3 is a non-metal, so you would expect them to bond ionic ally. But the compound has properties that are characteristic of covalent compounds, such as a low Figure 11 Polarized light micrograph melting point and high volatility. The electronegativity dierence between these of ammonium nitrate crystals. Ammonium + two elements (1.6) suggests they do not bond ionic ally. nitrate contains two polyatomic ions: NH 4 and NO. Its uses include fertilizers and 3 Polyatomic ions rocket propellants Some ionic compounds contain more than two elements. For instance, NH Cl, 4 + + which is made up of NH c ations and Cl anions. Ammonium ions, NH , are Name Formula 4 4 polyatomic ions. As their name suggests, polyatomic ions are ions that contain + ammonium NH 4 several atoms. hydroxide OH You are expected to know the names and formulas of the polyatomic ions shown nitrate NO 3 in table 4. hydrogenc arbonate HCO 3 2– c arbonate CO 3 2– ATL Self-management skills sulfate SO 4 3– phosphate PO 4 You will need to spend some time memorizing the names and formulas of the polyatomic ions in table 4. Some students like to use ashc ards, others make Table 4 Common polyatomic ions up mnemonics. What strategies will you use? How will you make sure you actively engage with them? Naming ionic compounds Name Formula Consider the list of ionic compounds shown in table 5. C an you notice any potassium uoride KF patterns in their names? magnesium uoride MgF 2 You should notice that, in the names of ionic compounds: c alcium c arbonate C aCO 3 the c ation name is given rst and is followed by the anion barium hydroxide Ba(OH) 2 iron(III) oxide Fe O c ations adopt the name of the parent atom and the name remains unchanged 2 3 silver(I) sulde Ag S monatomic anions adopt the rst part of the name of the parent atom, 2 followed by the sux -ide. If the anion is polyatomic, refer to table 4 Table 5 Names and formulas of some ionic compounds the name of the compound does not reect the number of ions in the formula. 104  Structure 2.1 The ionic model Practice questions Anions are conjugate bases of common acids. The strength of 9. State the name of each of the following compounds: acids and stability of their anions a. RbF d. Sr(OH) 2 c an be compared quantitatively b. Al S e. BaCO 2 3 3 using their dissociation constants, c. AlN f. NH HCO K , which will be introduced in 4 3 a Reactivity 3.1 The formulas of ionic compounds The name of an ionic compound tells you what elements it contains, but not the ratio of the ions in it. The basis for working out the formula of an ionic compound is remembering that the net charge of the compound is zero, so the positive charges and negative charges must c ancel out. First, determine the charge of the anion and the c ation, then work out how many of each ion you need to reach a total charge of zero. Worked example 4 Deduce the formulas of the following ionic compounds: a. c alcium oxide b. c alcium nitride c. sodium c arbonate d. aluminium nitrate Solution a. To deduce the formula of c alcium oxide, work through The second method is the criss-cross rule. Swap the following steps. the charges and turn them into the other ion’s subscript, ignoring the sign: Step 1: Determine the charges of the c ation and the anion Ca C alcium has an electron conguration of 2 2 6 2 6 2 1s 2s 2p 3s 3p 4s. C alcium atoms have two outer shell electrons, so they form ions with a 2+ charge. 2 2 4 Ca Oxygen has an electron conguration of 1s 2s 2p. Oxygen atoms have six outer shell electrons, so they Then, simplify the ratio: form ions with a 2 charge. 2+ 2 Therefore, calcium ions = Ca and oxide ions = O Ca O 1 1 Step 2: Determine how many of each ion are needed Step 3: Check that the net charge is zero in order to achieve a net charge of zero You check your working by adding up the charges of There are two methods you can use for this step. The rst e ach individual ion. If you did Step 2 correctly, the is the bar diagram method. Write out the ions as blocks charges will add to zero: equal to the number of charges on each individual ion: 2+ Ca + 2 Ca Total positive charge = 2+ 2 2– O O Total negative charge = 2 The bar diagram contains one c alcium ion and one Net charge = 2 2 = 0 oxide ion, so the ratio of c alcium to oxide in the compound is 1:1. Step 4: Write the formula This is a straightforward example where the Ca O 1 1 magnitude of charge is equal for the c ation and anion, and hence the formula is C aO.