The IUPAC accepted systematic name for potassium nitrate is potassium trioxonitrate(v)

Transcription

1 Topic 4 Bonding Answers 4.1 Exercises 1. Ionic bonding a) Ionic bonding occurs between what types of elements? A: non metal non metal B: metal metal : metal non metal Answer: b) What is the octet rule? Atoms have a tendency to gain or loose electrons in order to end up with a full outer shell of electrons. A full outer shell consists of eight electrons (for all but the period 1 elements) and, therefore, the tendency to reach this outer electron arrangement is called completing the octet. 2. Attraction a) Define electronegativity Electronegativity is the ability of an atom to attract electrons to itself in a bond. Electronegative elements tend to gain electrons and form negative ions, e.g. the halogens. There are various ways of assigning values for electronegativity. The IB hemistry Data Booklet contains values for the electronegativity of the elements using the Pauling scale, which is a dimensionless scale based on bond dissociation energies in which fluorine (the most electronegative element) has a value 4. b) What determines the electronegativity of an atom? Electronegativity is determined by how easily an atom will either lose or gain electrons. This, in turn, is determined by the number of protons in the nucleus and the size of the atom. Many protons in a small atom, e.g. F, means a high electronegativity. It has been proposed that the average of an atoms electron affinity and first ionisation energy is a good approximation of the atoms electronegativity. The electronegativity of an atom in a bond is most easily predicted by its position in the periodic table. Atoms high in a RS group have high electronegativity. Atoms low in a LS group have the lowest electronegativity. c) What determines how many electrons and element will lose or gain? An element will gain or lose electrons in order to satisfy the octet rule. By identifying an atom s position of the period table, it can quickly be determined how many electrons it will either lose or gain. For example, all atoms in group 2 of the periodic table need to lose two electrons in order to have a full octet, while elements in group 6 must gain two electrons for a full octet. 3. Predicting Ionic Bonds a) Predict, from their position on the periodic table, whether atoms of Rubidium, Rb and xygen,, will form ionic bonds when they react. Rubidium is low in group 1 and is a metal. xygen is high in group 6 and is a non-metal. They will therefore form an ionic bond due to the large difference in electronegativities between the elements. b) If an ionic compound was formed by these elements, which element would lose electron(s) and which would gain electron(s)? Rubidium would lose one electron and oxygen would gain two in order to have a stable octet of electrons. 1

2 c) With the help of Lewis dot diagrams, explain the bonding that occurs between rubidium and oxygen and state the final formula of the compound that is formed. Rb Rb 2- Rb + Rb + This electron transfer results in the formation of oppositely charged ions which bond via electrostatic attraction to form an ionic lattice structure. Since rubidium loses only one electron and oxygen requires two to obtain a stable octet, there must be two rubidium atoms per oxygen atom; the ionic formula is therefore Rb 2. d) Explain the nature of the forces that hold Rb and atoms together in their crystal lattice? Ionic bonding, i.e. the strong electrostatic force of attraction between oppositely charged species. The Rb + and 2- ions are held together by electrostatic attraction to form a three-dimensional, continuous crystal lattice. 4. Deduce the formulas and names of the ionic compounds that arise due to the combination of: Determine the charge on each ion then adjust with subscripts to balance charges. a) Na and Na 2 : Sodium oxide b) Mg and P Mg 3 P 2 : Magnesium phosphide A phosphide compound is one in which phosphorous has an oxidation state of 3, that is it has gained three electrons. c) K and Br KBr: Potassium bromide d) Li and S Li 2 S: Lithium sulphide A sulfide compound is one in which sulfur has an oxidation state of 2, that is it has gained two electrons. e) a and N a 3 N 2 : alcium nitride A nitride compound is one in which nitrogen has an oxidation state of 3, that is it has gained three electrons. f) Sr and F SrF 2 : Strontium fluoride 5. Describe the formation and structure of a three-dimensional ionic lattice using KBr as an example. Bromine is much more electronegative than potassium. Therefore, electrons are transferred from potassium to bromine when the two collide. Potassium has one electron in its valence shell and can therefore lose one electron easily to obtain a stable eight electron arrangement (full octet). Bromine has seven electrons in its valence shell and readily gains one electron to obtain a stable eight electron arrangement. Potassium donates an electron to bromine and the ions K + and Br - are formed. They are attracted to each other through electrostatic forces to form an ionic crystal lattice. 2

3 6. Name, and give the formulas for, three compounds with ionic bonding, one with an ion of single positive charge, the next with a double positive change, and then one with a triple positive change. Sodium; Na +. Lithium; Li +. Potassium; K + (Group 1 metals and transition metals) may form Nal, Lil, Kl. alcium; a 2+. Magnesium; Mg 2+. Strontium; Sr 2+ (Group 2 metals and transition metals) may form al 2, Mgl 2, Srl 2. Aluminium; Al 3+. Iron; Fe 3+. hromium, r 3+ (Group 3 metals and transition metals) may form All 3, Fel 3, rl Some elements in the transition metals are able to form more than one ion, an example being iron, Fe. a) Write the formula for iron(ii) sulfate and for iron(iii) sulfate. Iron(II) sulfate: FeS 4, iron(iii) sulfate: Fe 2 (S 4 ) 3 b) opper can form copper(i) and copper(ii) chlorides. Write formulas for these chlorides. opper(i) chloride: ul, copper(ii) chloride ul 2 8. What are some of the experimental tests you would carry out to decide if ionic bonding was present in a compound? ompounds containing ionic bonding have crystalline lattice structures. The crystalline lattice structures are hard, brittle and not malleable (cannot be moulded into other shapes) and normally have very high melting and boiling points. ompounds that contain ionic bonding do not conduct electricity in the solid state but will do so in an aqueous environment or molten state because the free ions are then free to move. Experimental tests therefore include: testing the melting and boiling points, testing for electrical conductivity in both the solid and molten state. 9. Write the formula, and show the ions present, for the following: a) iron(ii) chloride Fel 2, ions present: Fe 2+, l - b) iron(iii) chloride Fel 3, ions present: Fe 3+, l - c) magnesium nitride Mg 3 N 2, ions present: Mg 2+ N 3- d) aluminium oxide Al 2 3, ions present: Al 3+, 2- e) potassium nitrate KN 3, ions present: K +, N 3 - The IUPA accepted systematic name for potassium nitrate is potassium trioxonitrate(v) f) calcium fluoride af 2, ions present: a 2+, F - g) magnesium hydroxide Mg() 2, ions present: Mg 2+, - h) sodium carbonate Na 2 3, ions present: Na + 2-, 3 3

4 i) ammonium sulfate (N 4 ) 2 S 4, ions present N 4 +, S 4 2- j) sodium phosphate Na 3 P 4, ions present: Na +, P 4 3- k) sodium carbonate Na 2 3, ions present: Na +, 3 2- l) sodium hydrogencarbonate Na 3, ions present: Na +, 3 - The IUPA accepted common name for sodium hydrogencarbonate is sodium bicarbonate Now use a circle to indicate in each pair above which species was originally the less electronegative, i.e. which atomic species became the cation in the above pairs. The positive ion (cation) forms when a species (often a metal) loses electrons to a more electronegative species. In the majority of the examples above, the metal loses electrons and is, therefore, the less electronegative species. In part i) the ionic compound, N 4 S 4, has formed as the result of an acid-base reaction rather than as the result of differing atomic electronegativities. 10. Ionic compounds have many properties in common. utline the general properties associated with compounds that have ionic bonding. rystalline structures ard, brittle, high melting solids Do not conduct electricity in solid state but conduct when molten or in solution Examples include common table salt. ur bodies need these and other salts as a sort of electricity conductor in our bodies. In solution ionic salts are called electrolytes, because they conduct. 11. rystals of sodium chloride consist of a lattice of Na + and l -, in which each ion is surrounded by six of the other kind. a) Draw a diagram showing how the 6 l - surrounding a Na + are arranged. Please note: These diagrams only to show part of the structure of Na + l - (s). In reality there is not only one Na + surrounded by six l nor one l surrounded by six Na +. Instead there is always one Na + for every one l, in a repeating fashion as shown in the diagram on page 118. b) Draw a diagram showing how the 6 Na + surrounding a l - are arranged. Please see above diagram and note. c) Give the name for the structure of sodium chloride 4

5 ontinuous ionic lattice. d) Name the forces holding these ions in place. The electrostatic attraction between positive and negatively charged ions e) Explain how this type of bonding is formed. Sodium needs to lose one electron in order to end up with a full outer shell (octet). hlorine needs to gain one more electron to obtain a full outer shell. Sodium transfers one electron completely to chlorine, and the ions Na + and l - are formed. These are then held together by electrostatic attraction to form the ionic crystal lattice. 12. Why does chlorine exist as l 2 and not just as l like elemental argon, Ar? hloride is a group 7 non-metal that contains 7 valance electrons. In order to satisfy the octet rule, two chlorine atoms share one valence electron each with each other through a single bond. Argon on the other hand is a noble gas and has complete octet in its valence shell. Therefore, it does not need to form additional bonds to complete its octet. 4.2 Exercises 1. Draw Lewis (electron dot) structures for: Lewis structures, also known as electron dot structures, show all outer (valence) electrons, plus any charge. a) l - l b) + c) 2-2 d) 4 e) N 3 N f) 2 5

6 + g) N 4 N h) 2 4 or Both are correct Lewis structures. owever, the second structure shows the arrangement of the electron pairs due to their repelling affect, known as the VSEPR (Valence Shell Electron Pair Repulsion) theory. i) 3 2- j) 2 k) BF 3 F B F F l) S 2 S m) 3 + n) 2 2 6

7 2. Why are only valence (or outer shell) electrons drawn in Lewis dot structures? nly valence electrons participate in the formation of chemical bonds. 3. Use your Lewis structures to predict the shape of the molecules d) to n) above from question 1 above and explain your reasoning using the VSEPR Theory. Parts a) c) are all individual atoms, not molecules, so shape not applicable here. d) The central atom is carbon; it has four regions of electron density around it, these are the - single bonds. These are all bonding regions, and the electrons in each bond repel the others equally, so according to VSEPR the structure is tetrahedral. e) The central atom is nitrogen; it has four regions of electron density around it: three are bonding regions (N- single bonds) and one is a non-bonding region (the nitrogen lone pair). According to VSEPR the lone pair electrons will repel the electrons in the three bonds to a slightly greater extent, so the three bonding regions are pushed into a trigonal pyramidal shape. f) The central atom is oxygen; it has four regions of electron density around it: two are bonding regions (- single bonds) and two are non-bonding regions (the two oxygen lone pairs). According to VSEPR the lone pair electrons will repel the electrons in the two bonds to a slightly greater extent, so the two bonding regions are pushed into a bent shape. g) The central atom is nitrogen; it has four regions of electron density around it: all are bonding regions (N- single bonds), and the electrons in each bond repel the others equally, so according to VSEPR the structure is tetrahedral. h) There are two central atoms in this molecule, both are carbon. Each carbon atom has three regions of electron density; all are bonding regions (2 x - bonds and 1 x = double bond on each carbon). By VSEPR each of these three bonding regions will repel each other equally, therefore the shape around each carbon is trigonal planar. i) The central atom is carbon, it has three regions of electron density (2 x - single bonds and 1 x = double bond). By VSEPR each of these three bonding regions will repel each other equally, therefore the shape is trigonal planar. j) The central atom is carbon, it has two regions of electron density (2 x = double bonds). By VSEPR each of these two bonding regions will repel each other equally therefore the shape is linear. k) The central atom is boron, it has three regions of electron density (3 x B-F single bonds). By VSEPR each of these three bonding regions will repel each other equally, therefore the shape is trigonal planar. l) The central atom is sulfur, it has two regions of electron density (2 x S= double bonds). By VSEPR each of these two bonding regions will repel each other equally therefore the shape is linear. m) The central atom is oxygen, it has four regions of electron density around it: three are bonding regions (- single bonds) and one is a non-bonding region (the oxygen lone pair). According to VSEPR the lone pair electrons will repel the electrons in the three bonds to a slightly greater extent, so the three bonding regions are pushed into a trigonal pyramidal shape. n) The central atom is carbon, it has two regions of electron density (2 x triple bonds). By VSEPR each of these two bonding regions will repel each other equally therefore the shape is linear. 4. ovalent bonding in 2 7

8 a) Predict, with reasoning, whether a molecule consisting of carbon and oxygen, 2, will contain covalent bonding given the positions of the atoms in the periodic table. Both carbon and oxygen are non-metals, close to each other on the periodic table. Their electronegativities are therefore not different enough to enable a complete transfer of electrons between the two elements. arbon requires 4 electrons to complete its octet and oxygen requires two. ovalent bonding in the form of two = double bonds ensures that both carbon and oxygen complete their octet. b) Deduce the Lewis structure of the 2 molecule. c) Indicate the bond dipoles on the structure by considering the electronegativity of the atoms present. Show which atom(s) has the partial positive charge and which atom(s) has the partial negative charge. δ δ+ δ xygen is more electronegative therefore it will withdraw electron density away from carbon, leaving carbon with a partial positive charge and oxygen with a partial negative charge. Bond dipole moments are indicated by using d) Predict the shape and bond angles of 2 using VSEPR theory. ow many regions of electron density are around the central atom? ow many of these are bonding regions of electron density? The central atom is carbon, it has two regions of electron density (2 x = double bonds). By VSEPR each of these two bonding regions will repel each other equally at 180 degrees, therefore the shape is linear. Note that a double bond does not count as two regions of electron density despite the fact that there are two bonds present. A single region of electron density refers to any type of bond whether it s a single, double or triple bond or a lone pair of electrons. e) From (c) it is shown that the bonds are polar. Is the molecule polar overall? Give your reasons. verall the molecule is not polar because the bond moment vectors of the = bonds are equal and opposite to one another. They are equal in magnitude (size) and opposite in direction and, therefore, cancel each other out. This makes the compound overall non-polar despite its polar bonds. f) What force is actually holding the atoms within this molecule together? Electrostatic forces, i.e. the attraction between the positive nuclei and the negatively charged electrons within the bonds. 5. Explain, with the help of an example, the following terms: a) single bond A single bond is one pair of electrons shared between two atoms, for example the - bonds in 4. b) double bond A double bond is two pairs of electrons shared between two atoms, e.g. =, in 2. 8

9 c) triple bond A triple bond is three pairs of electrons shared between two atoms, e.g. N N, in N 2. d) polar bond A polar bond arises from the difference in electronegativity between two atoms covalently bonded. The electron pair is more localised around the atom that has the greatest electronegativity, creating a partial negative charge, δ-, on that atom and a partial positive charge, δ+, on the less electronegative atom in the bond. An example (δ+) l(δ-) e) coordinate covalent bond (dative bond) A coordinate covalent bond (or dative bond) is a type of covalent bond in which one atom donates both electrons. For example, ammonia can react with a proton (hydrogen ion) to form the positively charged ammonium ion. The bond is formed by the nitrogen using both the electrons in its lone pair in order to form the covalent bond with the proton: N N The proton, +, has no electrons which it can donate in order to form a bond. f) bond moment (dipole) A bond moment is a vector that indicates the magnitude and size of the polarity of a covalent bond, i.e. over which of the two atoms the electrons spend the majority of their time and, therefore, which atom is partially positive and which is partially negative. For example, in the molecules of dichloroethene shown below, there are bond moments indicated for the -l bonds: bond moment l l l l The chlorine atom is much more electronegative than carbon which results in a polar bond with partial charges distributed in this way: associated with them. δ l. The and = bonds are non-polar bonds, so have no bond moments arbon and hydrogen are different elements and so their electronegativities are not identical. owever, according to the Pauling scale, they are very close. Therefore, despite being different elements, the pull each atom has on the electrons within the bond is similar and therefore the bond is non-polar. g) dipole moment The dipole moment of a molecule is the sum of its individual bond moments. If the molecule has an overall dipole moment it is polar. If the dipole moment = 0 the molecule is non-polar, even if there are some polar bonds within the molecule. For example again considering the molecules of dichloroethene from the previous question: 9

10 dipole moment l l polar molecule l l no overall dipole moment non-polar molecule The molecule on the left had two bond moments which, by vector addition, give the overall dipole moment shown. Therefore the molecule is polar. The molecule on the right has two bond moments in equal and opposite directions so they cancel each other out and there is no net dipole moment. The molecule is non-polar. h) polar molecule A polar molecule has an overall dipole moment, the individual bond dipole vectors do not cancel each other out, e.g. l or 3 l. i) bond length A measurement of the distance between two nuclei. It varies in length depending on the nature of the bond. Single bonds are longer than double bonds, and double bonds are longer than triple bonds. j) bond strength A measurement of the amount of energy required to break the bond apart and, therefore, a reflection of how tightly the bonds hold the connected atoms together. Strong bonds, like primary bonds (ionic, metallic, covalent) require large amounts of energy to break. Weak bonds, like secondary bonds (dispersion forces) require small amounts of energy to break. k) bond angle The measurement of the angle at which two atoms sit apart from each other in space. For a tetrahedral molecule the bond angles are 109.5º. 6. Explain how Valence Shell Electron Pair Repulsion theory can be used to predict the shape and bond angles in molecules. Address how electron density affects the bond positions. The basis of the VSEPR theory considers how repulsive forces between valence electrons produces a geometric shape that results in the repulsive forces being as far apart from each other as possible. The negative charge density of electrons whether they are in bonds or lone pairs affects repulsion and hence the bond position. Lone pairs of electrons have a higher negative charge density than electrons in a single bond and therefore repel other regions to a greater extent. This is reflected in the differing bond positions when molecules contain lone pairs instead of all single bonds. For example, the electrons in the four single bonds around the carbon atom in methane ( 4 ) all repel each other equally producing bond angles of owever, the electrons in the three single bonds around the nitrogen atom in ammonia (N 3 ) do not repel the lone pair on the nitrogen atom equally. The electrons in the lone pair have a higher charge density (their double negative charge is more concentrated) and they repel the bonding electrons to a greater extent. This results in angles >109.5 between the lone pair and the single bonds and < between the single bonds. 7. ow is the type of bonding in the chlorides of potassium (Kl) and phosphorus (Pl 3 ) related to the position of K and P in the periodic table? K is a group 1 metal and can readily give up an electron to a chlorine atom (group 7 non-metal) to obtain a stable octet. This complete transfer of one electron results in ionic bonding. P is a group 5 non-metal and forms covalent bonds in order to share electrons with chlorine atoms and so obtain a stable octet. 8. Both N 2 and 2 contain covalent bonds, yet only one molecule is polar. Which is this and why? 10

11 2 is a polar molecule and N 2 is a non-polar molecule. N 2 is a diatomic molecule that contains two atoms of the same element which consequently have the same electronegativity. Therefore the electrons are equally shared between both N atoms. There is no bond moment and therefore no overall dipole moment. N 2 is nonpolar. 2 has bonds. xygen and hydrogen have different electronegativities and the electrons spend more time around the atom than the atom, creating a bond moment. The dipole vectors do not cancel each other out, so creating an overall dipole moment in the molecule. Therefore 2 is polar. 9. A, B and represent elements with atomic numbers 9, 16 and 19 respectively. a) Give the electron arrangement for each and name the elements A 2.7 fluorine B sulfur potassium b) What kind of bonding would you expect from: A and B covalent A and ionic B and iionic c) Draw Lewis structures for the compounds formed in (b) A and B F S F A and An ionic compound potassium fluoride, KF: continuous lattice of F - and K + ions. Ions: K F B and An ionic compound potassium sufide K 2 S: continuous lattice of S 2- and K + ions. 2 K S Ions: 10. Using an example, e.g. 4, outline the formation of covalent bonds. ovalent bonds are a result of electron sharing. Each atom must share enough electrons with other atoms in order to complete its octet. In 4, the carbon atom needs four electrons to complete its octet, so it forms four bonds with four hydrogen atoms sharing one electron with each. Each hydrogen atom needs one electron to fill its valence shell (which for hydrogen has only two electrons) so each gets one electron by forming a single bond with carbon. 11. Using an example, eg. 2, outline the general physical properties of covalent compounds. ovalent molecules, like 2, interact with each other through intermolecular forces. These forces are not very strong compared to covalent or ionic and metallic primary bonding and this is the reason why most covalent molecules have relatively low boiling and melting points. They therefore exist mostly as liquids and gases. Molecules that have covalent bonds do not conduct electricity in the solid or molten state as their outer electrons 11

12 are shared between the atoms and are not free to move between molecules. Water, 2, is a covalent molecule which exists as a liquid despite having the strongest of all intermolecular forces between its molecules hydrogen-bonding. There are some exceptions to these generalisations of course; continuous covalent lattices of Si, Si 2 and some allotropes of carbon for example are high melting solids. This is because rather than existing as single molecules they form three dimensional covalent lattices which require the breaking of strong covalent bonds to melt. Section 4.3 covers intermolecular forces in detail. 12. Which of the following is not an electrical conductor? A: KBr(aq) B: octane, 8 18 : l(aq) D: Nal(l) Answer: B ctane is a covalent liquid. There is no conduction because there are no free electrons nor free ions. KBr and Nal are ionic compounds and will conduct when molten or in solution. l(aq) contains + (aq) and l (aq) and so conducts when it ionizes. 13. Which of these group Vll hydrides has the smallest bond dipole? A: F B: l : Br D: I Answer: D Iodine is the lowest in group 7 and it therefore has the lowest electronegativity. Therefore iodine and hydrogen will have the smallest electronegativity difference of the group 7 hydrides, resulting in the smallest bond dipole. 14. The table below relates to the some chlorides of period 3 in the periodic table: Nal Mgl 2 Sil 4 Pl 3 Type of bonding Ionic Ionic ovalent ovalent Structure Lattice of Na + and l - ions Lattice of l - and Mg 2+ ions l l Si l l l P l l Boiling Point º a) Fill in the Type of Bonding row with either ionic or covalent. b) Name and draw the structure for each. Nal and Mgl 2 ionic lattices; silicon tetrachloride is tetrahedral; phosphorus trichloride is trigonal pyramidal. c) Explain, in terms of forces, the difference between the boiling points of Nal and Pl 3. 12

13 Nal is a compound that contains ionic bonding. This type of primary bonding requires vast amounts of thermal energy in order to break the ionic lattice and cause the compound to boil. Pl 3 is a compound that contains covalent bonding and the individual molecules are held together by intermolecular interactions: dispersion and dipole-dipole forces. Intermolecular interactions do not require as much thermal energy to break and therefore Pl 3 has a much lower boiling point. 15. The sequence that shows these molecules in increasing order of carbon-carbon bond strengths is: A: 2 2, 2 4, 2 6 B: 2 4, 2 6, 2 2 : 2 2, 2 6, 2 4 D: 2 6, 2 4, 2 2 Answer: D If you draw out the structural formulas of these molecules, you will find that 2 2 contains a carbon-carbon triple bond, 2 4 contains a carbon-carbon double bond, and 2 6 contains a carbon-carbon single bond. 16. Bond strengths. a) Draw a structural formula for ethanoic acid, 3. 3 b) omment on the bond lengths and strengths of the bond and the = bond present in ethanoic acid. The bond is weaker than the = double bond. There are more electrons in the double bond, and the atoms are drawn closer together because of this, making the = bond shorter and stronger than the single bond. 17. What is a difference between the bond formed between + and N 3 and the bond formed between and Br? Include an explanation as to how each bond is formed. The bond between + and N 3 is a coordinate (or dative) covalent bond, whereas the and Br bond is a covalent bond. In the coordinate covalent bond, an electron pair from the N is used to form the bond with +. ne atom donates both of the two electrons required to form the single bond. The -Br bond is a regular covalent bond, which is formed when the two atoms both donate an electron to complete their valence shells. Each atoms donates one of the two electrons required to form the single bond. 18. Is the bond considered to be a polar bond? Again explain your answer. Though the bond is made up of two different atoms, the bond is said to be non-polar because there is a small difference in electronegativity between these two elements. This small electronegativity difference means that the electrons are shared fairly equally between both the carbon and hydrogen atoms, resulting in a nonpolar bond. 19. Which statement correctly describes polar molecules? 13

14 A All molecules with polar bonds are polar. B Some molecules without polar bonds are polar. Polar molecules have bond moments which cancel out. D Polar molecules have bond moments that do not cancel each other out. Answer: D (Because in these molecules there is overall polarity) A: Not all molecules with polar bonds are polar, as the individual bond moments may cancel out, to give a nonpolar molecule overall. B: nly molecules with polar bonds can have a dipole moment and therefore be classified as being polar. : If the bond moments in a molecule cancel out, the molecule is non-polar. 20. onsider the following un-named elements: Element A B D E Electronegativity a) Which two elements would combine to form the most polar compound (which can include the extreme an ionic compound)? A and E would combine to form the most polar compound as they have the greatest difference in electronegativity. They may undergo a complete transfer of electron(s) to form an ionic bond. b) Which two elements would combine to form the least polar compound? A and B would combine to form the least polar compound because they have the smallest electronegativity difference. c) Suggest the most likely group and period for element E. Group 7, period 2, (the element is fluorine). 21. Allotropes a) What is an allotrope? An allotrope is the same pure element differing in the geometric structure, i.e. having a different physical forms. For example, oxygen has two allotropic forms, dioxygen ( 2 ) and ozone or trioxygen ( 3 ), while carbon has three. b) Describe and compare the structure and bonding present in the three carbon allotropes: diamond, graphite and fullerene ( 60 ) to account for properties such as hardness and electrical conductivity. Diamond is the hardest mineral known. This hardness results from the covalent crystal structure in which each carbon atom is linked to four others by covalent bonds to form a continuous tetrahedral structure. The carbon atoms are all sp 3 hybridised. The bond length is nm and the bond angel 109.5º and these properties are constant throughout the entire crystal lattice which results in no point of weakness in the structure. There are no free electrons in the structure since all four outer electrons of carbon are used in bonding, therefore diamond does not conduct electricity. Graphite is a soft black slippery substance. These properties are derived from the structure in which carbon atoms are bonded covalently to three others to form a flat sheet. The carbon atoms are sp 2 hybridised and sheets of carbon hexagons are are stacked and held together by weak van der Waals forces using delocalised electrons. The delocalised electrons can move between the layers, which explains the electrical conductivity of 14

15 graphite. The - length in the layers of graphite is nm and the distance between the layers is 0.34 nm. Longer bonds are weaker than shorter bonds, therefore there is a weakness between the layers and they slip over each other, resulting in the soft, slippery properties of graphite. 60 fullerene is a form of carbon that a bonded to form a cluster which resembles a sphere composed of sp 2 hybridised carbons joined to form hexagons and pentagons, to look like a soccer ball. Fullerene does not conduct electricity. owever the properties of fullerenes are still the subject of much scientific investigation. For diagrams see page 132 or refer to other sources. 22. ontinuous covalent network structures a) Explain the structures of silicon and silicon dioxide. Silicon can occur in a diamond like continuous covalent network structure, where each silicon atom is covalently bonded to another silicon atom: Si Si Si Si Si Silicon dioxide also forms a continuous covalent network structure, with the molecular formula Si 2 Si Si Si Si Si Si Si Si Si Because silicon is a larger atom than carbon, it is more stable with single bonds in continuous covalent networks rather than in forming double and triple bonds like carbon does. b) Use the bonding and structures to explain why silicon and silicon dioxide (quartz) both have melting points in excess of 1400º. Silicon and silicon dioxide both occur as crystalline structures made from a continuous covalent network of atoms. It is the continuous network structure made up of strong covalent bonds that gives these compounds their high melting points. Unlike other covalent molecules that have weak intermolecular forces between individual molecules, there are no separate molecules in silicon or silicon dioxide. Therefore, in order to convert between the solid and liquid states, the strong covalent bonds between atoms must be broken and this requires a lot of energy, reflected in the melting points in excess of 1400º. 4.3 Exercises 1. Explain how the English meaning of the word disperse relates to the use of the word in the term dispersion forces in chemistry. Disperse means to spread in different directions. Dispersion forces arise when a molecule collides or comes into close proximity to another molecule and their associated electron clouds are dispersed (displaced or spread) due to electron-electron repulsion. This creates areas of low and high electron density associated with the molecule, i.e. temporary dipoles within the molecules that may then become instantaneously attracted to one another. This attraction is only instantaneous. owever, as the electron cloud quickly disperses back, the molecules repel each other once more. This is why dispersion forces are such a weak intermolecular interaction. 15

16 See also: diagram question 3 answer below. 2. What is the difference between a permanent electric dipole and temporary electric dipole? Permanent electric dipole refers to a permanent partial positive and negative charge between molecules, for example a bond moment or overall dipole moment found in polar molecules. Induced electric dipole refers to one molecule inducing a partial negative or positive charge into another molecule when in close proximity, such as the case in a dispersion force or an ion-dipole interaction. nce the molecules separate, the partial charge is no longer present as the electron density redisperses. 3. Define the following terms with diagram to illustrate your answer: a) Dispersion force A weak force of interaction between temporary and induced dipoles. See question one for an explanation of dispersion forces as well as the following diagram: 1. ollision of two non-polar molecules 2. Associated electron clouds are dispersed due to electron-electron repulsions, resulting in induced dipoles 3. The molecules are then instantaneously attracted to each other due to their induced dipoles b) Dipole-dipole interaction Molecules which have a permanent dipole are attracted to one another For example molecules of l interact through dipole-dipole forces: δ l δ l There is electrostatic attraction through oppositely charged dipoles. c) Ion-dipole interaction The solvation (dissolving) of salt, Nal, is an example of ion-dipole interaction. The ion, either Na + or l -, is solvated by water molecules as shown below: 16

17 δ δ δ Na + δ δ d) ydrogen bond δ δ+ δ l - δ δ ydrogen bonding is a very strong type of dipole-dipole interaction. The strength of this electrostatic interaction is a result of the large bond moment that exists when hydrogen is bonded to the highly electronegative and small atoms, nitrogen, oxygen and fluorine (N,, F). ydrogen bonding occurring between molecules of water is shown: δ δ δ When drawing hydrogen bonds make them longer than the covalent bonds. They are an intermolecular interaction and therefore the "bond" is longer and weaker than a intramolecular interaction. 4. omment of the relative electronegativities of N and l and explain why molecules of N 3 can hydrogen bond to one another, but l molecules cannot. Nitrogen and hlorine have the same electronegativity value on the Pauling scale (see IB hemistry date booklet). owever, nitrogen (period 2) is smaller than chlorine (period 3). This smaller size allows the nitrogen atom to get closer to hydrogen atoms in adjacent N 3 molecules, so the electrostatic interaction between the hydrogen and δ - nitrogen atoms is much stronger and hydrogen bonding can occur. 5. Draw a diagram to show the hydrogen bonding that occurs between molecules of ethanol, δ 3 2 δ There is no hydrogen bonding with hydrogen atoms that are attached to carbon atoms. The bond is not polar enough to cause the charge on the hydrogen atom in order for it to take part in hydrogen bonding. 6. ydrogen bonding has been described as nature s glue. an you give some examples to support this statement? 17

18 The high Tb of water, the open structure of ice and the stickiness of honey depend upon hydrogen bonds. ydrogen bonding is important is maintaining biological structures for proteins and DNA fragments. DNA relies on hydrogen bonding between base pairs to maintain the double helical structure, and the unique three dimensional structure and folding of proteins is heavily reliant on the individual hydrogen bonds that can occur between amino acid side chains and water molecules. Life depends upon the reactions that complex structures, like DNA and proteins, undergo. ertain bonds in these structures must be able not only hold the units together strongly but also be able to break and reform readily. Two such types of interactions that occur in proteins and DNA are hydrogen bonding, and pi-stacking (an aromatic interaction-learn about aromatic/benzene rings in Topic 10). 7. Explain how intermolecular forces account for the following properties, use a diagram to illustrate your answer where possible. a) Water has a higher boiling point than ammonia xygen has two lone pairs, it can form two hydrogen bonds in the water ( 2 ) molecule, whereas nitrogen only has one lone pair and can only form one hydrogen bond per ammonia (N 3 ) molecule. b) exane 6 14 is immiscible with water exane is a non-polar molecule, water is polar. The two do not mix as the strong polar interactions between water molecules (hydrogen bonding) cannot be overcome by hexane as it can only have weak dispersion and induced dipole interactions with water. c) Solid water (ice) less dense than liquid water When water freezes an open structure is formed as the molecules are held apart and in an ordered network structure by hydrogen bonding. Each ice molecule is hydrogen bonded to four other ice molecules δ δ This structure is less dense than liquid water, which although it has hydrogen bonding, has too much energy for the molecules to be fixed in a rigid structure. ence the molecules are a lot closer to one another are therefore more dense in liquid water. When the kinetic energy of the water molecules is removed by freezing, they are able to resume the highly ordered, more open structure of ice. When ice melts some of the hydrogen bonds break and the water molecules come closer to each other. d) N 3 has a higher boiling point than 4 Ammonia, N 3 is capable of hydrogen bonding which is the strongest intermolecular attraction. This is a strong force that requires more energy to break, hence the boiling point of N 3 is higher than that of 4. 4 is a nonpolar molecule and only has weak dispersion forces. Therefore less energy is required to interrupt these forces so making the boiling point lower. e) Kl has a higher melting point than I 2 Kl is an ionic species, it is held together by electrostatic attraction between ions which results in a strong continuous crystalline lattice. Iodine, I 2, on the other hand, is a non polar molecule with only weak dispersion forces between molecules of I 2. More energy is required to break the strong electrostatic interactions of the continuous lattice of Kl than the weak dispersion forces of I 2. Therefore Kl has a higher melting point than I 2. 18

19 8. List the types of intermolecular forces that exist between molecules of each of the following species: a) 3 l Dipole-dipole interactions b) Nal Ionic bond - electrostatic attraction. c) S 2 Dispersion forces d) PF 3 Dipole-dipole interactions e) 3 ydrogen bonding f) 3 ( 2 ) 4 3 Dispersion forces 9. The following graph illustrates the boiling point (K) against period number for the following hydrides, 2, 2 S, 2 Se and 2 Te. Explain the deviation shown by water Boiling points of group 6 hydrides Boiling point ( o ) Period Since oxygen, sulfur, selenium and tellurium are all in group 6, one may expect that the properties of hydrides of this group would be similar to one another, or at least follow a trend. The graph shows that 2, the hydride of oxygen (period 2) deviates quite significantly from the increasing boiling point trend of the other group 6 hydrides. This is because water is capable of hydrogen bonding interactions, whereas all the other hydrides can only interact through weaker dipole-dipole interactions. You may be wondering why hydrides of sulfur, selenium and tellurium follow a slow steady increase in boiling point. This is because the size of the atoms of the group 6 elements increases down the group and this increases the dispersion forces at work between their molecules. The larger molecules are, the more electron density they 19

20 have and the stronger the dispersion forces become. These forces become significant, for example, in plastics. Plastics are made up of very long carbon chains that exist as solids at room temperature due to the presence of extensive dispersion forces between the very large molecules. 10. Nitrogen and phosphorus are in the same group on the periodic table. Account for the large difference in boiling points of the hydrides, N 3-33º and P 3-90º The difference in boiling points between N 3 and P 3 is due to the types of intermolecular interactions present between the different molecules. P 3 has dipole-dipole interactions, whereas N 3 is able to form hydrogen bonds, which is a much stronger interaction. ence more thermal energy is required to break intermolecular forces between N 3 molecules and therefore it has a higher boiling point than P Despite their similar molecular masses, propane, 3 8 and ethanol, 2 5 have boiling points of 231 K and 351 K respectively. Account for this seemingly large discrepancy. The difference in the boiling points between ethanol and propane is due to the types of intermolecular interactions present between the different molecules. Propane is a non-polar molecule and only contains dispersion forces, the weakest type of secondary interactions. Ethanol is a polar molecule and has hydrogen bonding as well as dispersion forces. ydrogen bonding is much stronger than dispersion forces, and hence requires more thermal energy to break and therefore ethanol has a higher boiling point. 12. The smell associated with decaying animal matter is, in part, due to the smelly gas mixture of ammonia, N 3, methane, 4, and hydrogen sulfide, 2 S. a) Draw Lewis structures for each, showing all valence electrons. N S b) Describe the geometry of the electron pairs and the atoms around the central atom in each molecule and predict the bond angles around the central atom in each molecule, using the valence shell electron pair repulsion (VSEPR) theory. In N 3 there are four areas of electron density around the central nitrogen atom, three bonding (N- bonds), and one non-bonding (nitrogen lone pair). The nitrogen lone pair repels the electrons in the N- bonds more than they repel each other, which results in the angles being slightly more or less than 109.5º as displayed on the diagram below. The structure is a trigonal pyramidal molecular shape. >109.5º N <109.5º In 4 there are four regions of electron density around the central carbon atom, all of which are bonding regions (4 x - bonds). The electrons in these bonds all repel each other equally so the result is a tetrahedral molecular shape. 20

21 all bond angles 109.5º In 2 S there are four regions of electron density around the central sulfur atom, two bonding (S- bonds) two non bonding (sulfur lone pairs). The lone pairs repel the bonding electrons more, which results in the angles being slightly less or more than 109.5º. The structure is a bent molecular shape. >109.5º S <109.5º c) The relative molecular masses, M r, and boiling points, T b/k, of these three substances are given below. Identify the intermolecular forces present in each substance and account for the boiling points in terms of these forces. ompound T b/k, M r Intermolecular forces ammonia ydrogen bonding, highest boiling point as it is the strongest type of intermolecular interaction methane Dispersion forces, lowest boiling point weakest type of intermolecular interaction hydrogen sulfide Dipole-dipole interactions, moderate boiling point due to moderate strength intermolecular interaction 13. Water is the most abundant compound on Earth. Much of the chemistry of water is influenced by its polarity and its ability to form hydrogen bonds. a) Why are water molecules polar? Use a diagram to illustrate your answer. overall dipole moment Water molecules contain two polar - bonds. The vector addition of these bond moments results in an overall dipole moment for the molecule, meaning it is polar. Although water is abundant on earth, only a very small amount (about 3%) is fresh water. Much of the water on Earth is trapped in glaciers or found as seawater. b) The high polarity of the - bond in water results in the formation of hydrogen bonds. Draw a diagram including dipoles and lone pairs of electrons, to illustrate hydrogen bonding between water molecules. 21

22 δ δ c) Why must many people living in anada drain their swimming pools before winter? Water expands when it freezes, for reasons outlined in question 7 c). The people in anada need to drain their pools to stop damage to the structure that the large volume of water expanding as it freezes would cause. 14. onsider the following hydrides: Na, N 3, 2 and l. Explain how the difference in the polar nature of these hydrides can lead to a difference in their acid character, and their ability to form hydrogen bonding. Give equations to support your answer. Na is an ionic material composed of Na + and - ions. It is extremely reactive and used as a base in organic syntheses as it readily removes a proton from even weak acids: Na + A Na + A N 3 is a covalent compound with polar N- bonds. The nitrogen is electronegative and has a lone pair. Thus ammonium acts as a base (accepts a proton) and reacts with water by the following equation: N N Water may act as an acid or a base: 2 + A 3 + A _ (water acts as BASE) 2 + A _ _ + A (water acts as AID) l is a polar covalent molecule. Due to its large polarity the l bond is quite weak and l is therefore acidic: l + 2 l nly water and ammonia are capable or -bonding. Na is ionic so will not hydrogen bond (the species are already ionised, the electrostatic interactions by far overcome any intermolecular interactions). l cannot be involved in hydrogen bonding as the l atom is too large (period 3), the atoms simply cannot get close enough to one another for the hydrogen bonds to form. 15. abon dioxide 2 and silicon dioxide Si 2 are oxides of group 4 elements. 2 is a gas at room temperature, while Si 2 is a hard, high melting solid. a) Draw a Lewis structure for 2. b) Predict the shape and polarity of 2 using VSEPR theory. There are two regions of electron density around the central carbon atom, and according to VSEPR theory the electrons in these bonds will repel one another equally at 180º. Therefore 2 will have a linear molecular shape. The bond moments of the = bonds are in equal and opposite so the 2 molecule is non-polar. c) Draw the structure of Si 2. 22

23 Si Si Si Si Si Si Si Si Si d) Why do 2 and Si 2 have such different structures? Because silicon is bigger than carbon, it is more stable with single bonds in continuous covalent networks rather than forming double and triple bonds like carbon does. Silicon cannot form double bonds with oxygen like carbon so it cannot form a discrete Si 2 molecule. Instead its octet must be satisfied by forming four individual covalent bonds in a continuous network structure. e) Explain how their structures relate to the large difference in boiling points and other physical properties. Silicon is a continuous covalent network of silicon and oxygen, which is very strong and gives silicon dioxide its hard, high melting properties. arbon dioxide is a small non-polar molecule that is held together by weak dispersion forces. At room temperature there is enough thermal energy to break the dispersion forces apart; hence carbon dioxide is a gas at room temperature. 16. onsider the following molecules, ethane 3 3 (bp K), 3 (bp 293 K) and 3 2 (bp 352 K). a) alculate the Mr of each. 3 3 M r = 30.08, 3 M r = 44.06, 3 2 M r = b) Explain the differences in their boiling points in terms of their intermolecular forces. The difference in the boiling point of ethane, ethanal and ethanol is due to the type of intermolecular interactions present between the different molecules. Ethane is a non-polar molecule with only weak dispersion forces, it also has a smaller mass than the other two, so the dispersion forces will be much less. Ethanal and ethanol have quite similar masses, so there must be differences in their intermolecular interactions to account for the large difference in their boiling points. Ethanal is a polar molecule (due to polar = bond) and has dipoledipole interactions. Ethanol is a polar molecule and is capable of hydrogen bonding. Dipole-dipole interactions are stronger than dispersion forces and hydrogen bonding is much stronger than dipole-dipole interactions. More thermal energy is required to overcome the stronger interactions, therefore the highest boiling point is ethanol ( 3 2 ) followed by ethanal ( 3 ) and ethane ( 3 3 ). 23