Attractive Forces Holding Atoms Together: Chemical Bonds - PDF

The attractive force which holds atoms, ions, and molecules together is called a chemical bond. Since these forces of attracstion are intramolecular forces, they have an effect on the chemical properties as well as the physical properties of the chemical. UNIT 2 67 CHEMISTRY GRADE 11 68 UNIT 2 Activ...

The attractive force which holds atoms, ions, and molecules together is called a chemical bond. Since these forces of attracstion are intramolecular forces, they have an effect on the chemical properties as well as the physical properties of the chemical. UNIT 2 67 CHEMISTRY GRADE 11 68 UNIT 2 Activity 2.2 From your knowledge of Grade 9 chemistry discuss the following questions in a group of four, then present your responses to the whole class. 1. Why do some atoms combine while others do not? Give some examples in your answer. 2. Why do different atoms form different types of bonding? All noble gases except helium (1s2) have ns2np6 electron configurations (where n indicates the highest occupied shell). The noble gases are quite unreactive because they have very stable electron configurations, as reflected by their high ionization energies and low electron affinities. Because all the noble gases (except helium) have outer shells with eight electrons, many atoms undergoing reactions also attain eight valence electrons (ns2np6). This rule has become known as the octet rule, as follows: Atoms tend to gain or lose electrons until they have achieved an outer shell that contains an octet of electrons (eight electrons). Do you know any compound whose central atom does not obey the octet rule? Activity 2.3 Discuss the following questions in pairs and share your ideas with the whole class. 1.Why do metals tend to form cations and why do non-metals tend to form anions? 2.Why do two non-metals combine to form compounds while no similar process exists for two metals to form a compound? In general, there is a gradual change from metallic to non-metallic character as you move from left to right across a period and from bottom to top within most groups in the periodic table. Accordingly, atoms can combine to form three types of bond: metal with non-metal (ionic bond), non-metal with non-metal (covalent bond), and metal with metal (metallic bond). Ionic Bonds 2.2 At the end of this section, you will be able to: ) define ionic bonding ) use Lewis electron dot symbols for main group elements ) describe ionic bonding using Lewis electron dot symbols ) list the favourable conditions for the formation of ionic bonds ) explain the formation of ionic bonding ) give examples of ionic compounds ) define lattice energy ) calculate lattice energy of ionic crystals from given data using the Born Haber cycle ) discuss the exceptions to the octet rule ) describe the properties of ionic bonding ) carry out an activity to demonstrate the effect of electricity on ionic compounds (PbI2 and NaCl) ) carry out an investigation into the melting point and solubility of some ionic compounds (NaCl and CuCl2 ). Activity 2.4 Form a group of four and discuss the following questions, then present your responses to the whole class. 1. Why do elements form ions in certain chemical reactions? 2. Which of the following pairs of elements are likely to form an ionic compound? a. sodium and chlorine b. nitrogen and fluorine c. lithium and oxygen An ionic bond is formed by the electrostatic attraction between positive and negative ions. The bond forms between two atoms when one or more electrons are transferred from the valence shell of one atom to the valence shell of the other. The atom that loses electrons becomes a cation (positive ion), and the atom that gains electrons becomes an anion (negative ion). Any given ion tends to attract as many neighboring ions of opposite charge as possible. UNIT 2 69 CHEMISTRY GRADE 11 When large numbers of ions gather together, they form an ionic solid. The solid normally has a regular, crystalline structure that allows for the maximum attraction of ions, given their particular sizes. Example 2.1 The formation of NaCl from sodium and chlorine can be explained as: Na → Na+ + e− \[Ne\]3s1 Cl + e− \[Ne\]3s23p5 \[Ne\] → Cl− \[Ne\]3s23p6 or \[Ar\] Na+ + Cl− → NaCl or Na+Cl− The bond formed, as a result of the electrostatic attraction between the positive and negative ions is called the electrovalent bond or ionic bond. Note: Ionic compounds are usually formed when metal cations bond with non-metal anions. What about an ionic compound containing an ammonium ion? Exercise 2.1 1. Explain the formation of bonds in the following pairs of elements: a. potassium and chlorine c. sodium and oxygen b. magnesium and oxygen 2. Which of the following elements will form an ionic bond with chlorine, why? a. calcium b. carbon c. oxygen d. silicon 3. Identify the species found in the following ionic compounds: CaCl2 , MgO, and Al2 O3. From this what can you conclude about the formation of these compounds? 4. Why do elements located on the opposite sides of the periodic table tend to form ionic bonds? 70 UNIT 2 Ionic Bonds 71 UNIT 2 The American Chemist Gilbert N. Lewis (1875--1946) developed a special set of symbols for his theory. A Lewis symbol consists of a chemical symbol to represent the nucleus and core (inner-shell) electrons of an atom, together with dots placed around the symbol to represent the valence (outer-shell) electrons. For example, the Lewis symbol for chlorine, which has the electron configuration, \[Ne\]3s23p5, is Cl A Lewis structure is a combination of Lewis symbols that represents either the transfer or the sharing of electrons in a chemical bond. For example, the formation of sodium chloride from sodium and chlorine atoms can be represented as: Cl Na + \[Na\] Cl x \[ \] Lewissymbols Lewisstructure x Activity 2.5 Discuss the following questions in pairs and present your responses to the whole class. 1. Sodium atom is highly reactive, but sodium ion is not. Explain. 2. Write the Lewis symbols of K+, S2--, Al3+, and F-- ions. Exercise 2.2 1. Use Lewis electron-dot symbols to depict the formation of lithium and fluoride ions from the atoms and determine the formula of the compound. 2. Apply the Lewis formula to illustrate the formation of calcium chloride from calcium and chlorine atoms. 3. Write Lewis structures for the following compounds: a. BaO b. potassium oxide c. aluminum oxide 4. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: a. K and S b. Ca and O c. Al and N CHEMISTRY GRADE 11 You may be surprised to learn that the electron-transfer process actually absorbs energy! So why does it occur? As you will see, the reason ionic substances exist at all is because of the enormous release of energy when the ions come together and form a solid. Consider just the electron-transfer process for the formation of lithium fluoride, which involves two steps: 1. A gaseous Li atom loses an electron (IE1 ): Li(g) → Li+(g) + e− IE1 = +520 kJ 2. A gaseous F atom gains an electron lost by Li atom (EA): F(g) + e-- → F--(g) EA = --328 kJ Note that the two-step electron-transfer process by itself requires energy: Li(g) + F(g) → Li+(g) + F--(g) IE1 + EA = +192 kJ The total energy needed for ion formation is even greater than this because metallic lithium and diatomic fluorine must first be converted to separate gaseous atoms, which also requires energy. Despite this, the standard heat of formation (∆Hf °) of solid LiF is −617 kJ/mol; that is, 617 kJ is released when 1 mol of LiF(s) forms from its elements. That is, there must be some exothermic energy component large enough to overcome the endothermic steps. These are: When 1 mol of Li+(g) and 1 mol of F−(g) form 1 mol of LiF (g): Li+(g) + F--(g) → LiF(g) ∆Η° = −755 kJ When the gaseous ions coalesce into a crystalline solid. That occurs because each ion attracts others of opposite charge: Li+(g) + F--(g) → LiF(s) ∆Η° = −1050 kJ The negative of this value, 1050 kJ, is the lattice energy of LiF. The lattice energy (U) is the enthalpy change that occurs when 1 mol of ionic solid separates into gaseous ions. It indicates the strength of ionic interactions, which influence melting point, hardness, solubility, and other properties. 72 UNIT 2 Ionic Bonds Lattice Energies from the Born--Haber Cycle So we need to know the lattice energy of solid lithium fluoride. Finding the lattice energy of an ionic solid by eperiment is difficult. However, this quantity can be found indirectly using the Born--Haber cycle. The reasoning is based on Hess's law, which states that an overall reaction's enthalpy change is the sum of the enthalpy changes for the individual reactions that make it up: ∆Htotal = ∆H1 + ∆H2 + ∆H3 + \... Consider the Born-Haber cycle for the formation of NaCl. We think that solid sodium chloride can be formed from the elements by two different routes, as shown in Figure 2.1. In one route, NaCl(s) is formed directly from Na(s) and ½Cl2 (g); ∆f Η° −411 kJ mol-1 Figure 2.1: Born--Haber cycle for NaCl The second route consists of the following five steps, along with the enthalpy change for each. Step1: Metallic sodium is vaporized to a gas of sodium atom: ∆H°step1 = + 108 kJ mol--1 Na (s) → Na (g) Step 2: Chlorine molecules are dissociated to atoms: ½Cl2 (g) → Cl (g) ∆H°step2 = ½ bond energy of Cl2 = ½ (240 KJ) = + 120 kJ mol--1 Step 3: Sodium atoms are ionized to Na+ ions: Na (g) → Na+ (g) + e ∆Η°step3 = IE1 = + 496 kJ mol--1 Step 4: Formation of chloride ion: Cl (g) + e-- → Cl-- (g) ∆Η°step4 = EA = --349 kJ mol--1 UNIT 2 73 CHEMISTRY GRADE 11 Step 5: Formation of NaCl(s) from ions. The ions Na+ and Cl− combine to give solid sodium chloride whose enthalpy changes (the lattice energy) is unknown: Na+(g) + Cl--(g) → NaCl(s) ∆H°step5 = U (lattice energy) =? We know the enthalpy formation (∆f Η°) of NaCl (Direct route) and equals −411 kJ mol-1. Therefore, we can calculate the lattice energy using Hess's low: Solving for UNaCl gives: UNaCl = ∆f Η°-- (∆Η°step1 + ∆Η°step2 ∆Η°step3 + ∆Η°step4 ) = --411 kJ mol--1 -- \[108 kJ mol--1 + 120 kJ mol--1 + 496 kJ mol--1 + (--349 kJ mol--1)\] = --786 kJ mol--1 Note: Ionic solids exist only because the lattice energy exceeds the energy required for the electron transfer. Exercise 2.3 1. Draw a Lewis electron-dot symbol for each atom: a. Rb b. As c. I 2. Give the group number and general electron configuration of an element with each electron-dot symbol: X a. b. a. X2 O3 X 3. How does the lattice energy of an ionic compound depend on the charges and sizes of the ions? 4. Use condensed electron configurations and Lewis electron dot symbols to depict the monatomic ions formed from each of the following atoms, and predict the formula of the compound the ions produce. a. Ba and Cl b. Sr and O c. Al and F d. Rb and O 5. Identify the main group to which X belongs to in each ionic compound formula: b. XCO3 c. Na2 X 6. For each pair, choose the compound with the lower lattice energy, and explain your choice: a. CaS or BaS b. NaF or MgO c. LiCl or CsCl d. BaS or CsCl 74 UNIT 2 Ionic Bonds Factors Affecting the Formation of Ionic Bonds Activity 2.6 Discuss each of the following questions and forward your responses to the whole class. 1. In general, how does the lattice energy of an ionic compound depend on the charges and sizes of the ions? 2. Why is an ionic compound unlikely to consist of ions of noble gases? 3. What characteristic charges do atoms in Groups 13 (IIIA) up to 17 (VIIA) of the periodic table have when they form ions? The formation of ionic bonding is influenced by various factors. Some of the major factors are presented below. Ionization energy (IE): The lesser the ionization energy, the greater is the ease of the formation of a cation. Thus, low ionization energy of metallic elements such as alkali and alkaline earth metals favours the formation of an ionic bond. Electron affinity (EA): A higher electron affinity favours the formation of an anion. Generally, the elements having higher electron affinity such as halogens favour the formation of an ionic bond. Thus, low ionization energy of a metal atom and high electron affinity of a non-metal atom facilitate the formation of an ionic bond between them. Lattice energy: When a cation and an anion come closer, they get attracted to each other due to the electrostatic (coulombic) force of attraction. The electrostatic force of attraction between oppositely-charged ions release a certain amount of energy and an ionic bond is formed. If the coulombic attraction forces are stronger, then more energy gets released and a more stable or a stronger ionic bond is formed. Larger lattice energy would favour the formation of an ionic bond. Lattice energy thus is a measure of coulombic attractive force between the combining ions. The lattice energy (U ) of an ionic compound depends directly on the product of the ionic charges (q1 × q2 ), and inversely on the distance (r) between them: ∝ × U q q 1 2 r UNIT 2 75 CHEMISTRY GRADE 11 76 UNIT 2 where q1 and q2 are the charges on +ve and --ve ions respectively, and r is the distance between the charges q1 and q2. Thus, small ions having a higher ionic charge will have a larger lattice energy. If the total energy released is more than that which is absorbed, then the formation of an ionic compound is favoured. Activity 2.7 Answer the following questions in groups, then present your answers to the whole class. 1. Given that the ions in LiF and in MgO are of similar size, which compound has the stronger ionic bonding? 2. The radii of the sodium and potassium ions are 102 pm and 138 pm, respectively. Which compound has stronger ionic attractions, sodium chloride or potassium chloride? Activity 2.8 Discuss the following issues in a group of four and share your responses with the whole class. 1. Comment on the possible formation of the K2+ ion. Why is its formation unlikely? 2. How many electrons does a Se atom have to gain to have a complete octet in its valence shell? 3. Identify the cations that obey the octet rule in the following compounds: a. LiF b. NaCl c. CuO d. FeCl3 Why does the octet rule work well only for the representative metals (Groups IA, IIA) and the non-metals, but not for the transition elements? There are certain exceptions to the octet rule. We will discuss it here with ionic compounds. Ionic Bonds Less than Octet (Central Atom is Deficient of Electrons) Ions of some elements which are near to helium in the periodic table do not obey the octet rule. The tendency of these atoms (H, Li, Be and B) is to attain an arrangement of two electrons like the noble gas He, which is also a stable configuration. The hydride ion (H--), lithium ion (Li+), beryllium ion (Be2+) and boron ion (B3+) are isoelectronic with He. Can you give some compounds that contain less than eight electrons around the central atom? Although atoms with less than an octet may be stable, they will usually attempt to form a fourth bond to get eight electrons. For example BF3 is stable, but it will form BF4 −when possible. More than Octet (18-Electron Rule) The ions of some transition and post-transition elements do not usually obey the octet rule. For transition metals, the 18-electron rule replaces the octet rule, due to the involvement of d orbitals of these atoms. Why don't the atoms of these elements lose electrons to achieve the noble-gas configurations of ns2np6? Note that when these atoms form positive ions, electrons are always lost first from the shells with the highest value of the principal quantum number (n). Consider the electron configurations of the ions of the transition elements iron and zinc and the post-transition elements gallium and tin. Electron Configurations of Iron: 26 Fe: 1s22s22p63s23p64s23d6 26 Fe2+: 1s22s22p63s23p63d6 26 Fe3+: 1s22s22p63s23p63d5 A stable ion of iron with valence shell electron configuration is 3s2 3p6 3d5 which is not isoelectronic with a noble gas. Fe2+ is a well-known stable ion with a valence shell electron configuration 3s2 3p6 3d6 which is not isoelectronic with any of the noble gases. UNIT 2 77 CHEMISTRY GRADE 11 Electron configurations of zinc: 30 Zn: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 = \[Ar\] 4s23d10 30 Zn2+: 1s2 2s2 2p6 3s2 3p6 3d10 = \[Ar\]3d10 Zn2+ is also not isoelectronic with any of the noble gases. Electron configurations of gallium: The post-transition element gallium (Ga) loses electrons first from the 4p orbital and then from the 4s orbital to from a Ga3+ ion as 31 Ga: 1s2 2s2 2p6 3s2 3p6 4s23d104p1 = \[Ar\]4s23d104p1 31 Ga3+: 1s2 2s2 2p6 3s2 3p63d10 = \[Ar\]3d10 On closely examining the electron configurations of Zn2+ and Ga3+, we will realize that ions have completely-filled outer subshells and a noble gas core. Their valance electron configuration can be generally represented as ns2np6nd10. Exercise 2.4 1. From which orbital (s) do the heavier post-transition elements such as Pb and Sn lose electrons? 2. Can you write the electron configurations of Sn, Sn2+, and Sn4+? Check if any of these tin ions are isoelectronic with any of the noble gases. 3. Which group of ions of elements, in general, show more than the octet (18-electron rule)? 78 UNIT 2 Ionic Bonds Experiment 2.1 Investigation of Solubility of Ionic Compounds Objective: To investigate the solubility of NaCl and CuCl2 in polar and non polar solvents Apparatus and Chemicals: Test tube, Bunsen burner, NaCl, CuCl2 , water, ethanol, cyclohexane, and petroleum ether Procedure: 1. Place about 0.5 g each of NaCl and CuCl2 in to two separate test tubes and add about 2.5 mL of water and shake both test tubes quickly. 2. If some residue is left in the test tubes, heat the solutions with a Bunsen burner. SAFETY: Turn off the Bunsen burners as these chemicals are HIGHLY FLAMMABLE! 3. Repeat Step 1 using fresh ethanol, cyclohexane, and petroleum ether solvents, separately. (SAFETY: Do not heat these, as they are highly flammable!) Observations and analysis: Prepare an observation table in your notebook for the solubility of NaCl and CuCl2 in all the four solvents at room temperature and on heating (SAFETY: only heat when water is the solvent) and record the observations. Solvent Water Ethanol Petroleum ether Cyclohexane NaCl CuCl2 Inference/Conclusion Interpret the observation table and give results. Make a generalized statement about the solubility of ionic compounds in polar and non-polar solvents. UNIT 2 79 CHEMISTRY GRADE 11 Experiment 2.2 Thermal behavior of ionic compounds Objective: To study the effect of heat on ionic compounds Apparatus and chemicals: Test tubes, Bunsen burner, test tubes holders, sodium chloride, and copper (II) chloride. Procedure: 1. Take two hard glass test tubes and label them as A and B. 2. Add 0.5 g each of dry sodium chloride crystals and copper (II) chloride in test tubes A and B, respectively. 3. Hold these test tubes with the help of test tube holders. 4. Heat the tubes at the same time in the Bunsen burner flame, first slowly and then strongly. Shake them a little when heating. Caution: Chlorine gas is produced, this would be best carried out in a fume cupboard. Observations and analysis: A. Do the crystals melt? B. Do they have high or low melting points? Experiment 2.3 Electrical Conductivity of Ionic Compounds Objective: To test the electrical conductivity of molten compounds Apparatus and chemicals: 9-volt battery, 6-watt bulb with a bulb holder, conducting wires include: beaker, tripod, wire gauze, clamp and stand, two carbon rods, lead (II) iodide or lead (II) bromide, copper (II) , sulphate or sodium chloride. Caution: Lead compounds are harmful. Copper sulphate is corrosive and an irritant. Procedure A: 1. Connect the circuit as shown in Figure 2.2 in which a 9 volt DC is connected via a bulb. 80 UNIT 2 Ionic Bonds 2. Using about 2 cm depth of lead (II) iodide or lead (II) bromide in a small beaker, test the conductivity of the crystals. Do not throw the crystal away and be careful not to contaminate the sample because you will reuse it later. (Note: A fairer test of the solid compound would be to use a lump of the compound rather than its powder). 3. Test the conductivity of large crystals of copper (II) sulphate and sodium chloride if any of them are available in your laboratory. Procedure B: 1. Now heat the same lead (II) iodide or lead (II) bromide (used in the above experiment) in a beaker on a tripod and wire gauze or in a boiling tube supported by a clamp and stand until it melts. 2. Test the conductivity of the molten compound by dipping the carbon electrodes (carbon rods) into the molten compound as shown in the Figure 2.2. 3. Repeat for the other compounds. Figure 2.2: Conductivity in the molten state Observations and analysis: A. What did you observe? B. Which of the compounds (molten or solid) conduct electricity? Why? C. All compounds contain at least two elements. Examine the names of those compounds which conduct electricity when in the molten state and decide which groups of the components of these compounds belong. Name the type of bonding that exists in the compounds used. UNIT 2 81 CHEMISTRY GRADE 11 Experiment 2.4 Electrical Conductivity of Ionic Compounds Objective:To test the electrical conductivity of the aqueous solutions of some common ionic compounds Apparatus and chemicals: 9-volt battery, 6-watt bulb with a bulb holder, conducting wires, two carbon rods, H2 O, lead (II) iodide, NaCl. Caution: Lead compounds are harmful. Procedure: 1. Dissolve the compound in 50 mL of water in two separate beakers. 2. Connect the same circuit you used in Experiment 2.3 and test the conductivity of each aqueous solution (Figure 2.3). Observations and analysis: A. Predict what happens at the electrodes based on the type of the compound used for the experiment. B. Do you expect the same product(s) at the electrodes when electricity passes through molten and aqueous solutions of the compounds? Figure 2.3: Electrochemical cell showing the conductivity of an aqueous solution 82 UNIT 2 Covalent Bonds and Molecular Geometry Group Assignment 2.1 2.3 Based on the observations and results of Experiments 2.1 to Experiment 2.4, give a summary of the properties of ionic compounds. At the end of this section, you will be able to: ) define covalent bonding ) explain the formation of covalent bonding using examples ) draw Lewis structures or electron dot formulas of some covalent molecules ) illustrate the formation of coordinate covalent bonding using examples ) draw resonance structures of some covalent molecules and polyatomic ions ) discuss the exceptions to the octet rule in covalent bonding ) distinguish between polar and non-polar covalent molecules ) describe the properties of covalent molecules ) carryout an activity to investigate the effects of heat, electricity and some solvents on covalent compounds (naphthalene, graphite, iodine and ethanol) ) describe the valence shell electron pair repulsion theory (VSEPR) ) distinguish between the bonding pairs and nonbonding pairs of electrons ) describe how electron pair arrangements and shapes of molecules can be predicted from the number of electron pairs ) explain why double bonds and lone pairs cause deviations from ideal bond angles ) explain the term dipole moment with the help of a diagram ) describe the relationship between dipole moment and molecular geometry ) describe how bond polarities and molecular shapes combine to give molecular polarity UNIT 2 83 CHEMISTRY GRADE 11 84 UNIT 2 ) predict the geometrical shapes of some simple molecules on the bases of hybridization and the nature of electron pairs ) construct models to represent shapes of some simple molecules ) define intermolecular forces ) name the different types of intermolecular forces ) explain dipole-dipole interactions ) give examples of dipole-dipole interaction ) define hydrogen bonding ) explain the effect of hydrogen bond on the properties of substances ) give reasons why hydrogen bonding is stronger than ordinary dipole-dipole interactions ) explain dispersion (London) forces ) give examples of dispersion forces ) predict the strength of intermolecular forces for a given pair of molecules. Formation of Covalent Bonding Activity 2.9 Discuss each of the following in your group and present your ideas to the whole class. 1.Can two non-metal atoms combine together? How? 2.Why doesn't a hydrogen atom lose an electron and form an ionic bond? 3.Give one main difference between ionic and covalent bonds. 4.Explain the formation of polar covalent and coordinate covalent bonds. Consider the formation of a hydrogen molecule (H2 ). When two isolated hydrogen atoms come close together, electrostatic interactions begin to develop between them. The two positively charged nuclei repel each other, and the two negatively charged electrons repel each other, but each nucleus attracts both electrons and each electron is attracted to both nuclei (Figure 2.4). If the attractive forces are stronger than the repulsive forces, a covalent bond is formed, with the two atoms held together and the two shared electrons occupying the region between the nuclei. Covalent Bonds and Molecular Geometry Figure 2.4: Attractive and repulsive forces in a covalent H-H bond formation The magnitudes of the various attractive and repulsive forces between nuclei and electrons in a covalent bond depend on how close together the atoms are. What happens if the hydrogen atoms are too far apart? What if the hydrogen atoms are too close together? There is an optimum distance between nuclei, called the bond length, where net attractive forces are maximized and the H−H molecule is most stable. In the H2 molecule, the bond length is 74 pm. On a graph of energy versus internuclear distance, the bond length is the H−H distance in the minimum energy. This is the most stable arrangement (Figure 2.5). Figure 2.5: Covalent bond formation in H2 UNIT 2 85 CHEMISTRY GRADE 11 Therefore, a covalent bond is formed when a pair of electrons is shared between two atoms. Formation of a bond always results in greater electron density between the nuclei. Some examples of covalent molecules are HCl, H2 S, C2 H4 , N2 , CCl4 , BCl3 , H2 O, NH3 , SO2 , PCl5 , O3 , etc. Generally, substances that contain covalent bonds are called molecules. Representation of Covalent Bonds (Drawing Lewis Structures) The representation of covalent bonding through Lewis symbols and shared electron pairs is called a Lewis structure. The Lewis structure for a hydrogen molecule formed from hydrogen atoms is: By sharing two electrons in a covalent bond, each hydrogen effectively has one electron pair and the stable, 1s2 electron configuration of helium. Similarly, fluorine has seven valence electrons, and an electron-dot structure for the F2 molecule shows how a covalent bond can form: The shared pairs of electrons in a molecule are called bonding pairs. The other electron pairs that stay with on atom and are not shared are called non-bonding pairs or lone pairs. Do the fluorine atoms in a fluorine molecule (F2 ) obey the octet rule? Note that in molecules such as O2 , N2 , and many others, the atoms share more than one pair of electrons, leading to the formation of multiple covalent bonds. 86 UNIT 2 Covalent Bonds and Molecular Geometry For example, consider the formation of oxygen molecule (O2 ) from the oxygen atoms. Can you give the atomic number, number of protons and electrons of an oxygen atom? The electronic configuration of oxygen atoms is 2,6. Now each oxygen atom needs two electrons to complete its octet. The two oxygen atoms share two electrons and complete their octet: The 4 electrons (or 2 pairs of electrons) which are shared between the two oxygen atoms. So these two pairs of shared electrons can be represented by two bonds between the oxygen atoms. An oxygen molecul can be sown as follow: The two oxygen atoms are said to be bonded together by two covalent bonds. Such a bond consisting of two covalent bonds is also known as a double bond. Steps to Writing a Lewis Formula: We can write the Lewis formula for a covalent compound of known geometry by using the following steps. 1. Determine the total number of valence electrons. The total number of electrons for a molecule is the sum of the valence electrons for each atom. For a polyatomic anion, which has one or more extra electrons, add one electron for each unit of negative charge. For a polyatomic cation, which is missing one or more electrons, subtract one electron for each unit of positive charge. 2. Write the skeletal structure. The most electropositive atom usually occupies the central position. Connect bonded atoms with an electron-pair bond (a dash). Hydrogen is an exception; it is always a terminal atom, even when bonded to a more electronegative atom. 3. Place electron pairs around terminal atoms so that each (except hydrogen) has an octet. Assign any remaining electrons as lone pairs around the central atom. 4. If at this stage, a central atom has fewer than eight electrons, a multiple bond(s) is likely. Move one or more lone-pair of electrons from a terminal atom(s) to a region between it and the central atom to form a double or a triple bond. UNIT 2 87 CHEMISTRY GRADE 11 Example 2.2 Write the Lewis structure for nitrogen trifluoride (NF3 ) in which all three F atoms are bonded to the N atom. Solution: We follow the steps for writing Lewis structures. Step 1: The outer-shell electron configurations of N and F are 2s22p3 and 2s22p5, respectively. Thus, there are 5 + (3 ´ 7), or 26, valence electrons to account for in NF3. Step 2: The N atom is less electronegative than F, so in the skeletal structure of NF3 , N goes in the center with the more electronegative F atoms bonded to it. F N F F Step 3: We draw a single covalent bond between N and each F, and complete the octets for the F atoms. We place the remaining two electrons on N: F N F F Because this structure satisfies the octet rule for all the atoms, step 4 is not required. 88 UNIT 2 Covalent Bonds and Molecular Geometry Example 2.3 Write the Lewis structure for the carbonate ion (CO32⁻). Solution: We follow the steps for writing Lewis structures and note that this is an anion with two negative charges. Step 1: The outer shell electron configurations of C and O are 2s22p2 and 2s22p4, respectively, and the ion itself has two negative charges. So, the total number of electrons is 4 + (3 ´ 6) + 2, or 24. Step 2: We can deduce the skeletal structure of the carbonate ion by recognizing that C is less electronegative than O. Therefore, it is most likely to occupy a central position as follows: O O C O Step 3: We draw a single covalent bond between C and each O and comply with the octet rule for the O atoms: O O C O This structure shows all 24 electrons. Step 4: Although the octet rule is satisfied for the O atoms, it is not for the C atom. Therefore, we move a lone pair from one of the O atoms to form another bond with C. Now the octet rule is also satisfied for the C atom: O O C 2 O UNIT 2 89 CHEMISTRY GRADE 11 90 UNIT 2 Exercise 2.5 1. Determine the total number of valence electrons for the following: a. CO2 b. SO4 2− c. NH4 + d. N2 O4 2. Write a Lewis structure of a. nitrogen trichloride, NCl3 b. chlorate ion, ClO3-- c. phosphonium ion, PH4 + 3. Draw a Lewis structure for SF4 and HCOOH (formic acid). 4. Which of the following atoms O, He, F, H, and P cannot be used as a central atom in a Lewis structure? Explain. 5. Write a Lewis structure for carbonyl sulphide, COS. Coordinate-Covalent Bonding Activity 2.10 Form a group and discuss the following issues, then present your responses to the whole class. 1. Consider ammonia and boron trifluoride molecules; identify which has an incomplete octet and which one has lone pair electrons on the central atom of the molecule. 2. Can these two molecules react with each other to complete the octet of an atom with an incomplete octet? How? Most covalent bonds form when two atoms each contribute one electron. However, bonds can also form when one atom donates both electrons (a lone pair) to another atom that has a vacant valence orbital. The ammonium ion (NH4 +), for example, forms when the lone-pair electrons from the nitrogen atom of ammonia, NH3 , bond to H+. Such kind of bonds is called coordinate covalent bonds or dative bonds. Some examples of molecules which contain coordinate-covalent bonds include NH3 , BCl3 , H3 O+ and AlCl4 −. Covalent Bonds and Molecular Geometry Resonance Structures Activity 2.11 Form a group and discuss the following questions and present your responses to the whole class. 1. How many structural formulas can you write for each of the following? a. H2 O b. NH4 + c. CO3 2− d. C6 H6 (Benzene) 2. What is the difference between a resonance structure and a resonance hybrid? Sometimes more than one valid Lewis structure is possible for a given molecule. For example, two Lewis structures are possible for ozone ( O3 ): O O O 1.278Å and O O O A B In formula A, the oxygen--oxygen bond on the left is a double bond and the oxygen-- oxygen bond on the right is a single bond. In formula B, the situation is just the opposite. These are not two different O3 molecules, just different Lewis structures for the same molecule. However, neither one of these two Lewis structures accounts for the known bond lengths in O3 (1.278 Å). O O O Since both the bonds are identical, which one is a double bond? Bond length and bond energy measurements indicate that the two oxygen-oxygen bonds in O3 are identical, with properties that lie between those of an O−O bond and an O=O bond, something like a "one-and-a-half" bond. The molecule is shown more correctly with two Lewis structures, called resonance structures (or resonance forms), and a two headed resonance arrow (↔) between them. O O O O O A B O 1.278Å UNIT 2 91 CHEMISTRY GRADE 11 Unfortunately, this notation can be misinterpreted. It does not mean that the ozone molecule flips back and forth between two forms. There is only one ozone molecule. The double-headed arrow means that you should form a mental picture of the molecule by joining the various resonance formulas. Resonance structures have the same relative placement of atoms but different locations of bonding and lone electron pairs. Our need for more than one Lewis structure to depict the ozone molecule is the result of electron-pair delocalization. You can convert one resonance form to another by moving lone pairs to bonding positions, and vice versa: O O O O O A B O O ≡ O O Aresonance hybrid This description of O3 is called resonance, in which two or more possible Lewis structures can be written and the true structure is a composite or hybrid of them. The different structures used to represent the molecule or ion are called resonance forms or resonance contributors or resonance structures. Example 2.4 Write two equivalent Lewis structures for the nitirite ion, NO2--. Describe its resonance hybrid structure. Solution: In NO2 −there are 5 + 2(6) + 1 = 18 valence electrons. Indicating the single bonds gives the structure N O O The remaining 14 electrons (18-4) can be distributed to produce these structures: N O O N O O The electronic structure of the molecule is correctly shown by the average of the two. There are two equivalent N\-\--O bonds, each one intermediate between a single and a double bond. 92 UNIT 2 Covalent Bonds and Molecular Geometry Exercise 2.6 1. Write three equivalent structures for the SO3 molecule that obey the octet rule. 2. Draw Lewis structures of all the important resonance forms of each of the following: a. NO2 F (N central) b. HNO3 Exceptions to the Octet rule in Covalent Bonding Activity 2.12 c. NO2 − Discuss the following issues in groups and present your answers to the whole class. 1. What requirements are needed for an atom to expand its valence shell? 2. Which of the following atoms can expand its valance shell: F, S, H, Al? 3. Identify the atoms in PCl3 , PCl5 , OF2 , BF3 , and SiF4 which do not obey octet rule. Although many Lewis structures follow the octet rule, there are exceptions. There can be categorized into three groups: i. Less than octet (central atom is deficient of electrons): Electron-deficient compounds are compounds in which an element has an incomplete octet. The central atoms of such molecules have fewer than eight electrons (below octet). This group consists of molecules containing central atoms from Group IIA and IIIA. For example, BeCl2 , BF3 and AlCl3 , whose Lewis formulas are shown: Cl Be Cl 4 electrons around Be F B F 6 electrons around B F Cl Al Cl Cl 6 electrons around Al ii. More than octet (central atom has excess of electrons): Elements with an expanded valence (also called an expanded octet) have more than 8 electrons (often 10 or 12) in a Lewis structure. Elements in the third period and below, such as phosphorus, sulfur, and bromine, often have an expanded valence because of their larger radii (when compared to the second-row elements) and the availability of UNIT 2 93 CHEMISTRY GRADE 11 empty d orbitals in the valence shell. For example PF5 , SF6 and XeF4 : iii. Molecules containing an odd number of electrons: Even if stable molecules of this kind (free radicals) are rare, they do exist. Free radicals can also be electron-deficient compounds. Some examples are ClO2 , NO and NO2 having 19, 11 and 17 valence electrons, respectively. For example, the best way to suggest the Lewis-like structure for the NO molecules is: N O Exercise 2.7 1. Suggest the Lewis-like structure for NO2 and ClO2 molecules. 2. The following do not obey the octet rule. Draw a Lewis structure for each one and state the type of octet rule exception: a. BH3 b. AsF4-- c. SeCl4 d. PF6-- Polar and Non-Polar Covalent Molecules Activity 2.13 each of the following: Discuss the following questions in groups and present your responses to the whole class. 1. Describe the trend in electronegativity as "increases" or "decreases" for a. from B to F b. from Mg to Ba c. from F to I 2. What does the word "dipole" refer to? 3. How does electronegativity relate to bond polarity? 94 UNIT 2 Covalent Bonds and Molecular Geometry When the atoms are alike in a covalent bond, as in the case of the H−H bond of H2 , the bonding electrons are shared equally. That is, the electrons spend the same amount of time near each nucleus. The bond formed is called a non-polar covalent bond. In a covalent bond between atoms of different electronegativities, there is an unequal sharing of an electron pair and the electrons spend more of their time around the more electronegative atom. Such a bond is said to be polar covalent bond. The H--F bond is a polar bond because fluorine is more electronegative than the hydrogen atom. The bonding electrons spend more time near the fluorine atom than the hydrogen atom. The polar nature of a bond is shown like this: Hδ+ -- Fδ In the representation the δ+ and δ-- (read "delta plus" and "delta minus") signify that one end (H) is partially positive and the other end (F) is partially negative. The term "partial charge" implies something less than the full charges of the ions that would result from complete electron transfer. The bond polarity can also be shown by drawing an arrow so that the head points toward the negative end (F) of the bond and the crossed tail indicates the positive end (H). H F For a diatomic molecule having a polar covalent bond, such as HF, we can describe a quantity called the dipole moment, which is a vector sum of the bond moments in a molecule. Bond moment is a measure of polarity of a diatomic covalent bond. The dipole moment (µ) is defined as the product of the magnitude of the charge (δ) at either end of the dipole multiplied by the distance (d) that separates the charge. µ= δ× d The SI-unit of dipole moment is coulomb-metre (C.m). Dipole moments are often expressed in the non-SI unit debye (D), where 1D = 3.33564 × 10--30 Cm. A very polar molecule is one with a large dipole moment, while a non-polar molecule will have a zero-dipole moment. Note: For a diatomic molecule, the bond moment is the dipole moment. The dipole moment of a polyatomic molecule (three or more atoms) depends on the geometry of the molecule. If the bond moments are equal in magnitude but opposite in direction, then they will cancel each other and the resultant dipole moment will be zero, provided that the vector sum of the bond moments is zero. UNIT 2 95 CHEMISTRY GRADE 11 Exercise 2.8 1. Both CO2 and BCl3 have zero dipole moments, but the C = O and B -- Cl bond moments are not zero. Explain. 2. Compare the dipoles in the covalent molecules HF, HCl, HBr, and HI. Explain your answer. 3. For each of the following bonds, indicate the positive end with δ+ and the negative end with δ−. Draw an arrow to show the dipole for each. a. N and F b. Si and Br c. C and O d. P and Br e. N and P Properties of Covalent Compounds Carbon tetrachloride is a covalent compound. Would you expect it to conduct electricity? Experiment 2.5 Investigating the melting point of covalent compounds Objective: To investigate the melting point of naphthalene. Apparatus and chemicals: Thermometer, stirrer, beaker, melting point tube, naphthalene, glycerine, Bunsen burner. Procedure: 1. Set up the apparatus shown in Figure 2.6. 2. Place a small amount of naphthalene in the melting point tube. 3. Attach the tube to the side of the thermometer (the liquid in the beaker will hold the tube in position). 4. Heat the beaker slowly. When the naphthalene melts, record the reading on the thermometer. 96 UNIT 2 Covalent Bonds and Molecular Geometry Figure 2.6: Determination of melting point of naphthalen Observations and analysis: Check your result by referring to standard melting point values. Discuss possible point. Experiment 2.6 reasons for the differences in melting point compared to the standard melting The effects of electricity and some solvents on covalent compounds Objective: To test the effects of electricity and some solvents on covalent compounds. Apparatus and chemicals: Beakers, stirrer, test tubes, table sugar, cyclohexane, petroleum ether, water, alcohol, naphthalene. Procedure A: Solubility test 1. Place about 0.5 g or 5 drops each of naphthalene, table sugar, and ethanol in three separate test tubes and add about 2.5 mL of distilled water and shake the test tubes vigorously. 2. See a change in the amount of naphthalene, and table sugar added in each case; if some residue remains in the test tube/s, heat it on a Bunsen burner. UNIT 2 97 CHEMISTRY GRADE 11 3. Observe if a separating layer is formed between the ethanol and water, it indicates ethanol is not soluble in water. 4. Repeat the experiment using cyclohexane and petroleum ether solvents, separately (Caution: do not heat it.) Procedure B: Conductivity test 1. Connect the circuit as shown in Figure 2.2, (Experiment 2.3) in which a 9-volt DC is connected via a bulb. 2. Test the conductivity of table sugar. 3. Dissolve a spoonful of table sugar in 50 mL of distilled water in a beaker. Test the resulting solution for electrical conductivity using the setup of of Figure 2.3, in Experiment 2.4. Observations and analysis: Record your observations on your notebook. Copy and complete the table below. Property Substance Naphthalene Table sugar Ethanol Solubility in water Solubility in non polar solvent Results and Discussion: Effect of electricity 1. Draw your conclusion from the above observations. 2. Do these substances conduct electricity either in the solid state or in the molten (liquid) state? 3. Is there any chemical reaction that takes place at the electrodes? 98 UNIT 2 Covalent Bonds and Molecular Geometry Activity 2.14 Form a group of four and dscuss the following questions. Then, present your responses to the whole class. 1. Why does ammonia, NH3 , dissolve in water but methane, CH4 , does not? 2. Distilled water is very poor electrical conductor, whereas rain water can conduct electricity. Explain why. 3. Unlike ionic compounds, covalent compounds do not conduct electricity in molten state. Why is that so? Unlike any ionic compound, many of the covalent compounds are found in gaseous state at room temperature. Consider methane (CH4 ), the simplest compound between carbon and hydrogen, which have comparable, intermediate electronegativity. It is a gas at room temperature. Cooling methane to a low temperature condenses it first to a liquid and then to a solid. Unlike melted ionic compounds, liquid covalent compounds do not conduct electricity. Therefore, liquid methane does not conduct electricity. Covalent compounds are molecular substances. They have low melting and boiling points. Most covalent compounds are soluble in non-polar solvents. Generally: Covalent compounds exist as separate molecules because electrically neutral atoms form them and the forces of attraction between these molecules are relatively weak. Due to weak intermolecular forces, many covalent molecules or covalent compounds are liquids or gases at room temperature. However, some covalent molecules like iodine are solids at room temperature. Liquid: H2 O, Br2 Gas: CO2 , H2 , Cl2 , NH3 Covalent compounds are volatile. Generally, they have low melting points and boiling points. Covalent compounds are generally insoluble in water. Most covalent compounds are soluble in non-polar solvents. Non-polar covalent compounds are non-electrolytes because they do not conduct electricity. UNIT 2 99 CHEMISTRY GRADE 11 Molecular geometry is the three-dimensional arrangement of atoms in a molecule. There is a simple procedure that enables us to predict the geometry of a molecule or ion if we know the number of electrons surrounding a central atom in its Lewis structure. The basis of this approach is the assumption that electron pairs in the valence shell of an atom repel one another. Valence Shell Electron Pair Repulsion (VSEPR) Theory In a covalent bond, a pair of electrons, often called the bonding pair, is responsible for holding two atoms together. However, in a polyatomic molecule, where there are two or more bonds between the central atom and the surrounding atoms, the repulsion between electrons in different bonding pairs causes them to remain as far apart as possible. The geometry that the molecule ultimately assumes (as defined by the positions of all the atoms) minimizes the repulsion. This approach to the study of molecular geometry is called the valence-shell electron-pair repulsion(VSEPR)model. The basic principle of VSEPR theory is that the pair of valence-shell electron pairs around the central atom arrange as far away from one another as possible to minimize electron-pair repulsions. The arrangements that best minimize repulsions naturally depend on the number of electron sets. Once we have counted the number of electron groups surrounding the central atom, we can determine its specific shape from the number of atoms bonded to the central atom. Electron Pair Arrangement and Molecular Shape The electron pair arrangement is defined by the orientation/distribution of pairs of electrons, both bonding and non-bonding (lone pair), around the central atom. On the other hand, the molecular shape is defined by the relative positions of the atomic nuclei. Molecular shapes that occur when all the surrounding electrons are bonding differ from molecular shapes when some of the electrons are non-bonding. Thus, the same electron set arrangement can give rise to different molecular shapes. For the prediction of geometrical shapes of molecules with the help of VSEPR theory, it is convenient to divide molecules into two categories as molecules in which the central atom has no lone pair and molecules in which the central atom has one or more lone pairs. 100 UNIT 2 Covalent Bonds and Molecular Geometry i. molecules in which the central atom has no lone pair Consider molecules that contain atoms of only two elements, A and B, of which A is the central atom. These molecules have the general formula ABx, where x is an integer 2, 3,.... (If x = 1, we have the diatomic molecule AB, which is linear by definition.) In the vast majority of cases, x is between 2 and 6. In the compounds of AB2 , AB3 , AB4 , AB5 and AB6 , the arrangements of electron pairs and the B atoms around the central atom A are: linear, trigonal planar, tetrahedral, trigonalbipyramidal and octahedral, respectively. Such arrangements can be seen in molecules such as BeCl2 (AB2 ), BF3 (AB3 ), CH4 (AB4 ) and PCl5 (AB5 ) as shown below (Figure 2.7) by their ball and stick models. Figure 2.7: The shapes of molecules in which the central atom has no lone pair In the case of multiple (double or triple) bonds the counting is not different, because a double bond is considered as one set and a triple bond as another set of electrons. For example, in the case of CO2 , O=C=O, there are two sets of electrons around the central atom, acetylene, C2 H2 , H--C≡C--H, two sets of electrons around each of the central atoms. ii. molecules in which the central atom has one or more lone pairs In such molecules there are three types of repulsive forces---those between bonding pairs, those between lone pairs, and those between a bonding pair and a lone pair. In general, according to the VSEPR model, the repulsive forces decrease in the following order: Lone pair vs lone pair \> lone pair vs bonding pair \> bonding pair vs bonding pair Why is that repulsions exerted by lone pair is greater than that of bonding pair? To keep track of the total number of bonding pairs and lone pairs, we show molecules with lone pairs as ABx Ey , where A is the central atom, B is a surrounding atom, and E is a lone pair on A. Both x and y are integers; x = 2, 3,... , and y = 1, 2,.... Thus, the values of x and y indicate the number of surrounding atoms and number of lone pairs on the central atom, respectively. The simplest such molecule would be a triatomic molecule with one lone pair on the central atom and the formula is AB2 E. UNIT 2 101 CHEMISTRY GRADE 11 AB2 E: Sulfur Dioxide (SO2 ) The Lewis structure of sulfur dioxide is: O S O Because VSEPR treats double bonds as though they were single, the SO2 molecule can be viewed as consisting of three electron pairs on the central S atom. Of these, two are bonding pairs and one is a lone pair. Thus, the overall arrangement of three electron pairs is trigonal planar. But because one of the electron pairs is a lone pair, the SO2 molecule has a "bent" shape. S O O Because the lone-pair versus bonding-pair repulsion is greater than the bonding-pair versus bonding-pair repulsion, the two sulfur-to-oxygen bonds are pushed together slightly and the OSO angle is less than 120°. AB3 E: Ammonia (NH3 ) The ammonia molecule contains of three bonding pairs and one lone pair: H N H H The overall arrangement of four electron pairs is tetrahedral. But in NH3 one of the electron pairs is a lone pair, so the geometry of NH3 is trigonal pyramidal. Because the lone pair repels the bonding pairs more strongly, the three NH bonding pairs are pushed closer together: N H H AB2 E2 : Water (H2 O) H A water molecule contains two bonding pairs and two lone pairs: The overall arrangement of the four electron pairs in water is tetrahedral, the same as in ammonia. However, unlike ammonia, the HOH angle is 104.5°. The geometry of H2 O is bent. Why do you think this is? 102 UNIT 2 Covalent Bonds and Molecular Geometry O H AB4 E: Sulfur Tetrafluoride (SF4 ) The Lewis structure of SF4 is: F H S F F F The central sulfur atom has five electron pairs whose arrangement is trigonal bipyramidal. In the SF4 molecule, however, one of the electron pairs is a lone pair, so the molecule must have one of the following geometries: In (a) the lone pair occupies an equatorial position, and in (b) it occupies an axial position. The axial position has three neighboring pairs at 90° and one at 180°, while the equatorial position has two neighboring pairs at 90° and two more at 120°. The repulsion is smaller for (a), and indeed (a) is the structure observed experimentally. Guidelines for Applying VSEPR Model Here are some guidelines for applying the VSEPR model: 1. Write the Lewis structure of the molecule, considering only the electron pairs around the central atom (that is, the atom that is bonded to more than one other atom). 2. Count the number of electron pairs around the central atom (bonding pairs and lone pairs). Treat double and triple bonds as though they were single bonds. If there is more than one central atom, treat each central atom separately. 3. Use the VSEPR geometry to predict the shape of the molecule. 4. In predicting bond angles, note that a lone pair repels another lone pair or a bonding pair more strongly than a bonding pair repels another bonding pair. UNIT 2 103 CHEMISTRY GRADE 11 104 UNIT 2 Molecular Shape and Molecular Polarity Activity 2.15 Discuss the following questions in a group of four and present your answers to the whole class. 1. Why is it useful to know the polarities of molecules? 2. Are all dissymmetric molecules polar? 3. Under what conditions do bonds of a compound become non-polar? Many aspects of molecule's chemical behaviour can be understood if we know the geometry (shape) of a substance. Molecular shape affects many properties of the molecule like molecular polarity, which in turn influence melting and boiling points, solubility, and even reactivity. Molecular polarity is created by molecules with a net imbalance of charge. In molecules with more than two atoms, both shape and bond polarity determine the molecular polarity. Bond Polarity and Dipole Moment Activity 2.16 Discuss the following concepts in groups, and share your ideas with the rest of the class. 1. How do we indicate a bond dipole when we draw the structure of a molecule? 2. Use a drawing to show why the SO2 molecule is polar. 3. Water has a resultant dipole moment of 6.23×10--30 C m (1.87D). Explain why this fact proves that the H2 O molecule must have a bent shape. Diatomic molecules containing atoms of different elements (for example, HCl, CO, and NO) have dipole moments and are called polar molecules. Diatomic molecules containing atoms of the same element (for example, H2 , O2 , and F2 ) are examples of non-polar molecules because they do not have dipole moments. For a molecule made up of three or more atoms, both the polarity of the bonds and the molecular geometry determine whether there is a dipole moment. Polar bonds do not necessarily lead to polar molecules. For example, the large electronegativity difference between carbon Covalent Bonds and Molecular Geometry and oxygen makes each C−O bond quite polar. However, because carbon dioxide (CO2 ) has a linear shape, and its bonds are directed 180° from each other, there is no net dipole moment (µ); µ= 0 D. Another molecule with identical atoms bonded to the central atom is water. Unlike carbon dioxide, water has a significant dipole moment (µ= 1.87 D). In each O--H bond, electron density is pulled toward the more electronegative O atom, but the bond polarities do not counter-balance each other, because the water molecule is V-shaped. Instead, the bond polarities partially reinforce each other, and the oxygen end of the molecule is considerably more negative than the other end. Predict whether ammonia molecule is polar or non-polar and show the direction of bond dipoles and the overall molecular dipole. Carbon dioxide and water demonstrate how molecular shape influences polarity. When two or more different molecules have the same shape, the nature of the atoms surrounding the central atom can have a major effect on the polarity of a molecule. Consider tetrachloromethane (CCl4 ) and trichloromethane (CHCl3 ), two AB4 type molecules: tetrahedral shape with different polarities. In CCl4 , the surrounding atoms are all Cl atoms. Although each C--Cl bond is polar, the molecule is non-polar because the individual bond polarities counter-balance each other. In CHCl3 , an H atom substitutes for one of the Cl atoms, disturbing the balance and giving chloroform a significant dipole moment. UNIT 2 105 CHEMISTRY GRADE 11 If you consider the two constitutional isomers of dichloroethene (C2 H2 Cl2 ), they have the same molecular formula. However, they have different physical and chemical properties. VSEPR theory predicts that all the nuclei lie in the same plane with a trigonal planar molecular shape around each carbon atom. resultant dipole moment Cl C C Cl H H cis-dichloroethylene µ =1.89 D Cl C C H Cl H trans-dichloroethylene µ =0D The trans isomer has no dipole moment (µ= 0 D) because the C -- Cl bond polarities balance each other. In contrast, the cis-isomer is polar (µ= 1.89 D) because the bond dipoles partially reinforce each other, with the molecular dipole pointing between the Cl atoms. Exercise 2.9 1. Determine the molecular shape and ideal bond angles in: a. COCl2 b. PCl3 c. SF2 d. SO4 2 e. CS2 2. When is the molecular shape and the electron-set arrangement the same? 3. Arrange the following AFn species in order of increasing F--A--F bond angles. BF3 , BeF2 , CF4 , NF3 , OF2 4. In the gas phase, phosphorus pentachloride exists as separate molecules. In the solid phase, however, the compound is composed of alternating PCl4 + and PCl6 − ions. How does the molecular shape change as PCl5 solidifies? How does the angle change? 5. For molecules of general formula ABn (where n \> 2), how do you determine if a molocule is polar? 106 UNIT 2 Covalent Bonds and Molecular Geometry Activity 2.17 Form a group of four and discuss the following questions. Then, present your responses to the whole class. 1. Why does a polar liquid generally have a higher normal boiling point than a non-polar liquid of the same molecular mass? 2. State the reason why CH4 is a gas at room temperature whereas H2 O is a liquid. 3. Only one of these substances is a solid at standard temperature and pressure (STP): C6 H5 COOH, CH3 (CH2 )8 CH3 , CH3 OH, (CH3 CH2 )2 O. Which do you think it is and why? 4. Only one of these substances is a gas at STP: NI3 , BF3 , PCl3 , and CH3 COOH. Which do you think it is and why? There are two types of force that hold matter together. These are intramolecular and intermolecular forces. Intramolecular force is a chemical bond (ionic, covalent or metallic) that exists within a particle (molecule or polyatomic ion) and affects the chemical property of the species. Intermolecular force are those bonds that hold particles (ions or molecules) together. A glass of water for example, contains many molecules of water. These molecules are held together by intermolecular forces, whereas the intramolecular forces hold the two hydrogen atoms to the oxygen atom in each molecule of water. Intermolecular forces are relatively weak as compared to intramolecular forces, because they typically involve lower charges that are farther apart. However, the strength of the intermolecular forces is important because they affect physical properties of the species such as melting point and boiling point. UNIT 2 107 CHEMISTRY GRADE 11 Activity 2.18 Discuss the following in a group of four, and share your answers with the rest of the class. 1. Name some intermolecular forces. 2. Explain why cis-1,2-dichloroethene has a higher boiling point than trans 1,2-dichloroethene. 3. Which type of compound exhibit hydrogen bonding? 4. Name the factors on which the strength of dispersion forces of the particles depends. 5. Ice is lighter than water. Why? Three types of attractive force are known to exist between neutral molecules: dipole dipole forces, London (or dispersion) forces, and hydrogen bonding forces. The term van der Waals forces are a general term for those intermolecular forces that include dipole--dipole and London forces. Van der Waals forces are the weak attractive forces in a large number of substances, including Cl2 , and Br2. Dipole-Dipole forces Dipole-dipole forces act between the molecules possessing permanent dipole. When polar molecules are brought near one another, their partial charges act as tiny electric f ields that orient them and give rise to dipole-dipole forces; the partially positive end of one molecule attracts the partially negative end of another. Ends of the dipoles possess "partial charges" and these charges are shown by Greek letter delta (δ). This interaction is stronger than the London forces but is weaker than ion-ion interaction because only partial charges are involved. The attractive force decreases with the increase of distance between the dipoles. 108 UNIT 2 Covalent Bonds and Molecular Geometry Hydrogen Bonding Polar molecules containing hydrogen atoms bonded to highly electronegative atoms of nitrogen, oxygen, or fluorine form especially strong dipole--dipole attractions. This type of attraction, called a hydrogen bond, occurs between the partially positive hydrogen atom in one molecule and the partially negative nitrogen, oxygen, or fluorine atom in another molecule. The atom sequence that leads to an H bond (dotted line) is indicated below. Hydrogen bonds are the strongest type of attractive forces between polar covalent molecules. What is the difference between hydrogen bonds and chemical bonds? Example 2.5 Some examples of H-bonding are given below. F H δ− δ− δ− δ− δ− δ− δ+ δ+ δ+ O O H N N H F Dispersion or London Forces Usually, the electrons in a non-polar covalent molecule are distributed symmetrically. However, the movement of the electrons may place more of them in one part of the molecule than another, which forms a temporary dipole. These momentary dipoles align the molecules so that the positive end of one molecule is attracted to the negative end of another molecule. UNIT 2 109 CHEMISTRY GRADE 11 These weak attractive forces between molecules resulting from the small, instantaneous dipoles that occur because of the varying positions of the electrons during their motion about nuclei are called London forces or dispersion forces. They are present between all particles (atoms, ions, and molecules). Although dispersion forces are very weak, they make it possible for non-polar molecules like CO2 , Cl2 , noble gases, etc., to form liquids and solids. London forces tend to increase with molecular weight, why do you think this is? For molecules of about the same molecular weight, the strength of the dispersion forces is affected by molecular geometry (shape). Shapes that allow more points of contact have more area over which electron clouds can be distorted, so stronger attractions result. This is probably the reason why n-pentane has a higher boiling point (36 °C) than its isomers 2-methylbutane (28 °C), and 2, 2-dimethylpropane (9.5 °C). Exercise 2.10 1. In which of the following substances do hydrogen bonds occur? Explain by using diagrams. a. CH4 b. CH3 CH2 OH c. R CH NH2 C O OH d. H3C O bonding) would you expect in the following substances? a. CHCl3 CH3 2. What kinds of intermolecular forces (London, dipole--dipole, hydrogen b. H2 O c. F2 d. HBr 3. Identify the dominant intermolecular force that is present in each of the following substance and select the substance with the higher boiling point in each pair: a. CH3 OH or CH3 CH2 OH b. Hexane or cyclohexane c. MgCl2 or PCl3 d. CH3 NH2 or CH3 F 4. Compare intermolecular forces with intramolecular bonding. 5. Compare the strength and polarity of intermolecular forces. 110 UNIT 2 Metallic Bonding 2.4 At the end of this section, you will be able to: ) explain how a metallic bond is formed ) explain the properties of metals related to the concept of bonding, and investigate the conductivity, malleability and ductility of some metals and non-metals (Al, Cu, Fe, Sn, Zn, S, C charcoal, C graphite and Si). Activity 2.19 Form a group and discuss the following questions. Then, share your responses to the whole class. 1. What are the two characteristics that lead to an element being classified as a metal? 2. Which types of bond exist in solid copper? 3. What is delocalization of electrons? Metallic bonds are the chemical bonds that hold atoms together in solid metals such as copper, iron, and aluminum. In these metals, each metal atom is bonded to several neighboring atoms. Metallic bonding is the sharing of free electrons (delocalized electrons) among a lattice of positively charged metal ions. The bonding electrons are relatively free to move throughout the three-dimensional structure. How is a metallic bond different from that of covalent and ionic bonds? The strength of the metallic bond depends on: 1. the number of electrons in the delocalized "sea" of electrons. More delocalized electrons result in a stronger bond and a higher melting point. 2. the packing arrangement of the metal atoms. The more closely packed the atoms are the stronger the bond is and the higher the melting point. UNIT 2 111 CHEMISTRY GRADE 11 The electron-sea model is a very simple model, which pictures the metal as an array of metal cations in a "sea" of electrons, as illustrated in Figure 2.8. It proposes that all the metal atoms in a sample pool their valence electrons to form an electron "sea" that is delocalized throughout the piece. The metal ions (nuclei plus core electrons) are submerged within this electron sea in an orderly array. Figure 2.8: The electron-sea model for the electronic structure of metals The metal ions are not held in place as rigidly as the ions in an ionic solid, and no two metal atoms are bonded through a localized pair of electrons as in a covalent bond. Rather, the valence electrons are shared among all the atoms in the sample, and the piece of metal is held together by the mutual attraction of the metal cations for the mobile, highly delocalized valence electrons. The mobile electrons, known as conduction electrons, can transfer thermal vibration from one part of the structure to another i.e., metals can conduct heat. They are good conductors of electricity also. Activity 2.20 Perform the following activity in groups, and share your results to the rest of the class. 1. Take a piece of metal and a piece of wood of the same size. 2. Let these remain on a table so that both reach room temperature. 3. Place your left hand on the metal and right one on the wood. Which hand feels colder? Why? 112 UNIT 2 Metallic Bonding The general properties of metals include malleability and ductility and most are strong and durable. They are good conductors of heat and electricity. Their strength indicates that the atoms are difficult to separate. To what do properties of metals like malleability and ductility refer? The regularity, but not rigidity, of the metal-ion array and the mobility of the valence electrons account for the physical properties of metals. We can explain a lot of the properties of metals by thinking about this model of metallic bonding. The delocalized electrons are not attached to atoms, so they can move through the metal easily. This makes metals good conductors of electricity and heat, even when they are solid. Why do not ionic compounds conduct electricity when they are in solid form? But conductor when they are dissolved in water? The electrostatic bonds between the metal ions and the delocalized electrons are hard to break, so metals are strong. Why do metals bend or dent, rather than crack or shatter as ionic solids do when they struck by a hammer? Many metals can be flattened into sheets (malleable) and pulled into wires (ductile). The combination of strength and malleability means that metals are easy to bend but hard to break. This makes them extremely useful for making things. Metals have moderate to high melting points because the attractions between the cations and the delocalized electrons are not broken during melting, but boiling points are very high because each cation and its electron(s) must break away from the others. Gallium provides a striking example: it melts in your hand (mp 29.8 °C) but doesn't boil until 2403 °C. The alkaline earth metals \[Group IIA(2)\] have higher melting points than the alkali metals \[Group IA(1)\] because of greater attraction between their 2+ cations and twice the number of valence electrons. UNIT 2 113 CHEMISTRY GRADE 11 114 UNIT 2 Activity 2.21 Form a group and discuss the following questions, then present your responses to the whole class. 1. List four physical characteristics of a solid metal. 2. Briefly account for the following relative values: a. Lithium boils at 1317 °C and melts at 179 °C. The boiling point is about 1138 °C higher than its melting point. b. The melting point of Li and Be are 180 °C and 1287 °C, respectively. c. The melting point of sodium is 89 °C, whereas that of potassium is 63 °C. 2.5 At the end of this section, you will be able to: ) name two chemical bond theories ) explain the valence bond theory ) distinguish the Lewis model and the valence bond model ) discuss the overlapping of orbitals in covalent bond formation ) explain hybridization ) show the process of hybridization involved in some covalent molecules ) draw hybridization diagrams for the formation of sp, sp2, sp3, sp3d and sp3d2 hybrids ) suggest the kind of hybrid orbitals on the basis of the electron structure of the central atom ) predict the geometrical shapes of some simple molecules on the basis of hybridization and the nature of electron pairs ) discuss the hybridization involved in compounds containing multiple bonds ) explain bond length and bond strength ) explain molecular orbital theory ) describe molecular orbital using atomic orbitals ) describe bonding and anti-bonding molecular orbitals Chemical Bonding Theories ) draw molecular orbital energy level diagrams for homonuclear diatomic molecules ) write the electron configuration of simple molecules using the molecular orbital model ) define bond order and determine the bond order of some simple molecules and molecule-ions ) determine the stability of a molecule or an ion using its bond order; and predict magnetic properties of molecules. The Lewis model, one of the earliest models of covalent bond formation, represents chemical bonds by an electron dot formula. However, it fails to explain the formation of chemical bonds. It gives no idea about the shapes of polyatomic molecules. Similarly, the VSEPR theory gives the geometry of simple molecules but, theoretically, it does not explain them and it has limited applications. Covalent bonds are more accurately explained using modern bonding theories: the valence bond theory and the molecular orbital theory. The valence bond (VB) theory assumes that the electrons in a molecule occupy atomic orbitals of the individual atoms. It give us a picture of individual atoms taking part in the bond formation. The molecular orbital (MO) theory, assumes the formation of molecular orbitals from the atomic orbitals. Which theory explains all aspects of bonding? The valence-bond approach is the most useful approach to answer questions such as: what is a covalent bond, and what characteristic gives it strength? How can we explain molecular shapes based on the interactions of atomic orbitals? The basic principle of valence bond theory is that a covalent bond forms when orbitals of two atoms overlap and the overlap region, which is between the nuclei, is occupied by a pair of electrons. By overlap, we mean that the two orbitals share a common region in space. UNIT 2 115 CHEMISTRY GRADE 11 For example, consider the formation of the H2 molecule from two hydrogen atoms. Each atom has the electron configuration 1s1. As the H atoms approach each other, their 1s orbitals begin to overlap and a covalent bond forms (Figure 2.9). Electron density is higher in the overlap region than anywhere else, and the build-up of negative electron charge between the positive nuclei provides the electrostatic attraction that holds the atoms together. Figure 2.9: Atomic orbital overlap and bonding in H2 How is the strength of the bond between any two atoms affected by the extent of the overlap between the two orbitals? When two atoms are brought closely together, the repulsion of the atomic nuclei becomes more important than the electron-nucleus attraction and the bond becomes unstable. There is a condition of optimal orbital overlap that leads to maximum bond strength (bond energy) at a particular internuclear distance (bond length). The bond strength depends on the attraction of the nuclei for the shared electrons, so the greater the orbital overlap, the stronger (more stable) the bond. The extent of overlap depends on the shapes and directions of the orbitals. An s orbital is spherical, but p and d orbitals have more electron density in one direction than in another. Thus, whenever possible, a bond involving p or d orbitals will be oriented in the direction that maximizes overlap. In the HF bond, for example, the 1s orbital of H overlaps the half-filled 2p orbital of F along the long axis of that orbital (Figure 2.10). Any other direction would result in less overlap and, thus, a weaker bond. Note that a single bond consists of two electrons of opposite spin: hence, the two atomic orbitals that give rise to a single bond can have no more than two electrons in total. 116 UNIT 2 Chemical Bonding Theories Figure 2.10: Orbital and spin pairing in the formation of a HF molecule Similarly, in the F--F bond of F2 , the two 2p orbitals interact end-to-end, that is along the orbital axes, to attain maximum overlap (Figure 2.11). Exercise 2.11 Figure 2.11: Orbital and spin pairing in the formation of a F2 molecule When two atomic orbitals overlap to build up electron density along the two axis between the two nuclei, the resulting localized bond is called a sigma bond (σ). Such an overlap is sometimes referred to as a head on, end-to-end, end on or linear overlap. In Figures 2.9, 2.10 and 2.11, head-to-head overlap resulted in sigma bonds. 1. Sketch the sigma bonds in Cl2 and HCl molecules. 2. Identify the overlapping orbitals that form the single bond in a BrCl molecule. So far we have considered only sigma bonds, bonds in which the highest electron density lies along the bond axis (an imaginary line joining the nuclei) in between the bonded nuclei. The shape of p atomic orbitals allows them to overlap not only in a head-to-head fashion. If they are oriented in a parallel position, they can overlap sideways or laterally as shown in Figure 2.12. A bond formed in this way is called pi-bond (π). UNIT 2 117 CHEMISTRY GRADE 11 Figure 2.12: Formation of a pi bond (π bond) by the overlap of two half-filled p orbitals that are perpendicular to the internuclear axis Note that the directional nature of p orbital allows to overlap in two ways: end-to end or sideways. These two modes give rise to the two types of covalent bond: sigma (σ) and pi (π) bond, respectively. However, s-s and s-p will always overlap along the nuclear axis, resulting only in sigma bonds. Pi bonds are present in molecules containing multiple bonds. Can you give examples of compounds contain multiple bonds? All of the electron density in a π bond is found in lobes above and below the line connecting the atomic centers: one above and one below the sigma bond axis. One π bond holds two electrons that occupy both regions of the bond. A double bond always consists of one σ and one π bond. The double bond increases electron density between the nuclei. A triple bond consists of one σ and two π bonds. Activity 2.22 Answer the following questions in groups, and share your answers with the rest of the class. 1. Which bond is stronger: sigma (σ) or pi (π)? Why? 2. Compare the bond strength and length of single, double, and triple bonds. 118 UNIT 2 Chemical Bonding Theories Hybridization of Orbitals How many covalent bonds is a carbon atom supposed to form? How many covalent bonds does carbon actually form? This discrepancy and molecular structure, specifically bond angle, can be explained using the idea of hybridization. It is an imaginary mixing process in which the orbitals of an atom rearrange themselves to form new orbitals called hybrid orbitals. Consider covalent bond formation of carbon atom. From its ground state configuration, two unpaired electrons in the 2p subshell are observed. Ground state C: 1s 2s 2p One can predict the simplest hydrocarbon molecule to be CH2 , by overlapping the two unpaired electrons from two H atoms with the two unpaired electrons of carbon. However, experiment shows that CH2 is not a stable molecule. The simplest stable hydrocarbon is methane, CH4. To account for this, you need an orbital diagram that shows four unpaired electrons in the valence shell of carbon, requiring four bonds (and therefore four atoms of hydrogen). To get such a diagram imagine that one of the 2s electrons is promoted to the empty 2p orbital. To excite the 2s electron to a higher energy sublevel, energy must be absorbed. The resulting electron configuration is that of an excited state having energy greater than the configuration in the ground state. Excited state C: 1s 2s 2p Valence bond theory proposes that the one 2s and all three 2p orbitals of the carbon atom mix to produce four new orbitals ( sp3 ) that are equivalent to each other in energy and in shape and pointing in different directions with equal H--C--H bond angles. This blending is called hybridization and the resultant orbital as hybrid orbitals. The symbols used for hybrid orbitals identify the kinds and numbers of atomic orbitals used to form the hybrids. UNIT 2 119 CHEMISTRY GRADE 11 The number of new hybrid orbitals is equal to the total number of atomic orbitals that are combined. Hybridization provides a useful and convenient method of predicting the shapes of molecules. Note that it does not explain the reason for the shape. Exercise 2.12 Which of the following statements are false? Give a reason for your answer. 1. Two σ bonds comprise a double bond. 2. A triple bond consists of one π bond and two σ bonds. 3. Bonds formed from atomic s orbitals are always σ bonds. 4. A π bond consists of two pairs of electrons. 5. End-to-end overlap results in a bond with electron density above and below the bond axis. sp hybrid orbital sp type of hybridization involves the mixing of one s and one p orbital resulting in the formation of two equivalent sp hybrid orbitals. Each sp hybrid orbitals has 50% s-character and 50% p-character. Such a molecule in which the central atom is sp hybridized and linked directly to two other central atoms possesses linear geometry. This type of hybridization is also known as diagonal hybridization. For example, Be in BeCl2 molecule has an sp hybridization. BeCl2 : The ground state electronic configuration of Be is 1s22s2. In the exited state one of the 2s-electrons is promoted to vacant 2p orbital to account for its bivalency Figure 2.13. One 2s and one 2p-orbital gets hybridized to form two sp hybridized orbitals as shown in Figure 2.14. Figure 2.13: Excitation of Be 2s electrons and sp hybrid orbitals formation 120 UNIT 2 Chemical Bonding Theories Figure 2.14: The formation of two equivalent sp hybride orbitals in Be These two sp hybrid orbitals face opposite directions, forming an angle of 180°. Each of the sp hybridized orbitals overlaps with the 2p-orbital of chlorine axially and forms two BeCl sigma bonds. This is shown in Figure 2.15. Figure 2.15: The sp hybrid orbitals in gaseous BeCl2 sp2 hybrid orbitals sp2 hybridization In this hybridization there is involvement of one s and two p-orbitals in order to form three equivalent sp2 hybridized orbitals. For example, in BCl3 molecule, the ground state electronic configuration of central boron atom is 1s22s22p1. In the excited state, one of the 2s electrons is promoted to vacant 2p orbital (Figure 2.16) as a result boron has three unpaired electrons. These three orbitals (one 2s and two 2p) hybridize to form three sp2 hybrid orbitals as shown in Figure 2.16. Figure 2.16: Excitation of boron 2s electrons and sp2 hybrid orbitals formation UNIT 2 121 CHEMISTRY GRADE 11 The three equivalent hybrid orbitals formed are oriented in a trigonalplanararrangement and overlap with 2p orbitals of chlorine to form three B-Cl bonds. Therefore, in BCl3 (Figure 2.17), the geometry is trigonal planar with ClBCl bond angle of 120°. What is the percent of s-character and p-character in each sp2 hybrid orbital? Figure 2.17: sp2 hybrid orbitals and bonding in BCl3 sp3 hybridization sp3 type of hybridization can be explained by taking the example of CH4 molecule in which there is mixing of one s-orbital and three p-orbitals (Figure 2.18) of the valence shell to form four sp3 hybrid orbital of equivalent energies and shape. Can you determine the percent of s-character and p-character in each sp3 hybrid orbital? The four sp3 hybrid orbitals so formed are directed towards the four corners of the tetrahedron. The angle between sp3 hybrid orbital is 109.5° as shown in Figure 2.19. Figure 2.18: Excitation of carbon 2s electrons and sp3 hybrid orbitals formation 122 UNIT 2 Chemical Bonding Theories Figure 2.19: sp3 hybrid orbitals and bonding in methane The structure of NH3 and H2 O molecules can also be explained with the help of sp3 hybridization. In NH3 , the valence shell (outer) electronic configuration of nitrogen in the ground state is 2s22p x12p y12pz1. In the formation of NH3 , one 2s orbital and three 2p orbitals of nitrogen are mix up forming four hybrid orbitals (sp3) of equivalent energy. One of the sp3 hybrid orbitals has a lone pair of electrons and the other three sp3 orbitals are half-filled. These three half-filled hybrid orbitals overlap with 1s orbitals of hydrogen atoms to form three N--H sigma (σ) bonds. We know that the force of repulsion between a lone pair and a bond pair is more than the force of repulsion between two bond pairs of electrons. So the molecule gets distorted and the bond angle is reduced to 107° from 109.5°. The geometry of such a molecule will be pyramidal as shown in Figure 2.20. Figure 2.20: sp3 hybrid orbitals and bonding in NH3 In the case of H2 O molecule, the four oxygen orbitals (one 2s and three 2p) undergo sp3 hybridization forming four sp3 hybrid orbitals out of which two contain one electron each and the other two contain a pair of electrons. These four sp3 hybrid orbitals acquire a tetrahedral geometry, with two corners occupied by hydrogen atoms while UNIT 2 123 CHEMISTRY GRADE 11 the other two by the lone pairs. The bond angle in this case is reduced to 104.5° from 109.5° (Figure 2.21) and the molecule thus acquires a V-shape or angular geometry. Figure 2.21: sp3 hybrid orbitals and bonding in H2 O Why is the HOH bond angle in water less than the HNH bond angle in NH3 ? Hybridization of elements involving d orbitals The elements present in the third period contain d orbitals in addition to s and p orbitals. The energy of the 3d orbitals is comparable to the energy of the 3s and 3p orbitals. The energy of 3d orbitals is also comparable to those of 4s and 4p orbitals. As a consequence, the hybridization involving either 3s, 3p and 3d or 3d, 4s and 4p is possible. Do you expect hybrid orbitals formation by mixing 3p, 3d and 4s orbitals? Explain it. i. Group Assignment 2.2 Summarize in a table the shapes of molecules and ions, types of hybridization involving s, p and d orbitals, the number of atomic orbitals engaged in the hybridization, and give examples. Formation of PCl5 (sp3d hybridization): The ground state and the excited state outer electronic configurations of phosphorus (Z=15) are represented below in Figure 2.22a. 124 UNIT 2 Chemical Bonding Theories Figure 2.22 ( a ): sp3d hybrid orbitals filled by electron pairs donated by five Cl atoms Now the five orbitals (i.e., one s, three p and one d orbitals) are available for hybridization to yield a set of five sp3d hybrid orbitals which are directed towards the f ive corners of a trigonal bipyramidal as in Figure 2.22b. Figure 2.22 ( b ): Trigonal bipyramidal geometry of PCl5 molecule Note that all the bond angles in trigonal bipyramidal geometry are not equivalent. In PCl5 the five sp3d orbitals of phosphorus overlap with the singly occupied p orbitals of chlorine atoms to form five P--Cl sigma bonds. Three P--Cl bonds lie in one plane and make an angle of 120° with each other; these bonds are termed as equatorial bonds. The remaining two P--Cl bonds--one lying above and the other lying below the equatorial plane, make an angle of 90° with the plane. These bonds are called axial bonds. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds; which makes PCl5 molecule more reactive. UNIT 2 125 CHEMISTRY GRADE 11 ii. Formation of SF6 (sp3d2 hybridization): In SF6 the central sulphur atom has the ground state outer electronic configuration 3s23p4. In the exited state the available six orbitals i.e., one s, three p and two d orbitals are singly occupied by electrons (Figure 2.23a). These orbitals hybridize to form six new sp3d2 hybrid orbitals, which are projected towards the six corners of a regular octahedron in SF6. These six sp3d2 hybrid orbitals overlap with singly occupied orbitals of fluorine atoms to form six S--F sigma bonds. Thus, SF6 molecule has a regular octahedral geometry as shown in Figure 2.23b. Figure 2.23 ( a ): sp3d2 hybridization Figure 2.23 ( b ): Octahedral geometry of SF6 molecule 126 UNIT 2 Chemical Bonding Theories In hybridization schemes, one hybrid orbital is produced for every single atomic orbital involved. In a molecule, each of the hybrid orbitals of the central atom acquires an electron pair, either a bond-pair or a lone-pair. And the hybrid orbitals have the same orientation as the electron-set arrangement predicted by VSEPR theory. When describing the probable hybridization scheme for a structure, follow the following four steps: 1. Write a reasonable Lewis structure for the chemical. 2. Use VSEPR theory to predict the electron-set arrangement of the central atom. 3. Select the hybridization scheme that corresponds to the VSEPR prediction. 4. Describe the orbital overlap. Example 2.6 Describe a hybridization scheme for the central atom of iodine pentafluoride, IF5. Solution: a. The reasonable Lewis structure for IF5 is F F I F F F b. VSEPR predicts an octahedral electron-set arrangement for six electron pairs (AB5 E type) c. The hybridization scheme corresponding to octahedral electron arrangement is sp3d2 d. The six sp3d2 hybrid orbitals are directed to the corners of an octahedron, but one of the orbitals is occupied by a lone pair of electrons. The resulting molecular shape is that of a square pyramid with bond angles of approximately 90°. UNIT 2 127 CHEMISTRY GRADE 11 Exercise 2.13 1. List the three types of hybrid orbital that may be formed by an atom with only s and p orbitals in its valence shell. Draw the shapes of the hybrid orbitals produced. 2. Describe a hybridization scheme for the central atom and molecular geometry of the triiodide ion, I3--. Multiple bonds and hybridization The hybrid orbital model accounts well for the geometry of molecules involving single covalent bonds. Is it also capable of describing molecules containing double and triple bonds? We have already discussed that multiple bonds consist of σ and π bonds. Now, we will consider how we visualize these components and how they relate to hybrid orbitals. We will discuss here the ethene, C2 H4 , and acetylene, C2 H2 molecules, however pi bonds are also present in other molecules such as N2 , SO2 , NO3 , etc. The Lewis structure of ethene, C2 H4 , shows us that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms. H H C C H H The three bonding regions form a trigonal planar electron-pair geometry. Thus we expect the σ bonds from each carbon atom are formed using a set of sp2 hybrid orbitals that result from hybridization of two of the 2p orbitals and the 2s orbital (Figure 2.24). These orbitals form the C--H single bonds and the σ bond in the C = C double bond (Figure 2.25). The π bond in the C= C double bond results from the overlap of the third (remaining) 2p orbital on each carbon atom that is not involved in hybridization. This unhybridized p orbital (lobes shown in red and blue in Figure 2.25 is perpendicular to the plane of the sp2 hybrid orbitals. Thus, the hybridized 2p orbitals overlap in a side by-side fashion, above and below the internuclear axis and form a π bond. 128 UNIT 2 Chemical Bonding Theories Figure 2.24: sp2 hybridization of each carbon atom in ethene Figure 2.25: Bond formation in the ethene molecule, C2 H4 : (a) five σ bonds and (b) one π bond In an ethene molecule, the four hydrogen atoms and the two carbon atoms are all in the same plane. If the two planes of sp2 hybrid orbitals tilted relative to each other, the p orbitals would not overlap enough to create the π bond. Why does rotation around single (σ) bond occur easily, but is much more difficult for multiple bonds? In molecules with sp hybrid orbitals, two unhybridized p orbitals remain on the atom (Figure 2.26). We find this situation in acetylene, H−C≡C−H, which is a linear molecule. The two linear sp hybrid orbitals of a carbon atom which lie in a straight line, and the two unhybridized p orbitals at perpendicular angles (Figure 2.26). The sp hybrid orbitals of the two carbon atoms overlap end to end to form a σ bond between the carbon atoms (Figure 2.27). The remaining sp orbitals form σ bonds with two hydrogen atoms. The two unhybridized p orbitals per carbon are positioned such that they overlap side by side and, hence, form two π bonds. The two carbon atoms of acetylene are bound together by one σ bond and two π bonds, giving a triple bond. UNIT 2 129 CHEMISTRY GRADE 11 Figure 2.26: sp hybridization of each carbon atom and the two linear sp hybrid orbitals of a carbon atom in acetylene Figure 2.27: (a) The formation of σ bonds in an acetylene molecule, C2 H2 , and (b) π bonds Exercise 2.14 1. Describe the hybridization of the central atoms N and S in molecules and the molecular geometry of: a. NO2 b. SO3 2. Discuss the bonding in ammonium and sulphate ions: predict the ideal bond angles, bond length, shape of the ions, and the number of σ and π bonds. 3. Prepare sketches of the overlaps of the following atomic orbitals: a. s with s; b. s with p along the bond axis; c. p with p along the bond axis (head-on overlap); d. p with p perpendicular to the bond axis (side-on overlap). 130 UNIT 2 Chemical Bonding Theories The molecular orbital theory assumes that when atoms come together, their orbitals not only overlap, but are also simultaneously transformed into new orbitals. These new orbitals, called molecular orbitals (MOs), play the same role for molecules that atomic orbitals play for atoms. Note that atomic orbitals are the allowed states for an electron moving in the field of one nucleus, whereas molecular orbitals are the allowed states for an electron moving in the field of several nuclei (atoms present in the molecules). Which theory (VBT or MOT) perfectly explains all aspects of bonding? Bonding and Anti-Bonding Molecular Orbitals In contrast to VB theory, in which one bonding orbital is formed as a result of the interaction of two atomic orbitals, in MO theory two molecular orbitals result from the combination of two atomic orbitals. In mathematical terms, the two molecular orbitals result from the addition of the atomic orbitals that overlap. The molecular orbitals that results from the addition of the atomic orbitals that overlap is called a bonding orbital. In a bonding molecular orbital, because the electron density is concentrated between the two nuclei, the energy of the system is lowered compared with that of isolated atomic orbitals. The molecular orbital that results from the subtraction of atomic orbitals that overlap is called an antibonding molecular orbital. In antibonding molecular orbitals, the electron density is concentrated away from the region between the two nuclei. That is, anti-bonding molecular orbitals have a region of zero electron density (a node) between the nuclei. The net effect of having a low electron density between the nuclei is that the two nuclei repel each other. Therefore, energy of an antibonding orbital is increased compared with that of the two isolated atomic orbitals. When two atomic orbitals overlap end-to-end, they form two σ MOs. Consider the H2 molecule, which has two H atoms and therefore two 1s AOs. The two 1s atomic orbitals combine (Figure 2.28) to produce two σ MOs, which differ in energy and location. One of the σ MOs is a bonding orbital, denoted σ1s , the other is an anti bonding orbital denoted σ\* 1s. The relative energy levels of these two MOs are different. The σ1s MO has a lower energy than the original 1s AOs, while the σ\* 1s MO has a higher energy. What is the advantage of VBT compared with that of MOT? UNIT 2 131 CHEMISTRY GRADE 11 Electron Configuration of Diatomic Molecules Electrons occupy molecular orbitals following the same rules that apply to the filling of atomic orbitals: the aufbau principle and the Pauli Exclusion Principle and Hund's rule are obeyed. The electronic structure of the molecule is obtained by feeding the appropriate number of electrons into the new molecular orbitals. For example, both electrons in H2 will go into the lower energy orbital denoted by (σ1s )2. There are no more electrons in H2 so the σ\* 1s orbital remains empty in the ground state. A molecular orbital diagram shows the relative energy and number of electrons in each MO, as well as the atomic orbitals from which they form. Figure 2.28 is the molecular orbital diagram for H2. Figure 2.28: Molecular orbital diagram and bonding in the H2 molecule Note: The number of molecular orbitals formed must equal the number of atomic orbitals available for combination. Unfilled molecular orbitals are considered to be there, even when there are no electrons in them. MO theory uses the term "bond order" to indicate whether a covalent bond is single (bond order = 1), double (bond order = 2) or triple (bond order = 3). Bond order is defined as the number of electrons in bonding MOs minus the number in antibonding MOs, divided by two: B ond order = 1 2  Numberof in − −   Numberof in   e e      bonding MOs −     − anti bondingMOs       Thus, for H2 , the bond order is ½ (2 − 0) = 1. A bond order greater than zero indicates that the molecule is stable relative to the separate atoms, whereas a bond order of zero implies no net stability. In general, the higher the bond order, the stronger the bond. Why is MOT better than VBT in explaining covalent bond formation? 132 UNIT 2 Chemical Bonding Theories Exercise 2.15 1. Draw molecular orbital energy diagrams for: a. C2 − b. C2 c. C2 + The order of energy of molecular orbitals has been determined mainly from spectroscopic data. a. In simple homonuclear diatomic molecules (total electrons = 14 or less) the order is: = σσσ σ π π σ π π σ \* \* \* \* \* 1 1 2 2 2 2 2 2 2 2 ( ) ( ) = s s s s p p p p p p y z x y z x b. For simple homonuclear diatomic molecules (total electrons greater than 14) the ( ) 2 2 π π = comes after 2 x p p p y z σ and the order is: σσσ σ σ π π π π σ = \* \* \* \* \* 1 1 2 2 2 2 2 2 2 2 ( )( ) = s s s s p p p p p p x y z y z x Note that the 2py atomic orbital give π bonding and π antibonding (π\*) MOs of the same energy as those produced from 2pz orbitals. The 2 y p to be double degenerate, and similarly \* 2 y p π and \* 2 z p π and 2 z p π orbitals are said π are double degenerate. A similar arrangement of MOs exists from Example 2.7 to \* 3 3 x s p σ σ , but energies are known with less certainty. Use the molecular orbital theory to derive the electron configuration of: a. H2 + b. Li2 c. He2 + Which of these molecules or molecular ions exist? Solution: d. O2 a. H2 + molecular ion. This may be considered as a combination of H atom and a H+ ion, giving one electron to be accommodated in a MO. The bond order is ½(1− 0) = ½, so we predict that H2 + does exist. The electron configuration is σ1s 1. b. Li2 molecule. Each Li atom has two electrons in its inner shell, and one in its outer shell, making a total of six electrons arranged: (σ1s )2 (σ\* 1s )2 (σ2s )2. We ignore the inner electrons here because, in general, only outer (valence) orbitals interact enough to form molecular orbitals. The bond order is ½(2− 0) = 1, so we predict that Li2 does exist. UNIT 2 133 CHEMISTRY GRADE 11 c. He2 + Molecular ion. The electron configuration is ( σ1s )2 (σ\*1s )1. The f illed bonding orbital gives stabilization whilst the half-filled gives destabilization. The bond order is ½(2− 1) = ½, there is some stabilization, so the He2 + should exist, and it has been observed spectroscopically. d. O2 molecule. Each oxygen has eight electrons, making a total of sixteen for the molecule. These are arranged:    ( ) ( ) π π 2 \* 1 2 2  s s s s p  ( ) ( ) ( ) ( ) ( ) ( ) ( ) σ σ σ σ σ π π 2 \* 2 2 \* 2 2 1 1 2 2 2 2 \* 1 2 2 x The antibonding \* 2 y p π and \* 2 z p      π orbitals are singly occupied in accordance with Hund's rule. As in previous examples, the inner shell does not participate in bonding and the 2 2 and p p y z each other. A σ bond results from the filling of arise form bonding and antibonding. Therefore, 1 1 1 σ π π 2 2 Exercise 2.16  p p  y y  p p   z z π π bonding and antibonding 2s orbitals cancel σ 2 ( ) 2 2 + + = bonds are formed. p. Two half π bonds x 1. Use the molecular orbital theory and derive the electron configuration of the following molecules. Identify those which exist and those which do not exist. a. He2 b. Be2 c. B2 d. C2 e. N2 2. Use the MO theory to predict the bond order and the number of unpaired electrons in O2 2 , O2--, O2 +, NO and CO. Magnetic Properties Molecules with unpaired electrons exhibit paramagnetic properties. The species is attracted by an external magnetic field. When all the electrons are paired, there is diamagnetism. Such species are not attracted (and, in fact, are slightly repelled) by a magnetic field. Can you predict magnetic properties of diatomic molecules by employing VBT? 134 UNIT 2 Types of Crystal The antibonding π2py and π2pz orbitals for O2 are singly occupied in accordance with Hund's rule. Unpaired electrons give rise to paramagnetism. Since there are two unpaired electrons with parallel spins, this explains why oxygen is paramagnetic. The MO theory successfully predicts the paramagnetism of O2 , a fact not even thought of with the VB representation of the oxygen molecule (O = O). Exercise 2.17 1. What is the bond order for CN--, CN and CN+? 2. Which homonuclear diatomic molecules from the second period elements, besides O2 , should be paramagnetic? 2.6 At the end of this section, you will be able to: ) define a crystal ) name the four types of crystalline solid and give examples ) mention the types of attractive force that exist within each type of crystalline solid ) describe the properties of each type of crystalline solid ) build a model of sodium chloride crystal structure. Solids can be crystalline or amorphous. A crystalline solid is composed of one or more crystals. Each crystal has a well-defined, ordered structure in three dimensions. Sodium chloride (table salt) and sucrose (table sugar) are examples of crystalline substances. Are metals crystalline or amorphous solids? An amorphous solid has a disordered structure; it lacks the well-defined arrangement of basic units (atoms, molecules, or ions) found in a crystal. A glass is an amorphous solid obtained by cooling a liquid rapidly enough that its basic units are "frozen" in random posit