Chemical Bonding PDF

4.1 Types of chemical bonding IMPORTANT When elements form compounds, they either gain, lose or share electrons to get to the nearest stable noble gas electronic configuration. There are exceptions to this rule for some Group 15, 16 and 17 elements. Ionic bonding is the electrostatic attr...

4.1 Types of chemical bonding IMPORTANT When elements form compounds, they either gain, lose or share electrons to get to the nearest stable noble gas electronic configuration. There are exceptions to this rule for some Group 15, 16 and 17 elements. Ionic bonding is the electrostatic attraction between positive ions (cations) and negative ions (anions) in an ionic crystal lattice. Covalent bonds are formed when the outer electrons of two atoms are shared. The ionic or covalent bonds formed are usually very strong: it takes a lot of energy to break them. There is also a third form of strong bonding: metallic bonding. Although the atoms within molecules are kept together by strong covalent bonds, the forces between molecules are weak. We call these weak forces between molecules van der Waals’ forces. The term van der Waals’ forces is a general term used to describe all intermolecular forces. There are several types of van der Waals forces: dipole (instantaneous dipole–induced dipole (id–id) forces. These are also called London dispersion forces. permanent dipole–permanent dipole (pd–pd) forces. hydrogen bonding, which is a stronger form of permanent dipole–permanent dipole force. If you understand these different types of chemical bonding and understand intermolecular forces, this will help you to explain the structure and some physical properties of elements and compounds. 4.2 Ionic bonding How are ions formed? One way of forming ions is for atoms to gain or lose one or more electrons. Positive ions are formed when an atom loses one or more electrons. Metal atoms usually lose electrons and form positive ions. Negative ions are formed when an atom gains one or more electrons. Non-metal atoms usually gain electrons and form negative ions. The charge on the ion depends on the number of electrons lost or gained (see Section 2.4). When metals combine with non-metals, the electrons in the outer shell of the metal atoms are transferred to the non-metal atoms. Each non-metal atom usually gains enough electrons to fill its outer shell. As a result of this, the metal atoms and non-metal atoms usually end up with outer electron shells that are complete: they have the electronic configuration of a noble gas. In Figure 4.2 we can see that: the sodium ion has the same electronic structure as neon: [2,8]+ the chloride ion has the same electronic structure as argon: [2,8,8]−. The strong force of attraction between the positive ions and negative ions in the ionic crystal lattice results in an ionic bond. An ionic bond is sometimes called an electrovalent bond. In an ionic structure such as sodium chloride crystals (Figure 4.3), the ions are arranged in a regular repeating pattern (see Chapter 5). As a result of this, the force between one ion and the ions of opposite charge that surround it is very great. In other words, ionic bonding is very strong. Figure 4.2: The formation of a sodium ion and chloride ion by electron transfer. Figure 4.3: These crystals of salt are made up of millions of sodium ions and chloride ions. Dot-and-cross diagrams You will notice that in Figure 4.2 we use dots and crosses to show the electronic configuration of the chloride and sodium ions. This helps us keep track of where the electrons have come from. It does not mean that the electron transferred is any different from the others. Diagrams like this are called dot-and- cross diagrams. When drawing a dot-and-cross diagram for an ionic compound it is usually acceptable to draw the outer electron shell of the metal ion without any electrons. This is because it has transferred these electrons to the negative ion. Figure 4.4 shows the outer shell dot-and-cross diagram for sodium chloride. A dot-and-cross diagram shows: the outer electron shells only the charge of the ion is spread evenly, by using square brackets the charge on each ion, written at the top right-hand corner. Figure 4.4: Dot-and-cross diagram for sodium chloride. Some examples of dot-and-cross diagrams Magnesium oxide When magnesium reacts with oxygen to form magnesium oxide, the two electrons in the outer shell of each magnesium atom are transferred to the incompletely filled orbitals of an oxygen atom. By losing two electrons, each magnesium atom achieves the electronic configuration [2,8] (Figure 4.5). By gaining two electrons, each oxygen atom achieves the electronic configuration [2,8]. [2,8] is the electronic configuration of neon. It is sometimes called a noble-gas configuration. When ions or atoms have 8 electrons in their outer shell like this it is called an octet of electrons (the prefix oct means 8). Figure 4.5: Dot-and-cross diagram for magnesium oxide. Calcium fluoride Each calcium atom has two electrons in its outer shell, and these can be transferred to two fluorine atoms. By losing two electrons, each calcium atom achieves the electronic configuration [2,8,8] (Figure 4.6). The two fluorine atoms each gain one electron to achieve the electronic configuration [2,8]. [2,8] is the electronic configuration of neon; it is a ‘noble-gas configuration’. Figure 4.6: Dot-and-cross diagram for calcium fluoride. Question 1 Draw dot-and-cross diagrams for the ions in the following ionic compounds. Show only the outer electron shells. a Potassium chloride, KCl b Sodium oxide, Na2O c Calcium oxide, CaO d Magnesium chloride, MgCl2 4.3 Covalent bonding Single covalent bonds When two non-metal atoms combine, they share one or more pairs of electrons. A shared pair of electrons is called a single covalent bond, or a bond pair. A single covalent bond is represented by a single line between the atoms: for example, Cl─Cl. You can see that when chlorine atoms combine not all the electrons are used in bonding. The pairs of outer- shell electrons not used in bonding are called lone pairs. Figure 4.7: a Bromine and b iodine are elements. They both have simple covalent molecules. Each atom in a chlorine molecule has three lone pairs of electrons and shares one bonding pair of electrons (Figure 4.8). Figure 4.8: Atoms of chlorine share electrons to form a single covalent bond. IMPORTANT A dot-and-cross diagram shows the arrangement of the electrons in the main energy levels. We generally only show the outer energy level when we draw dot-and-cross diagrams. Remember that we only draw the electrons as dots and crosses to show us which atoms the electrons come from. When drawing the arrangement of electrons in a molecule: use a ‘dot’ for electrons from one of the atoms and a ‘cross’ for the electrons from the other atom if there are more than two types of atom, use additional symbols such as a small circle or a small triangle draw the outer electrons in pairs, to show the number of bond pairs and the number of lone pairs. Some examples of dot-and-cross diagrams for simple covalently bonded molecules are shown in Figure 4.9. Figure 4.9: Dot-and-cross diagrams for some covalent compounds: a hydrogen, H2, b methane, CH4, c water, H2O, d ammonia, NH3, e hydrogen chloride, HCl and f ethane, C2H6. There are some cases in which the electrons around a central atom may not have a noble gas configuration (an octet of electrons). For example: boron trifluoride, BF3, has only six electrons around the boron atom. We say that the boron atom is ‘electron deficient’. sulfur hexafluoride, SF6, has twelve electrons around the central sulfur atom. We say that the sulfur atom has an ‘expanded octet’ (Figure 4.10). phosphorus(V) chloride, PCl5, has 10 electrons around the central phosphorus atom. Phosphorus has expanded its octet (Figure 4.10). sulfur dioxide, SO2, the sulfur atom forms a double bond with each oxygen atom. This leaves a pair of non-bonding electrons on the sulfur atom. There are 10 electrons around the sulfur atom. Sulfur has expanded its octet (Figure 4.10). Figure 4.10: Dot-and-cross diagrams for a boron trifluoride, BF3, b sulfur hexafluoride, SF6, c phosphorus(V) chloride, PCl5 and d sulfur dioxide, SO2. Some elements in Periods 15, 16 and 17 have more than 8 electrons in their outer shell when they form compounds. According to one theory they use unfilled p-orbitals or d-orbitals in the third main energy level for the extra electrons. Some common examples of this expansion of the octet are seen in compounds of phosphorus, sulfur and chlorine with oxygen or fluorine. Question 2 Draw dot-and-cross diagrams for the following covalently bonded molecules. Show only the outer electron shells. Note that in part d the beryllium atom is electron deficient and in part e the antimony atom has an expanded octet. a tetrachloromethane, CCl4 b phosphorus(III) chloride c bromine, Br2 d beryllium chloride, BeCl2 e antimony pentafluoride, SbFl5 Multiple covalent bonds Some atoms can bond together by sharing two pairs of electrons. We call this a double covalent bond. A double covalent bond is represented by a double line between the atoms: for example, O═O. The dot- and-cross diagrams for oxygen, carbon dioxide and ethene, all of which have double covalent bonds, are shown in Figure 4.11. In order to form an oxygen molecule, each oxygen atom needs to gain two electrons to complete its outer shell so two pairs of electrons are shared and two covalent bonds are formed. For carbon dioxide, each oxygen atom needs to gain two electrons as before. But the carbon atom needs to gain four electrons to complete its outer shell. So two oxygen atoms each form two bonds with carbon, so that the carbon atom has eight electrons around it. In ethene, two hydrogen atoms share a pair of electrons with each carbon atom. This leaves each carbon atom with two outer shell electrons for bonding with each other. A double bond is formed between the two carbon atoms. Figure 4.11: Dot-and-cross diagrams for a oxygen, O2, b carbon dioxide, CO2, and c ethene, C2H4. Atoms can also bond together by sharing three pairs of electrons. We call this a triple covalent bond. Figure 4.12 shows a dot-and-cross diagram for the triple-bonded nitrogen molecule. In order to form a nitrogen molecule, each nitrogen atom needs to gain three electrons to complete its outer shell. So three pairs of electrons are shared and three covalent bonds are formed. Figure 4.12: Dot-and-cross diagram for a nitrogen molecule, N2. Question 3 Draw dot-and-cross diagrams for the following covalently bonded molecules. Show only the outer electron shells: a carbonyl chloride O═CCl2 b carbon disulfide, CS2. Co-ordinate bonding (dative covalent bonding) A co-ordinate bond (or dative covalent bond) is formed when one atom provides both the electrons needed for a covalent bond. For dative covalent bonding we need: one atom to have a lone pair of electrons a second atom to have an unfilled orbital to accept the lone pair. In other words, an electron-deficient compound. An example of this is the ammonium ion, NH4+, formed when ammonia combines with a hydrogen ion, H+. The hydrogen ion is electron deficient. It has space for two electrons in its shell. The nitrogen atom in the ammonia molecule has a lone pair of electrons. The lone pair on the nitrogen atom provides both electrons for the bond (Figure 4.13). Figure 4.13: The formation of a co-ordinate bond in the ammonium ion. In a displayed formula (which shows all atoms and bonds), a co-ordinate bond is represented by an arrow. The head of the arrow points away from the lone pair that forms the bond. Another molecule that has co-ordinate bonds is aluminium chloride. At high temperatures aluminium chloride exists as molecules with the formula AlCl3. This molecule is electron deficient. It still needs two electrons to complete the outer shell of the aluminium atom. At lower temperatures two molecules of AlCl3 combine to form a molecule with the formula Al2Cl6. The AlCl3 molecules are able to combine because lone pairs of electrons on two of the chlorine atoms form co-ordinate bonds with the aluminium atoms, as shown in Figure 4.14. Figure 4.14: A dot-and-cross diagram for an aluminium chloride molecule, Al2Cl6.  IMPORTANT A displayed formula shows all of the atoms and all of the bonds. You do not have to show the correct bond angles in a displayed formula. Question 4 a Draw dot-and-cross diagrams to show the formation of a co-ordinate bond between the following: i boron trifluoride, BF3, and ammonia, NH3, to form the compound F3BNH3 ii phosphine, PH3, and a hydrogen ion, H+, to form the ion PH4+. b Draw the displayed formulae of the products formed in part a. Show the co-ordinate bond by an arrow. Bond length and bond energy In general, double bonds are shorter than single bonds. This is because double bonds have a greater quantity of negative charge between the two atomic nuclei. The greater force of attraction between the electrons and the nuclei pulls the atoms closer together. This results in a stronger bond. We measure the strength of a bond by its bond energy. This is the energy needed to break one mole of a given covalent bond in the gaseous state (see also Chapter 6). The distance from the nucleus of one atom to another depends on the two atoms forming the bond. The internuclear distance between two covalently bonded atoms is called the bond length. Table 4.1 shows some values of bond lengths and bond energies. Bond Bond energy / kJ mol−1 Bond length / nm C─C 350 0.154 C═C 610 0.134 C─O 360 0.143 C═O 740 0.116 Table 4.1: Examples of values for bond energies and bond lengths. Bond strength can influence the reactivity of a compound. The molecules in liquids and gases are in random motion so they are constantly colliding with each other. A reaction only happens between molecules when a collision occurs with enough energy to break bonds in either or both molecules. Nitrogen is unreactive because it has a triple bond, N≡N. It takes a lot of energy to break the nitrogen atoms apart. The bond energy required is 994 kJ mol−1. Oxygen is much more reactive. Although it has a double bond, it only takes 496 kJ to break a mole of O═O bonds. However, bond strength is only one factor that influences the reactivity of a molecule. The polarity of the bond (see Section 4.7) and whether the bond is a σ bond (sigma bond) or a π bond (pi bond) (see Section 4.5) both play a large part in determining chemical reactivity. Question 5 The table lists bond lengths and bond energies of some hydrogen halides. Hydrogen halide Bond length / nm Bond energy / kJ mol−1 HCI 0.127 431 HBr 0.141 366 HI 0.161 299 Table 4.2: Information table for Question 5. a Describe the relationship between the bond length and the bond energy for these hydrogen halides. b Suggest why the bond energy values decrease in the order HCl > HBr > HI. c Suggest a value for the bond length in hydrogen fluoride, HF. 4.4 Shapes of molecules Valence shell electron pair repulsion theory (VSEPR) Because all electrons have the same (negative) charge, they repel each other when they are close together. So, a pair of electrons in the bonds surrounding the central atom in a molecule will repel other electron pairs. This repulsion forces the pairs of electrons apart until the repulsive forces are minimised. The shape and bond angles of a covalently bonded molecule depend on: the number of pairs of electrons around each atom whether these pairs are lone pairs or bonding pairs. IMPORTANT VSEPR: The valence shell electrons are the electrons in the main outer shell. Pairs of electrons repel each other because they have the same charge. A lone pair of electrons repel each other more than a bonded pair of electrons. Repulsion between multiple and single bonds is treated the same as for repulsion between single bonds. Repulsions between pairs of double bonds are greater. The shape of a molecule can be deduced using this theory, with the most stable shape being that which minimises the forces of repulsion. Lone pairs of electrons have a more concentrated electron charge cloud than bonding pairs of electrons. Their cloud charges are wider and slightly closer to the nucleus of the central atom. This results in a different amount of repulsion between different types of electron pairs. The order of repulsion is lone pair– lone pair (most repulsion) > lone pair–bond pair > bond pair–bond pair (least repulsion). Figure 4.15 shows the repulsions between lone pairs (pink) and bonding pairs (white) in a water molecule. Figure 4.15: Repulsion between lone and bonding electron pairs in water. Working out the shapes of molecules The differences in electron-pair repulsion determine the shape and bond angles in a molecule. Figure 4.16 compares the shapes and bond angles of methane, ammonia and water. Space-filling models of these molecules are shown in Figure 4.17. Each of these molecules has four pairs of electrons surrounding the central atom. Note that in drawing three-dimensional diagrams, the triangular ‘wedge’ is the bond coming towards you and the dashed black line is the bond going away from you. Methane has four bonding pairs of electrons surrounding the central carbon atom. The equal repulsive forces of each bonding pair of electrons results in a structure with all H─C─H bond angles being 109.5°. We call this a tetrahedral structure. In ammonia and water, the tetrahedral arrangement of the electron pairs around the central atom becomes distorted. Ammonia has three bonding pairs of electrons and one lone pair. As lone pair–bond pair repulsion is  greater than bond pair–bond pair repulsion, the bonding pairs of electrons are pushed closer together. This gives the ammonia molecule a pyramidal shape. The H─N─H bond angle is about 107°. We call this a pyramidal structure. Water has two bonding pairs of electrons and two lone pairs. The greatest force is between the two lone pairs. This results in the bonds being pushed even closer together. The shape of the water molecule is a non-linear V shape. The H─O─H bond angle is 104.5°. Figure 4.16: The bond angles in a methane, b ammonia, and c water depend on the type of electron-pair repulsion. Figure 4.17: Shapes of molecules. These space-filling models show the molecular shapes of a methane, CH4, b ammonia, NH3, and c water, H2O. Question 6 a Predict, using VSEPR theory, the shapes of the following molecules, which you drew in question 2 in Section 4.3: i tetrachloromethane, CCl4 ii beryllium chloride, BeCl2 iii phosphorus(III) chloride. b Draw dot-and-cross diagrams for the following molecules and then predict their shapes: i hydrogen sulfide, H2S ii phosphine, PH3. WORKED EXAMPLE 1 Predict the shape and bond angles in a molecule of stibine, SbH3. Explain how you arrived at your conclusions.  Solution Step 1: Use the Periodic Table to find to which groups the atoms belong. Sb = 15 Step 2: Deduce the number of electrons in the outer principal quantum shell. Sb = 5, H = 1 Step 3: Draw a dot-and-cross diagram for the molecule. In this case you should recognise the similarity to structure of ammonia. So there will be 3 bonding pairs of electrons and 1 lone pair. Step 4: Use VSEPR theory to work out approximate bond angles. Repulsion between lone pair and bond pair electrons is greater than repulsions between a bonding pair and a bonding pair. So for four pairs of electrons, the H─Sb─H bond angle is reduced from the tetrahedral bond angle to lower than 109.5° (the tetrahedral bond angle). Any suitable value lower than about 107° (the value of ammonia is a suitable estimate). But the value should be above 90°. (The actual value is 91.7°. The low value is related to the large size of the Sb atom). Step 5: Deduce the shape. You should recognise the similarity with the shape of ammonia which is pyramidal. More molecular shapes We can work out the shapes of other molecules by following the rules for electron-pair repulsion. Boron trifluoride Boron trifluoride is an electron-deficient molecule. It has only six electrons in its outer shell. The three bonding pairs of electrons repel each other equally, so the F─B─F bond angles are 120° (Figure 4.18). We describe the shape of the molecule as trigonal planar. Trigonal means having three angles. Figure 4.18: Boron trifluoride. Carbon dioxide Carbon dioxide has two carbon–oxygen double bonds and no lone pairs. The four electrons in each double bond repel other electrons in a similar way to the two electrons in a single bond (Figure 4.19). So, the O═C═O bond angle is 180°. We describe the shape of the carbon dioxide molecule as linear. Figure 4.19: Carbon dioxide. Phosphorus pentafluoride Phosphorus pentafluoride has five bonding pairs of electrons and no lone pairs. The repulsion between the electron pairs results in the most stable structure being one where two pyramids with bases of three angles are joined (Figure 4.20). We call this structure trigonal bipyramidal. Three of the fluorine atoms lie in the same plane as the phosphorus atom. The bond angles F─P─F within this plane are 120°. Two of the fluorine atoms lie above and below this plane, at 90° to it. Figure 4.20: Phosphorus pentafluoride. Sulfur hexafluoride Sulfur hexafluoride has six bonding pairs of electrons and no lone pairs. The equal repulsion between the electron pairs results in the structure shown in Figure 4.21. All F─S─F bond angles are 90°. We describe the shape as octahedral. Figure 4.21: Sulfur hexafluoride. Questions 7 a Draw a dot-and-cross diagram for a molecule of selenium hexafluoride, SeF6. A single selenium atom has six electrons in its outer shell. b Predict the shape of selenium hexafluoride. c Draw the shape of the phosphorus(V) chloride molecule that you drew as a dot-and-cross diagram in question 2 in Section 4.3. 8 In trimethylamine, (CH3)3N: three CH3 groups are attached to the nitrogen by C─N bonds. The repulsion of the CH3 groups is slightly greater than the repulsion of the H atoms in ammonia. Which one of these bond angles for the C─N─C bond in trimethylamine, is correct? A about 90° B about 107° C about 109.5° D about 120° 4.5 σ bonds and π bonds A single covalent bond is formed when two non-metal atoms combine. Each atom that combines has an atomic orbital containing a single unpaired electron. In the formation of a covalent bond the atomic orbitals overlap so that a combined orbital is formed, containing two electrons. We call this combined orbital a molecular orbital. The amount of overlap of the atomic orbitals determines the strength of the bond: the greater the overlap, the stronger the bond. Figure 4.22 shows how the s atomic orbitals of two hydrogen atoms overlap to form a covalent bond. IMPORTANT Covalent bonds are formed when atomic orbitals overlap. Figure 4.22: Two 1s atomic orbitals in hydrogen overlap to form a covalent bond. The greater the overlap, the stronger the bond. The mixing of atomic orbitals is called hybridisation. Mixing an s with three, two or one p-type orbitals forms sp3, sp2 and sp hybrid orbitals. Sigma bonds (σ bonds) are formed from end-on overlap of atomic orbitals. Pi bonds (π bonds) are formed from sideways overlap of atomic orbitals. The p atomic orbitals can also overlap linearly (end-on) to form covalent bonds. When p orbitals are involved in forming single bonds, they become modified to include some s orbital character. The orbital is slightly altered in shape to make one of the lobes of the p orbital bigger. The process of mixing atomic orbitals (for example, one s orbital and three p orbitals) in this way is called hybridisation. The hybrids are called sp3 hybrids. In sp3 hybrids, each orbital has s character and p character. When one s orbital and two p orbitals are hybridised, the hybrids are called sp2 hybrids. When one s orbital and one p orbital are hybridised, the hybrids are called sp orbitals. When hybridised orbitals overlap linearly (end-on) we call the bond a σ bond (sigma bond). Figure 4.23 shows the formation of σ bonds. Figure 4.23: Bonds are formed by the linear (end-on) overlap of atomic orbitals. The electron density of each σ bond is symmetrical about a line joining the nuclei of the atoms forming the bond. Bonds formed by the sideways overlap of p orbitals are called π bonds (pi bonds). A π bond is not symmetrical about the axes joining the nuclei of the atoms forming the bond. Figure 4.24 shows how a π bond is formed from two p orbitals overlapping sideways. We often draw a single π bond as two electron clouds, one arising from each lobe of the p orbitals. You must remember, though, that the two clouds of electrons in a π bond represent one bond consisting of a total of two electrons. Figure 4.24: π bonds are formed by the sideways overlap of atomic orbitals. The shape of some organic molecules We can explain the shapes of molecules in terms of the patterns of electron density found in σ bonds and π bonds. Ethane The displayed formula for ethane is: All the bonds in ethane are formed by linear overlap of atomic orbitals. They are all σ bonds. Figure 4.25 shows the electron density distribution in ethane formed by these σ bonds. All the areas of electron density repel each other equally. This makes the H─C─H bond angles all the same (109.5°). Figure 4.25: The electron density distribution in ethane. Ethene The displayed formula for ethene is: Each carbon atom in ethene uses three of its four outer electrons to form σ bonds. Two σ bonds are formed with the hydrogen atoms and one σ bond is formed with the other carbon atom. The fourth electron from each carbon atom occupies a p orbital, which overlaps sideways with a similar p orbital on the other carbon atom. This forms a π bond. Figure 4.26 shows how this occurs. Figure 4.26: Overlap of p orbitals to produce a π bond in ethene. The electron density distribution of both the σ and π bonds in ethene is shown in Figure 4.27. Figure 4.27: The electron density distribution in ethene. Ethene is a planar molecule because this ensures the maximum overlap of the p orbitals that form the π bond. You will notice that the electron clouds that make up the π bond lie above and below the plane of the carbon and hydrogen nuclei. We would expect the H─C─H bond angle in ethene to be about 120° because the three areas of electron density of the σ bonds are equally distributed. However, because of the position of the π bond, this bond angle is actually 117°. This minimises the repulsive forces. Triple bonds In the nitrogen molecule, N≡N, hydrogen cyanide, H─C≡N and ethyne, H─C≡C─H, there are triple bonds. The triple bond is formed from two π bonds (pi bonds) and one σ bond (sigma bond). The two π bonds are at right angles to each other. In hydrogen cyanide one σ bond is formed between the hydrogen atom and the carbon atom (by overlap of an sp carbon hybrid orbital with the 1s hydrogen orbital). A second σ bond is formed between the carbon and the nitrogen atom (by overlap of an sp carbon hybrid orbital with a p orbital on the nitrogen). This leaves the remaining p orbitals on each nitrogen atom to form two π bonds at right angles to each other. Figure 4.28: Overlap of two sets of p orbitals of nitrogen and carbon to produce two π bonds in hydrogen cyanide. 4.6 Metallic bonding What is a metallic bond? In a metal, the atoms are packed closely together in a regular arrangement called a lattice. Metal atoms in a lattice tend to lose their outer shell electrons and become positive ions. The outer shell electrons occupy new energy levels and are free to move throughout the metal lattice. We call these electrons delocalised electrons (mobile electrons). Delocalised electrons are electrons that are not associated with any one particular atom or bond. Metallic bonding is strong. This is because the ions are held together by the strong electrostatic attraction between their positive charges and the negative charges of the delocalised electrons (Figures 4.29 and 4.30). Figure 4.29: Metals: a sodium, b gold, c mercury, d magnesium and e copper. Figure 4.30: Metallic bonding: there are strong attractive forces between the positively charged ions and the delocalised electrons. This electrostatic attraction acts in all directions. The strength of metallic bonding increases with: increasing positive charge on the ions in the metal lattice decreasing size of metal ions in the lattice increasing number of mobile electrons per atom. Metallic bonding and the properties of metals We can use our model of metallic bonding to explain many of the properties of metals. Most metals have high melting points and high boiling points It takes a lot of energy to weaken the strong attractive forces between the metal ions and the delocalised electrons. These attractive forces can only be overcome at high temperatures. However, mercury is a liquid at room temperature (Figure 4.31). This is because some of the electrons in a mercury atom are bound more tightly than usual to the nucleus, weakening the metallic bonds between atoms. Figure 4.31: Mercury is a liquid at room temperature. Metals conduct electricity When a voltage is applied to a piece of metal, an electric current flows in it because the delocalised electrons (mobile electrons) are free to move. Metallic bonding is the only type of bonding that allows us to predict reliably that a solid will conduct electricity. Covalent solids cannot conduct electricity because none of their electrons are free to move throughout the structure, although graphite is an exception to this. Ionic solids cannot conduct because neither their electrons nor their ions are free to move from place to place. Question 9 Answer the following, giving a full explanation in terms of metallic bonding. a Explain why aluminium has a higher melting point than sodium. b The thermal conductivity of stainless steel is 82 W m−1 K−1. The thermal conductivity of copper is 400 W m−1 K−1. Why do some stainless steel saucepans have a copper base? c Why does aluminium conduct electricity better than sodium? Metals conduct heat The conduction of heat is partly due to the movement of the delocalised electrons but mainly due to the vibrations passed on from one metal ion to the next. 4.7 Intermolecular forces The forces within molecules due to covalent bonding are strong. However, the forces between molecules are much weaker. We call these forces van der Waals’ forces (intermolecular forces). There are two types of van der Waals’ (intermolecular) force: instantaneous (temporary) dipole–induced dipole forces (id–id) forces. These are also called London dispersion forces. permanent dipole–permanent dipole (pd–pd) forces (including hydrogen bonding). A special case of permanent dipole–permanent dipole forces is hydrogen bonding. This occurs between molecules where hydrogen is bonded to a highly electronegative atom. IMPORTANT Remember that ‘van der Waals’ forces’ is a general term used for all intermolecular forces including id– id forces, pd–pd forces and hydrogen bonding. Table 4.2 compares the relative strength of these intermolecular forces and other bonds. Type of bond or force Bond strength / kJ mol−1 ionic bonding in sodium chloride 760 O─H covalent bond in water 464 hydrogen bonding 20–50 permanent dipole–permanent dipole force 5–20 id–id forces 1–20 Table 4.2: Strengths of different types of bond and intermolecular force. In order to understand how intermolecular forces work, we first have to know about electronegativity and bond polarity. Electronegativity Electronegativity is the power of a particular atom that is covalently bonded to another atom to attract the bonding pair of electrons towards itself. The greater the value of the electronegativity, the greater the power of an atom to attract the electrons in a covalent bond towards itself. For Groups 1 to 17 the pattern of electronegativity is: electronegativity increases across a period from Group 1 to Group 17 electronegativity increases up each group. This means that fluorine is the most electronegative element. For the most electronegative elements, the order of electronegativity is: Carbon and hydrogen have electronegativities that are lower than those of most other non-metallic elements. Factors influencing electronegativity The value of the electronegativity depends on:  nuclear charge: atoms in the same period with a greater (positive) nuclear charge are more likely to attract the bonding pair of electrons. atomic radius: atoms in the same group in which the outer electrons are further from the nucleus are less likely to attract the bonding pair of electrons because the pull of the positive nucleus on the electron pair is lower. shielding: the greater the number of inner electron shells and sub-shells, the lower the effective nuclear charge on the bonding electrons. Ionic or covalent? The most commonly used scale of electronegativity values is called the Pauling electronegativity scale, NP. Table 4.3 shows some Pauling electronegativity values. Li Be B C N O F 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Na Mg Al Si P S Cl 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Table 4.3: Some Pauling electronegativity values. Note: carbon (NP = 2.5) and hydrogen (NP = 2.1) have electronegativities that are lower than those of many other non-metallic elements Pauling electronegativity has no units. Differences in electronegativity values can be used to predict whether a simple compound has ionic or covalent bonds. If the electronegativity difference is high, e.g. 2.0 or more, the compound is likely to be ionic. E.g. sodium chloride: Na = 0.9, Cl = 3.0. Difference = 2.1. If the electronegativity difference is lower, e.g. below 1.0, the compound is likely to be covalent. E.g. methane: C = 2.5, H = 2.1. Difference = 0.4. A zero value shows that there is no ionic character in the bond, e.g. Cl─Cl. Some compounds are not entirely covalent and have some ionic character in them. These have intermediate electronegativity differences, e.g. 1.0. Polarity in molecules When the electronegativity values of the two atoms forming a covalent bond are the same, the pair of electrons is equally shared. We say that the covalent bond is non-polar. For example, hydrogen (H2), chlorine (Cl2) and bromine (Br2) are non-polar molecules. When a covalent bond is formed between two atoms having different electronegativity values, the more electronegative atom attracts the pair of electrons in the bond towards it. As a result: the centre of positive charge does not coincide with the centre of negative charge we say that the electron distribution is asymmetric the two atoms are partially charged we show: the less electronegative atom with the partial charge δ+ (‘delta positive’) the more electronegative atom with the partial charge δ– (‘delta negative’) we say that the bond is polar (or that it has a dipole). Figure 4.32 shows the polar bond in a hydrogen chloride molecule. Figure 4.32: Hydrogen chloride is a polar molecule. As the difference in electronegativity values of the atoms in a covalent bond increases, the bond becomes more polar. The degree of polarity of a molecule is measured as a dipole moment. The direction of the dipole is shown by the sign. The arrow points to the partially negatively charged end of the dipole. In molecules containing more than two atoms, we have to take into account: the polarity of each bond the arrangement of the bonds in the molecule. Trichloromethane, CHCl3, is a polar molecule. The three C─Cl dipoles point in a similar direction. Their combined effect is not cancelled out by the polarity of the C─H bond. This is because the C─H bond is virtually non-polar. The electron distribution is asymmetric. The molecule is polar, with the negative end towards the chlorine atoms. This is shown in Figure 4.33a. Some molecules contain polar bonds but have no overall polarity. This is because the polar bonds in these molecules are arranged in such a fashion that the dipole moments cancel each other out. An example is tetrachloromethane, CCl4 (Figure 4.33b). Tetrachloromethane has four polar C─Cl bonds pointing towards the four corners of a tetrahedron. The dipoles in each bond cancel each other, so tetrachloromethane is non-polar. Figure 4.33: The polarity of a trichloromethane and b tetrachloromethane. IMPORTANT Remember that not all molecules with polar bonds are polar molecules. Sometimes the polar bonds cancel each other out because the same types of bonds have dipoles ‘pulling’ in opposite directions. We can determine the charge distribution in molecules and ions by a method called X-ray spectroscopy. One method involves firing X-rays at molecules and measuring the energy of the electrons given off. Using this method, scientists have found that in a sulfate ion, the sulfur atom has a charge of +1.12 units and the four oxygen atoms each have a charge of –0.78 units. Question 10 Are the following molecules polar or non-polar? In each case give a reason for your answer. (Electronegativity values: F = 4.0, Cl = 3.0, Br = 2.8, S = 2.5, C = 2.5, H = 2.1) a chlorine, Cl2 b hydrogen fluoride, HF c the V-shaped molecule, sulfur dichloride, SCl2 d the tetrahedral molecule, chloromethane, CH3Cl e the tetrahedral molecule, tetrabromomethane, CBr4 Polarity and chemical reactivity Bond polarity influences chemical reactivity. For example, both nitrogen, N≡N, and carbon monoxide, C≡O, have triple bonds requiring a similar amount of energy to break them. Nitrogen is a non-polar molecule and is fairly unreactive. But carbon monoxide is a polar molecule, and this explains its reactivity with oxygen and its use as a reducing agent. Many chemical reactions are started by a reagent attacking one of the electrically charged ends of a polar molecule. For example, chloroethane, C2H5Cl, is far more reactive than ethane, C2H6. This is because reagents such as OH− ions can attack the delta-positive carbon atom of the polarised C─Cl bond (see also Section 16.2). Such an attack is not possible with ethane because the C─H bond is virtually non-polar. This helps to explain why alkanes such as ethane are not very reactive. Instantaneous dipole–induced dipole (id–id) forces Noble gases such as neon and argon exist as isolated atoms. Noble gases can be liquefied, but at very low temperatures, so there must be very weak forces of attraction between their atoms. These weak forces keep the atoms together in the liquid state. Bromine is a non-polar molecule that is liquid at room temperature. The weak forces of attraction are keeping the bromine molecules together at room temperature. These very weak forces of attraction are called instantaneous dipole–induced dipole (id–id) forces, or London dispersion forces. These forces exist between all atoms or molecules. So, how do these forces arise? The electron charge clouds in a non-polar molecule (or atom) are constantly moving. It often happens that more of the charge cloud is on one side of the molecule than the other. This means that one end of the molecule has, for a short time, more negative charge than the other. A temporary dipole is set up. This dipole can set up (induce) a dipole on neighbouring molecules. As a result of this, there are forces of attraction between the δ+ end of the dipole in one molecule and the δ– end of the dipole in a neighbouring molecule (Figure 4.34). These dipoles are always temporary because the electrons clouds are always moving. Instantaneous dipole–induced dipole forces are sometimes called temporary dipole–induced dipole forces. Instantaneous dipole–induced dipole forces increase with: increasing number of electrons (and protons) in the molecule increasing the number of contact points between the molecules: contact points are places where the molecules come close together. Differences in the size of the instantaneous dipole–induced dipole (id–id) forces can be used to explain the trend in the enthalpy change of vaporisation and boiling points of the noble gases. Figure 4.35 shows how these vary with the number of electrons present. (The enthalpy change of vaporisation is the energy required to convert a mole of liquid into a mole of gas.) Figure 4.34: How instantaneous dipole–induced dipole forces arise. Figure 4.35: a Enthalpy changes of vaporisation and b boiling points of the noble gases plotted against the number of electrons present. You can see that both the enthalpy change of vaporisation and the boiling points of the noble gases increase as the number of electrons increases (Figure 4.35). This is because the id–id forces between the atoms are increased with an increasing number of electrons. So, more energy is needed to change the liquid into vapour and the boiling point is higher. The effect of increasing the number of contact points can be seen by comparing the boiling points of pentane (boiling point 36 °C) and 2,2-dimethylpropane (boiling point 10 °C) (Figure 4.36). These compounds have equal numbers of electrons in their molecules. The molecules in pentane can line up beside each other so there is a large number of contact points. The id–id forces are higher, so the boiling point is higher. The molecules of 2,2-dimethylpropane are more compact. The surface area available for coming into contact with neighbouring molecules is smaller. The id–id forces are relatively lower, so the boiling point is lower. The id–id forces between individual atoms are very small. However, the total id–id forces between very long non-polar molecules such as poly(ethene) molecules (see Section 15.5) can be much larger. That is why poly(ethene) is a solid at room temperature. Figure 4.36: The difference in boiling points of pentane and 2,2-dimethylpropane can be explained by the strength of the instantaneous dipole–induced dipole forces. Two types of poly(ethene) are low-density poly(ethene), LDPE, and high-density poly(ethene), HDPE. Both have crystalline and non-crystalline regions in them (Figure 4.37). Figure 4.37: Crystalline and non-crystalline regions in poly(ethene). HDPE has more crystalline regions where the molecules are closer together than LDPE. The total id–id forces are greater, so HDPE is the stronger of the two. Question 11 a The boiling points of the halogens are: fluorine –188 °C chlorine –35 °C bromine +59 °C iodine +184 °C i Describe the trend in these boiling points going down Group 17. ii Explain the trend in these boiling points. b The table lists the formulae and boiling points of some alkanes. Explain the trend in terms of instantaneous dipole–induced dipole forces. Alkane Structural formula Boiling point / °C methane CH4 –164 ethane CH3CH3 –88 propane CH3CH2CH3 –42 butane CH3CH2CH2CH3 0 Permanent dipole–permanent dipole (pd–pd) forces In some molecules, the dipole is permanent. Molecules with a permanent dipole are called polar molecules. A fine jet of polar molecules will be attracted towards an electrically charged plastic rod or comb. (The rod can be charged by rubbing it with a woollen cloth.) Figure 4.38 shows the result of this experiment. Figure 4.38: The deflection of water by an electrically charged nylon comb. The molecules are always attracted to the charged rod, whether the rod is positively or negatively charged. This is because the molecules have both negatively and positively charged ends. The forces between two molecules having permanent dipoles are called permanent dipole–permanent dipole (pd–pd) forces. The attractive force between the δ+ charge on one molecule and the δ– charge on a neighbouring molecule causes a weak attractive force between the molecules (Figure 4.39). Figure 4.39: Permanent dipole–permanent dipole forces in propanone. For small molecules with the same number of electrons, permanent dipole–permanent dipole forces are often stronger than instantaneous dipole–induced dipole forces. For example, propanone (CH3COCH3, Mr = 58) has a higher boiling point than butane (CH3CH2CH2CH3, Mr = 58). This means that more energy is needed to break the intermolecular forces between propanone molecules than between butane molecules) (Figure 4.40). Figure 4.40: The difference in the boiling points of propanone and butane can be explained by the different types of intermolecular force between the molecules. The permanent dipole–permanent dipole forces between propanone molecules are strong enough to make this substance a liquid at room temperature. There are only instantaneous dipole–induced dipole forces between butane molecules. These forces are comparatively weak, so butane is a gas at room temperature. Question 12 Bromine, Br2, and iodine monochloride, ICl, have the same number of electrons. But the boiling point of iodine monochloride is nearly 40 °C higher than the boiling point of bromine. Explain this difference. 4.8 Hydrogen bonding Note: Learners are only expected to describe hydrogen bonding in relation to molecules containing N─H and O─H. The material in this section relating to the F─H group is extension material, and is not on the syllabus. Hydrogen bonding is the strongest form of intermolecular bonding. It is a type of permanent dipole– permanent dipole bonding. For hydrogen bonding to occur between two molecules we need: one molecule to have a hydrogen atom covalently bonded to F, O or N (the three most electronegative atoms) a second molecule to have an F, O or N atom with an available lone pair of electrons. For hydrogen bonding to happen the molecules must have: a H atom covalently bonded to a highly electronegative atom, e.g. N, O or F another highly electronegative atom with a lone pair of electrons. When a hydrogen atom is covalently bonded to a very electronegative atom, the bond is very highly polarised. The δ+ charge on the hydrogen atom is high enough for a bond to be formed with a lone pair of electrons on the F, O or N atom of a neighbouring molecule (Figure 4.41). Figure 4.41: Hydrogen bonding between two ammonia molecules. A hydrogen bond is represented by a line of dots. The force of attraction is about one-tenth of the strength of a normal covalent bond. For maximum bond strength, the angle between the covalent bond to the hydrogen atom and the hydrogen bond is usually 180°. The average number of hydrogen bonds formed per molecule depends on: the number of hydrogen atoms attached to F, O or N in the molecule the number of lone pairs present on the F, O or N. Water has two hydrogen atoms and two lone pairs per molecule (Figure 4.42). So water is extensively hydrogen bonded with other water molecules. It has an average of two hydrogen bonds per molecule. Figure 4.42: Water can form, on average, two hydrogen bonds per molecule. Ammonia is less extensively hydrogen bonded than water (see Figure 4.41). It can form, on average, only one hydrogen bond per molecule. Although each ammonia molecule has three hydrogen atoms attached to the nitrogen atom, it has only one lone pair of electrons that can be involved in hydrogen bond formation. REFLECTION Take an A4 sheet of paper and divide it into two columns. In the left-hand column write down, one  under another, the key words, phrases and definitions that you have come across in this chapter in any order. There are at least 15! Do this without looking at the book. Work with another learner: a Ask another learner to define the first term that you have chosen. b Write the definition next to the term in the second column. c After each turn, check your explanation or definition with that given in the book and rewrite if necessary. d Take it in turns to ask each other to explain the terms or give definitions. How many of these key words, phrases and definitions did you know before studying this chapter? Questions Note: Questions 13 c and 14 are extension material. 13 Draw diagrams to show hydrogen bonding between the following molecules: a ethanol, C2H5OH, and water b ammonia and water c two hydrogen fluoride molecules. 14 The boiling points of the hydrogen halides are shown in the table. Hydrogen halide HF HCl HBr HI Boiling point / °C +20 –85 –67 –35 Table 4.4: Information table for Question 14. a Explain the trend in boiling points from HCl to HI. b Explain why the boiling point of HF is so much higher than the boiling point of HCl. How does hydrogen bonding affect boiling point? Some compounds have higher boiling points than expected. This can be due to hydrogen bonding. Figure 4.43 shows a graph of the boiling points of the hydrogen halides, HF, HCl, HBr and HI, plotted against the position of the halogen in the Periodic Table. Figure 4.43: The boiling points of the hydrogen halides. The rise in boiling point from HCl to HI is due to the increasing number of electrons in the halogen atoms as we go down the group. This leads to increased instantaneous dipole–induced dipole forces as the molecules get bigger. If hydrogen fluoride only had instantaneous dipole–induced dipole forces between its molecules, we would expect its boiling point to be about –90 °C. However, the boiling point of hydrogen fluoride is +20 °C, which is much higher. This is because of the stronger intermolecular forces of hydrogen bonding between the HF molecules. Question 15 The table lists the boiling points of some Group 15 hydrides. Hydride Boiling point / °C ammonia, NH3 –33 phosphine, PH3 –88 arsine, AsH3 –55 stibine, SbH3 –17 Table 4.5: Information table for Question 15. a Explain the trend in the boiling points from phosphine to stibine. b Explain why the boiling point of ammonia does not follow this trend. The peculiar properties of water Enthalpy change of vaporisation and boiling point Water has a much higher enthalpy change of vaporisation and boiling point than expected. This is due to its extensive hydrogen bonding. Figure 4.44 shows the enthalpy changes of vaporisation of water and other Group 16 hydrides. The rise in enthalpy change of vaporisation from H2S to H2Te is due to the increasing number of electrons in the Group 16 atoms as we go down the group. This leads to increased instantaneous dipole–induced dipole forces as the molecules get bigger. If water only had instantaneous dipole–induced dipole forces between its molecules, we would expect its enthalpy change to be about 17 kJ mol−1. But the enthalpy change of vaporisation of water is much higher. This is because water is extensively hydrogen bonded. The boiling point of water is also much higher than predicted by the trend in boiling points for the other Group 16 hydrides. This also indicates that we need much more energy to break the bonds between water molecules compared with other hydrides of Group 16 elements. Figure 4.44: Enthalpy changes of vaporisation for Group 16 hydrides plotted against number of electrons present. Surface tension and viscosity Water has a high surface tension and high viscosity. Hydrogen bonding reduces the ability of water molecules to slide over each other, so the viscosity of water is high. The hydrogen bonds in water also exert a significant downward force at the surface of the liquid. This causes the surface tension of water to be higher than for most liquids. Ice is less dense than water Most solids are denser than their liquids. This is because the molecules are more closely packed in the solid state. But this is not true of water. In ice, there is a three-dimensional hydrogen-bonded network of water molecules. This produces a rigid lattice in which each oxygen atom is surrounded by a tetrahedron of hydrogen atoms. This ‘more open’ arrangement, due to the relatively long bond lengths of the hydrogen bonds, allows the water molecules to be slightly further apart than in the liquid (Figures 4.45 and 4.46). So the density of ice is less than that of liquid water. Figure 4.45: Ice floats on water. Figure 4.46: A model of ice. Oxygen atoms are red, hydrogen atoms are white, hydrogen bonds are lilac. This hydrogen- bonded arrangement makes ice less dense than water. Question 16 Which pair of liquids when mixed has the strongest van der Waals’ forces? A H3C─CO─CH3 and H3C─CO─CH2─CH3 B H3C─CH2─Cl and H3C─O─H C H3C─CH2─CH2─CH2─CH2─CH2─CH3 and H3C─CH2─CH2─CH2─CH2─CH3 D H3C─O─H and H3C─CH2─CH2─NH2 4.9 Bonding and physical properties The type of bonding between atoms, ions or molecules influences the physical properties of a substance. IMPORTANT Giant structures such as ionic structures, metals and carbon (graphite and diamond) generally have high melting points but this does not give evidence about the type of bonding. If a substance has a low melting point, it is a better indication that the bonding is covalent within the molecule and there are weak van der Waals’ forces of attraction between the molecules (intermolecular forces). Physical state at room temperature and pressure Ionic compounds Ionic compounds are solids at room temperature and pressure. This is because: there are strong electrostatic forces (ionic bonds) holding the positive and negative ions together the ions are regularly arranged in a lattice (see Chapter 5), with the oppositely charged ions close to each other. Ionic compounds have high melting points, high boiling points and high enthalpy changes of vaporisation. It takes a lot of energy to overcome the strong electrostatic attractive forces. Metals Metals, apart from mercury, are solids. Many metals have high melting points, high boiling points and high enthalpy changes of vaporisation. This is because it takes a lot of energy to overcome the strong attractive forces between the positive ions and the ‘sea’ of delocalised electrons. IMPORTANT Remember that not all metals have high melting points. Group 1 metals and mercury have lower boiling points than some non-metals, so do not use high melting point as a characteristic of metals unless you are only referring to transition elements. Covalent compounds Covalently bonded substances with simple molecular structures, for example, water and ammonia, are usually liquids or gases. This is because the intermolecular forces are weak. It does not take much energy to overcome these intermolecular forces. So these substances have low melting points, low boiling points and low enthalpy changes of vaporisation compared with ionic compounds. Some substances that have covalently bonded molecules may be solids at room temperature: for example, iodine and poly(ethene). These are usually molecules where the total instantaneous dipole–induced dipole forces are considerable. However, the melting points of these substances are still fairly low compared with ionic compounds or most metals. Solubility Ionic compounds Most ionic compounds are soluble in water. This is because water molecules are polar and they are attracted to the ions on the surface of the ionic solid. These attractions are called ion–dipole attractions (see Section 19.4). These attractions replace the electrostatic forces between the ions and so the ions go into solution. Metals Metals do not dissolve in water. The force of attraction between the ions and the delocalised electrons is too great for water molecules to disrupt the structure and form bonds with the ions. However, some metals, for example sodium and calcium, react with water rather than dissolving in water. Covalent compounds Covalently bonded substances with simple molecular structures fall into two groups. Those that are insoluble in water. Most covalently bonded molecules are non-polar. Water molecules are not attracted to them so they are insoluble. An example is iodine. Those that are soluble in water. Small molecules that can form hydrogen bonds with water are generally soluble. An example is ethanol, C2H5OH. Some covalently bonded substances react with water rather than dissolving in it. For example, hydrogen chloride reacts with water to form hydrogen ions and chloride ions, and the ions are soluble. Silicon chloride reacts with water to form hydrogen ions, chloride ions and silicon dioxide. This reaction is called a hydrolysis reaction. Electrical conductivity IMPORTANT Remember that electrical conductivity in molten or aqueous ionic compounds is due to mobile ions. Electrical conductivity in metals is due to mobile electrons. Ionic compounds Ionic compounds do not conduct electricity when in the solid state. This is because the ions are fixed in the lattice and can only vibrate around a fixed point. When molten, an ionic compound conducts electricity because the ions are mobile. Metals Metals conduct electricity both when solid and when molten. This is because the delocalised electrons move throughout the structure when a voltage is applied. Covalent compounds Covalently bonded substances with simple molecular structures do not conduct electricity. This is because they have neither ions nor electrons that are mobile. PRACTICAL ACTIVITY 4.1 Ionic, covalent or metallic? SAFETY: Only carry out this activity in the presence of a teacher after safety aspects have been explained. You are given several solid substances and have to decide whether the bonding is ionic, covalent or metallic. Carry out these investigations and record your observations. Procedure 1 Set up an electrical circuit containing batteries, connecting leads, a lamp (bulb) and crocodile clips. This may already be set up for you. 2 Place one of the solids between the crocodile clips. Does the lamp (bulb) light? 3 Add a small amount of the solid to 20 cm3 of water in a beaker and stir with a glass rod.  Does the solid dissolve? If the solid dissolves, keep the solution. 4 Dip a clean conductivity probe into the solution from step 3. Does the solution conduct electricity? Which solids are ionic, which are covalent and which are metallic? Explain your answers. Question 17 Explain the following differences in terms of the type of bonding present. a Aluminium oxide has a melting point of 2980 °C but aluminium chloride changes to a vapour at 178 °C. b Magnesium chloride conducts electricity when molten but not when solid. c Iron conducts electricity when solid but the ionic solid iron(II) chloride does not conduct when solid. d Sodium sulfate dissolves in water but sulfur does not. e Propanol, CH3CH2CH2OH, is soluble in water but propane, CH3CH2CH3, is not. f A solution of hydrogen chloride in water conducts electricity. REFLECTION Work with another learner to make a mind map (spider diagram) of the different types of bonding that you have learned about in this chapter. On your mind map include information about how different types of bonding affect the physical properties of substances. Reflect on any ideas you both might have on how you could improve your mind map. SUMMARY Ionic bonding is the electrostatic attraction between oppositely charged ions. Covalent bonding is the electrostatic attraction between the nuclei of two atoms and a shared pair of electrons. Electronegativity is the power of a particular atom that is covalently bonded to another atom to attract the bonding pair of electrons towards itself. Bond energy is the energy required to break one mole of a particular covalent bond in the gaseous state. The units of bond energy are kilojoules per mole, kJ mol−1. The bond angles within molecules and the shapes of molecules can be predicted by VSEPR theory (lone pairs of electrons repel each other more than bonding pairs of electrons). Bonding can be described in terms of hybridisation of atomic orbitals and sigma (σ) bonds and pi (π) bonds. Van der Waals’ forces can be classified as instantaneous dipole–induced dipole forces (very weak), permanent dipole–permanent dipole forces (slightly stronger) and hydrogen bonding (the strongest). Van der Waals’ forces are weaker than covalent, ionic or metallic bonding. Metallic bonding is the electrostatic attraction between positive ions and delocalised electrons. Some of the physical properties of ionic, simple covalent molecules and metals can be related to the type of bonding present. EXAM-STYLE QUESTIONS 1 The table shows the atomic number and boiling points of some noble gases. Gas helium neon argon krypton xenon Atomic number 2 10 18 36 54 Boiling point / °C –269 –246 –186 –153 –107 a Use ideas about forces between atoms to explain this trend in boiling points. b Xenon forms a number of covalently bonded compounds with fluorine. i Define the term covalent bond. ii Draw a dot-and-cross diagram for xenon tetrafluoride, XeF4 iii Suggest a shape for XeF4. Explain your answer. c The structure of xenon trioxide is shown below. i By referring to electron pairs, explain why xenon trioxide has this shape. ii Draw the structure of xenon trioxide to show the partial charges on the atoms and the direction of the dipole in the molecule. [Total: 11] 2 Aluminium chloride, AlCl3, and ammonia, NH3, are both covalent molecules. a i Draw a diagram of an ammonia molecule, showing its shape. Show any lone pairs of electrons. ii State the bond angle in the ammonia molecule. b Explain why ammonia is a polar molecule. c An ammonia molecule and an aluminium chloride molecule can join together by forming a co-ordinate bond. i Explain how a co-ordinate bond is formed. ii Draw a dot-and-cross diagram to show the bonding in the compound formed between ammonia and aluminium chloride, H3NAlCl3. (Use for a nitrogen electron, o for an aluminium electron and × for the hydrogen and chlorine electrons.) d Aluminium chloride molecules join together to form a compound with the formula Al2Cl6. Draw a displayed formula (showing all atoms and bonds) to show the bonding in one Al2Cl6 molecule. Show the dative covalent bonds by arrows. [Total: 12] 3 Electronegativity values can be used to predict the polarity of bonds. a Define the term electronegativity. b The electronegativity values for some atoms are given below: H = 2.1, C = 2.5, F = 4.0, Cl = 3.0, I = 2.5 Use these values to predict the polarity of each of the following bonds by copying the bonded atoms shown below and adding δ+ or δ− above each atom. i H─I  ii F─I iii C─Cl c The shape of iodine trichloride, ICl3, is shown below. i Use the electronegativity values above to explain how you know that iodine trichloride is a covalent compound. ii Describe the shape of this molecule. iii Explain why the ICl3 molecule has this shape. iv Suggest a value for the Cl─I─Cl bond angle. d Tetrachloromethane, CCl4, is a non-polar molecule. i Draw a diagram to show the shape of this molecule. ii Explain why this molecule is non-polar. [Total: 15] 4 The diagram below shows part of a giant metallic structure. a Use this diagram to explain the main features of metallic bonding. b Explain why metals are good conductors of electricity. c Explain why, in general, metals have high melting points. d Suggest why potassium is a better conductor of electricity than lithium. [Total: 11] 5 Methane, CH4, is a gas at room temperature. a Explain why methane is a gas at room temperature. b Draw a diagram to show the shape of a molecule of methane. On your diagram show a value for the HCH bond angle. c Perfumes often contain molecules that have simple molecular structures. Explain why. d When a negatively charged rod is held next to a stream of propanone, CH3COCH3, the stream of propanone is attracted to the rod. Draw the full structure of a molecule of propanone and use your diagram to explain why the stream of propanone is attracted to the rod. [Total: 10] 6 Sodium iodide and magnesium oxide are ionic compounds. Iodine and oxygen are covalent molecules. a Draw dot-and-cross diagrams for: i magnesium oxide ii oxygen b Describe how sodium iodide and iodine differ in their solubility in water. Explain your answer.  c Explain why molten sodium iodide conducts electricity but molten iodine does not. d The boiling point of sodium iodide is 1304 °C. The boiling point of iodine is 184 °C. Explain this difference. e The Pauling electronegativities of Na and iodine are shown. sodium = 0.9, iodine = 2.5 Use these electronegativity values to explain why sodium iodide is an ionic compound and not a covalent compound. [Total: 14] 7 Hydrogen sulfide, H2S, is a covalent compound. a Draw a dot-and-cross diagram for hydrogen sulfide. b Draw a diagram of a hydrogen sulfide molecule to show its shape. Show on your diagram: i the value of the HSH bond angle ii the partial charges on each atom as δ+ or δ– iii an arrow showing the exact direction of the dipole in the molecule as a whole. c Oxygen, O, sulfur, S, and selenium, Se, are in the same group in the Periodic Table. i Explain why hydrogen selenide, H2Se, has a higher boiling point than hydrogen sulfide, H2S. iiExplain why the boiling point of water is so much higher than the boiling point of hydrogen sulfide. [Total: 13] 8 The table shows the type of bonding in a number of elements and compounds. Element or compound Type of bonding Fe, Na metallic NaCl, MgCl2 ionic CO2, Br2 covalent within the molecules a Draw a labelled diagram to show metallic bonding. b Explain why magnesium chloride has a high melting point but bromine has a low melting point. c Explain why solid sodium conducts electricity but solid sodium chloride does not conduct electricity. d i Draw a dot-and-cross diagram for carbon dioxide. ii Describe the shape of the carbon dioxide molecule. iii Explain why a carbon dioxide molecule has this shape. e Bromine is a liquid at room temperature. Weak instantaneous dipole–induced dipole forces hold the bromine molecules together. Describe how these forces arise. [Total: 18] 9 a Water is extensively hydrogen bonded. This gives it anomalous (peculiar) properties. i Explain why ice is less dense than liquid water. ii State two other anomalous properties of water. b Propanone has the structure shown below.  When propanone dissolves in water, it forms a hydrogen bond with water. Describe the features water and propanone molecules have in order to form a hydrogen bond. c Draw a diagram to show a propanone molecule and a water molecule forming a hydrogen bond. d Propanone has a double bond. One of the bonds is a σ bond (sigma bond). The other is a π bond (pi bond). Explain the difference between a σ bond and a π bond in terms of how they are formed. e Copy the diagram, then complete it to show the shapes of the electron clouds in the σ bond and the π bond between the carbon atoms in ethene. Label your diagram. [Total: 15] SELF-EVALUATION After studying this chapter, complete a table like this: See Needs Almost Ready to I can section... more work there move on define electronegativity and explain the factors influencing the 4.7 electronegativity values of the elements explain the trends in electronegativity across a period and 4.7 down a group in the Periodic Table use differences in the Pauling electronegativity values to 4.7 predict if a compound has ionic or covalent bonds define ionic bonding and describe ion bonding in a sodium chloride 4.2 b magnesium oxide c calcium fluoride define covalent bonding and describe covalent bonding in a hydrogen b oxygen c nitrogen d chlorine e hydrogen chloride 4.3 f carbon dioxide g ammonia h methane i ethane j ethene describe how some atoms in Period 3 can expand their octet of electrons to form compounds such as sulfur dioxide, 4.3 phosphorus pentachloride and sulfur hexafluoride describe co-ordinate bonding (dative covalent bonding) in ions 4.3 such as NH4+ and molecules such as Al2Cl6 use dot-and-cross diagrams to show the arrangement of electrons in compounds with ionic, covalent and co-ordinate 4.3 bonding describe covalent bonding in terms of orbital overlap giving 4.5 sigma (σ) and pi (π) bonds describe how sigma and pi bonds form in a H2 b C 2H 6 4.5 c C 2H 4 d HCN e N2 describe hybridisation of atomic orbitals to form sp, sp2 and 4.5 sp3 orbitals define the terms bond energy and bond length and use these 4.3 to compare the reactions of covalent molecules using ‘valence shell electron pair repulsion’ (VSEPR) theory, describe and explain the shapes and bond angles in: a BF3 b CO2 c CH4 4.4, 4.5 d NH3 e H 2O f SF6 g PF5 predict the shapes and bond angles in other molecules and 4.4 ions similar to those above describe hydrogen bonding 4.8 explain, in terms of hydrogen bonding, the unusual properties of water (relatively low density of ice, relatively high melting 4.8 point, relatively high surface tension) describe and understand the different types of intermolecular forces (van der Waals’ forces) as either instantaneous dipoles 4.7 or permanent dipoles describe metallic bonding 4.6 describe the relative bond strengths of ionic, covalent and 4.9 metallic bonds compared with intermolecular forces. images Chapter 5 States of matter LEARNING INTENTIONS In this chapter you will learn how to: explain the origin of pressure in a gas in terms of collisions between gas molecules and the walls of the container explain that ideal gases have zero particle volume and no intermolecular forces of attraction use the ideal gas equation pV = nRT in calculations, including the determination of relative molecular mass describe the lattice structure of a crystalline solid to include: giant ionic structures including sodium chloride and magnesium oxide simple molecular structures including iodine, buckminsterfullerene and ice giant molecular structures including silicon(IV) oxide, graphite and diamond giant metallic structures including copper describe, interpret and predict the effect of different types of structure and bonding on the physical properties of substances, e.g. effect on melting point, boiling point, electrical conductivity and solubility deduce the type of structure and bonding present in a substance from given information. BEFORE YOU START 1 Describe the states of matter (solids, liquids and gases) in terms of the proximity (closeness), arrangement and motion of particles. Compare your answers with those of another learner. How well did they compare? 2 How does the volume of a gas change when the temperature is changed at constant pressure and the pressure is changed at constant temperature? Discuss with another learner the reasons for these changes, using ideas about moving particles. 3 Draw a diagram of a giant covalent structure such as diamond or graphite. Compare your drawing with one in a textbook. How could your drawing be improved? 4 Discuss these questions with another learner. a Why does diamond have a very high melting point but sulfur has a low melting point? b Why does graphite conduct electricity but diamond does not? c Explain how metals are able to conduct electricity. d Use your knowledge of the structure of metals to explain why metals are malleable. Be prepared to share your ideas with the rest of the class. 5 Try this problem by yourself and then discuss with another learner. A substance, Y, is soluble in water. Y does not conduct electricity either when solid or in solution. Y has a melting point of 185°C. Describe the structure and bonding in Y, giving reasons for your answer. 6 Phosphorus has a simple molecular structure. Work with another learner to make a list of the physical properties of phosphorus. LIQUID CRYSTALS: THE FOURTH STATE OF MATTER? Figure 5.1: A liquid crystal display in action in a mobile phone. Liquid crystal displays (LCDs) in mobile phones and computers are now part of everyday life (Figure 5.1). But how do they work? When you heat crystals of cholesteryl benzoate to 147 °C, they melt to form a cloudy liquid. At 180 °C the liquid becomes transparent. The cloudy state between the liquid and solid is called a liquid crystal metaphase. Many substances with rod-shaped molecules form a metaphase. In the metaphase, the molecules can move slightly. They have lost their regular arrangement but there is still some structure. For example, they point in about the same direction or are twisted into a helix. The direction in which the molecules face can be changed by small electric fields. LCDs have a thin film of liquid crystal between two transparent plates. The plates are treated chemically so that the liquid crystals arrange themselves in a particular way. The plates have an electrically conducting coating on their inner surface. Light that enters the display unit goes through a polariser, which makes the light vibrate in one plane. When the polarised light passes through the liquid crystal it turns through 90°. The light then passes through a second polariser set at 90° to the first. The display unit appears bright (Figure 5.2). Figure 5.2: The bright and dark states of a liquid crystal (LCD) display. When you apply a small voltage. the direction of the molecules changes so they no longer rotate in the plane of the polarised light. The unit appears dark. When the voltage is switched off the molecules go back to their original arrangement. By making the conductive layer as separate small areas any pattern can be displayed and switched on or off by a small voltage. Questions for discussion Discuss with another learner or group of learners: Compare the structures of a metaphase with a liquid and a solid. Is a metaphase another state of matter or not? How many hours a day do you spend looking at LCD screens (phone, computer, etc)? Do you think this is good for you or not? Why do some people think that too much time spent on looking mobile phone screens is a bad thing? 5.1 States of matter Gases have no fixed shape or volume. Gas particles: are far apart, therefore gases can be compressed are randomly arranged can move freely from place to place, in all directions. Liquids take the shape of the container they occupy. Liquid particles: are close together, so liquids have a fixed volume and can only be compressed slightly are mostly arranged randomly have limited movement from place to place, in all directions. Solids have a fixed shape and volume. Solid particles: are touching each other, so solids cannot be compressed are usually in a regular arrangement cannot change positions with each other: they can only vibrate. Question 1 Describe the changes that occur in the closeness and motion of the particles when: a a solid changes to a liquid b a liquid changes to a gas. The state of a substance at room temperature and pressure depends on its structure and bonding. Four types of structure are found in elements and compounds: simple molecular or simple atomic, e.g. carbon dioxide, argon giant ionic, e.g. sodium chloride giant metallic, e.g. iron, copper giant molecular, e.g. silicon(IV) oxide. The simple atomic structures found in the noble gases generally have similar physical properties to simple molecular gases. Although noble gases exist as isolated atoms, they can be thought of as having a molecular structure. 5.2 The gaseous state The kinetic theory of gases The idea that molecules in gases are in constant movement is called the kinetic theory of gases. This theory makes certain assumptions: the gas molecules move rapidly and randomly the distance between the gas molecules is much greater than the diameter of the molecules, so the volume of the molecules is negligible (hardly any volume). there are no intermolecular forces of attraction or repulsion between the molecules all collisions between particles are elastic: this means no kinetic energy is lost in collisions (kinetic energy is the energy associated with moving particles) the temperature of the gas is related to the average kinetic energy of the molecules. A theoretical gas that fits this description is called an ideal gas. In reality, the gases you are learning about don’t fit this description exactly, although they may come very close. We call these gases real gases. Noble gases with small atoms, such as helium and neon, come close to ideal gas behaviour. This is because the intermolecular forces are so small. Question 2 Explain why the intermolecular forces in a sample of helium and neon are very small. IMPORTANT Ideal gases have zero particle volume and no intermolecular forces of attraction. Ideal gases The volume that a gas occupies depends on: its pressure; we measure pressure in pascals, Pa its temperature; we measure temperatures of gases in kelvin, K. The kelvin temperature equals the Celsius temperature plus 273. For example, 100 °C is 100 + 273 = 373 K. Question 3 a Convert the following temperatures into the kelvin temperature: i 245 °C ii −45 °C b How many pascals are there in 15 kPa? IMPORTANT The pressure of a gas is due to collisions between the gas molecules and the walls of the container. Gases in a container exert a pressure. This is because the gas molecules are constantly hitting the walls of the container. If we decrease the volume of a gas (at constant temperature) the molecules are squashed closer together and hit the walls of the container more often. So the pressure of the gas increases (Figure 5.3a). A graph of volume of gas plotted against gives a proportional relationship (as shown by the straight line in Figure 5.3b). We say that the volume is inversely proportional to the pressure. Figure 5.3: a As the volume of a gas decreases, at constant temperature, its pressure increases due to the increased frequency of the gas molecules hitting the walls of the container. b For an ideal gas a plot of the volume of gas against shows a proportional relationship IMPORTANT Two variables are proportional if there is a constant ratio between them. Look at the figures for temperature and gas volume in the table below: Temperature / K 136.5 273 546 Gas volume / dm3 11.2 22.4 44.8 Table 5.1: Data table showing temperature against gas volume. When a gas is heated at constant pressure its volume increases (Figure 5.4a). This is because the particles move faster and hit the walls of the container with greater force. For the pressure to be constant, the molecules must get further apart. The volume of a gas at constant pressure is proportional to its temperature measured in kelvin (Figure 5.4b). Figure 5.4: a As the temperature increases, at constant pressure, the volume of a gas increases. Molecules hit the walls with increased force. b For an ideal gas, the volume of a gas is proportional to its kelvin temperature. An ideal gas will have a volume that varies exactly in proportion to its temperature and exactly in inverse proportion to its pressure. Question 4 Some chemical reactions involving gases are performed in sealed glass tubes that do not melt at high temperatures. If the tubes have walls that are too thin, the tubes can easily break. Use the kinetic theory of gases to explain why these tubes should not be heated to high temperatures. Limitations of the ideal gas laws Scientists have taken accurate measurements to show how the volumes of gases change with temperature and pressure. These show us that gases do not always behave exactly as we expect an ideal gas to behave. This is because real gases do not always obey the kinetic theory in two ways: there is not zero attraction between the molecules we cannot ignore the volume of the molecules themselves. These differences are especially noticeable at very high pressures and very low temperatures. Under these conditions: the molecules are close to each other the volume of the molecules is not negligible compared with the volume of the container there are instantaneous dipole−induced dipole forces or permanent dipole−permanent dipole forces of attraction between the molecules attractive forces pull the molecules towards each other and away from the walls of the container the pressure is lower than expected for an ideal gas the effective volume of the gas is smaller than expected for an ideal gas. Question 5 a What is meant by the term ideal gas? b Under what conditions do real gases differ from ideal gases? Explain your answer. The general gas equation IMPORTANT The ideal gas equation pV = nRT shows the relationship between pressure, volume, temperature and number of moles of gas. Make sure that you can rearrange this equation to make any of these values the subject of the expression, e.g. For an ideal gas, we can combine the laws about how the volume of a gas depends on temperature and pressure. We also know from Section 3.8 that the volume of a gas is proportional to the number of moles present. Putting all these together, gives us the ideal gas equation: pV = nRT p is the pressure in pascals, Pa V is the volume of gas in cubic metres, m3 (1 m3 = 1000 dm3) n is the number of moles of gas R is the gas constant, which has a value of 8.31 J K−1 mol−1 T is the temperature in kelvin, K. Calculations using the general gas equation If we know any four of the five physical quantities in the general gas equation, we can calculate the fifth. IMPORTANT To change °C to kelvin, add 273 to the Celsius (°C) temperature. A pressure of 101 325 pascals = 1.0 atmosphere pressure. WORKED EXAMPLES 1 Calculate the volume occupied by 0.500 mol of carbon dioxide at a pressure of 150 kPa and a temperature of 19 °C. (R = 8.31 J K−1 mol−1) Solution Step 1: Change pressure and temperature to their correct units: 150 kPa = 150 000 Pa; 19 °C = 19 + 273 = 292 K Step 2: Rearrange the general gas equation to the form you need: pV = nRT so V= Step 3: Substitute the figures: V = = = 8.09 × 10−3 m3 = 8.09 dm3 2 A flask of volume 5.00 dm3 contained 4.00 g of oxygen. Calculate the pressure exerted by the gas at a temperature of 127 °C. (R = 8.31 J K−1 mol−1; Mr oxygen = 32.0) Solution Step 1: Change temperature and volume to their correct units and calculate the number of moles of oxygen. 127 °C = 127 + 273 = 400 K 5 dm3 = m3 = 5.00 × 10−3 m3 n = = = 0.125 mol Step 2: Rearrange the general gas equation to the form you need: pV = nRT so p= Step 3: Substitute the figures: p= p= p = 8.31 × 104 Pa Note: If you are given the pressure in kPa and the volume in dm3, you can insert these figures directly into the equation without having to convert into Pa and m3. This is because by changing kPa to Pa you are multiplying by 1000 and by changing dm3 to m3 you are dividing by 1000. Because p and V are  on the same side of the equation pV = nRT, the × 1000 and ÷ 1000 cancel each other. IMPORTANT Remember that if you substitute the pressure in kPa in the equation pV = nRT, the volume of gas calculated will be in dm3. If you substitute the pressure in Pa in the equation pV = nRT, the volume of gas calculated will be in m3. Calculating relative molecular masses An accurate method of finding the relative molecular mass of a substance is to use a mass spectrometer (see Chapter 3). A less accurate method, but one that is suitable for a school laboratory, is to use the general gas equation to find the mass of gas in a large flask. As the number of moles is the mass of a substance divided by its relative molecular mass, we can find the relative molecular mass of a gas by simply substituting in the general gas equation. Although weighing gases is a difficult process because they are so light and you have to consider the buoyancy of the air, the method can give reasonable results. You can apply this method to find the relative molecular mass of a volatile liquid. The volatile liquid is injected into a gas syringe placed in a syringe oven (Figure 5.5). The liquid vaporises and the volume of the vapour is recorded. WORKED EXAMPLE 3 A flask of volume 2.00 dm3 was found to contain 5.28 g of a gas. The pressure in the flask was 200 kPa and the temperature was 20 °C. Calculate the relative molecular mass of the gas. (R = 8.31 J K−1 mol−1) Solution Step 1: Change pressure, temperature and volume to their correct units. 200 kPa = 2.00 × 105 Pa 20 °C = 20 + 273 = 293 K 2.00 dm3 = m3 = 2.00 × 10−3 m3 Step 2: Rearrange the general gas equation to the form you need: pV = nRT and n = so pV = RT, which gives Mr = Step 3: Substitute the figures: Mr = = = 32.14 = 32 g mol−1 Question 6 a Calculate the volume occupied by 272 g of methane at a pressure of 250 kPa and a temperature of 54 °C. (R = 8.31 J K−1 mol−1; Mr methane = 16.0) b The pressure exerted by 0.25 mol of carbon monoxide in a 10 dm3 flask is 120 kPa. Calculate the  temperature in the flask in kelvin. PRACTICAL ACTIVITY 5.1 Determining relative molecular mass Figure 5.5: The relative molecular mass of a volatile liquid can be found using a syringe oven. SAFETY: Only carry out this activity in the presence of a teacher after safety aspects have been explained. 1 Put a gas syringe in the syringe oven and leave until the temperature is constant. 2 Record the volume of air in the gas syringe. 3 Fill a hypodermic syringe with the volatile liquid and find its total mass. 4 Inject a little of the liquid into the gas syringe then find the total mass of the hypodermic syringe again. 5 Allow the liquid to vaporise in the gas syringe. 6 Record the final volume of vapour + air in the gas syringe. 7 Record the atmospheric temperature and pressure. The calculation is carried out in the same way as Worked example 3. The volume of vapour produced is: final gas syringe volume − initial gas syringe volume The mass used in the calculation is: initial mass of hypodermic syringe + liquid − final mass of hypodermic syringe + liquid Questions 7 When 0.08 g of liquid X was vaporised at 100 °C, 23 cm3 of vapour was formed. The atmospheric pressure was 1.02 × 105 Pa. Calculate the relative molecular mass of liquid X. (R = 8.31 J K−1 mol−1) 8 A flask of volume 2 × 10−3 m3 contains 4.19 g of a gas. The pressure in the flask is 300 kPa and the temperature is 20 °C. R = 8.31 J K−1 mol−1. You can find the relative molecular mass using the expression: Mr = Which one of the following gives the correct value for the relative molecular mass of the gas? A B C D 5.3 The liquid state The behaviour of liquids When we heat a solid: the energy transferred to the solid makes the particles vibrate more vigorously the forces of attraction between the particles weaken the solid changes to a liquid at its melting point. IMPORTANT The melting point of a solid is the temperature at which it changes to a liquid at 1 atmosphere pressure. We call this change of state melting. For ionic compounds, we need a high temperature to melt the substance because ionic bonding is very strong. For molecular solids, we need a lower temperature, just enough to overcome the weak intermolecular forces between the particles. The particles in a liquid are still close to each other but they have enough kinetic energy to keep sliding past each other in a fairly random way. They d