42 Chemistry Term 1 STPM CHAPTER 2 4 The single electron in the hydrogen atom can be located in any of the orbits (depending on its energy) but not in spaces between the orbits. + n = 7 n = 6 n = 5 n = 4 n = 3 n = 2 n = 1 5 Each orbit is also known as a principle shell and is represented by an alphabet. Orbit, n Principle shell 1 K 2 L 3 M 4 N 5 O 6 P 7 Q 6 When the electron is in the orbit closest to the nucleus (i.e. at orbit n = 1), it is in its ground state and is stable. It will not lose energy as it revolves round the nucleus. (This is in contradiction to the classical law of physics which states that an electron revolving around a positive nucleus must lose energy continuously and will eventually fall into the nucleus). 7 The electron can be excited to higher energy levels by absorbing the required energy. This is called the excited state. 8 The electron in the excited state will ‘drop’ back to lower levels by emitting the excess energy in the form of electromagnetic radiations. Quantised energy levels can be visualised as the shelves in a book shelf. The books can be on any shelf, but not the space between the shelves. In the ground state, the electron is stable. Contradiction to classical Physics Excitation of the electron Returning to the ground state Shelf Book Energy absorbed Nucleus Energy released Nucleus
43 Chemistry Term 1 STPM CHAPTER 2 9 The frequency of the radiation emitted is given by Plank’s equation: ∆E = hf where ∆E = energy difference between the two levels concerned 10 All transitions from higher energy levels to level n = 1 emit radiations with frequencies in the ultraviolet region, giving rise to Lyman Series. 11 Transitions from higher levels back to level n = 2 emit radiations with frequencies in the visible region, giving rise to Balmer Series. This is summarised in the diagram below: Paschen series Balmer series Frequency Lyman series n = 6 n = 5 n = 3 n = 2 n = 1 n = 4 Example 2.2 The difference in energy between the first and second levels in the hydrogen atom is 985 kJ mol–1. Calculate the frequency of the radiation emitted when an electron falls from the second level to the first. Solution ∆E = 985 kJ mol–1 Using the equation: ∆E = hf 985 = (3.99 10–13) f ∴ f = 2.47 1015 Hz Exam Tips Exam Tips Exam Tips Exam Tips Exam Tips Exam Tips This frequency corresponds to the ultraviolet region. Since ∆E is given in terms of kJ mol –1, h will have the value of 3.99 × 10–13 kJ mol –1s and not 6.63 × 10–34 Js. The intervals between levels/ orbits gets smaller with increasing value of n. 2009/P1/Q3
44 Chemistry Term 1 STPM CHAPTER 2 Quick Check 2.2 1 The difference in energy between two levels in the hydrogen atom is 3.03 10–19 J. (a) Calculate the frequency of light emitted when transition takes place between these two levels. (b) In what part of the electromagnetic spectrum is the light emitted? 2 The line spectrum of sodium consists of a line with wavelength of 590 nm. (a) Calculate its frequency. (b) Calculate the difference in energy (in kJ mol–1) between the two levels where the transition occurs. 3 The energies of the orbits in the hydrogen atom are as shown below: 1277 n = 6 n = 5 n = 3 n = 2 n = 1 n = 4 1261 1231 1168 985 0 y x Calculate the frequencies of the lines marked x and y. Definition of the ionisation energy of hydrogen Ionisation Energy of Hydrogen 1 Ionisation energy of hydrogen is the minimum energy required to remove the lone electron in the hydrogen atom per mole of gaseous hydrogen atom under standard conditions (298 K and 101 kPa). H(g) → H+(g) + e 2 For the hydrogen atom, the ionisation energy corresponds to the difference in energy between the levels, n = 1 and n = ∞. n = ∞ ∆E = ionisation energy n = 1 3 The electronic transition between these two levels will result in a line with maximum frequency in the Lyman series of the hydrogen line spectrum. 2013/P1/Q18(b) The line with maximum frequency is known as the convergence limit or convergence frequency.
45 Chemistry Term 1 STPM CHAPTER 2 n = 1 3.29 1015 Hz f (maximum) Frequency n = ∞ 4 Using the formula: ∆E = hf = (3.99 10–13)(3.29 1015) kJ mol–1 = 1312.7 kJ mol–1 Example 2.3 The frequencies of the first four lines in the Lyman Series are given below: 2.56; 2.92; 3.08; 3.16 ( 1015 Hz) (a) Plot a suitable graph to determine the frequency of the last line in the Lyman Series. (b) Use your answer to question (a), calculate the ionisation energy of hydrogen. Solution Using the equation: The Rydberg’s equation for the Lyman series is: 1 λ = RH 1 12 – 1 n2 Or, f = CRH 1 12 – 1 n2 Rearranging: f = CRH – CRH 1 n2 A graph of frequency, f, against 1 n2 will give a straight line where the maximum frequency can be obtained by the intercept at the frequency axis (where n = ∞) and 1 n2 = 0 f (× 1015)/s–1 2.56 2.92 3.08 3.16 n 2 3 4 5 1 n2 0.25 0.11 0.063 0.040 Graphical method to calculate the ionisation energy of hydrogen
46 Chemistry Term 1 STPM CHAPTER 2 f(× 1015)/s–1 1 n2 3.29 The maximum frequency is 3.29 1015 Hz. (b) Using the equation: ∆E = hf ∴ ∆E = (3.99 10–10)(3.29 1015) 10–3 = 1312.7 kJ mol–1 Alternative method: The convergence frequency of the Lyman series occurs when the difference in frequency of successive lines (∆f) is zero. Thus, if we plot a graph of frequency (f) against ∆f, the intersection on the f axis (where ∆f = 0) would give the value of the convergence frequency. f(× 1015)/Hz 2.56 2.92 3.08 3.16 ∆f(× 1015)/Hz – 0.36 0.16 0.08 3.29 ƒ (× 1015) / Hz Δƒ (× 1015) / Hz Using the equation: ∆E = hf = (3.99 × 10–10)(3.29 × 1015) × 10–3 = 1312.7 kJ mol–1 Ionisation Energies and Electronic Shells 1 Evidence of energy levels or electronic shells is evident when one studies the successive ionisation energies of an atom. 2 A graph of ionisation energy against order of electrons removed provides information on the number of principle shells or energy levels and the distribution of the electrons in the shells. 3 The table below lists the successive ionisation energies of carbon atom, 6 C. No. of electrons removed 1 2 3 4 5 6 Ionisation energy/kJ mol–1 1090 2350 4610 6220 37 830 42 280 4 A plot of the successive ionisation energies gives the graph on the left. Ionisation energy Order of electron removed 1 2 3456
47 Chemistry Term 1 STPM CHAPTER 2 5 The first and second ionisation energies correspond to the process: C(g) → C+(g) + e ∆H = 1090 kJ mol–1 C+(g) → C2+(g) ∆H = 2350 kJ mol–1 6 The ionisation energy increases with the number of electrons removed. This is because successive species have ever increasing proton electron ratio. This causes the remaining electrons to be more tightly held by the nucleus. 7 The graph shows that there is a gradual increase in the first four ionisation energies. However, there is a very large increase in energy in removing the fifth electron. 8 This suggests that the first four electrons and the fifth electron occupy different energy levels or shells. 9 The ionisation energies of the last two electrons are very large compared to the first four, suggesting that the last two electrons are closest to the nucleus in energy level or orbit n = 1, while the first four electrons are in the outermost energy level or orbit, n = 2. 10 The distribution of the five electrons in the boron atom is: Orbit, n Principle shell No. of electrons 1 K 2 2 L 3 11 A similar study shows that silicon, 14Si, has the following electronic distribution. Orbit, n Principle shell No. of electrons 1 K 2 2 L 8 3 M 4 Sucessive ionisation energy increases because electrons are removed from cations of increasing charge. Quick Check 2.3 1 Use appropriate equations to represent the following ionisation: (a) the 5th ionisation energy of sodium. (b) the 12th ionisation energy of aluminium. 2 The following graph shows the successive ionisation energies of an element X. Ionisation energy Order of electrons removed
48 Chemistry Term 1 STPM CHAPTER 2 (a) How many electrons are there in one atom of X? (b) How many energy levels are filled with electrons? (c) State the distribution of the electrons in X. 3 Element W is in Group 15 of the Periodic Table. Sketch a graph of successive ionisation energies for the first 8 electrons. Short-comings of Bohr’s model. 3 n 2 1 s p d Energy 2.2 Atomic Orbitals: s, p and d 1 Although Bohr’s model can satisfactorily explain the formation of the line spectrum of hydrogen or one-electron species such as He+ and Li2+. It cannot explain the formation of the line spectra of multielectron atoms. 2 Bohr’s model also cannot explain the appearance of extra lines in the emission spectra of elements when a magnetic field is applied. 3 Modern theories still embrace the concept of quantised energy levels but totally reject the circular orbit concept. Energy Subshell 1 A detailed study of the line spectrum of atoms shows that each principle shell is further made up of subshells with slightly different energies. 2 The number of subshells in a principle shell is equal to the quantum number of the shell concerned. Principle shell n No. of subshells First, K 1 1 Second, L 2 2 Third, M 3 3 Fourth, N 4 4 Fifth, O 5 5 3 In any one subshell, the level with the lowest energy is assigned the letter s, followed by p, d and f. For example, Principle shell n No. of subshells Symbol First, K 1 1 1s Second, L 2 2 2s, 2p Third, M 3 3 3s, 3p, 3d Fourth, N 4 4 4s, 4p, 4d, 4f [In all cases, the s orbital has the lowest energy in each shell.] 2015/P1/Q5
49 Chemistry Term 1 STPM CHAPTER 2 4 The total number of electrons that can occupy any principle shell (with quantum number n) is 2n2 . Principle shell n No. of subshells First, K 1 2 Second, L 2 8 Third, M 3 18 Fourth, N 4 32 5 Further experiments on the line spectra shows that each subshell is further made up of orbital where the electrons are placed. 6 The number of orbital depends on the type of subshell as shown below: Subshell No. of orbitals Symbol s 1 s p 3 px , py , pz d 5 dxy, dyz, dxz, dx 2– y 2 , dz 2 7 The three p orbitals have equivalent energies. In other words, they degenerate. The same goes for the five d orbitals. Shapes of the Orbitals 1 The concept of orbitals arises from the fact that an electron has dual nature. It is a particle as well as a wave. 2 As a wave, it extends in all directions in space. 3 However, we can define the region in space within which the ‘wave density’ of a particular electron is maximum. 4 An orbital is defined as the region (or volume) in space around the nucleus where the probability of finding a particular electron is maximum ( 95%). This is opposed to the idea of orbit which refers to a fixed circular path where the electron moves. (a) (b) (c) d Energy Subshell Orbital p s Definition of orbital VIDEO Shapes of the Orbitals
50 Chemistry Term 1 STPM CHAPTER 2 5 The s orbital is spherical and non-directional. Y X Z The shaded area represents the region in which the chance of finding the s electron is more that 95%. The size of the orbital increases in the order 1s 2s 3s ..... 6 The p orbitals have dumb-bell shapes and are directional. The diagram below shows the distribution of three p orbitals around the nucleus. y z x
51 Chemistry Term 1 STPM CHAPTER 2 7 The shapes of the five d orbitals are: z z z z z y y y y y x x x x x dxz dyz dxy dx2 dx2 y2 – 8 Note that in the set of d orbitals, two of them, dx 2 – y 2 and dz 2, have their ‘lobes’ along the x, y and z axes while the other three have their 'lobes' in between the axes. 2.3 Electronic Configurations 1 There are three rules that govern the filling of electrons in an atom: The Aufbau Principle, the Pauli’s exclusion Principle and Hund’s rule. 2 The Aufbau Principle states that in the ground state of an atom, the electrons must occupy orbitals in order of increasing energy. The diagrams below show the arrangement of the energy levels and the order they are filled. 2008/P2/Q1(a) 2013/P1/Q3, Q18(c) 2011/P1/Q3, Q5 2014/P1/Q18(b) 2016/P1/Q4 2010/P1/Q4 2014/P1/Q3 2015/P1/Q18 2011/P2/Q5(a) 2015/P1/Q6 2018/P1/Q19(b)
52 Chemistry Term 1 STPM CHAPTER 2 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 4f 6s 5p 4d 5s 4p 3d 4s 3p 3s 2p 2s 1s Energy 3 The Pauli’s Exclusion Principle states that an orbital can accommodate a maximum of two electrons only. Both the electrons must have opposite spins. (Otherwise they will repel one another). 4 The total number of electrons in an orbital or a set of degenerate orbital is given below: Orbital Maximum number of electrons s 1 2 = 2 p 3 2 = 6 d 5 2 = 10 f 7 2 = 14 5 Hund’s Rule states that in a set of degenerate orbitals, electrons must occupy the orbital singly (with parallel spins) first before pairing. 6 An example of Hund’s rule is illustrated below: If the p orbitals are occupied by only two electrons, the distribution is: and not: 7 In the above diagram, each box represents an orbital. The boxes that are joined together represent degenerate orbitals. Electrons with opposite spins attract one another. Exam Tips Exam Tips Two electrons occupying the same orbit experience repulsion due to interaction of their electron clouds. Exam Tips Exam Tips Another statement of Hund's rule: 'If two or more orbitals with equal energy are available, the orbitals are filled with one electron, with the electrons having parallel spins, before a second electron of the opposite spin is added.' Info Chem The spins of the electrons are represented by ‘ ’ and ‘ ’.
53 Chemistry Term 1 STPM CHAPTER 2 8 Following the three rules introduced, the electronic configurations of the first 30 elements in the Periodic Table are given below: Element Proton number Electronic configuration Condensed configuration 1 H 1 1s1 1 2 He 2 1s 2 2 3 Li 3 1s 2 2s 1 2.1 4 Be 4 1s 2 2s 2 2.2 5 B 5 1s 2 2s 2 2p1 2.3 6 C 6 1s2 2s2 2p2 2.4 7 N 7 1s 2 2s2 2p3 2.5 8 O 8 1s2 2s2 2p4 2.6 9 F 9 1s2 2s2 2p5 2.7 10Ne 10 1s2 2s2 2p6 2.8 11Na 11 1s2 2s2 2p6 3s1 2.8.1 12Mg 12 1s2 2s2 2p6 3s2 2.8.2 13Al 13 1s2 2s2 2p6 3s2 3p1 2.8.3 14Si 14 1s2 2s2 2p6 3s2 3p2 2.8.4 15P 15 1s2 2s2 2p6 3s2 3p3 2.8.5 16S 16 1s2 2s2 2p6 3s2 3p4 2.8.6 17Cl 17 1s2 2s2 2p6 3s2 3p5 2.8.7 18Ar 18 1s2 2s2 2p6 3s2 3p6 2.8.8 19K 19 1s2 2s2 2p6 3s2 3p6 4s1 2.8.8.1 20Ca 20 1s2 2s2 2p6 3s2 3p6 4s2 2.8.8.2 21Sc* 21 1s2 2s2 2p6 3s2 3p6 3d1 4s2 2.8.9.2 22Ti* 22 1s2 2s2 2p6 3s2 3p6 3d2 4s2 2.8.10.2 23V* 23 1s2 2s2 2p6 3s2 3p6 3d3 4s2 2.8.11.2 24Cr* 24 1s2 2s2 2p6 3s2 3p6 3d5 4s1 2.8.13.1 25Mn* 25 1s 2 2s 2 2p6 3s 2 3p6 3d5 4s 2 2.8.13.2 26Fe* 26 1s 2 2s 2 2p6 3s 2 3p6 3d6 4s 2 2.8.14.2 27Co* 27 1s 2 2s 2 2p6 3s 2 3p6 3d 7 4s 2 2.8.15.2 28Ni* 28 1s 2 2s 2 2p6 3s 2 3p6 3d 8 4s 2 2.8.16.2 29Cu* 29 1s 2 2s 2 2p6 3s 2 3p6 3d 10 4s 1 2.8.18.1 30Zn* 30 1s 2 2s 2 2p6 3s 2 3p6 3d 10 4s 2 2.8.18.2 [* denotes d-block elements] Exam Tips The outermost shell cannot have more than 8 electrons.
54 Chemistry Term 1 STPM CHAPTER 2 9 Note that for potassium, the 19th electron, occupies the 4s orbital and not the 3d because in an empty atom, the 4s is of lower energy than the 3d. 10 For elements with proton number 21 to 30 (known as the d-block elements), the energy level of the 4s and 3d are reversed. This is because once the 3d orbital is/are filled, the 3d electrons repel the 4s electrons to a higher energy level. 3d 4s 4s 3d Empty 3d orbitals Filled 3d orbitals Energy 11 Orbitals that are fully filled (e.g. s 2 , p6 or d10) or are half-filled (e.g. p3 or d5 ) have extra stability due to their symmetrical charge distribution. This is evident in the electronic configuration of 24Cr and 29Cu. 12 The configuration of chromium is [Ar] 3d5 4s 1 and not [Ar] 3d4 4s 2 . Similarly, the configuration of copper is [Ar] 3d10 4s 1 and not [Ar] 3d9 4s 2 . 3d 4s Cr: 3d 4s Cu: Electronic Configuration of Ions 1 In the formation of cations, electrons are lost in the reverse order from electron filling. That is, last to be in, first to be out. 2 The electronic configuration of aluminium is 2.8.3 or 1s 2 2s 2 2p 6 3s 2 3p 1 . 1s 2s 2p 3s 3p Reversal of the 4s and 3d orbitals s2 , p6 , d,10 p3 and d5 are more stable because of the even distribution of charge
55 Chemistry Term 1 STPM CHAPTER 2 In the formation of Al+ , the electron removed is from the 3p orbital: The configuration of Al+ is 2.8.2 or 1s 2 2s 2 2p 6 3s 2 . 3 In the formation of anions, electrons are added in the same manner as filling of electrons. 4 The electronic configuration of fluorine is 2.7 or 1s 2 2s 2 2p 5 . 1s 2s 2p The additional electron is added to one of the 2p orbitals which have only one electron. 1s 2s 2p The electronic configuration of F – is 2.8 or 1s 2 2s 2 2p 6 . Quick Check 2.4 1 Write the electronic configurations for the following atoms: (a) 32Ge (c) 82Pb (b) 35Br 2 Iron (proton number 26) is a d block element. It can form two stable ions, Fe2+ and Fe3+. Write the electronic configuration for: (a) the iron atom, and (b) the Fe2+ and Fe3+ ions. (c) Based on the electronic configuration in (b), state and explain which oxidation state of iron is more stable? 2.4 Classification of Elements in the Periodic Table 1 In the modern Periodic Table, the elements are arranged in order of increasing proton numbers. 2 In 1984, the International Union of Pure and Applied Chemistry (IUPAC) formally recommended the format as shown: 2017/P1/Q3 2018/P1/Q3
56 Chemistry Term 1 STPM CHAPTER 2 1 H 3 Li 11 Na 19 K 37 Rb 55 Cs 87 Fr 13 14 15 16 17 2 He 1 18 1 H 2 5 B 6 C 7 N 8 O 9 F 10 Ne 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar 4 Be 12 Mg 20 Ca 38 Sr 56 Ba 88 Ra 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 58 Ce 59 Pr 60 Nd 61 Pm 62 Sm 63 Eu 64 Gd 65 Tb 66 Dy 67 Ho 68 Er 69 Tm 70 Yb 71 Lu 90 Th 91 Pa 92 U 93 Np 94 Pu 95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lw 57 La 72 Hf 73 Ta 74 W 75 Re 76 Os 77 Ir 78 Pl 79 Au 80 Hg 81 T1 82 Pb 83 Bi 84 Po 85 At 86 Rn 89 Ac 104 Rf 105 Ha [106] [107] [108] [109] Proton number Symbol 3 4 5 6 7 8 9 10 11 12 d-block p-block s-block f-block 3 The Table consists of 7 horizontal rows called Periods and 18 vertical columns called Groups. 4 It is interesting to note that only 11 of all the known elements are gases under room conditions. They are H2 , He, Ne, Ar, Kr, Xe, Rn, F2 , Cl2 , N2 and O2 . 5 Only 2 elements are liquids at room conditions: Br2 and Hg (mercury). Electronic Configuration and the Periodic Table 1 The elements in the Periodic Table can be divided into four blocks depending on their valence shell electronic configurations. 2 They are the s block, p-block, d-block and f-block elements. 3 The s and p-blocks are called the representative elements, the d-blocks are the transition elements, and the f-block is called the inner transition elements. 4 All elements in the same group have the same number of valence electrons with similar electronic configurations. The s-block Elements 1 The s-block elements consist of elements from Group 1 and Group 2. 2 Group 1 elements (also called the alkaline metals) have one valence electron in the outermost orbital with the configuration of s 1 . Info Chem 7 Periods 18 Groups 4 Blocks
57 Chemistry Term 1 STPM CHAPTER 2 Element Valence shell configuration 1 H 1s1 11Na 3s1 37Rb 5s1 87Fr 7s1 6s1 55Cs 3 Li 2s1 4s1 19K 3 Group 2 elements (also called the alkaline earth metals) have two electrons in the outermost orbital with the configuration of s 2 . Element Valence shell configuration 4 Be 2s 2 20Ca 4s 2 56Ba 6s 2 7s 2 88Ra 12Mg 3s 2 5s 2 38Sr The p-block Elements 1 Elements of Group 13 to Group 18 are known as the p-block elements because the outermost orbitals that are filled with electrons are the p orbitals. [The Group 17 elements are also called the halogens, and the Group 18 elements are called the noble gases.] 2 They have valence shell electronic configurations of s 2 p1 to s 2 p6 . Group Representative element Valence shell configuration 13 13Al 3s 2 3p1 14 14Si 3s 2 3p2 15 15P 3s 2 3p3 16 16S 3s 2 3p4 17 17Cl 3s 2 3p5 18 18Ar 3s 2 3p6 The d-block Elements 1 Elements of Group 3 through 12 are known as d-block elements. 2 They have valence shell electronic configurations of d1 s 2 to d 10s 2 .
58 Chemistry Term 1 STPM CHAPTER 2 Group Element Valence shell configuration 3 21Sc 3d1 4s 2 4 22Ti 3d 2 4s 2 5 23V 3d 3 4s 2 6 24Cr 3d 5 4s 1 7 25Mn 3d 5 4s 2 8 26Fe 3d 6 4s 2 9 27Co 3d 7 4s 2 10 28Ni 3d 8 4s 2 11 29Cu 3d10 4s 1 12 30Zn 3d 10 4s 2 3 They are typical metals with very high melting points and boiling points. The f-block Elements 1 Elements with proton numbers 58 – 71 (known as the lanthanides) and from 90 – 103 (also known as the actinides) are f-block elements. 2 They have valence shell electronic configuration of f 1 d 10s 2 to f 14d 10s 2 . Valence Shell Configuration from the Periodic Table 1 Most chemical recations involve only the valence shell electrons. Hence, the valence shell electronic configuration of an element is of utmost importance in predicting the chemical properties of an element. 2 The valence shell configuration of the representative elements (Group 1, 2, 13 to 18) can be deduced from its position in the Periodic Table. 3 The period in which the atom is found corresponds to the Principle Quantum Number of the valence shell. 4 The group in which the atom is in corresponds to the number of valence electrons. 5 For example, aluminium (proton number 13) is in Period 3 and Group 13 of the Periodic Table. This means that the outermost energy level that is filled with electrons is the 3rd shell, and it has 3 (not 13) valence electrons. Hence, its valence shell configuration is 3s 2 3p 1 . 6 Lead (proton number 82) is in the 6th Period and Group 14. Its valence shell configuration is 6s 2 6p 2 . Exam Tips Exam Tips The number of valence shell electrons of Group 13 to 18 elements is given by: (Group number – 10)
59 Chemistry Term 1 STPM CHAPTER 2 SUMMARY SUMMARY Example 2.4 Write the valence shell electronic configuration of the following elements. Element Period Group A 4 17 B 6 1 C 2 18 Solution A . . . . . 4s 2 4p 5 B . . . . . 6s 1 C . . . . . 2s 2 2p 6 1 Electromagnetic radiations are characterised by their wavelengths (λ) and frequencies (f ). 2 The relationship between wavelength and frequency is: λ = c f 3 The energy associated with a electromagnetic radiation is given by: E = hf 4 The line spectrum of hydrogen: (a) shows that the energy levels in the hydrogen atom are quantised, (b) shows the electron in the hydrogen atom can have certain fixed values only and not any values, (c) shows that electronic transitions between energy levels are possible, (d) can be used to calculate the ionisation energy of the hydrogen atom. 5 The lines in the Lyman series (ultra-violet region) is a result of transitions between higher energy levels and level n = 1. 6 The lines in the Balmer series (visible region) is a result of transitions between higher energy levels and level n = 2. 7 The wavelengths of the lines in the hydrogen spectrum can be calculated using the equation: 1 λ = RH 1 n1 2 – 1 n2 2 8 Evidence of energy levels or shells is supported by the successive ionisation energy graphs of elements. 9 Aufbau’s Principle states that electrons will fill orbitals of lower energy before those of higher energies are filled. 10 Pauli’s exclusion Principle states that an orbital can accommodate a maximum of two electrons with opposite spins only. 11 Hund’s Rule states that electrons must first fill a set of degenerate orbitals singly before pairing occurs. 12 An orbital is the region/volume in space around the nucleus where the probability of finding an electron is maximum ( 95%).
60 Chemistry Term 1 STPM CHAPTER 2 Objective Questions 1 Which of the following statements is true regarding the line spectrum of hydrogen? A The first line in each series has the longest wavelength. B The ionisation energy of hydrogen can be calculated using the line with the lowest frequency in the ultraviolet region. C The lines are all equally spaced. D The lines converge when the wavelength increases. 2 What is the maximum number of atomic orbitals with the principle quantum number 3? A 3 C 9 B 6 D 12 3 Which of the following statements is not true of the line spectrum of hydrogen atom? A The lines become closer as frequency increases B There are no more lines after the line with maximum frequency in the Lyman series C The spectrum consists of only two series, Lyman series and Balmer series D The lines are produced when the electron absorbed energy and gets ‘promoted’ to higher energy levels 4 What is the maximum number of electronic transitions possible between the first 7 energy levels in the hydrogen atom? A 7 B 14 C 21 D 5040 5 Which of the following statements explains why the emission spectrum of hydrogen is a line spectrum? A The hydrogen has only one electron. B The lone electron in the hydrogen atom can revolve round the nucleus in fixed orbits only. C The energy levels in the hydrogen atom are quantised. D Electrons have dual nature. 6 An orbital always contains A 0, 1 or 2 electrons C 1 or 2 electrons B 2 electrons D 8 electrons 7 A d subshell contains how many degenerate orbitals? A 1 C 5 B 3 D 7 8 In the building up of the electronic configuration of atoms, electrons are added to the 4s subshell before the 3d subshell because A the 3d subshell can hold more electrons than the 4s subshell. B the 3d subshell is closer to the nucleus. C the 4s subshell is spherical. D the 3d subshell is of higher energy. 9 Which of the following correctly represents the Balmer series in the line spectrum of hydrogen? A Wave number Wave number Wave number Wave number B C D 10 Tungsten (74W) is used as the element in incandescent bulbs. Which of the following is not true about the element? A Tungsten oxide is amphoteric B Tungsten exhibits variable oxidation states in its compound C The maximum oxidation state of tungsten is +6 D Tungsten compounds are used as catalyst STPM PRACTICE 2
61 Chemistry Term 1 STPM CHAPTER 2 11 What is the proton number of an element that has 6 unpaired electrons in its ground state? A 14 C 24 B 22 D 30 12 The lines in the emission spectrum of hydrogen are the results when A electrons are removed from the atom. B electrons absorb energy. C electrons lose energy. D the hydrogen molecules split to form atoms. 13 Part of the lines in the Lyman series of an emission spectrum of atomic hydrogen is shown below. X Which of the following statements is not true? A The lines have frequencies corresponding to ultra-violet radiation B The wavelength of line X can be used to calculate the ionisation energy of hydrogen C Each line corresponds to electronic transition between higher energy levels and level n = 1 D The hydrogen atom has 6 electrons 14 In a hydrogen atom, electron in which orbital can absorb energy but cannot emit energy? A 1s C 3s B 2p D 3d 15 Which of the following electronic transitions would emit visible light of highest frequency? A n4 → n2 B n2 → n6 C n5 → n1 D n1 → n2 16 Which statement is not true about the s, p and d orbitals? A The five d orbitals have the same shape and same energy B The three p orbitals have the same shape and same energy C In any energy shell, the energy of the orbitals increases in the order s < p < d D d orbitals are directional whereas s orbitals are non-directional 17 Which of the following elements has 5 unpaired electrons in its ground state? A Nitrogen B Phosphorous C Manganese D Iron 18 Which of the following valence configuration is for an atom that can form positive ions of different oxidation state? A 3d10 4s1 C 3d10 4s2 B 3s2 3p1 D 3d1 4s2 19 Which of the following species would have a line spectrum similar to that of the hydrogen atom? A He2+ C Na+ B Li2+ D K+ 20 Which of the following statements is not true? A 3d orbitals have the same energy as the 4s orbital. B p orbitals are filled with electrons according to Hund’s rule. C The principle quantum number of p orbitals start with n = 2. D An s orbital is spherical. 21 An atom X has six valence electrons and forms a stable X3+ ion. In which group of the Periodic Table does X belong to? A 3 C 13 B 6 D 16 22 The proton number for species X is 8 and its nucleon number is 18. Species X has two electrons less than the number of neutron. What is the charge on X? A 0 C +1 B –1 D +2 23 Which of the following electronic arrangements violates Hund’s rule? A B C D
62 Chemistry Term 1 STPM CHAPTER 2 Structured and Essay Questions 1 In the stratosphere, chlorine molecules absorb electromagnetic radiation and dissociate to produce chlorine atoms which are responsible for the destruction of ozone. Cl—Cl(g) → 2Cl(g) Given that the bond energy of chlorine is 242 kJ mol–1, calculate the wavelength of the light absorbed. 2 Part of the Lyman series in the hydrogen emission spectrum is as shown below: 8.22 10.27 10.62 10.97 Wave number ( 106 ) m–1 (a) Draw an energy level diagram to show how the lines are produced. (b) (i) Define ionisation energy of hydrogen. Illustrate your answer with an appropriate equation. (ii) Calculate the ionisation energy (in kJ mol–1) of hydrogen from the above spectrum. 3 The diagram below shows the lines in the Balmer series in the emission spectrum of hydrogen atom. (a) Draw a labelled energy diagram to show how the lines marked X and Y are formed. (b) Name two species that would produce similar emission spectrum as that of the hydrogen atom. Explain your reasoning. X Y Frequency 4 The frequencies of the first five lines in the Lyman series of the hydrogen atom are given below: 2.47, 2.92, 3.08, 3.16, 3.20 ( 105 s–1) Plot a suitable graph to determine the ionisation energy of the hydrogen atom (a) in J, and (b) kJ mol–1. 5 The first seven ionisation energies (kJ mol–1) of an element D are as follows: 496, 4563, 6913, 9544, 13 352, 16 611, 20 115 (a) What do you understand by the term ionisation energy? (b) Explain why the ionisation energy increases with the number of electrons removed. (c) Plot a suitable graph or otherwise, determine the Group number of D in the Periodic Table. Explain how you arrived at the answer. 6 (a) Sketch the energy level diagram for the orbitals of an atom with the principle quantum numbers of n = 1 to n = 3 inclusive of the 4s orbital. (b) Using arrows to represent electrons, show the electronic configuration of (i) a carbon atom in its ground state. (ii) a carbon atom in its excited state of lowest energy. (iii) a carbon atom in its excited state with the highest energy.
63 Chemistry Term 1 STPM CHAPTER 2 7 Explain the following observations: (a) An orbital can accommodate a maximum of only two electrons with opposite spins. (b) The electronic configuration of chromium is [Ar]3d 5 4s 1 instead of [Ar]3d 4 4s 2 . 8 (a) State Hund’s rule and Pauli’s exclusion principle. (b) Draw an energy level diagram showing the electron distribution in an N2– ion. 9 The number of electrons occupying the different orbitals of atom X is shown in the following table. Orbital s p d Number of electrons 7 12 10 Write the electronic configuration of X, and explain how each of these orbitals is filled with electrons. 10 (a) The lines in the emission line spectrum of hydrogen are the result of electronic transitions. (i) Draw an energy level diagram to show the transitions that give rise to the first and last lines in both the Lyman series and the Balmer series and state in which part of the electromagnetic spectrum that each of the series belongs to. (ii) The energy difference between two of the levels in the hydrogen atom is 987.3 kJ mol–1. Calculate the frequency and the wavelength of the line produced when electronic transition occurs between these two levels. [h = 6.63 × 10–34 Js; C = 3.00 × 108 ms–1; L = 6.02 × 1023 mol–1] (b) An atom X has 24 electrons in its nucleus. (i) Draw an energy diagram to show how the electrons are arranged in its atom. (ii) Explain any abnormality in the electronic arrangement.
CHAPTER CHEMICAL BONDING 3 Concept Map Learning earning Outcomes Students should be able to: Ionic bonding • describe ionic (electrovalent) bonding as exemplified by NaCl and MgCl2. Covalent bonding • draw the Lewis structure of covalent molecules (octet rule as exemplified by NH3, CCl4, H2O, CO2, N2O4 and exception to the octet rule as exemplified by BF3, NO, NO2, PCl5, SF6); • draw the Lewis structure of ions as exemplified by SO4 2−, CO3 2−, NO3 − and CN−; • explain the concept of overlapping and hybridisation of the s and p orbitals as exemplified by BeCl2, BF3, CH4, N2, HCN, NH3 and H2O molecules; • predict and explain the shapes of and bond angles in molecules and ions using the principle of valence shell electron pair repulsion, e.g. linear, trigonal planar, tetrahedral, trigonal bipyramid, octahedral, V-shaped, T-shaped, seesaw and pyramidal; • explain the existence of polar and non-polar bonds (including C−Cl, C−N, C−O, C−Mg) resulting in polar or/and non-polar molecules; • relate bond lengths and bond strengths with respect to single, double and triple bonds; • explain the inertness of nitrogen molecule in terms of its strong triple bond and nonpolarity; • describe typical properties associated with ionic and covalent bonding in terms of bond strength, melting point and electrical conductivity; • explain the existence of covalent character in ionic compounds such as Al2O3, AlI3 and LiI; • explain the existence of coordinate (dative covalent) bonding as exemplified by H3O+, NH4 +, Al2Cl6 and [Fe(CN)6] 3−. Metallic bonding • explain metallic bonding in terms of electron sea model. Intermolecular forces: van der Waals forces and hydrogen bonding • describe hydrogen bonding and van der Waals forces (permanent, temporary and induced dipole); • deduce the effect of van der Waals forces between molecules on the physical properties of substances; • deduce the effect of hydrogen bonding (intermolecular and intramolecular) on the physical properties of substances. Covalent Bonding • Electron sharing theory • Multiples bonds • Coordinate bonds • Exception of the octet rule • Strength of covalent bonds • Drawing of Lewis diagram • Resonance structure Ionic Bonding • Electron transfer theory • Strength of ionic bond • Properties of ionic compounds • Factors affecting ionic compounds formation Intermolecular Forces • Polar and non-polar bonds • Polar and non-polar molecules • Polarisation in ionic bonds • Hydrogen bonding • Force of attraction between covalent molecules Chemical Bonding Metallic Bonding • Strength of metallic bonds
Exam Tips Exam Tips 65 Chemistry Term 1 STPM CHAPTER 3 3.1 Ionic Bonding Lewis Diagram 1 As most chemical reactions involve only the redistribution of the valence electrons of the atoms concerned, G. N. Lewis devised a simple method to indicate the number of valence shell electrons in an atom and how they are utilised in bond formation. 2 The valence shell electrons of an atom are represented either by a cross () or a dot (·). The Lewis diagram is also called a ‘dot-and-cross diagram’. 3 The number of valence electrons of Group 1 and Group 2 elements corresponds to the Group number of the element concerned. For elements of Group 13 to 18, the number of valence electrons = Group number – 10. 4 The table below lists the Lewis diagram for the main group elements. Group no. Example Electronic configuration Lewis diagram 1 Sodium 2.8.1 • Na 2 Magnesium 2.8.2 •Mg• 13 Aluminium 2.8.3 • •AI• 14 Silicon 2.8.4 • •SI• • 15 Phosphorus 2.8.5 •• •P• • 16 Sulphur 2.8.6 •• •S• •• 17 Chlorine 2.8.7 •• •CI •• 18 Argon 2.8.8 •• Ar •• Example 3.1 Draw the Lewis diagram for the following atoms, showing their valence shell electrons. (a) 20Ca (b) 54Xe (c) 50Sn (d) 35Br (e) 34Se Solution • •• • (a) •Ca (b) Xe (c) •Sn• • • • • • • • (d) Br• (e) •Se• • • • • Lewis diagrams show only the number of electrons in the valence shell. 2017/P1/Q4 2013/P1/Q4
66 Chemistry Term 1 STPM CHAPTER 3 Na+ Cl– Ionic bond Electron Transfer Theory 1 Ionic bonds are electrostatic attractions between opposite charged ions. 2 In 1916, Kossel suggested that the ions are formed by the complete transfer of electrons from atoms of one element to atoms of another element. 3 Ionic bonds are usually formed between metallic elements (Groups 1, 2 and 13) and non-metallic elements (Groups 15, 16 and 17). 4 The element that loses the electrons forms a positive ion (cation), while the atom that accepts the electrons forms a negative ion (anion). The opposite charged ions attract one another and an ionic bond is formed. 5 The compounds formed are called ionic compounds. 6 An example of ionic compound is sodium chloride. The formation of ionic bond in sodium chloride is explained as follows: (a) The electronic configurations of sodium and chlorine are: Na: 1s 2 2s 2 2p 6 3s 1 (or 2.8.1) Cl: 1s 2 2s 2 2p 6 3s 2 3p 5 (or 2.8.7) (b) The sodium atom has one electron in excess of an octet while the chlorine atom has one electron short of an octet. (c) During the formation of sodium chloride, the sodium atom transfers its lone electron in the 3s orbital to a 3p orbital (which contains one electron) of chlorine. 1s 3p Cl: Na: 2s 3s 3p (d) In the process, sodium becomes Na+ with a configuration of neon (2.8), and chlorine becomes Cl– with a configuration of argon (2.8.8). This can be represented by the Lewis diagram: x • • • •• Na + •Cl → [Na]+[xCl ] – • • •• 2.8.1 2.8.7 2.8 2.8.8 (e) The opposite charged ions attract one another to form sodium chloride, Na+Cl– . 7 Other examples of ionic compounds are: (a) Magnesium oxide x • • • • • • xMg + •O• → [Mg]2+[xOx] 2– • • • • 2.8.2 2.6 2.8.8 2.8 Atom → cation + electron Atom + electron → anion The valence shell of sodium in Na+ is empty. Ionic bond is the electrostatic attraction between opposite charged ions formed by the complete transfer of electrons from an atom to another atom.
67 Chemistry Term 1 STPM CHAPTER 3 (b) Magnesium chloride x •• • •• xMg + 2 Cl• → [Mg]2+ 2[xCl ] – •• •• 2.8.2 2.8.7 2.8 2.8.8 Example 3.2 Draw the Lewis diagram for the following ionic compounds: (a) Potassium oxide, K2 O (b) Aluminium oxide, Al2 O3 (c) Magnesium fluoride, MgF2 Solution • • • • • •• (a) 2[K]+ [xO x] 2– (c) [Mg]2+ 2[ xF ] – • • •• • • • • (b) 2[Al]3+ 3[xO x] 2– • • Strength of Ionic Bonds 1 The strength of an ionic bond is a measure of the electrostatic attraction between the ions. 2 The force of attraction between oppositely-charged ions is proportional to the charge on the ions and inversely proportional to the square of the distance between the ions. E ∝ (Q +) (Q – ) d2 where Q + = charge on the cation Q – = charge on the anion d = distance between the ions = (radius of cation + radius of anion) d + – 3 The smaller the ions and/or the higher the charge on the ions, the stronger is the attraction between the ions. This makes the bond stronger. 4 For example, Compound NaCl NaBr Melting point/°C 801 750 The ionic bond in NaCl is stronger than that in NaBr. 2015/P1/Q10
68 Chemistry Term 1 STPM CHAPTER 3 (a) The melting point of sodium chloride is higher than that of sodium bromide. This shows that the ionic bond in NaCl is stronger than that in NaBr. (b) This is because the Cl– ion is smaller than that of the Br– ion. Ion Na+ CI– Br– Ionic radius/nm 0.095 0.181 0.195 (c) The interionic distance in NaCl is 0.276 nm, whereas the interionic distance in NaBr is 0.290 nm. As a result, the electrostatic attraction between Na+ and Cl– is stronger. 0.276 + Cl– + Br– 0.290 5 The melting points of sodium chloride and magnesium chloride are: Compound NaCI MgCI2 Melting point/°C 801 987 Cationic radius/nm 0.095 0.065 (a) The Mg2+ ion is smaller in size than the Na+ ion. On top of that, the Mg2+ ion has a higher charge. (b) As a result, the ionic bond in MgCl2 is stronger than that in NaCl. This accounts for the higher melting point of MgCl2 . Properties of Ionic Compounds 1 Due to the strong ionic bonds, all ionic compounds have giant ionic structures and are solid at room conditions with relatively high melting points and boiling points. Compound Al2 O3 BaF2 CaO MgCl2 Melting point/°C 2345 1628 2887 987 Boiling point/°C 3252 2410 3123 1685 2 Most ionic compounds are soluble in water because the ions can be solvated by water molecules through the formation of iondipole attraction. They are insoluble in non-polar solvents. Br– is larger than Cl– . The ionic bond in MgCl2 is stronger than that in NaCl. Mg2+ is smaller than Na+. Ionic compounds are all solid. 2016/P1/Q19(b)
69 Chemistry Term 1 STPM CHAPTER 3 δ+ δ– δ+ δ– + – 3 In the solid state, ionic compounds do not conduct electricity because the ions are immobile. When molten or in aqueous solutions, the ions are set free and can conduct electricity. 4 Ionic compounds are hard and brittle. They shattered when knocked. This is due to the displacement of the layers of ions which brings like-charged particles adjacent to one another causing repulsion between the layers. Applied force + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – + – – + – + + – + – – + – + Displacement of layers + – + – Repulsion Plane of cleavage Factors Affecting the Formation of Ionic Compounds 1 During the formation of the ionic bond, one atom loses electron(s) and another atom accepts the electron(s). For example, Na Electron Cl Na+ ............................. Cl– Ionic bond 2 Hence, the ease of formation of ionic bonds depends on how easily an atom can lose electron(s) and how readily another atom can accept electron(s). 3 The larger the size and the lower the nuclear charge of an atom, the lower the ionisation energy and the tendency to form cation is higher. 4 The smaller the size and the higher the nuclear charge of an atom, the higher the tendency to attract electron(s) to form anions. Ion-dipole attraction No mobile ions in the solid state. Ionic solids are hard but brittle.
70 Chemistry Term 1 STPM CHAPTER 3 5 Thus, ionic bonds are usually formed between metallic elements (Group 1, 2, 13 and the transition elements) and non-metallic elements (Group 15, 16 and 17). 3.2 Covalent Bonding Electron Sharing Theory 1 Certain compounds are formed without the transfer of electrons from one atom to the other. 2 Instead, the atoms concerned share their valence electrons with one another so that each atom achieves octet configuration. 3 Take the fluorine molecule, F2 , as an example: (a) The electronic configuration of fluorine is: 1s 2 2s 2 2p 5 F: (b) The fluorine atom is one electron short of an octet. (c) In the F2 molecule, each fluorine atom shares one unpaired electron with one another as illustrated below to achieve octet configurations: F F or (d) There is now electrostatic attraction between the two nuclei of the two fluorine atoms and the pair of electrons that are shared between them. (e) This type of bond is called a covalent bond. (f) A covalent bond is defined as the force of attraction between two adjacent nuclei and the electrons that are shared between them. (g) A covalent bond is represented by a straight line joining the two bonding atoms. Cl —Cl Covalent bond Info Chem The two circles are added for clarity. 2009/P1/Q42 2017/P1/Q18(a)(b)
71 Chemistry Term 1 STPM CHAPTER 3 4 Another example is ammonia, NH3 . H N H H Bond pair Lone pair Note that in the ammonia molecule, there is a pair of electrons on the nitrogen atom that is not used in bond formation. This is called a lone pair. The other three pairs that bond nitrogen to hydrogen are called the bond pair. 5 Other examples are: CI C CI CI CI Carbon tetrachloride, CCl4 H O H Water, H2 O [Note that in the water molecule, there are two lone pairs and two bond pairs on the oxygen atom.] Multiple Bonds 1 In some molecular species, some atoms are forced to share more than one electron with another atom in order to achieve octet configuration giving rise to either a double covalent bond or a triple covalent bond. 2 There are no quadruple or higher covalent bonds. Double Bonds 1 A double bond is formed when atoms share two electrons each with one another. An example is the carbon dioxide molecule, CO2 . O C O 2 A double bond is represented by joining the atoms with two parallel lines. O == C == O Info Chem The lone pair electrons of the terminal atoms are omitted for clarity.
72 Chemistry Term 1 STPM CHAPTER 3 3 Other examples are the oxygen molecule, O2 , and dinitrogen tetraoxide molecule, N2 O4 . O O N O O O O N ∆ ∆∆ ∆ ∆ Triple Bonds 1 In a triple bond, each bonding atoms shares three electrons with one another. 2 An example is the nitrogen molecule, N2 : X• X• X• • • X X N N 3 A triple bond is represented by three lines joining the bonding atoms. N ≡ N 4 Another example is the hydrogen cyanide molecule, H—C ≡ N. NCH Lewis Diagrams of Ions 1 We can similarly draw the Lewis diagram for polyatomic ions, such as SO4 2–, CO3 2–, NO3 – and CN– . 2 In the Lewis diagram for polyatomic anions, the negative charge will be carried by the more electronegative atom(s). 3 For ions with a charge of –2 and above, the negative charge will be shared among as many other electronegative atoms as possible to help disperses the charge and stabilises the anion. 4 Take the example of SO4 2–: (a) The negative charge will be carried by oxygen, which is more electronegative than sulphur. 2016/P1/Q19(a)(ii)
73 Chemistry Term 1 STPM CHAPTER 3 (b) The two negative charges will be shared between two oxygen atoms in the form of O– ions. O O– x x • • x x • • • x • x O– O S 5 The Lewis diagrams for CO3 2–, NO3 – and CN– are shown below: O– O– O C O– O O N NC– Exception to the Octet Rule Molecular species that do not follow the octet rule fall under two categories: (a) Octet deficient species (b) Expanded octet species Octet Deficient Species 1 In octet deficient or incomplete octet species, the central atom has less than eight electrons (4 pairs) surrounding it. 2 Examples are: (a) Boron trifluoride, BF3 F B F F Boron is surrounded by only 6 electrons. (b) Nitrogen monoxide, NO N O • x • x • x • • x x x Nitrogen is surrounded by only 7 electrons.
74 Chemistry Term 1 STPM CHAPTER 3 (c) Nitrogen dioxide, NO2 O N O Nitrogen is surrounded by only 7 electrons. [Note: The two electrons in the nitrogen-oxygen single bond come from nitrogen alone. This is an example of coordinate (dative covalent) bond.] Expanded Octet Species 1 Atoms of the Period 2 (Li, Be, B, C, N, O and F) cannot have more than 8 valence electrons surrounding them. This is because the outermost shell contains only the 2s and 2p subshell which can accommodate a maximum of 8 electrons only. 2 However, atoms of Period 3 and higher do not have such restriction. Their atoms can accommodate more than 8 electrons in their valence shells. 3 These atoms have empty d orbitals in their respective valence shells which can be used to expand their valency. 4 Consider the sulphur hexafluoride molecule, SF6 . The sulphur atom in SF6 has 12 electrons (6 bonding pairs) surrounding it. This is made possible because sulphur ([Ar]3s 2 3p 4 3d 0 ) can promote one of its 3s 2 electrons and one of its paired 3p electrons to two empty 3d orbitals, resulting in the formation of 6 unpaired electrons which are then used to form 6 covalent bonds with 6 fluorine atoms. 3s Promotion 6 unpaired electrons 3p The Lewis diagram of SF6 is shown below: F S x • • x • x F F F F F x • • x x • F S x • • x • x F F F F F x • • x x •
75 Chemistry Term 1 STPM CHAPTER 3 5 Another example is phosphorus pentachloride, PCl5 . By promoting one electron from the 3s orbital to an empty 3d orbitals, phosphorus now has five electrons available for covalent bond formation. Cl P x• • x • Cl x Cl Cl Cl x • • x Promotion 5 unpaired electrons Drawing of the Lewis Diagram 1 The Lewis diagrams of polyatomic species show how the atoms are connected to one another and the type of bonds (single, double or triple) that are formed between them. 2 When drawing the Lewis diagram for a covalent compound or molecular ion, we first have to determine which is the central atom, and which is/are the terminal atoms. For example, in the ammonia molecule, NH3 , nitrogen is the central atom while hydrogen atoms are terminal atoms. 3 During the sharing of electrons, the terminal atoms must attain octet configuration (duplex for the case of hydrogen), but not necessarily so for the central atom. 4 If the central atom is from the second period in the Periodic Table (Li, Be, B, C, N, O and F), the total number of electrons surrounding it cannot exceed eight (although it can be less than eight). 5 Atoms will endeavor to form multiple bond (double or triple bonds) between them if possible. This is because multiple bonds are stronger than single bonds. 6 In a molecular anion, the charge will be shared among as many other atoms as possible. This is to decrease the charge density on any particular atom and to stabilise the ion. For example, in the CO3 2– ion, the two negative charges are distributed between two oxygen atoms, and in the PO4 3– ion, the three negative charges are shared among three oxygen atoms. 7 For molecular cations, the positive charge will be carried by the less electronegative atom in the molecular ion. For example, in the PCl4 + ion, the positive charge is on the phosphorus atom (which is less electronegative than the chlorine atom). The negative charge in a molecular anion is shared by as many other atoms as possible. H | H —N— H Terminal atom Central atom
76 Chemistry Term 1 STPM CHAPTER 3 8 If the terminal atom already has an octet (for example, Cl–), then it will contribute two electrons to the central atom to form a coordinate covalent bond. Drawing the Lewis Diagram for NH3 The following explanation outlines the steps in the drawing of the Lewis diagram for the NH3 molecule: 1 Write the symbol of the central atom. Do not assign any electrons to the nitrogen atom yet. 2 The nitrogen atom is bonded to three hydrogen atoms (the terminal atoms). 3 Each hydrogen atom (which is one electron short of a duplet configuration) shares one electron with the nitrogen atom in order to achieve the configuration of helium. 4 The nitrogen atom in turn shares one electron with each of the hydrogen atoms to form covalent bonds. 5 Nitrogen is in Group 15 of the Periodic Table. After sharing three electrons with three hydrogen atoms, it still has two electrons which are not used in the bond formation process. These are the lone pair electrons. 6 Thus, there are three bonding pairs of electrons and one lone pair in the NH3 molecule. H | H—N—H • • 7 The circles can be omitted if necessary. They are put there for clarity. Drawing the Lewis Diagram of AlCl4 – 1 The central atom is aluminium, Al. 2 Aluminium is bonded to three chlorine atoms and one Cl– ion. 3 Each chlorine atom (which is one electron short of an octet) shares one electron with aluminium. However, the Cl– ion already has an octet. Thus, it will not share any electrons with aluminium (otherwise it would have more than 8 electrons surrounding it). Instead it contributes two electrons to aluminium to form a coordinate bond, giving rise to the Lewis structure as shown in the diagram. (1) (2) (3) (4) (5) N H H H N H H H N x x x H H N H x x • • x • H H N H x x • • x • •• (1) (2) (3) Al Cl Cl Cl Cl– Al Cl Cl Cl Cl– Al x • • x x x • x
77 Chemistry Term 1 STPM CHAPTER 3 Drawing the Lewis Diagram of CH3 + 1 The central atom is carbon. 2 Carbon is bonded to two hydrogen atoms and one H+ ion. 3 Each hydrogen atom shares one electron with carbon. However, since there are no electrons in the H+ ion, it will accept two electrons from carbon to form a coordinate bond. This gives the Lewis structure of CH3 + as shown in the diagram. Quick Check 3.1 1 Draw the Lewis structure of the following species (You may use either one of the methods). (a) SO4 2– (b) SO3 2– (c) I 3 – (d) PO4 3– (e) PCl4 + (f) ICl4 – (g) O3 (h) CO (i) S2 Cl2 (j) HCN 2 (a) Draw the Lewis diagram of aluminium chloride, AlCl3 . (b) At room temperature, aluminium chloride exists in as a dimer of AlCl3 . 2AlCl3 Al2 Cl6 The two AlCl3 molecules are joined by two coordinate bonds. Draw the Lewis structure of the dimer. 3 Beryllium chloride, BeCl2 , is a covalent molecule. It reacts with ammonia to form an addition compound: BeCl2 + 2NH3 → BeCl2⋅2NH3 (a) Draw the Lewis structure of BeCl2 . (b) What features are present in the BeCl2 and NH3 molecules that lead to the formation of the addition compound? (c) Draw the Lewis diagram of the addition compound. 4 (a) Nitrogen dioxide, NO2 is a brown gas. Draw the Lewis structure of nitrogen dioxide. (b) When cooled in ice, the brown colour of the gas fades due to dimerisation to form the colourless dinitrogen tetroxide, N2 O4 . 2NO2 N2 O4 Draw the Lewis structure of dinitrogen tetroxide. Covalent Bonds and Overlapping of Atomic Orbitals 1 Earlier on in the chapter we learnt that a covalent bond is formed when electrons are shared between two atoms. 2 For example, there is the sharing of electrons to form the hydrogen molecule, H2 . H x • H How atoms share their electrons (1) (2) (3) C H H C H+ H H C H+ X• X X X •
78 Chemistry Term 1 STPM CHAPTER 3 3 However, no mention was made as to how the electrons are shared. What happens to the electrons both before and after sharing? 4 The valence bond theory attempts to explain how covalent bonds are formed using the concept of overlap of atomic orbitals. 5 Let us take the hydrogen molecule, H2 . The Lewis theory explains the formation of the covalent bond in terms of pairing of the two valence electrons. 6 In the valence bond theory, the covalent bond in the H2 molecule is formed by the overlap of the two 1s orbitals in the hydrogen atom. 7 When two hydrogen atoms are far apart, there is no interaction (no attraction or repulsion) between the two. The potential energy of the system is almost zero. 8 As the atoms approach one another, they start to attract one another and the potential energy of the system decreases because each electron is attracted by the nucleus of the other hydrogen atom. The attraction continues until the potential energy of the system reaches a minimum. 9 If the distance between the atoms were to decrease further, the potential energy would increase sharply because of electronelectron and nucleus-nucleus repulsion. 10 At the point of minimum potential energy, there is substantial overlap of the two 1s orbital of the two hydrogen atom and the system is the most stable. This corresponds to a bond length of 0.074 nm and the formation of a stable H2 molecule. H + Maximum overlap (0.074 nm) H 1s orbitals 11 In the H2 molecule, the two atomic orbitals merge to form a molecular orbital. The two electrons are free to move throughout the new orbital. 12 There is a large build up of electron density between the two hydrogen nuclei. 13 The attraction between the two hydrogen nuclei and the electrons in the electron cloud holds the two hydrogen atoms together and a covalent bond is formed. + +
79 Chemistry Term 1 STPM CHAPTER 3 14 The variation of potential energy of the bond formation process is summarised in the energy diagram below. 0 Potential energy Distance of separation H: H: 1s 1s H H H H 15 Using the same model, we can explain the formation of the Cl—Cl bond in the Cl2 molecule. Each chlorine atom has an unpaired electron in one of their 3p orbitals. Overlap of the two 3p orbitals of the two chlorine atoms gives rise to the Cl2 molecule. Cl Cl + + Cl Cl 16 The formation of the H—Cl bond in the HCl molecule is due to the overlap of a 1s orbital of hydrogen with the 3p orbital of chlorine. Cl + + H Cl H
80 Chemistry Term 1 STPM CHAPTER 3 Sigma and Pi Bond 1 Unlike the s orbital which is spherical, p orbitals are directional. 2 As a result, two p orbitals can overlap in two different ways to produce two different types of covalent bonds, a sigma (σ) bond and a pi (π) bond. 3 A sigma bond (σ) is a result of end-to-end overlap of two p orbitals, as in the case of the Cl2 molecule. + + In a sigma bond, the electron density is concentrated in the region between the nuclei. A σ bond is also formed when an s orbital overlaps with a p orbital, as in the case of the HCl molecule. + + 4 A pi bond (π), on the other hand, is a result of sideway overlap of two p orbitals. + + In a pi bond, the electron density is concentrated above and below the plane of the nuclei. 5 π bonds are weaker than σ bonds because the overlap is not as extensive as that of σ bonds. On top of that, the repulsion between the two ‘unshielded’ nuclei weakens the π bond. 6 A π bond can only be formed after the two atoms are joined by a σ bond. 7 σ bonds involve overlap of hybridised orbitals while π bonds involve overlap of unhybridised orbitals. Hybridisation 1 Let us take a look at the Lewis diagram of beryllium chloride molecule, BeCl2 . 2 In the Lewis diagram shown, the beryllium atom shares two unpaired electrons with two chlorine atoms to form two Be—Cl bonds. The adjacent nuclei are not shielded from one another. The two nuclei must be bonded by a σ bond first before a π bond can be formed between them. Cl Be Cl x • x • 2016/P1/Q19(a)(i) 2018/P1/Q4
81 Chemistry Term 1 STPM CHAPTER 3 3 However, the electronic configuration of beryllium shows that all the electrons in the orbitals are paired. Be: 1s Be: 2s This configuration shows that beryllium could not form any covalent bonds through electron sharing. 4 The same happens when we consider the electronic configuration of carbon: C: 1s 2p C: 2s Carbon has only two unpaired electrons. Thus, carbon can form only two covalent bonds with other atoms. However, compounds of carbon show that carbon is tetravalent. Examples are CH4 and CH3 Cl. 5 Another example involves oxygen. The electronic configuration of oxygen is: O: 1s 2p O: 2s Oxygen has two unpaired electrons in the 2p orbitals. Overlap of these two singly occupied 2p orbitals with the 1s orbitals of two hydrogen atoms would produce the water molecule, H2 O. 6 Such overlap would give a bond angle of 90° as the two 2p orbitals are perpendicular to one another. However, experiment shows that the bond angle in the water molecule is 104.5° (which is closer to the angle of a tetrahedron of 109.5°). H S H 2py 2pz 2px H S H 7 To account for these discrepancies, the hybridisation theory was introduced. 8 Hybridisation is the mixing of two or more non-equivalent atomic orbitals to produce a set of equivalent hybrid orbitals. 9 In our syllabus, we are concerned with hybridisation involving the s and p orbitals only. Hybridisation involving d orbitals is not required. Beryllium has no unpaired electrons in its valence shell. Carbon has only 2 unpaired electrons in its valence shell. Definition of hybridisation 90° H O H
82 Chemistry Term 1 STPM CHAPTER 3 10 The number of hybrid orbitals produced is the same as the number of atomic orbitals that takes part in the hybridisation process. 11 There are three types of hybridisation involving the s and p orbitals. • sp hybridisation (pronounce as s-p) • sp2 hybridisation (pronounced as s-p-two and not s-p-square) • sp3 hybridisation (pronounced as s-p-three) 12 Hybridisation of an s and a p orbital produces two sp hybrid orbitals of equivalent energy. Hybridise Two sp hybrid orbitals sp hybrid orbitals around the nucleus s orbital p orbital 13 The energy profile for the hybridisation is Atomic orbitals Two sp hybrid orbitals p orbital s orbital Energy 14 For example, the mixing of two orbitals with energies of 40 and 60 kJ mol–1 respectively produces two hybrid orbitals with energy of 50 kJ mol–1 each. New hybrid orbitals Energy (kJ mol–1) 60 50 40 15 Hybridisation does not alter the number of valence shell electrons. So hybridisation can be ignored when working out the Lewis structure of a species. 16 However, one must take note that hybridisation is not an actual physical process. The theory was suggested to account for the discrepancies in the shape and number of covalent bonds in a molecule. VIDEO Hybridisation
83 Chemistry Term 1 STPM CHAPTER 3 sp3 Hybridisation 1 sp3 hybridisation is exemplified by the methane molecule, CH4 . 2 The valence shell electronic configuration of carbon is C: 2p C: 2s 3 During the formation of the CH4 molecule, one of the 2s electron is promoted (or excited) to one of the empty 2p orbitals. As a result, the excited carbon atom (C*) now has 4 unpaired electrons. C*: 2p C*: 2s 4 The four orbitals (one 2s and three 2p) then ‘mix’ with one another to produce four equivalent sp3 hybrid orbitals with the same shape and energy. C*: 2p C*: 2s C*: sp3 sp3 sp3 sp3 Mixing C*: 2p Energy 2s sp3 hybridisation The methane molecule: H | H—C—H | H
84 Chemistry Term 1 STPM CHAPTER 3 5 The four sp3 hybrid orbitals are directed towards the corners of a tetrahedron. 6 Overlap between the four sp3 hybrid orbitals with the 1s orbitals from four hydrogen atoms produces four C—H sigma-bonds with bond angles of 109.5°. C α α α α 109.5° H H H H sp3 sp3 1s sp3 1s sp3 1s 1s + z y x s z y x px z y x py z y x pz z y x z y x – z y x z y x
85 Chemistry Term 1 STPM CHAPTER 3 The H2 O molecule 1 The valence shell electronic configuration of oxygen is [He] 2s 2 2p 4 . O: 2p O: 2s 2 Since the oxygen atom already has two unpaired electrons, ‘promotion’ of electrons is not necessary. 3 The 2s and the three 2p orbitals ‘mix’ to produce four equivalent sp3 hybrid orbitals. 2p O: 2s sp3 sp3 sp3 sp3 ‘Mixing’ 4 The four sp3 hybrid orbitals are directed towards the corners of a tetrahedron. 5 Overlap of the two of sp3 hybrid orbitals, each containing an unpaired electron with the 1s orbital of two hydrogen atom, gives rise to two O—H sigma bond, with a bond angle of 104.5°. Example 3.3 Draw a diagram to show the type of orbitals that are involved in the formation of the chloromethane molecule, CH3 Cl. Solution Cl C H H H s orbital sp3 orbital p orbital sp3 hybridisation 2p Energy 2s sp3 hybridisation
86 Chemistry Term 1 STPM CHAPTER 3 sp2 Hybridisation 1 sp 2 hybridisation involves the mixing of one s orbital with two p orbitals to produce three equivalent sp 2 hybrid orbitals. 2 An example of sp 2 hybridisation is found in boron trihydride, BH3 . 3 The valence shell electronic configuration of the boron atom is 2s 2 2p 1 . B: 2p B: 2s 4 One electron from the 2s orbital is ‘promoted’ to one of the empty 2p orbital of boron to produce three unpaired electrons. B*: 2p B*: 2s 5 The excited boron atom then undergoes sp 2 hybridisation to produce three equivalent sp 2 hybrid orbitals. B*: 2p B*: 2s sp2 sp2 sp2 2p ‘Mixing’ BB: *: 6 The three sp 2 hybrid orbitals are arranged in a trigonal planar arrangement. + z y x s z y x px z y x z y x py z y x z y x z y x 120° sp2 hybridisation Energy 2p 2s sp2 hybridisation
87 Chemistry Term 1 STPM CHAPTER 3 7 Overlap between the three sp 2 hybrid orbitals and the 1s orbital of three hydrogen atoms gives rise to three B—H sigma bonds with bond angle of 120°. 8 However, there is still a 2p orbital in the boron atom which is not involved in the hybridisation or bond forming process. This unhybridised 2p orbital is at right angle to the plane containing the three sp 2 hybrid orbitals. Example 3.4 Draw the orbital overlap diagram for the formation of aluminium chloride, AlCl3 . Solution Al Cl p orbital sp2 orbital Cl Cl sp Hybridisation 1 sp hybridisation involves the mixing of one s and one p orbitals to produce two equivalent sp hybrid orbitals. 2 An example of sp hybridisation is found in beryllium hydride, BeH2 . 3 The valence shell electronic configuration of the beryllium atom is: Be: 2p Be: 2s 4 ‘Promotion’ of a 2s electron to an empty 2p orbital produces two unpaired electrons. This is followed by the mixing of the s and p orbitals (each containing an unpaired electron) to produce two sp hybrid orbitals. Be*: 2s 2p sp sp 2p ‘Mixing’ Be*: 120° sp2 p sp2 sp2 Empty unhybridised p orbital Planar B H H H 120° Energy 2p 2s sp hybridisation sp hybridisation
88 Chemistry Term 1 STPM CHAPTER 3 5 The two sp hybrid orbitals arranged themselves in a straight line. + z y x s z y x px z y x z y x z y x 180° 6 Overlap between the two sp hybrid orbitals with the 1s orbital of two hydrogen atoms produces two Be—H sigma bonds with a bond angle of 180°. 7 Note that there are two unhybridised 2p orbitals on the beryllium atom. These two p orbitals are perpendicular to one another, and also perpendicular to the plane containing the two sp hybrid orbitals. H σ σ Be Py Pz H Hybridisation and Multiple Bonds The formation of double bonds and triple bonds are also explained by the hybridisation theory. The Double Bond 1 The Lewis diagram of ethene, C2 H4 , shows that two carbon atoms are joined by a double bond which consists of four electrons shared between the two carbon atoms. 2 In the formation of the ethene molecule, each carbon atom undergoes sp2 hybridisation to produce three sp2 hybrid orbitals with one unhybridised p orbital perpendicular to the plane containing the three sp2 hybrid orbitals. H–Be–H • • • x x • x x C C H H H H
89 Chemistry Term 1 STPM CHAPTER 3 2pz sp2 sp2 sp2 z y x 3 The two carbon atoms then use their sp2 hybrid orbitals to form four C—H sigma bond with four hydrogen atoms and a C—C sigma bond with one another. C C Η Η Η σ σ σ σ σ 4 The two p orbitals then overlap sideways, giving rise to a π-bond as shown below. C C H H H H 2pz 2pz H H H H C C π π 5 Thus, the two electron-pairs in a double bond is distributed with 1 pair in a σ-bond and the other pair in a π-bond. π σ C H C H H H π π σ σ σ σ σ Η Η Η Η Overlap to form sigma bonds Overlap to form π bonds
90 Chemistry Term 1 STPM CHAPTER 3 The Triple Bond 1 The Lewis diagram of nitrogen molecule N2 is: N N It is a linear molecule with a nitrogen-nitrogen triple bond. 2 During the formation of the N2 molecule, each nitrogen atom undergoes sp hybridisation to produce two sp hybrid orbitals with two unhybridised 2p orbitals in the following arrangement. 180° p p sp sp C 3 Each of the nitrogen atoms uses one of its sp hybrid orbitals to form a sigma bond with one another. The remaining four unhybridised 2p orbitals then overlap sideways to produce two π bonds. N p π N π p sp fi Lone pair electrons A triple bond is made up of one σ bond and two π bonds. N ≡≡ N 1 σ-bond and 2 π-bonds 4 Another example of a triple bond molecule is hydrogen cyanide, HCN with the following Lewis diagram: NCH H—C ≡ N
91 Chemistry Term 1 STPM CHAPTER 3 5 Both the carbon and nitrogen atoms undergo sp hybridisation. The orbital overlap diagram is shown below. C p π N π p s sp H fi Valence-shell Electron-pair Repulsion Theory 1 The valence-shell electron-pair repulsion theory was proposed by N. V. Sedgwick and H. W. Powell to predict the general shape of molecular species. 2 The theory states that electron-pairs (bonding pairs and lone pairs) will repel one another and orientate themselves to be as far apart as possible. 3 The force of repulsion increases in the order: Weakest * bonding pair – bonding-pair * lone pair – bonding pair Strongest * lone pair – lone pair [This is because lone pair electrons are closer to the nucleus and hence exert a greater repulsive force.] 4 Multiple bonds (double and triple bonds) are treated like single bonds and are assumed to occupy the position as single bond. 5 The table below lists the general shape of molecular species together with the number and type of electron pairs surrounding the central atom. Shape Total of electron pairs No. of bonding pairs No. of lone pairs Linear 2 2 0 Bent •• 3 2 1 Trigonal planar 3 3 0 Tetrahedron 4 4 0 2015/P1/Q16(a) 2014/P1/Q6, Q19(a) 2016/P1/Q5