Pre-U STPM Chemistry Term 1 CC039342a

92 Chemistry Term 1 STPM CHAPTER 3 Shape Total of electron pairs No. of bonding pairs No. of lone pairs Trigonal pyramid •• 4 3 1 Bent •• •• 4 2 2 Trigonal bipyramidal 5 5 0 See-saw shape •• 5 4 1 T-shape •• •• 5 3 2 Octahedron 6 6 0 Squrare planar •• •• 6 4 2 Squarebased pyramid •• 6 5 1 Predicting the Shapes of Molecules from the Lewis Diagram 1 The steps below outline the procedure for determining the general shapes of molecular species. (a) Draw the Lewis diagram of the species. (b) Determine the number of bond pairs and lone pairs surrounding the central atom. (c) A double bond or a triple bond occupies the same space as a single bond. (d) Determine the general shape of the species (Refer to the table). A multiple bond occupies the same space as a single bond. 2008/P1/Q6 2009/P1/Q6 2008/P2/Q5(a) 2011/P1/Q43 2013/P1/Q5, Q19(b)

93 Chemistry Term 1 STPM CHAPTER 3 2 Take the methane molecule, CH4 , as example. (a) The Lewis diagram of CH4 is C H H H H x • x • x • x • (b) The central carbon atom is surrounded by 4 bond pairs (with no lone pairs). (c) Thus, the shape is tetrahedral, with bond angles of 109.5°. C H H H H 3 Another example is the water molecule, H2 O. (a) The Lewis diagram of H2 O is x • H x • x x x x O H (b) There are two bond pairs and two lone pairs surrounding the central oxygen atom. (c) Thus, it has a bent structure. (d) Note that the bent structure is derived from the tetrahedron. Lone pair electrons O H • • • • H 4 Let us look at the carbon dioxide molecule, CO2 . (a) The Lewis diagram is O C O x • x • x • x • A double bond is counted as one double bond pair. Similary this applies for a triple bond.

94 Chemistry Term 1 STPM CHAPTER 3 (b) The central carbon atom is surrounded by 2 bond pairs (i.e. 2 double bonds) and not 4 bond pairs. (c) Thus, its shape is linear with a bone angle of 180°. O == C== O 5 The structure of the nitrate ion, NO3 – : (a) The Lewis diagram is shown on the left. (b) There are 3 bond pairs (make up of 2 single bonds and 1 double bond) surrounding the nitrogen atom. This gives rise to a trigonal planar shape with bond angles of 120°. Predicting the Shapes of Molecules without Drawing the Lewis Structure An alternate way to determine the number of electrons pairs surrounding the central atom is as follows. (This method is suitable in cases where the Lewis diagram is not needed, such as in multiplechoice questions). (a) Determine the number of valence electrons of the central atom (this is the same as the group number of the element). (b) Add to it the number of electrons per atom shared between the terminal atoms and the central atom. (c) For positive ions, subtract one electron for every one positive charge. (d) Determine the number of electron pairs surrounding the central atom by dividing the total number of electrons surrounding the central atom by two. (e) Determine the number of bond pairs (remember that a multiple bond is counted as one bond pair) and lone pairs surrounding the central atom. (f) Refer to the table to obtain the shape of the species. The Shape of H2 O 1 O 6 (Valence electrons) H 1 (No. of electrons shared) H 1 (No. of electrons shared) Total electrons: 8 Total electron pairs: 4 2 But, oxygen is bonded only to two hydrogen atoms. Hence, the 4 electron pairs are made up of 2 bond pairs and 2 lone pairs. Thus, the H2 O molecule has a bent structure. O H • • • • H N O O O

95 Chemistry Term 1 STPM CHAPTER 3 3 The following format can also be adopted. Atom Number of valence electrons/shared Type of bond Oxygen, O 6 Hydrogen, H 1 O—H Hydrogen, H 1 O—H Total electrons 8 Total electrons pairs 8 2 = 4 4 But, oxygen is bonded only to two hydrogen atoms by two single bonds. Hence, the 4 electron pairs are made up of 2 bond pairs and 2 lone pairs, giving rise to a bent structure. The Shape of NH4 + Atom Number of valence electrons/shared Type of bond Nitrogen, N 5 Hydrogen, H 1 N—H Hydrogen, H 1 N—H Hydrogen, H 1 N—H Hydrogen, H 1 N—H Positive charge (+1) –1 Total electrons 8 Total electrons pairs 8 2 = 4 The 4 electron pairs consist of 4 single bond pairs, giving rise to a tetrahedral shape. The Shape of CO3 2– 1 Atom Number of valence electrons/shared Type of bond Carbon, C 4 Oxygen, O 2 C=O Oxide, O– 1 C—O– Oxide, O– 1 C—O– Total electrons 8 Total electrons pairs 4 N H H H H O– with the electronic configuration of 2.7 needs to share one electron only to achieve octet configuration.

96 Chemistry Term 1 STPM CHAPTER 3 2 The 8 electrons are made up of 3 bond pairs (1 double bond and 2 single bonds). 3 This gives rise to a trigonal planar shape with bond angles of 120°. The Shape of SO4 2– 1 Atom Number of valence electrons/shared Type of bond Sulphur, S 6 Oxygen, O 2 S=O Oxygen, O 2 S=O Oxide, O– 1 S—O– Oxide, O– 1 S—O– Total electrons 12 Total electrons pairs 6 2 The 6 electron pairs are made up of 4 bond pairs (2 double bonds, 2 single bonds). 3 The SO4 2– ion has a tetrahedral shape. The Shape of I3 – 1 Atom Number of valence electrons/shared Type of bond Iodine, I 7 Iodine, I 1 I—I Iodide, I– 2 I—I Total electrons 10 Total electrons pairs 5 2 The 5 electron pairs are made up of 2 bond pairs and 3 lone pairs. 3 The shape is linear. O– C O O– O– S O O– O I •• •• •• I I – 2011/P2/Q2(b) Info Chem The I– ion already has an octet. It donates two electrons to form a coordinate bond. Quick Check 3.2 Predict the shape of the following species. 1 NH3 6 PCl 4 + 11 SbF5 2– 2 BeCl2 7 CH3 – 12 SO2 3 PbCl4 8 HCN 13 SO3 2– 4 PCl5 9 ICl 4 – 14 ClO2 – 5 SF6 10 SnCl2

97 Chemistry Term 1 STPM CHAPTER 3 Bond Angle The bond angle is the angle between any two adjacent covalent bonds. Effect of Lone Pairs on Bond Angle 1 The Lewis diagrams of methane, ammonia and water are shown below. C H H H H x • x • x • x • N H H H x • x • x • • • H O H x • x • x x x x 2 These three molecules have one thing in common that is there are four electron pairs surrounding the central atom. The electron pair distribution around the central atom is that of a tetrahedron. 3 However, their bond angles are different as shown below. H C H H H 109.5° H N H H 107° •• •• O H H 104.5° •• 4 The difference arises because of the presence of lone-pair electrons as shown in the table below. The four electron pairs are divided as: Molecule Total electron pairs No. of bond pairs No. of Ione pairs Methane 4 4 0 Ammonia 4 3 1 Water 4 2 2 5 In CH4 , all the bond pairs are identical. Repulsion between the bonds is the same. Thus, CH4 has a perfect tetrahedral structure with bond angles of 109.5°. 6 In H2 O, the repulsion between the two lone pairs is much stronger than the repulsion between the two bond pairs. The two bond pairs are thus pushed closer to one another. This decreases the bond angle from 109.5° to 104.5°. 7 In NH3 , there is only one lone pair. The force of repulsion on the three bond pairs is lesser. This causes the bond angle to be larger than that in H2 O but smaller than in CH4 . 2008/P1/Q44 2018/P2/Q5 O H • • • • H Greater repulsion N H • • H H

98 Chemistry Term 1 STPM CHAPTER 3 Bond Angle and Electronegativity 1 Water and the hydrogen sulphide molecule have the same general shape with the same number of bond pairs and lone pairs as shown by their Lewis diagrams: H x • O H x • • • • • H x • S H x • • • • • 2 However, the bond angle in water is 104.5° while the bond angle in hydrogen sulphide is only 92.5°. 3 The difference is due to the difference in electronegativities between the oxygen atom and the sulphur atom. 4 Electronegativity is a measure of the ability of an atom to attract electrons in a covalent bond to which it is bonded. 5 Oxygen (E.N. = 3.5) is more electronegative than sulphur (E.N. = 2.5). The bonding electrons in the O—H bonds are closer to the oxygen atom than the bonding electrons in the S—H bond. H O H 104.5° x • x • H S H 92.5° x • x • • • • • • • • • 6 This results in a greater repulsion between the two O—H bonds pushing them further apart than the S—H bonds. 2009/P2/Q3(c) Exam Tips Exam Tips The higher the electronegatvity of the central atom, the larger is the bond angle. Quick Check 3.3 1 Explain the difference in bond angles in the following molecules. Molecule Bond angle NH3 107° PH3 94° 2 Between H2 S and H2 Se, which molecule is expected to have a larger bond angle? Explain your answer.

99 Chemistry Term 1 STPM CHAPTER 3 Polar and Non-polar Bonds 1 Consider the hydrogen chloride molecule below. H Cl 2 In the above representation, the electron cloud in the molecule is evenly distributed throughout the molecular orbital containing the hydrogen and chlorine atom. 3 In other words, we say that the two electrons in the covalent bond are shared equally between the two atoms. However, in practice, this is not the case. 4 A covalent bond is formed due to the attraction of the hydrogen and chlorine nuclei with the electrons in the molecular orbital. 5 However, their relative attractive force towards the shared electrons is not equal. The chlorine nucleus, which has 17 protons compared to only 1 proton in the hydrogen atom, exerts a greater attractive force on the electrons. That means chlorine is more electronegative than hydrogen. 6 As a result, the electron density around the chlorine atom is higher than that of hydrogen. The ‘chlorine end’ of the molecule will now have a higher electron density compared to the ‘hydrogen end’ of the molecule. This can be represented as H Cl 7 The ‘chlorine end’ of the molecule acquires a partial negative charge, while the ‘hydrogen end’ acquires a partial positive charge. δ+H—Clδ– 8 The covalent bond in the hydrogen chloride molecule is polarised, giving rise to a small amount of ionic character. 9 Most heteroatom molecules have polar covalent bonds due to the difference in the electronegativity between the bonding atoms. Examples are: H2 O, NH3 , CHCl3 , HF, AlCl3 and others. 10 The percentage of ionic character (i.e. the extent of polarisation) depends on the difference in the electronegativity of the bonding atoms. 11 On the other hand, the covalent bond in the chlorine molecule, Cl2 , is non-polar. This is because the electronegativity of both the bonding atoms is the same. Unequal sharing of the bonding electrons 2008/P1/Q5

100 Chemistry Term 1 STPM CHAPTER 3 12 The graph below shows the variation in percentage of ionic character in a covalent bond with the difference in electronegativity of the atoms. 100 Difference in electronegativity 80 60 40 20 1 2 3 % ionic character 13 The larger the difference in the electronegativity of the bonding atoms, the greater is the polarisation of the bond. Difference in electronegativity  2.0  2.0 and  0.4  0.4 0 Ionic character Ionic Polar covalent Mostly covalent Non-polar covalent bond Example 3.5 The following bonds involving carbon are polarised. Determine the polarity of the bonds. (a) C—Cl (b) C—N (c) C—O (d) C—Li (e) C—Mg Solution (a) δ+C—Clδ– (b) δ+C —Nδ– (c) δ+C—Oδ– (d) δ– C—Liδ+ (e) δ– C —Mgδ+ Polar and Non-polar Molecules 1 When a charged rod is brought near to a fine stream of running water from a burette, the path of the water will be deflected towards the charged rod. 2 However, if we replace the water in the burette with tetrachloromethane (also known as carbon tetrachloride), the stream is not deflected. 3 Water, which is affected by an electrostatic field, is called a polar molecule. Tetrachloromethane, which is not affected by an Charged rod electrostatic field, is a non-polar molecule. Water

101 Chemistry Term 1 STPM CHAPTER 3 4 Polar molecules when placed in an electric field will align themselves such that the negative end of the molecules faces the positive terminal of the field, and the positive end faces the negative terminal. 5 A non-polar molecule will have a random arrangement in the electric field. 6 For a polar molecule, the covalent bonds in the molecule must be polarised. However, there are molecules whose bonds are polarised but they are non-polar (i.e. they are not affected by electric fields). 7 Take the example of hydrogen chloride. The bond is polarised as shown. δ+H—Clδ– The molecule is said to have a dipole moment (μ) and it is represented as H—Cl µ ≠ 0 8 Now let’s look at the beryllium chloride molecule, BeCl3 (which is linear). The bonds in the molecule are polarised as chlorine is more electronegative than beryllium. Cl—Be—Cl µ = 0 However, in this case, the two dipole moments cancel out one another. Even though the bonds are polarised, the molecule as a whole does not have a net dipole moment. There is no distortion to the electron density in the bonds. Beryllium chloride is non-polar. 9 On the other hand, water, H2 O, is a polar molecule. The water molecule has a bent structure. The dipole moments in the bonds do not cancel out one another. Thus, the water molecule has a net dipole moment as shown on the right. 10 A molecule is polar provided, (a) the covalent bonds are polarised, and (b) the dipole moments of the bonds do not cancel out one another. Presence of net dipole moment Dipole moments cancel out O H H Net dipole moment μ ≠ 0

102 Chemistry Term 1 STPM CHAPTER 3 Quick Check 3.4 Determine whether the following molecules are polar or non-polar. 1 H2 O 4 CH4 2 NH3 5 SO3 3 AlCl3 6 CO2 Polarisation in Ionic Bonds 1 Polarisation also occurs in ionic bonds. 2 A pure ionic bond is visualised as consisting of perfectly spherical cations and anions, where their respective charges are uniformly distributed. + – Uniform distribution of charge density 3 However, in practice, this is not always the case. This is because the positive electric field around the cation tends to attract the electron cloud around the anion towards it. 4 As such, the electron cloud of the anion is distorted. This gives rise to a certain amount of covalent character in the ionic bond. + – Distortion of the electron cloud of the anion 5 The percentage of covalent character in an ionic bond depends on: (a) the polarising power of the cation, (b) the polarisibility of the anion. 6 The polarising power of a cation is directly proportional to its charge density ( Charge Ionic radius ). 2015/P1/Q7

103 Chemistry Term 1 STPM CHAPTER 3 Exam Tips Exam Tips 7 The table below lists the charge densities of some common cations. Cation Charge Ionic radius/ nm Charge Ionic radius Relative polarising power K+ 1 0.133 7.5 1.0 Na+ 1 0.095 10.5 1.4 Mg2+ 2 0.065 30.8 4.1 Be2+ 2 0.031 64.5 8.6 AI3+ 3 0.050 60.0 8.0 8 The polarisibility of an anion is the ease with which the electron cloud around an anion can be distorted by a cation. 9 The polarisibiltiy increases with increasing size of the anion The bigger the anion, the weaker is the attraction between the nucleus and the electrons. Example 3.6 Arrange the following compounds in order of increasing covalent character. Explain your answer. NaCl, NaBr, NaI Solution NaCl  NaBr  NaI The size of the anion increases in the order: Cl–  Br–  I– Hence, the larger I– is more easily to be polarised by Na+ compared to Br– and Cl– . Example 3.7 Which of the following oxide has a higher degree of covalent character? MgO, Al2 O3 Solution The charge density of Al3+ is higher than that of Mg2+. Thus, its polarising power towards the O2– ion is higher, giving aluminium oxide more covalent character than magnesium oxide. Strength of Covalent Bonds 1 The strength of a covalent bond is measured by its bond energy and bondlength. The bigger the anion, the easier it is polarised by cations. The following compounds also show covalent character: Al2 O3 , AlI3 , LiI 2016/P1/Q19(a)(iii)

104 Chemistry Term 1 STPM CHAPTER 3 2 The bond energy is defined as the energy required (in kJ mol–1) to break a covalent bond, per mole of the bond. The higher the bond energy, the stronger the bond. A—B(g) → A(g) + B(g) ∆H = positive 3 The bondlength is defined as the distance between two adjacent nuclei of the two atoms that are joined by the covalent bond. Bond length 4 The bondlength decreases as the bond energy increases. 5 The table below lists the bondlength and bond energy of some common covalent bonds. Bond Bondlength/nm Bond energy/kJ mol–1 H—H 0.074 436 H—F 0.092 567 H—CI 0.127 431 H—Br 0.141 366 H—I 0.161 299 C—C 0.154 348 C=C 0.134 612 C≡C 0.120 804 C—O 0.143 360 C=O 0.116 740 C—F 0.138 484 C—CI 0.177 338 C—Br 0.194 276 C—I 0.214 238 O=O 0.121 496 N≡N 0.110 944 CI—CI 0.200 242 Br—Br 0.228 193 I—I 0.266 151 6 Generally, multiple bonds (double and triple bonds) are stronger than single bonds. General Properties of Covalent Compounds 1 The bonds that hold the atoms together in a covalent molecule are the strong covalent bonds. 2 However, the bond that holds individual molecules together is the weak van der Waals force. Exam Tips Exam Tips The strong N≡N bond and the non-polar nature of the N2 molecule are the reasons for nitrogen's inertness. Note that all the species involved are in the gaseous state. 2014/P1/Q4 2016/P1/Q19(b)

105 Chemistry Term 1 STPM CHAPTER 3 3 Little energy is required to weaken or to break these weak intermolecular forces. Thus, covalent molecules are usually liquids or gases at room conditions while some are solids with low melting points. Molecule Melting point/°C Boiling point/°C H2 O 0 100 CO2 –56 –78 NH3 –78 –33 CHCI3 –63 62 C2 H4 –169 –103 4 However, there are some covalent compounds with exceptionally high melting/boiling points. An example is silicon dioxide, SiO2 (melting point = 1710 °C). These compounds do not exist as simple molecules but have giant covalent structures with strong covalent bonds holding all the atoms together. 5 Due to the strong covalent bonds in the structure, a lot of energy is required to weaken or to break the bonds. This accounts for their high melting/boiling points. 6 Generally, covalent compounds are insoluble in water but are soluble in non-polar solvents such as benzene and tetrchloromethane. 7 However, some covalent compounds such as alcohols and carboxylic acids dissolve in water because they can form hydrogen bonds with water molecules. 8 There are also some covalent compounds which can react with water to form water-soluble products. Examples are hydrogen chloride and ammonia. HCl(g) + aq → H+(aq) + Cl– (aq) NH3 (g) + H2 O(l)  NH4 +(aq) + OH– (aq) 9 Generally, covalent compounds are non-conductors of electricity either in the solid state or liquid state except those that dissolve in water to form aqueous ions. The Coordinate Bond 1 A coordinate bond or dative bond is a special type of covalent bond where both the electrons shared between two atoms are contributed by one of the atoms only. It can be represented as follows: A B + A B 2 Atom A is called the donor atom and atom B is called the acceptor atom. 3 The donor atom must have a lone pair electrons and the acceptor atom must have an empty orbital to accommodate the lone pair electrons. Simple molecular structure Giant covalent structure ‘Sand’ is an impure form of silicon dioxide. Some covalent compounds are soluble in water. 2008/P1/Q7 2009/P2/Q3(b) 2017/P1/Q5

106 Chemistry Term 1 STPM CHAPTER 3 4 A coordinate bond is usually represented by an arrow from the donor atom to the acceptor atom. 5 A coordinate bond is identical to a ‘normal’ covalent bond after it has been formed. It only differs from the way the bond is formed. 6 An example of coordinate bond is the ammonium ion, NH4 +, formed when ammonia reacts with a H+ ion: NH3 + H+ → NH4 + (a) The ammonia molecule has a lone pair of electrons on the nitrogen atom. H | H— N• • | H (b) The H+ ion on the other hand has an empty 1s orbital. (c) During the reaction, nitrogen shares its lone pair electrons with H+ by forming a coordinate bond between the nitrogen atom and the H+ ion: H H N HH • • + (d) The Lewis diagram of the NH4 + ion is: H N •x • •x x • • H H H+ 7 Another example is the hydroxonium ion, H3 O+, formed when a water molecule combines with an H+ ion. H2 O + H+ → H3 O+ H H H+ O 8 Aluminium chloride exists in the form of AlCl3 molecules with the following structure.

107 Chemistry Term 1 STPM CHAPTER 3 Cl X X• • X • Cl Al Cl [Note: Non-bonding electrons on the terminal atoms are omitted for clarity.] It is an octet deficient molecule as the aluminium atom is surrounded by only 6 electrons. 9 However, the aluminium atom can achieve octet if two AlCl3 combines to form a ‘dimer’ with the molecular formula of Al2 Cl6 through coordinate bond formation. Cl Cl Cl Cl Al Al Cl Cl Cl Cl Cl Cl Al Cl Cl Al 10 Another example is the hexacyanoferrate(III) ion [Fe(CN)6 ]3–. Fe3+ CN– CN– CN– CN– CN– CN– Note that the [FeCN)6 ]3- ion is an expended octet species. 3.3 Metallic Bonding 1 Metals have an unique property that is not shared by other substances (ionic or covalent), in that they conduct electricity in the solid state. 2 On top of that, metals are hard, with high melting/boiling points and are malleable (can be knocked into shapes) and ductile (can be pulled into thin wires). 3 Metals are also good conductors of heat. A metal can conduct electricity in the solid state. Electronic conduction is due to electrons. 2011/P1/Q7 2013/P1/Q6 2014/P1/Q5/Q19(b)(i)

108 Chemistry Term 1 STPM CHAPTER 3 4 X-ray crystallography shows that the atoms in metals are packed very tightly against one another in the crystal lattice. 5 The forces that hold the atoms together in the solid lattice is called metallic bonds. How are the bonds formed? 6 Let us take sodium as an example. In the solid state, the sodium atoms are packed so close together that the 3s orbitals (containing one unpaired electron) of the atoms overlap with one another to form a giant ‘molecular orbital’. 7 The valence electron of each sodium atom is then free to move throughout the whole crystal structure and is no more bound to the outer shell of any sodium atom. This denotes that, the electrons are ‘ionised’ and are delocalised throughout the whole solid. 8 The attraction between the ‘cations’ and the delocalised electrons constitute the metallic bond. The delocalised electrons are sometimes referred to as a ‘sea of delocalised mobile electrons’. 9 When a electric potential is applied to the metal, the mobile electrons will move towards the positive terminal and thus conducts electricity. 10 When one end of the metal is heated, the kinetic energy of the electrons increases. This is then transmitted to the other end by the system of delocalised mobile electrons. 11 Metals have relatively low ionisation energy. Their valence electrons are easily excited to higher energy levels. When these electrons returned to the ground state, the excess energy is liberated as electromagnetic radiation, which makes the solids shine. 12 When a piece of metal is knocked or pulled, the layers of cations simply slide over one another but are still held together by the ‘flexible glue’ of the delocalised electrons. Push Strength of the Metallic Bonds 1 The strength of the metallic bonds depends on (a) the number of electrons in the sea of delocalised electrons, (b) the size of the atom. Overlap of atomic orbitals in solid sodium. + + + + + + + + + e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - e - + + + + + + + + + + + + + + + + + + + + + + + + sea of delocalised mobile electrons VIDEO Metallic Bonding

109 Chemistry Term 1 STPM CHAPTER 3 2 The melting points and boiling points of three metals are given below. Metal Sodium Magnesium Aluminium No. of valence electrons 1 2 3 Melting point/°C 98 649 660 Boiling point/°C 883 1107 2467 The melting/boiling point increases in the order Na  Mg  Al. This is because the number of electrons per atom donated to the sea of delocalised electrons increases from 1 for sodium, 2 for magnesium and 3 for aluminium. As a result, the metallic bonding in aluminium is the strongest and that in sodium is the weakest. 3 Consider the melting points of beryllium and magnesium. Element Beryllium Magnesium No. of valence electrons 2 2 Atomic radius/nm 0.11 0.16 Melting point/°C 1280 649 The bigger size of the magnesium atom causes the attraction between the nucleus and the sea of delocalised electrons to be weaker. As a result, magnesium has a lower melting point than beryllium. 4 However, the strength of the metallic bond also depends on the packing of the atoms in the solid crystal. Melting points of magnesium and calcium are shown below: Element Magnesium Calcium No. of valence electrons 2 2 Atomic radius/nm 0.16 0.20 Melting point/°C 649 840 Lattice structure Hexagon Face-centred cube Even though calcium is larger than magnesium, yet its melting point is higher than that of magnesium. This is because calcium has a more close-packed structure than magnesium. This close-packed structure brings the calcium atom closer to one another compared to that of magnesium. Solid lattice structure also affects the strength of metallic bonds. Generally, the bigger the size of an atom, the weaker is the metallic bond. The more the valence electrons, the stronger is the metallic bond. Quick Check 3.5 The melting points of magnesium and chromium are 649 °C and 1875 °C respectively. Explain the difference in terms of metallic bonding.

110 Chemistry Term 1 STPM CHAPTER 3 3.4 Intermolecular Forces: van der Waals Forces and Hydrogen Bonding 1 The magnitude of intermolecular forces is reflected in the physical properties such as melting point, boiling point and enthalpy of vaporisation of a substance. 2 At room conditions, ammonia is a gas. When the temperature of ammonia is reduced, it will condense to a liquid when the temperature falls below –80 °C. 3 This suggests that there are some sort of attractive forces (although a weak one) that hold the ammonia molecules together in the liquid state. 4 What are these forces? What is their origin? The force that holds covalent molecules together either in the liquid state or solid state is called the van der Waals force. 5 The van der Waals force is made up of two major components: (a) Permanent dipole-permanent dipole attraction (b) Dispersion force (Temporary dipole-induced dipole attraction) Permanent Dipole-Permanent Dipole Attraction 1 Consider the hydrogen chloride molecule discussed above. The molecule is polar. It has a permanent separating of charges. δ+H—Clδ– 2 As a result, the polar molecules will attract one another, giving rise to dipole-dipole attractive force or sometimes it is called permanent dipole-permanent dipole attractive force. δ + δ – δ – δ + δ + δ – δ – δ + δ + δ – δ – δ + δ + δ – δ – δ + δ + δ – δ – δ + δ + δ – δ – δ + δ + δ – δ – δ + δ + δ – δ – δ + δ + δ – δ – δ + δ + δ – δ – δ + 3 Other examples of dipole-dipole attraction are found in trichloromethane, nitrogen dioxide and hydrogen sulphide. Temporary Dipole-Induced Dipole Forces 1 Oxygen, O2 , is a non-polar molecule. When cooled, oxygen will liquefy at –183 °C and eventually freeze at –219 °C. The same applies to other non-polar molecules such as the noble gases He, Ar, Ne and others. Physical properties are determined by intermolecular forces. INFO Van der Waals Forces 2010/P1/Q6 2011/P1/Q6, Q8 2015/P1/Q8 2011/P2/Q2(a) 2017/P1/Q6

111 Chemistry Term 1 STPM CHAPTER 3 Exam Tips Random movement of electrons in an atom or molecule gives rise the temporary dipoles. 2 The nature of the intermolecular forces between non-polar molecules was investigated by Fritz London in 1932. He suggested that the intermolecular forces arise as a result of the random movement of electrons in a molecule, giving rise to temporary dipoles. 3 Take neon (proton number 10) as an example. Neon is a monatomic ‘molecule’. Since neon is non-polar, the distribution of electrons in the atom, on the average, is uniform. 4 However, when these 10 electrons revolve round the nucleus of the neon atom, there is a high possibility that at any given instant, the electron density might be higher in one part of the atom compared to the other parts. 5 This results in the formation of a temporary dipole or instantaneous dipole because it lasts for just a tiny fraction of time. Higher electron density Lower electron density 6 In the next instance, the distribution of the electron density will change and the atom has a new instantaneous dipole. 7 The temporary dipole can distort the electron clouds of neighboring atoms, giving rise to induced dipoles. 8 The attraction between temporary dipoles and induced dipoles gives rise to the intermolecular force. δ+ δ– δ+ δ– δ+ δ– 9 The polarity of the temporary dipoles changes with time. However, the net attraction between them is permanent. δ– – δ+ δ+ δ– δ+ δ– δ δ+ δ– δ+ δ– δ+ 10 The dipole-dipole forces and the temporary dipole-induced dipole forces are collectively called the van der Waals forces, named after the Dutch physicist, Johanne Diderik van der Waals. Uneven distribution of electron density. The temporary dipole-induced dipole attraction is sometimes called the dispersion force. Ne Even distribution of electron cloud

112 Chemistry Term 1 STPM CHAPTER 3 11 Temporary dipole-induced dipole forces also exist between polar molecules. In many cases, it is the major contributor to the van der Waals forces as shown by the table below. Species Contribution to the van der Waals force Dipole-dipole attraction Temporary dipole-induced dipole force Ar 0 100 CO2 0 100 HCI 20 80 12 Attractive force Occurrence Dipole-dipole Between polar molecules only Temporary dipole-induced dipole Between all covalent molecules Factors Affecting the Strength of the van der Waals force 1 The strength of the van der Waals force depends on (a) the size of the atom or molecule, (b) the relative atomic/molecular mass, (c) the total number of electrons in the atom or molecule, (d) the total surface area of the atom or molecule. 2 The boiling points of the Group 18 elements are given in the table below. Element Atomic radius/nm Relative molecular mass Total number of electrons Boiling point/°C He 0.050 4.0 2 –269 Ne 0.065 20.0 10 –246 Ar 0.095 40.0 18 –186 Kr 0.110 83.8 36 –152 Xe 0.130 131.0 54 –108 3 Moving from He to Xe, the size and the total number of electrons in the atoms increases. The attraction between the nucleus and the electrons gets weaker. As a result, the electron clouds become progressively easier to be distorted. This causes the van der Waals force to get stronger as reflected in their boiling points. 4 Consider the boiling points of three isomeric alkanes with the molecular formula of C5 H12. Compound Structural formula Boiling point/°C Pentane CH3 CH2 CH2 CH2 CH3 36 2-methylbutane CH3 —CH—CH2 —CH3 | CH3 28 2,2-dimethylpropane CH3 | CH3 —C—CH3 | CH3 10 Dispersion force is the major contributor to the van der Waals force. The larger the molecule, the stronger is the van der Waals force. 2016/P1/Q16(b)

113 Chemistry Term 1 STPM CHAPTER 3 5 The pentane molecule is linear. It has a larger surface area. This increases its polarisibility. As a result, the van der Waals force is the strongest and pentane has the highest boiling point. CH3 CH2 CH2 CH2 CH3 CH3 CH2 CH2 CH2 CH3 van der Waals force (Larger surface area of contact) 6 2,2-dimethylpropane is a branched chain alkane. The two CH3 side groups make the molecule somewhat more spherical. This decreases the area of contact between dimethylpropane molecules. Hence, the van der Waals force is the weakest and it has the lowest boiling point. CH3 C CH3 CH3 CH3 CH3 C CH3 CH3 CH3 van der Waals force (Less surface area of contact) 7 2-methylbutane is also a branched chain alkane. However, it is less spherical than 2,2-dimetylpropane. Thus, its boiling point is intermediate between that of pentane and 2,2-dimethylpropane. 8 Considering the boiling point of the hydrogen halides. Molecule HCl HBr HI Relative molecular mass 36.5 81.0 128.0 Total no. of electrons 18 36 54 % ionic character 19 12 4 Boiling point/°C –85 –66 –35 (a) Of the three hydrogen halides, HCl has the highest percentage of ionic character. This would mean that the permanent dipole-permanent dipole attraction is the strongest. (b) However, the boiling point of HCl is the lowest. (c) This shows that the contribution of permanent dipolepermanent dipole attraction towards the van der Waals force is much less than that from the temporary dipole-induced dipole force. (d) HI with its largest size and the most number of electrons have stronger van der Waals forces because the dispersion force (the major contributor) is the strongest. 9 However, when comparing molecules having the same size and the same number of electrons, the molecule which is more polar would have a stronger van der Waals force. Branched chain isomers are more spherical. The H—Cl bond is more polarised than the H—Br or H—I bonds. HI molecule has the largest size.

114 Chemistry Term 1 STPM CHAPTER 3 For example: Molecule Br2 ICI Relative molecular mass 162 162.5 Total no. of electrons 70 70 % ionic character 0 6 Boiling point/°C 59 98 The van der Waals force between ICl molecules is stronger than that of between Br2 molecules because ICl is polar while Br2 is non-polar. Example 3.8 Between fluorine and chlorine, which would you expect to have a higher boiling point? Solution The chlorine molecules are larger and have more electrons than fluorine. The electrons in chlorine are less tightly held by the nucleus. The chlorine molecules are more polarisable than the fluorine molecules. As a result, the van der Waals force between chlorine molecules is stronger. Hence, chlorine would have a higher boiling point than fluorine. Example 3.9 Arrange the following halopropanes in order of increasing boiling point. Explain your answer. CH3 CH2 CH2 Cl, CH3 CH2 CH2 Br, CH3 CH2 CH2 I Solution CH3 CH2 CH2 Cl  CH3 CH2 CH2 Br  CH3 CH2 CH2 I The size of the molecules and the total number of electrons in the molecules increase from 1-chloropropane to 1-iodopropane. The strength of the van der Waals forces increases and subsequently the boiling point also increases from 1-chloropropane to 1-iodopropane. δ+I—Cl δ–

115 Chemistry Term 1 STPM CHAPTER 3 Example 3.10 Explain why chlorine is a gas while iodine is a solid. Solution Molecule Total number of electrons CI2 34 I 2 106 The van der Waals forces between I2 molecules are stronger than those between Cl2 molecules. This is due to the larger size of the I2 molecule and the large number of electrons in the molecule. Quick Check 3.6 1 Arrange the following compounds in order of increasing boiling point. Explain your answer. H2 , O2 , Cl2 , CH4 2 The graph below shows the boiling points of the Group 14 hydrides from CH4 to SnH4 . Relative molecular mass Boilling point Offer an explanation for the trend observed. 3 The table below lists the boiling points of silane, SiH4 and hydrogen sulphide, H2 S: Compound Relative molecular mass Total number of electrons Boiling point/°C Silane –112 Hydrogen sulphide –61 (a) Complete the table. (b) Explain the difference in their boiling point. 4 Which of the two, propane (C3 H8 ) or ethanenitrile (CH3 CN), is expected to have a higher boiling point? Explain your answer.

116 Chemistry Term 1 STPM CHAPTER 3 Hydrogen Bonding 1 The Group 14 elements form tetrahydrides with the general formula of MH4 , where M = C, Si, Ge, Sn and Pb. 2 The tetrahydrides are all simple covalent molecules with covalent bonds holding the atoms together in a tetrahedral geometry. 3 They are all colourless liquid at room conditions. 4 The graph below shows the variation of the boiling points of the tetrahydrides with relative molecular mass. – 150 – 100 – 50 – 0 CH4 SiH4 GeH4 SnH4 PbH4 – 200 °C 5 The trend in the boiling points can be explained as follows. Molecule CH4 SiH4 GeH4 SnH4 PbH4 No. of electrons 10 18 36 54 83 (a) The intermolecular forces between the molecules are the van der Waals forces. H M H H H H M H H van der Waals force H (b) Going down the group from carbon to lead, the size of the molecule and the total number of electrons in the molecule increase. (c) As a result, the van der Waals forces get stronger and cause an increase in the boiling points. 6 The graph of the variation of the boiling points of the hydrides of the Group 17 elements (HF, HCl, HBr and HI) is shown on next page. 7 The general increase in the boiling point from HCl to HI is due to the increase in the strength of the van der Waals forces. This is because the size of the molecules and the total number of electrons in the molecules increase down the group. Molecular size increases from CH4 to PbH4 . No. of electrons increases from CH4 to PbH4 . H M H H H H M H H H INFO Hydrogen Bond 2009/P2/Q7(c) 2010/P1/Q8

117 Chemistry Term 1 STPM CHAPTER 3 H F Molecule HF HCI HBr HI No. of electrons 10 18 36 54 Boiling point/°C 20 –85 –67 –35 Temparature / °C 20 10 0 -10 -20 -40 -60 -80 0 10 20 30 40 50 60 No. of electrons HF HI HBr HCl 8 However, the boiling point of hydrogen fluoride is exceptionally high for its small size and the small number of electrons in the molecule. 9 This indicates that there must be some additional intermolecular forces of attraction other than the van der Waals force. What is this force? What is its origin? 10 Due to the very high electronegativity and the small size of the fluorine atom, the H—F bond is greatly polarised. Since hydrogen has no inner electrons, this polarisation causes the nucleus of hydrogen to be partly exposed giving it a high partial positive charge. δ++H—Fδ– – 11 The partial positive hydrogen end can get very close to the lone pair electrons of another fluorine atom without being repelled by the electron cloud around the other fluorine atom. δ++H—Fδ– – ---δ++H—Fδ– – ---δ++H—Fδ– – This dipole-dipole force is significantly stronger than the ‘normal’ van der Waals forces. It is given a special name, hydrogen bond. 12 The same pattern is also observed for the hydrides of the Group 15 and Group 16 elements. HF has exceptionally high boiling point. d++ d– – FH FH 2018/P1/Q6

118 Chemistry Term 1 STPM CHAPTER 3 Compound NH3 PH3 AsH3 SbH3 B.P./°C –33 –88 –55 –17 B. pt / °C -20 -10 -30 -40 -50 -60 -70 -80 10 20 30 40 50 60 NH3 AsH3 SbH3 PH3 Total no. of electrons in the molecule Compound H2 O H2 S H2 Se B.P./°C 100 –61 –41 100 80 60 40 20 0 -20 -40 -60 -80 B-p / °C 0 10 20 30 40 50 H2S H2Se H2O No. of electrons in the molecule

119 Chemistry Term 1 STPM CHAPTER 3 13 This indicates that NH3 , H2 O and HF can form intermolecular hydrogen bonds with their own molecules. 14 The boiling point of water is the highest among the three hydride because the oxygen atom in water has two lone pair electrons, and hence, can form a total of four hydrogen bonds per molecule. H O •• •• H O •• •• •• O •• H H H H •• O •• H H H •• O •• H 15 On the other hand, the nitrogen atom in ammonia has only one lone pair electrons, and can form only two hydrogen bonds per molecule. As a result, the boiling point of ammonia is lower than that of water. 16 Although the fluorine atom in hydrogen fluoride has three lone pair electrons, it makes use of only one pair to form hydrogen bond. 17 However, the higher electronegativity of fluorine compared to nitrogen makes the intermolecular hydrogen bonds in HF stronger than that in NH3 . As a result, the boiling point of HF is higher than that of NH3 but lower than that of H2 O. 18 The hydrogen bond is defined as the electrostatic attraction between a hydrogen atom (which is covalently bonded to a small and highly electronegative atom) and the lone pair of electrons of another small and highly electronegative atom. 19 One of the conditions for hydrogen bond to form is that a hydrogen atom must be bonded directly to either a nitrogen atom, an oxygen atom or a fluorine atom. 20 The relative average bond energy of a covalent bond, hydrogen bond and van der Waals force is given in the table below. Type of bonds van der Waals Hydrogen Covalent (single) Bond strength/kJ mol–1 22 65 450 Water molecules can form more extensive hydrogen bonds. The hydrogen bond between HF molecules is stronger than that between NH3 molecules. Hydrogen bond is about 3 times stronger than van der Waals force. H H •• H N H H •• H N •• H F •• •• •• H F •• •• Quick Check 3.7 Which of the following compounds can form intermolecular hydrogen bonds? (a) CH3 CH2 OH (e) HCOOCH3 (b) CHCl3 (f) CH3 COCH3 (c) HCl (g) H2 S (d) CH3 COOH

120 Chemistry Term 1 STPM CHAPTER 3 Effect of Hydrogen Bonding on the Physical Properties of Substances Boiling Points of Organic Compounds 1 Organic compounds that exist as simple molecules usually have low melting point and boiling point because the intermolecular forces are the weak van der Waals force. 2 However, organic compounds that can form intermolecular hydrogen bonds have significantly higher melting/boiling points than those compounds with comparable molecular mass which cannot form intermolecular hydrogen bonds. 3 Examples are propane and ethanol. Compound Ethanol Propane Formula C2 H5 OH CH3 CH2 CH3 Molecular mass/g mol–1 46 44 No. of electrons 26 26 Boiling point/°C 78.5 –42.0 The boiling point of ethanol is significantly higher than propane although both have almost the same molecular mass and the same number of electrons in the molecule. This is because ethanol can form intermolecular hydrogen bonds but propane cannot. Intermolecular hydrogen bonds increase the boiling point of ethanol. O•• •• H C2H5 O C2H5 H O C2H5 H Quick Check 3.8 1 Which of the following compounds can form intermolecular hydrogen bonding? (a) C6 H5 COOH (d) CH3 CONH2 (b) C6 H5 CH2 OH (e) N2 H4 (c) CH3 COOC2 H5 (f) NH2 OH 2 Explain in terms of intermolecular forces the difference between the boiling point of ethanol (78.5 °C) and dimethyl ether, CH3 OCH3 (–23 °C). 3 The boiling point of methanoic acid, HCOOH is 100.7 °C, while that of ethanol is 78.5 °C. Explain. 4 Which would you expect to have a higher boiling point, ethanol or 1-propanol? Explain your answer. Solubility of Organic Compounds in Water 1 In order for a substance to dissolve in water, the intermolecular force that holds the molecule together in the substance, as well as the intermolecular forces between water molecules, has to be first broken. This requires absorption of energy.

121 Chemistry Term 1 STPM CHAPTER 3 2 New bonds are then formed between water molecules and the molecules of the substance concern and energy is released. 3 Most organic molecules are attracted to each other by weak van der Waals force that needs little energy to break. 4 However, water molecules are attracted to each other by strong hydrogen bonds. More energy is required to break the hydrogen bonds. 5 For most organic compounds, the energy required to disrupt the hydrogen bonds in water is too high compared to the energy released when weak van der Waals force formed between water molecules and the organic molecules. As a result they are insoluble in water. 6 However, polar organic compounds having the —OH, —NH2 and —COOH groups are soluble in water because they can form hydrogen bonds with water molecules. R H H N O R H H O H H H O R O H O C O H H 7 The energy released from the hydrogen bonds formed in the solution is enough to compensate for the energy required to break the intermolecular forces in water and the polar organic molecules. 8 However, their solubility decreases with the increasing size of the non-polar R group. Quick Check 3.9 1 Which is more soluble in water: ethanoic acid, CH3 COOH or benzoic acid, C6 H5 COOH? Explain your answer. 2 The solubility of 1,2,3-propantriol, CH2 OH—CH(OH)—CH2 OH, is higher than that of ethanol. Why is it so? 3 Which is more soluble in water: ethanol, C2 H5 OH or ethanethiol, C2 H5 SH? 4 Explain why ethylamine, C2 H5 NH2 , is more soluble in water than N,N,N-triethylamine, (C2 H5 )3 N. 5 The solubility of ammonia and phosphine in water is given in the table below. Compound NH3 PH3 Solubility (mol/100 g) 3.2 8.9  10–4 Explain their difference in terms of bonding.

122 Chemistry Term 1 STPM CHAPTER 3 Anomalous Relative Molecular Mass 1 Ethanoic acid, CH3 COOH, has a molar mass of 60 g mol–1. 2 However, when ethanoic acid is dissolved in a non-polar solvent (such as benzene) and its molecular mass determined, a value of 120 g mol–1 is obtained. 3 This molar mass suggests a molecular formula of (CH3 COOH)2 . 4 This happens because in a non-polar solvent, two molecules of CH3 COOH would join, through the formation of hydrogen bonds, to form a dimer as shown below. 2CH3 COOH  (CH3 COOH)2 O CH3 O H O C O C H Hydrogen bonding CH3 5 Dimerisation also occurs in the gaseous phase at temperatures slightly higher than the boiling point of the acid. O R O H O C O C R H Hydrogen bonding Abnormal Behaviour of Water 1 One of the abnormal behaviour of water is that its melting point decreases with increasing pressure. 2 This is because when ice melts, there is a contraction in volume. H2 O(s)  H2 O(l) Volume decreases In other words, ice is less dense than water. 3 In ice, each water molecule is rigidly bonded to four other water molecules by hydrogen bonds in a tetrahedral arrangement. 4 In this arrangement, each oxygen atom is rigidly bonded to four hydrogen atoms, two via covalent bonds and two via hydrogen bonds. 5 This arrangement leads to an ‘open structure’ with many empty spaces between the molecules. 6 When ice melts, some of the hydrogen bonds break and the ‘rigid’ structure collapses, causing the volume to decrease and its density to increase. 7 This is the ‘common’ phenomenon of ice floating on water. This phenomenon is of utmost importance in the living world. 8 When the water in a pond or lake freezes during winter, the ice formed will float on top of the water. Ice is less dense than water. Water Ice Empty space 2014/P1/Q19(b)(ii)

123 Chemistry Term 1 STPM CHAPTER 3 9 As this layer of ice gets thicker, it acts as an insulating layer and shields the water underneath it from the cold temperature outside, making subsequent freezing more difficult. 10 Unless the pond/lake is shallow, or the outside temperature is extremely low, the pond or lake will not be completely frozen. This prevents the aquatic life-forms found in the pond/lake from perishing during winter. Intramolecular and Intermolecular Hydrogen Bonds 1 Hydrogen bonds formed between different molecules are called intermolecular hydrogen bonds. 2 On the other hand, hydrogen bonds formed in the same molecule are called intramolecular hydrogen bonds. 3 A classical example of the two types of hydrogen bonds is illustrated by 2-nitrophenol and 4-nitrophenol. 2-nitrophenol 4-nitrophenol OH NO2 OH NO2 4 The boiling point of 2-nitrophenol is 216 °C while that of 4-nitrophenol is 279 °C. 5 The difference in melting point is explained as follows. (a) Due to the close proximity of the —NO2 and —OH groups, 2-nitrophenol can form intramolecular hydrogen bonding as shown below. O N O H O (b) As a result, the hydrogen atom on the —OH group cannot be utilised to form intermolecular hydrogen bonds with other 2-nitrophenol molecules. (c) However, intramolecular hydrogen bonding is not possible in 4-nitrophenol because the —OH and —NO2 groups are too far apart. Instead, they form intermolecular hydrogen bonds. O N H O H O N O O O Hydrogen bond (d) This accounts for the higher boiling point of 4-nitrophenol. 2-nitrophenol cannot form intermolecular hydrogen bonds. Ice has a hexagonal lattice structure 4-nitrophenol can form intermolecular hydrogen bonds. H2O(l) Ice Insulating layer

124 Chemistry Term 1 STPM CHAPTER 3 SUMMARY SUMMARY 1 Ionic bond is the electrostatic attraction between oppositely-charged ions that are formed by the complete transfer of electrons from one atom to another. 2 Atoms with larger size and lower nuclear charge readily form cations. Atoms with smaller size and higher nuclear charge readily form anions. 3 Covalent bond is the electrostatic attraction between adjacent nuclei and the electrons that are shared between them. 4 Bond energy is the energy required to break a particular covalent bond, per mole of the bond. 5 Generally, the strength of the covalent bond increases in the order: Single < double < triple 6 A sigma (σ) bond has high electron density in the region between the nuclei of the two bonding atoms. In a pi (π) bond, the electron density is concentrated in the regions above and below the plane that contains the two nuclei. 7 Hybridisation is the mixing of two or more non-equivalent atomic orbitals to form a set of equivalent hybrid orbitals. 8 The shape of a molecular species is determined by the angles between the covalent bonds in the molecule. 9 Polar molecules are molecules that have a permanent dipole. 10 The van der Waals force is the force of attraction between covalent molecules. It is made up of two components: (a) Permanent dipole-permanent dipole attraction (b) Temporary dipole-induced dipole attraction 11 The strength of the van der Waals force depends on: (a) the size of the atom/molecule (b) the number of electrons in the atom/ molecule (c) the total surface area of the atom/ molecule 12 The hydrogen bond is the electrostatic attraction between a hydrogen atom that is covalently bonded to a small and highly electronegative atom (such as N, O and F) and the lone pair electrons of another small and highly electronegative atom (N, O and F). 13 Metallic bond is the electrostatic attraction between the cations of metals and the delocalised electrons in the sea of mobile electrons.

125 Chemistry Term 1 STPM CHAPTER 3 Objective Questions 7 Which of the following best explains why the boiling point of H2 O is higher than HF? A H2 O is a liquid but HF is a gas B Fluorine is more electronegative than oxygen C The H2 O molecule has a bent structure while that of HF is linear D The intermolecular forces in H2 O is stronger than HF 8 Which compound has the highest covalent character? A BeF2 B CaO C Al2 O3 D SO3 9 Which molecule has the strongest intermolecular force? A C6 H5 Cl B SiO2 C (CH3 )3 N D CH3 COOH 10 At room conditions, hydrogen sulphide is a gas and water is a liquid. A correct explanation of this phenomenon is A hydrogen sulphide molecule is bent whereas water is linear. B the water molecule is larger than the hydrogen sulphide molecule. C there is no lone pair of electrons in the hydrogen sulphide molecule. D oxygen is more electronegative than sulphur. 11 Which of the following statements regarding π-bond is correct? A It consists of four electrons. B It is formed by the overlap of unhybridised orbitals. C It is stronger than a σ-bond. D It cannot be formed unless the two atoms are already joined by a double bond. 1 Which of the following compounds has a boiling point that is influenced by hydrogen bonding? A C2 H5 OC2 H5 B CH3 COOC2 H5 C C6 H5 NH2 D CH3 COCH3 2 Which of the following species does not have the same shape as H2 O? A H2 S C SiO2 B SO2 D NO2 3 Bromobenzene has the following structure. Br How many atoms undergo the following hybridisation in a molecule of bromobenzene? sp2 sp3 A 6 0 B 6 1 C 3 6 D 0 7 4 Which of the following compounds has van der Waals forces? A Sodium C Ammonium nitrate B Sulphur D Silicon 5 Which of the following solids has more than one kind of bonds in its structure? A Bronze C Ice B Alnico D Silicon dioxide 6 In which sequence are the following species arranged in order of increasing bond angles? A CCl4 AlCl3 NO C CH4 NH3 H2 S B CO2 HCN H2 O D HCl CO3 2– NO2 – STPM PRACTICE 3

126 Chemistry Term 1 STPM CHAPTER 3 12 Which of the following most accurately characterised elements with metallic bonds? A They have high melting points. B They are conductors of electricity in the solid and liquid states. C They are hard. D They have shiny surfaces. 13 Which of the following is the property of all ionic solids? A They are soluble in water. B They have low enthalpy of vaporisation. C They conduct electricity in the solid state. D They are electrolytes in the molten state. 14 The structure of the I3 – ion is A linear C trigonal planar B bent D pyramidal 15 In which of the following pairs do the species have the same shape? A AlCl3 and PH3 B CH3 + and NH4 + C CO and HF D CO2 and SO2 16 Which of the following processes is responsible for the identical sulphur-oxygen bonds in the SO4 2– ion? A Hybridisation B Delocalisation C Electron sharing D Isomerism 17 Which species has the largest bond angle? A AlCl3 C SO4 2– B HCN D NH2 – 18 What type of bond is responsible for the intermolecular forces in liquid silicon dioxide? A Covalent bond B Coordinate bond C van der Waals force D Ionic bond 19 Which of the following molecules has the largest bond angle? A AlCl3 C CH4 B H2 O D CO 20 What is the nature of the intermolecular force that is responsible for iodine being a solid at room conditions? A Covalent bond B Hydrogen bonding C Ionic bond D Van der Waals force 21 Why is CH3 + planar whereas CH4 is not? A CH4 is a neutral molecule whereas CH3 + is a cation. B There is a lone pair of electrons in CH3 +. C The carbon atom in CH3 + is surrounded by only three electron pairs whereas four electron pairs surround the carbon atom in CH4 . D CH3 + undergoes sp3 hybridisation whereas CH4 undergoes dsp3 hybridisation. 22 What is the nature of the forces that makes iodine a solid at room temperature even though it consists of simple molecules? A van der Waals forces B Hydrogen bonding C Covalent bonds D Coordinate bonds 23 Ammonia reacts with aluminium chloride to form an addition compound. Which is not true about this reaction? A Ammonia acts as a Lewis base B Coordinate (dative) bonds are formed between ammonia and chlorine C The reaction takes place in the gaseous phase D The reaction is reversible 24 Which of the following species is square planar? A CH4 B SiCl4 C XeF4 D C2 H4 25 Which of the following compounds contains more than four bonding electrons? A BeCl2 B H2 O C SnCl2 D CO2

127 Chemistry Term 1 STPM CHAPTER 3 26 What type of nitrogen-nitrogen bonds is present in the following species? N2 , N2 H4 and N2 (C2 H5 )2 N2 N2 H4 N2 (C2 H5 )2 A Double Double Single B Triple Double Single C Triple Single Double D Single Single Triple 27 Which species has a linear shape? I I3 – II CO2 III N2 A I B II C II and III D I, II and III 28 Coordinate bonds do not exist in A Al2 Cl6 B PCl4 + C [Cu(NH3 )4 (H2 O)2 ]2+ D CO 29 The bond angles in methane, ammonia and water are given in the table below: Molecule CH4 NH3 H2 O Bond angle 109° 107° 104° The difference in the bond angle in the three molecules is due to A the hybridisation of the central atom B the electronegativity of the central atom C the number of lone pair electrons in the central atom D the strength of hydrogen bonding between the molecules 30 Which species and its geometry is incorrect? Species Geometry A NH3 Pyramidal B I 3 – Linear C CN– Bent D [Fe(CN)6 ] 3– Octahedral 31 Which of the following statements about the bonding in ammonia and water are correct? I The central atom of both molecules undergo sp3 hybridisation. II The O—H bond is stronger than the N—H bond. III Both molecules are pyramidal. IV There are more lone pair electrons in the water molecule. A I and III B I, II and IV C II, III and IV D I, II, III and IV 32 Copper(I) oxide is a reddish-brown solid. In which orbitals are the valence electrons of copper(I) ion found? A 3s B 4s C 3d D 4s and 3d 33 The strength of intermolecular forces determines the boiling point of a substance. Which substance has the highest boiling point? A HF B H2 O C NaCl D SiO2 34 Glucose, C6 H12O6 , readily dissolves in water because A it is ionic B it is a polar molecule C it exists in the more stable cyclic form in aqueous solution D it can form hydrogen bonds with water molecules 35 Which statement is true of metallic bond? I The strength of the bond increases with the size of the metal atoms. II The valence electrons can move freely throughout the metallic lattice. III The bond is formed between an electropositive element and an electronegative atom. A I only B II only C III only D I and III

128 Chemistry Term 1 STPM CHAPTER 3 36 The boiling point of the halogens increases in the order: Cl2 < Br2 < I2 . Which property(ies) influence(s) the trend? I Covalent bond strength II Hydrogen bonding III Molecular size A I only B III only C II and III D I, II and III 37 The boiling points of CCl4 and SiCl4 are 76 °C and 57 °C respectively. This is because A carbon is more electronegative than silicon B the size of CCl4 is larger than that of SiCl4 C the intermolecular forces in CCl4 is stronger than in SiCl4 D CCl4 molecules can be more closely packed compared to SiCl4 38 Which of the following compounds is ionic? A HF B F2 O C AlCl3 D [Cu(H2 O)6 ]SO4 39 Consider the three molecules: HCl, H2 O and NH3 . Which of the following about the molecules is not true? A HCl is ionic while H2 O and NH3 are covalent. B The bond angles increase in the order of NH3 < H2 O < HCl. C The number of lone pair electrons in the molecules are 3, 2 and 1 respectively. D H2 O and NH3 can form intermolecular hydrogen bonding, but HCl cannot. 40 Consider the four elements, W, X, Y and Z below: Element W X Y Z Proton number 11 13 24 30 Which element has the highest melting point? A W B X C Y D Z 41 Which of the following statements is not true with regards to the nitrogen molecule? A Nitrogen exists as a diatomic molecule. B The nitrogen molecule is non-polar. C The triple bond in nitrogen molecule consists of two σ-bonds and one π-bond. D The nitrogen molecule is chemical inert. 42 Which of the following is not true with regards to the C2 Cl2 molecule? A It is linear. B The carbon atoms undergo sp hybridization. C It is polar. D The carbon atoms are joined by a triple bond. 43 Which statement about the Lewis structure of CO3 2- and NO3 - is not true? A Both ions have one coordinate bond in their structures. B Both ions have the same shape. C The central atoms in both ions have octet configuration. D Both ions have three σ-bonds and one π-bond.

129 Chemistry Term 1 STPM CHAPTER 3 Structured and Essay Questions 1 Explain what is meant by (i) ionic bonding, and (ii) covalent bonding. Illustrate your answer with a suitable example each. 2 (a) Explain what is meant by hydrogen bonding using hydrogen fluoride as an example. (b) Explain the trend in the boiling point of the following compounds. Compound CH4 NH3 H2 O Boiling point/°C –165 –33 100 (c) Glucose, C6 H12O6 , is soluble in water because it can form hydrogen bonding with water. Draw a diagram to show one such bond between water and glucose. 3 The boiling points and solubility of hydrogen chloride, carbon dioxide and ammonia are given below. Compound Hydrogen chloride Carbon dioxide Ammonia –85 –78 –33 23 0.03 18 Solubility in water/mol dm–3 Boiling point/°C at 298 K (a) Explain in terms of intermolecular forces the difference in boiling points between (i) hydrogen chloride and carbon dioxide, (ii) ammonia and hydrogen chloride. (b) Explain why the solubility of ammonia and hydrogen chloride is so much higher than that of carbon dioxide. 4 Aluminium chloride, AlCl3 , beryllium chloride, BeCl2 and ammonia, NH3 are covalent compounds. (a) Draw the Lewis diagram and state the shapes of the three compounds. (b) Using your diagrams in (a), explain what is meant by an electron-deficient compound. (c) Aluminium chloride and ammonia can form a 1 : 1 addition compound through coordinate bond formation. (i) Give the formula of the addition compound. (ii) Draw a dot-and-cross diagram to illustrate the formation of the coordinate bond between the two molecules. (d) Beryllium chloride, on the other hand, forms a 2 : 1 addition compound with ammonia. (i) Give the formula of the addition compound. (ii) Draw the Lewis structure of the addition compound. (iii) Explain why beryllium does not form a 1 : 1 compound with ammonia. 5 Draw labelled diagrams to show the overlapping of orbitals in the following molecular species. (b) H2 O (b) NH3 (c) HCN 6 Nitromethane, CH3 NO2 , is used as a dry-cleaning solvent as well as fuel for power car racing. (a) Draw the Lewis diagram of nitromethane and state the type of hybridisation involved in the molecule. (b) What is the geometry around the nitrogen atom in nitromethane? (c) Draw the Lewis structure of CN– ion. (d) The carbon-nitrogen bond length in CN– is 116 pm while that in CH3 NO2 is 147 pm. Explain the difference in the carbon-nitrogen bond length in CN– and CH3 NO2 .

130 Chemistry Term 1 STPM CHAPTER 3 7 (a) What do you understand by hybridisation? (b) Use the hybridisation theory to explain the formation of the following compounds. (i) Ammonia (ii) Trichloromethane 8 In liquid ammonia, the following equilibrium exists. 2NH3 (l) : NH4 +(l) + NH2 – (l) (a) Explain the above reaction with reference to the Bronsted-Lowry’s definition of acids and bases. (b) Draw the Lewis structures of NH3 , NH4 + and NH2 – and predict their shapes according to the ‘Valence shell electron-pair repulsion’ theory. (c) Nitrogen and phosphorous form trichlorides with the molecular formula of NCl3 and PCl3 . State the type of hybridisation of the nitrogen atom and phosphorous atom in their trichlorides. Which molecule is expected to have a larger bond angle? Explain your answer. 9 (a) Ammonia, NH3 , is a hydride of nitrogen. Ammonia reacts with beryllium chloride, BeCl2 , to form an addition compound. (i) Write an equation for the reaction. (ii) Draw the Lewis structure for the compound formed in (a)(i). (iii) State the change in the hybridisation of beryllium in the reaction. (b) Ammonia and phosphine, PH3 , have the same molecular shape. (i) Draw the shape for the phosphine molecule. (ii) Which of the two molecules is expected to have a larger bond angle? Explain your answer. 10 The halogens (F to I) react with hydrogen, under suitable conditions, to produce hydrogen halides. (a) Write a balanced equation for the reaction of chlorine with hydrogen. (b) Sketch a graph to show the variation of the boiling points of the hydrogen halides with the molar mass of the hydrogen halides. (c) Explain the trend in the boiling points. 11 (a) The melting points of lead(II) chloride and silicon dioxide are 498 °C and 1610 °C respectively. Explain, in terms of structure and bonding why the melting point of silicon dioxide is higher than that of lead(II) chloride. (b) Sulphur trioxide, SO3 , dissolves in water to form sulphuric acid, H2 SO4 . (i) Draw the Lewis structures of SO3 and the SO4 2– ion. (ii) Draw the shapes and state the O—S—O bond angles of both species. 12 (a) In terms of structure and bonding, explain the following observations. (i) The melting point of ice decreases with increasing pressure. (ii) Liquid hydrogen chloride does not conduct electricity. However, aqueous hydrogen chloride does. (b) Chlorine is a Group 17 elements. It forms a series of oxoanions: ClO2 – , ClO3 – and ClO4 – . (i) Determine the oxidation state of chlorine in each of the oxoanions. (ii) Draw the Lewis diagram for each of the oxoanions and predict their shapes.

CHAPTER STATES OF MATTERS 4 Students should be able to: Gases • explain the pressure and behaviour of ideal gas using the kinetic theory; • explain qualitatively, in terms of molecular size and intermolecular forces, the conditions necessary for a gas approaching the ideal behaviour; • define Boyle’s law, Charles’ law and Avogadro’s law; • apply the pV = nRT equation in calculations, including the determination of the relative molecular mass, Mr ; • define Dalton’s law, and use it to calculate the partial pressure of a gas and its composition; • explain the limitation of ideality at very high pressures and very low temperatures. Liquids • describe the kinetic concept of the liquid state; • describe the melting of solid to liquid, vaporisation and vapour pressure using simple kinetic theory; • define the boiling point and freezing point of liquids. Solids • describe qualitatively the lattice structure of a crystalline solid which is: (a) ionic, as in sodium chloride, (b) simple molecular, as in iodine, (c) giant molecular, as in graphite, diamond and silicon(IV) oxide, (d) metallic, as in copper; • describe the allotropes of carbon (graphite, diamond and fullerenes), and their uses. Phase diagrams • sketch the phase diagram for water and carbon dioxide, and explain the anomalous behaviour of water; • explain phase diagrams as graphical plots of experimentally determined results; • interpret phase diagrams as curves describing the conditions of equilibrium between phases and as regions representing single phases; • predict how a phase may change with changes in temperature and pressure; • discuss vaporisation, boiling, sublimation, freezing, melting, triple and critical points of H2O and CO2; • explain qualitatively the effect of a nonvolatile solute on the vapour pressure of a solvent, and hence, on its melting point and boiling point (colligative properties); • state the uses of dry ice. The Gas Laws • Boyle's law • Ideal gas equation • Charles' law • Dolton's law • Avogadro's law Liquid State • Vapour pressure • Melting • Boiling • Freezing Effect of Non-volatile Solutes on Vapour Pressure • Lowering of the vapour pressure • Elevation of boiling point • Depression of freezing Allotropy • Carbon MaxwellBoltzmann Distribution Curve The Kinetic Theory of Gas Deviation of Real Gases from Ideal Behaviour Solid • Ionic • Simple molecular • Giant molecular • Metallic Phase Diagram • Water • Carbon dioxide States of Matters Liquids, Solids and Phase Diagrams Learning earning Outcomes Gases Concept Map

132 Chemistry Term 1 STPM CHAPTER 4 4.1 Gases The Kinetic Theory of Gases 1 The kinetic theory of gases is introduced to explain the behaviours of gases. 2 The theory is based on the following assumptions: (a) The gas particles are considered as ‘point particles’. They have mass but negligible volume. (b) The gas particles are spaced far apart and are in a state of continuous random motion. (c) The gas particles travel in straight lines unless when colliding with one another or with the walls of the container. (d) There are no intermolecular forces (attractive or repulsive) between gas particles. (e) Collisions between the gas particles with one another or with the walls of the container are perfectly elastic, i.e. there is no lost in energy during collisions. (f) Collisions between the gas particles and the walls of the container give rise to gas pressure. (g) The average kinetic energy of the gas particles is directly proportional to its absolute temperature. 3 As there are a lot of empty spaces between gas particles, a gas is easily compressed. 4 A gas has no fixed shape or volume. It takes the shape of the container in which it is placed and always fills it. 5 Gases are completely miscible with one another. 6 The volume occupied by a gas is affected by (a) the pressure on the gas, (b) the temperature of the gas, (c) the number of moles of the gas. The Gas Law The gas laws govern the bahaviour of a gas under different conditions: Boyle’s Law 1 In the year 1662, Robert Boyle investigated the relationship between the volume occupied by a gas and the pressure on the gas, at constant temperature. 2 The result he obtained is illustrated by the graph shown on the left. 3 From the results, he formulated Boyle’s Law which states that the volume occupied by a fixed mass of gas (i.e. fixed number of moles) is inversely proportional to the pressure of the gas, at constant temperature. V ∝ 1 P Collisions give rise to the gas pressure. Kinetic energy ∝ T. A statement of Boyle’s Law • Exam Tips The volume of the gas molecules is negligible compared to the volume of the container where the gas is placed. Volume (V) Pressure (P) 2010/P1/Q7

133 Chemistry Term 1 STPM CHAPTER 4 4 This can be explained using the kinetic theory. When the pressure on the gas is increased, the gas particles are pushed closer to one another, and the volume they occupy decreases. 5 Conversely, if the volume occupied by the same amount of gas is decreased, more particles are now closer to the walls of the container. This causes an increase in the number of collisions per unit time leading to an increase in the gas pressure. 6 Boyle’s Law can also be expressed as: PV = constant or P1 V1 = P2 V2 = . . . . . . . . . = Pn Vn 7 This means that when the pressure of the gas increases, the volume must decrease accordingly such that the product of PV is always a constant. 8 Boyle’s Law can be represented by the following plots: (a) P 1 V (d) P PV (b) P 1 V (e) PV V (c) V 1 P 9 One consequence of Boyle’s Law is that when the pressure on the gas increases to infinity (i.e. when 1 P = 0), the volume occupied by the gas would be zero. Increasing pressure pushes the gas particles closer. The amount of empty spaces between the particles decrease.

134 Chemistry Term 1 STPM CHAPTER 4 10 However, in practice, this does not happen because at very high pressure, the gas particles would liquefied as shown by the graph on the left. 11 All real gases do not obey Boyle’s Law perfectly, especially under high pressures. A gas that obeys Boyle’s Law perfectly under any pressures is known as an ideal gas or perfect gas, a non-existing gas. 12 For a gas to obey Boyle’s Law perfectly, the gas particles should have zero volume so that the volume occupied by the gas would be zero at infinite pressure. 13 Furthermore, there should not be any intermolecular forces between the gas particles so that they do not ‘stick together’ when they are close to one another. Otherwise the gas will condense to liquid. 14 However, for most of our calculations, we assumed that the gas obeys Boyle’s Law. Example 4.1 The volume of a gas at 298 K and 100 kPa pressure is 76.0 cm3 . Calculate the volume it would occupy at 265 kPa at the same temperature. Solution Using: P1 V1 = P2 V2 100  76.0 = 265  V2 V2 = 28.7 cm3 Example 4.2 The graph shows the pressure/volume graph for a sample of carbon dioxide gas at 298 K. Explain the shape of the curve. Solution At low pressures, along AB, carbon dioxide follows Boyle’s Law where its volume decreases with increasing pressure. At point B, the carbon dioxide molecules are so closed to one another that condensation takes place. This is accompanied by a large decrease in the volume (along BC), after which the volume does not change significantly with pressure (along CD) because liquid carbon dioxide is not easily compressed. V P A B C D Exam Tips Exam Tips Definition of an ideal gas We can use any appropriate units for pressure and volume provided that the same units are used on both sides of the equation. V 1 P Liquefaction occurs Gas 0 Liquid

135 Chemistry Term 1 STPM CHAPTER 4 Quick Check 4.1 1 A 1.0 dm3 of a sample of nitrogen is compressed from 100 kPa to 560 kPa at 298 K. What is the final volume of the gas? 2 A sample of helium occupies a volume of 358 cm3 at 100 kPa and 300 K. At what pressure would the volume be reduced to 250 cm3 at the same temperature? 3 12.5 dm3 of air at 100 kPa is allowed to expand to 45.0 dm3 at constant temperature. What is the final pressure of the air sample? 4 Sketch on the same graph the variation of the volume occupied by a sample of oxygen with pressure at two different temperatures of 300 K and 400 K. Assuming that oxygen shows ideal behaviour. Explain the shape of your curves. 5 Sketch the graph of pressure (P) against 1 V for an ideal gas and a real gas such as carbon dioxide. Explain the shape of your curves. Charles’ Law 1 Charles’ Law relates the volume occupied by a gas with temperature. 2 Charles’ Law states that the volume occupied by a fixed mass of gas, at constant pressure, is directly proportional to its absolute temperature. 3 The absolute temperature scale has a unit of Kelvin (K). It is related to the Celsius scale by: x °C = (273 + x) K 4 Zero Kelvin (0 K) is known as absolute zero. It is equal to –273 °C (or more accurately –273.15 °C). 0 K = –273 °C 5 A plot of Charles’ Law yields the following graph: V T (K) 6 Mathematically, Charles’ Law is expressed as: V ∝ T or V T = constant i.e. V1 T1 = V2 T2 = ....... = Vn Tn A statement of Charles’ Law The volume occupied by the gas is zero at zero Kelvin. The gas has disappeared! Info Chem The unit for the Kelvin scale is Kelvin (K) and not degree Kelvin (°K). 2014/P1/Q8

136 Chemistry Term 1 STPM CHAPTER 4 7 Charles’ Law can be represented by the following plots: (a) V –273 0 T (°C) (c) 0 V T T (b) V V T 0 8 According to Charles’ Law, the volume occupied by a gas at absolute zero or –273 °C would be zero. 9 However, in practice this does not happen because the gas would liquefied before this temperature is reached as shown by the sketch. Example 4.3 Sketch the graph of volume occupied by an ideal gas against temperature (in K) at two different pressures. Explain the shape of your graph. Solution V T (K) P1 P2 P2 fi P1 The two lines pass through the origin because the volume of an ideal gas is zero at zero Kelvin. The line at pressure P1 is steeper than the line at pressure P2 . This is because the volume occupied by the gas will be larger at lower pressure, for any given temperature. V Liquefaction occurs Liquid Gas T (K)

137 Chemistry Term 1 STPM CHAPTER 4 Example 4.4 The volume occupied by a sample of helium at 25 °C is 350 cm3 under a pressure of 101 kPa. What will be its volume at 10 °C under the same pressure? Solution Using: V1 T1 = V2 T2 350 (25 + 273) = V2 (10 + 273) V2 = 332.4 cm3 Exam Tips Remember to change the temperature into Kelvin. Quick Check 4.2 1 Calculate the volume which 2.4 dm3 of a gas at 0 °C would occupy at 85 °C of the same pressure. 2 A sample of oxygen occupies a volume of 350 dm3 at 25 °C and 200 kPa. At what temperature would the gas occupy 700 dm3 at 200 kPa? 3 A certain amount of gas at 28 °C occupies a volume of 500 cm3 . At what temperature would its volume be doubled at the same pressure? 4 A given mass of chlorine occupies 25 dm3 at 87.6 °C. What volume would it occupy at 0 °C of the same pressure? Avogadro’s Law 1 Avogadro’s Law relates the volume occupied by a gas with the amount of gas. 2 Avogadro’s Law states that equal volume of all gases, at the same temperature and pressure, contains the same number of particles (or same number of moles of gas). 3 Conversely, equal number of moles of all gases will occupy the same volume at the same temperature and pressure. It does not depend on the nature of the gases. V ∝ n [where n = number of moles of the gas] Or: n V = constant Or: n1 V1 = n1 V2 = ....... = nn Vn 4 Experimentally, it was found that 1 mol of any gas would occupy 22.4 dm3 at 273 K (0 °C) and 1 atm (or 101 kPa). This volume is known as the molar volume of a gas. A statement of Avogadro’s Law

138 Chemistry Term 1 STPM CHAPTER 4 5 101 kPa and 273 K are known as standard temperature and pressure (s.t.p.). 6 The molar volume of a gas at room conditions is usually stated as 24.0 dm3 . 7 Consider the following gaseous reaction: N2 (g) + 3H2 (g) → 2NH3 (g) The stoichiometric equation shows that 1 mol of N2 will react with 3 mol of H2 to produce 2 mol of NH3 . 8 Applying Avogadro’s Law: 1 volume of N2 will react with 3 volume of H2 to produce 3 volume of NH3 . All volumes are measured under the same conditions. Example 4.5 Calculate the volume occupied by the following gas at s.t.p. (a) 9.8 g of nitrogen, (b) 70.4 g of carbon dioxide. Solution (a) No. of moles of N2 = 9.8 28 = 0.35 mol ∴ Volume at s.t.p. = 0.35  22.4 = 7.84 dm3 (b) No. of moles of CO2 = 70.4 44 = 1.6 mol ∴ Volume at s.t.p. = 1.6  22.4 = 35.84 dm3 s.t.p. Quick Check 4.3 1 Ethane, C2 H6 , burns in excess oxygen to produce carbon dioxide and water. (a) Calculate the volume of oxygen that is required to completely burn 2.5 dm3 of ethane. (b) Find the volume of carbon dioxide produced when 25 dm3 of ethane are completely burned in oxygen. [All volumes are measured under the same conditions.] 2 TNT is used as an explosive. It decomposes according to the following equation: 2C7 H5 N3 O6 (s) → 7C(s) + 7CO(g) + 3N2 (g) + 5H2 O(g) (a) Calculate the relative molecular mass of TNT. (b) What volume of gas (measured at s.t.p.) will be produced by the decomposition of 2 kg of TNT? 3 50 cm3 of an alkene W burns in 250 cm3 of oxygen to produce 150 cm3 of carbon dioxide and 150 cm3 of steam. 50 cm3 of oxygen is left behind. Determine the molecular formula of W. [All volumes are measured under the same conditions.] 4 25 cm3 of a hydrocarbon were exploded with excess oxygen, there was a contraction of 75 cm3 . There was a further contraction of 75 cm3 when the product was treated with aqueous sodium hydroxide. Deduce the molecular formula of the hydrocarbon. [All volumes are measured under the same conditions.]

139 Chemistry Term 1 STPM CHAPTER 4 The Ideal Gas Equation 1 Boyle’s Law deals with the variation of the volume occupied by a gas with pressure. Charles’ Law deals with the variation of the volume occupied by a gas with temperature. Avogadro’s Law deals with the variation of the volume occupied by a gas with the number of moles of the gas. 2 All the three laws above assumed ideal bahaviour for the gas concerned. 3 The three laws can be combined to yield a single equation that relates the volume occupied by an ideal gas with pressure, temperature and the number of moles of gas. 4 From: Boyle’s Law: PV = constant Charles’ Law: V T = constant Avogadro’s Law: V n = constant 5 Hence, V ∝ 1 P , V ∝ T and V ∝ n Combining the three relationships, we get: V ∝ nT P Rearranging: PV ∝ nT Or: PV = nRT [where R = constant, known as the gas constant] The final expression is called the ideal gas equation. 6 For a fixed mass of gas (i.e. n = constant): PV T = constant Or: P1 V1 T1 = P2 V2 T2 7 In calculations involving the ideal gas equation, the following units are adopted. Quantity P V T R Unit Pa or N m–2 m3 K 8.31 J mol–1 K–1 Example 4.6 Calculate the volume occupied by the following gases at the conditions stated. (a) 0.23 mol of nitrogen at 25 °C and 300 kPa (b) 1.5 mol carbon dioxide at 200 °C and 10 atm (c) A mixture containing 19.2 g of oxygen and 10.56 g of carbon dioxide at 45 °C and 250 kPa 2013/P1/Q2 2015/P1/Q9 2016/P1/Q7 2017/P1/Q1(a)(b) Exam Tips Exam Tips 1 atm ≈ 101 kPa, 1 cm3 = 1  10–3 dm3 ; 1 dm3 = 1  10–3 m3 ; 1 cm3 = 1  10–6 m3

140 Chemistry Term 1 STPM CHAPTER 4 Solution (a) Using the ideal gas equation: PV = nRT (300  103 )(V) = 0.23  8.31  (25 + 273) V = 1.90  10–3 m3 (b) Using: PV = nRT (10  101  103 )(V) = 1.5  8.31  (200 + 273) V = 5.84  10–3 m3 (c) No. of moles of O2 = 19.2 32 = 0.60 mol No. of moles of CO2 = 10.56 44 = 0.24 mol Total number of moles of gas = (0.60 + 0.24) = 0.84 mol Using: PV = nRT (250  103 )(V) = 0.84  8.31  (45 + 273) V = 8.88  10–3 m3 Quick Check 4.4 1 Calcium carbonate decomposes according to the equation: CaCO3 (s) → CaO(s) + CO2 (g) Calculate the volume of carbon dioxide liberated at 28 °C and 125 kPa on the decomposition of 25.0 g of calcium carbonate. 2 2.45 g of a gas occupies 10.0 dm3 at 30 °C and 22.6 kPa. Calculate the relative molecular mass of the gas. 3 Hydrogen peroxide decomposes according to the equation: 2H2 O2 (l) → 2H2 O(l) + O2 (g) Calculate the volume of oxygen (measured at 45 °C and 101 kPa) produced on the complete decomposition of 45.0 g of hydrogen peroxide. 4 Solid carbon dioxide, also known as ‘dry ice’, sublimes at room conditions. Given that the density of ‘dry ice’ is 1.60 g cm–3, what is the volume of carbon dioxide gas formed when 35.6 cm3 of ‘dry-ice’ sublimes at room conditions? [1 mole of gas occupies 24.0 dm3 at room conditions.] Dalton’s Law of Partial Pressure 1 Consider a mixture of two gases, X and Y in a closed container at a certain temperature. What is the total pressure of the mixture? 2 The answer is provided by the Dalton’s Law of partial pressure which states that the total pressure exerted by a mixture of gases which do not react with one another, is given by the sum of the partial pressures of each constituent gases in the mixture at the same temperature. [where PA P is the partial pressure of gas A in the mixture.] total = PA + PB + PC A, B, C

141 Chemistry Term 1 STPM CHAPTER 4 3 The partial pressure of a gas A in a mixture of gases is the pressure that will be exerted by gas A, if gas A alone was to occupy the same volume as the mixture under the same temperature. 4 For example, a 2 dm3 container contains 0.5 mol of oxygen and 1.2 mol of carbon dioxide at 298 K. The pressure exerted by oxygen in the mixture (i.e. its partial pressure) is the pressure that will be exerted by 0.5 mol of O2 in the same container without the presence of CO2 . Using: PV = nRT P(O2 )  (2  10–3) = 0.5  8.31  298 P(O2 ) = 619.1  103 Pa = 619.1 kPa Similarly for carbon dioxide: P(CO2 )  (2  10–3) = 1.2  8.31  298 P(CO2 ) = 1485.8  103 Pa = 1485.8 kPa ∴ Total pressure of the mixture = (619.1 + 1485.8) kPa = 2104.9 kPa 5 The partial pressure of a gas in a mixture of gases can also be calculated using the following formula: PA= (mole fraction of A in the mixture)  (total pressure of the mixture) PA = XAPT Example 4.7 A 1 dm3 vessel containing 0.50 mol of hydrogen gas at 28.0 °C is connected to a 2 dm3 vessel containing 0.20 mol of carbon dioxide at 28.0 °C. The valve between them is opened and the gases are allowed to mix. Calculate (a) the partial pressure of the gases, (b) the total pressure of the mixture. Solution The initial pressure of H2 can be calculated as follows: Using: PV = nRT P(1  10–3) = 0.5  8.31  (28 + 273) P = 1.25  106 Pa The initial pressure of CO2 : P(2  10–3) = 0.2  8.31  (28 + 273) P = 2.50  105 Pa 0.5 mol O2 1.2 mol CO2 Mole fraction of A = no. of moles of A ————————————– total no. of moles of gas