CHEM 101_L4b Atomic Structure & Atomic Properties of the Elements PDF

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Botswana International University of Science & Technology

Dr Disah Mpadi

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atomic structure atomic properties electromagnetic radiation chemistry

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These lecture notes cover atomic structure and atomic properties, including electromagnetic radiation and its interactions with matter. The document details concepts like electromagnetic radiation, interference, and diffraction, as well as important discoveries in quantum theory.

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Atomic Structure & Atomic Properties of the Elements Dr Disah Mpadi Office: Block 1, Room 16 Email: [email protected] 1 Part II 2 Outline Electrons in Atoms: ▪ Electromagnetic Radiation ▪ Impo...

Atomic Structure & Atomic Properties of the Elements Dr Disah Mpadi Office: Block 1, Room 16 Email: [email protected] 1 Part II 2 Outline Electrons in Atoms: ▪ Electromagnetic Radiation ▪ Important Discoveries Leading to Quantum Theory ▪ Two Ideas Leading to Quantum Mechanics ▪ Quantum Theory of the Hydrogen Atom: The Schrodinger Equation ▪ A Fourth Quantum Number: Electron Spin ▪ Multielectron Atoms ▪ Electron Configurations & The Periodic Table of Elements 3 Overview The electronic structure of atoms is better understood by gaining knowledge on the interactions of electromagnetic radiation and matter. 4 The Electron Movement & Electromagnetic Radiation Note: The description of the position of electrons relative to the atomic nucleus is closely related to absorption & emission of electromagnetic radiation. Note: Energy can be transmitted or transported by electromagnetic radiation as waves. 5 Cont’d ▪ Types Electromagnetic (EM) radiation: radio waves, microwaves, infrared, visible light, ultraviolet light, X-rays, & gamma rays. ▪ EM radiation consist of energy propagated by transverse oscillating electric & magnetic fields that are perpendicular to each other, & alternatively increase & decrease in intensity (watts/cm2) as they move through space. 6 Cont’d The Electromagnetic Spectrum: 7 Cont’d ▪ White light from the Sun consists of light with all wavelengths in the visible spectrum; & light with different wavelengths have different colours. ▪ This white light can be diffracted with a prism or by a raindrop to form a continuous spectrum & contains all visible colours, with no breaks in between the colours. ▪ In contrast to a continuous spectrum from the Sun, when samples of elements are burned, light is emitted, which is diffracted into a so-called line spectrum when it passes through a prism, characteristic of an element. ▪ The colours we see in nature results from an object absorbing some wavelengths of visible light & reflecting others back to our eyes. 8 Cont’d Frequency, Wavelength, & Amplitude: The following three variables describe the wave properties of EM radiation: ▪ Wavelength (𝝀, lambda) — the distance from one wave peak/crest to the next (or from one wave trough to the next) (units, m). ▪ Frequency (𝝂, nu) — the number of wave peaks that pass by a given point per unit of time (units of cycles per second, s-1, Hz [Hertz]). ▪ Amplitude — the height of the wave, measured from the center line between peak and trough; the intensity or brightness of the light is proportional to the square of the amplitude. 9 Cont’d The Speed of Light: Relating 𝜈 & 𝜆 ▪ All EM radiation travel at the same speed in a vacuum: about 2.998 x 108 ms-1 (speed of light, c). ▪ The frequency (𝝂) of EM radiation is inversely proportional to its wavelength (𝝀) through the equation, where c is constant: 𝒄 = 𝝂𝝀. ▪ Types of EM radiation differ by their wavelength & frequency. ▪ Visible light extends from wavelengths of 750 nm (red) to 400 nm (violet); these wavelengths make white light. 10 Example: Relating frequency & wavelength of electromagnetic radiation Most of the light from a sodium vapour lamp has a wavelength of 589 nm. Calculate the frequency of this radiation. Analyse: Rearrange the formula, 𝑐 = 𝜈𝜆, to solve for 𝜈. Solve: Change the units of 𝜆 from nanometers (nm) to metres (m): 1 𝑥 10−9 𝜆 = 589 nm 𝑥 = 5.89 𝑥 10−7 m 1 nm 2.998 𝑥 108 ms −1 14 s −1 = 𝟓. 𝟎𝟗 𝒙 𝟏𝟎𝟏𝟒 𝐇𝐳 𝜈= = 5.09 𝑥 10 5.89 𝑥 10−7 m 11 Cont’d An important characteristic of electromagnetic waves/light: Interference: ▪ Waves can interfere with each other constructively (wave amplitudes add, in-phase); ▪ resultant wave with increased intensity. ▪ or destructively (wave amplitudes cancel, out-of-phase); ▪ resultant wave with decreased intensity. Adding two waves: amplitude vs spatial coordinate. ▪ Many wave phenomena involve Waves have + & - amplitudes. interference effects. 12 Cont’d Diffraction (a consequence of interference): ▪ A wave bends around an object (grating & pinhole) that is comparable in size to its wavelength, producing a new set of waves in the process, e.g., XRD instrument. ▪ In case of a stream of particles, they would either be blocked by the barrier or go through the opening without being diffracted. 13 Cont’d Refraction (prism & lens): Is another property of light that makes light a wave. Reflection (bouncing off solid objects): Is also another property of light that makes it a wave. 14 Important Discoveries Leading to Quantum Theory The development of quantum theory was driven by numerous experiments, each involving the interaction of light (electromagnetic radiation) and matter. ▪ The following are some discoveries that demonstrated that light not only has wave properties, but also particle properties: ▪ Blackbody radiation. The photoelectric effect Atomic line spectra (Energy quantisation) 15 Cont’d ▪ When iron is heated, it first glows a dull red color, becomes orange as the Blackbody radiation: Light emission by temperature increases, then a blinding heated solids (classical physics). white glare, & eventually a dull blue. ▪ The longer wavelengths (red) have a lower intensity, & the shorter wavelengths (white, blue) have a higher intensity. ▪ If this trend were to continue, the intensity would be even higher in the UV range; ▪ instead, the intensity reaches a maximum, & then falls off with shorter wavelengths. 16 Cont’d ▪ Classical electromaganetic theory (light is purely wave, with energy continuously variable) couldn’t explain the variation in radiation emitted at different temperatures by the Blackbody. ▪ From Quantum Mechanics (QM) (particle model), Max Planck concluded that the energy radiated by a blackbody can’t be continuously variable, ▪ instead, the energy is emitted/absorbed only in discrete amounts or quanta (singular- quantum). ▪ Thus, each change in the energy of the atoms in a heated substance results from the gain or loss of these ‘packets’ of energy. 17 Cont’d ▪ This amount of ‘packets’ of energy, 𝜠, associated with each quantum of energy is given by the Planck’s equation: 𝒉𝒄 𝜠photon = 𝒉𝝂 = ; ℎ = 6.626 x 10-34 J s 𝝀 𝒏𝒉𝒄 or 𝜠photons = 𝒏𝒉𝝂 = ; where 𝒏 is an integer, equivalent to the number of 𝝀 photons. 18 Example: Using Planck’s equation to calculate the energy of photons of light For radiation of wavelength 242.4 nm, the longest wavelength that will bring about the photodissociation of O2, compute the energy of (a) one photon, & (b) a mole of photons of this light. Analyse: 𝒉𝒄 Use Planck’s equation, 𝑬 = 𝒉𝝂 = since 𝒄 = 𝝂𝝀, which is written for one photon of light. Once we have energy per photon, we can multiply it by the Avogadro constant to convert to a per-mole basis. 𝝀 Solve: Don’t forget to change the units of 𝜆 from nanometers (nm) to metres (m): ℎ𝑐 (6.626 𝑥 10−34 J.s)(2.998 𝑥 108 ms−1 ) (a) 𝐸 = ℎ𝜈 = = = 8.196 𝑥 10−19 J. mol−1 𝜆 (242.4 𝑥 10−9 m) 6.022 𝑥 1023 photons (a) Energy of a mole of photons: 𝐸 = 8.196 𝑥 10 −19 J. mol −1 𝑥 = 4.936 𝑥 105 J. mol−1 1 mol 19 ▪ The photoelectric effect occurs when light Cont’d shines/impinges on a flat metal surface, causing it to emit electrons. The Photoelectric effect: ▪ Albert Einstein explained the photoelectric effect by proposing that light behaves as if it were; ▪ ▪ a stream of small, massless discrete packets of energy, called photons, having an energy given by the Planck equation: ▪ 𝜠𝒑𝒉𝒐𝒕𝒐𝒏 = 𝒉𝝂 = 𝒉𝒄Τ𝝀 ▪ The energy of a photon of EM radiation is directly proportional to its frequency (color, wavelength); while the intensity (brightness) of the light is due to the number of photons. 𝑲𝒎𝒂𝒙 = 𝒉(𝝂 − 𝝂𝟎 ) = 𝒉𝝂 − 𝝓; 𝝓 = 𝒉𝝂𝟎 20 Cont’d ▪ If the frequency of light is below a specific threshold frequency, 𝝂𝟎 , no electrons are emitted, no matter how bright the light is. ▪ If the frequency is above 𝝂𝟎 , the number of electrons emitted increases with the intensity of the light. The kinetic energy of the electrons increases with the frequency. 𝑲𝒎𝒂𝒙 (excess energy) = 𝒉(𝝂 − 𝝂𝟎 ) = 𝒉𝝂 − 𝝓; 𝝓 = 𝒉𝝂𝟎 ▪ When a low-energy (low 𝝂) photon strikes the metal, nothing happens, but a high- energy (high 𝝂) photon strikes with enough force to knock an electron loose. ▪ 𝝓 is a work function of metal & is different for each metal. 21 Cont’d Atomic Line Spectra: ▪ Early clues to the arrangement of electrons inside the atom came from the discovery of line spectra; ▪ & this gave chemists insights in understanding the atomic structure of matter. ▪ If an electric current is passed through a sample of an element in the gas phase, the sample emits light (e.g., hydrogen gas, neon or fluorescent lights). ▪ And if this light is passed through a prism, a line spectrum is observed, consisting of a series of discrete lines separated by blank areas. 22 Cont’d Atomic Fingerprints- Atomic Emission & Absorption Spectra: Prism/grating diffraction Each element in the periodic table has its unique set of wavelengths of radiation; (i.e., UV, Vis, IR) that it emits when energetically excited, & the spectral lines it exhibits can be used as a ‘fingerprint’ for the identification of that element. Also, elements absorb the same wavelengths of light that they emit or radiate when they are heated; if white light is passed through a sample & observing what wavelengths of light are subtracted from it, then the element can be characteristically be identified in the sample. Flame Tests & Pyrotechnics: Metal salts are used in fireworks to produce different colors; & the colours of a flame can be used to identify what metal ions are present in a solution (Flame test). 23 Cont’d The frequencies of light in the atomic line spectra correspond to the energy QM & Atomic Line Spectra: differences between the orbitals in the ▪ Each 𝝀 in an atomic line spectrum corresponds atom. to an energy transition between orbitals. ▪ If an atom in its ground state absorbs energy (heat or light), an electron in a lower energy level is promoted to higher-energy orbital & that’s an excited state; ▪ An excited atom is unstable & electron goes back to the ground state by emitting this excess energy as a photon of EM radiation. ▪ The energy of the photon (𝒉𝝊) which is emitted corresponds to the energy difference between the excited state & ground state. 24 Cont’d What does the atomic line spectrum tell us? ▪ The spectroscopic observations of atoms suggest that an electron inside an atom can occupy only certain energy levels &; ▪ that the emission of discrete frequencies of EM radiation occurs when an electron makes a transition between these levels. ▪ The energy levels in atoms are quantised just like the energy of photons. ▪ The existence of line spectra is related to the structure of the atom; ▪ how electrons are arranged & the energy they possess. 25 Cont’d ▪ This spectroscopic observation led to the development of the Bohr model of the atom, & then to quantum mechanics. ▪ Johann Rydberg found that the position & all the wavelengths of any of the spectral lines can be described by the expression: 𝟏 𝟏 𝟏 = 𝝂෤ = 𝑹𝑯 − , 𝒏 are integers, with 𝝀 𝒏𝟐𝒇 𝒏𝟐𝒊 𝒏𝒇 &...,2 ,1 = 𝒏𝒊 = 𝑛𝑓 ,1+𝑛𝑓... ,2+ ▪ The Balmer-Rydberg equation is an empirically derived equation with an empirical constant, 𝑹𝑯 1.097 = x 107 m-126. Cont’d The Bohr Model of the Atom: ▪ The Rutherford model of the atom: the model treated an atom like a miniature solar system, where electrons orbit the nucleus just like planets orbiting the Sun; ▪ Limitation: moving charges emit EM radiation & would continuously lose energy, suggesting that an electron would crash into the nucleus in no time. ▪ The Bohr model: Based on the observed line spectrum of hydrogen atom, Bohr suggested that the energy levels in atoms are quantised, having only certain allowed energy levels associated with fixed electron orbits (or stationary states). 27 Energy Absorption in the Bohr Model: Cont’d Energy Emission in the Bohr Model: ▪ An atom absorbs an amount of energy equal to the energy difference between two fixed orbits, thus, an electron transition from low energy orbit to a higher energy orbit: ▪ An atom releases energy when an electron falls from a higher-energy orbit (an excited state) to the low energy orbit (the ground state): ▪ Note: the atomic line spectra arise becoz the atom’s energy has only certain allowed orbits & atoms can absorb or emit energy only in these ‘chunks’ or ‘packets’. 28 Cont’d ▪ Neils Bohr also stated that energy is absorbed or emitted only when an electron moves from one stationary state (orbit) to another & the energy change is given by: ∆𝜠 = 𝜠𝒏𝟐 - 𝜠𝒏𝟏 =𝒉𝝂; where 𝒏𝟏 & 𝒏𝟐 are the principal quantum numbers referring to the energy levels 𝜠𝒏𝟏 & 𝜠𝒏𝟐 ,respectively. ▪ Energy of an electron can also be obtained by combining: 𝒉𝒄 1 𝟏 𝟏 ∆𝜠 = ℎ𝜈 = and = 𝑅𝐻 ( − ); to get: 𝝀 𝜆 𝒏𝟐𝒇 𝒏𝟐𝒊 𝟏 𝟏 ∆𝜠 = 𝒉𝒄𝑅𝐻 ( − ) 𝒏𝟐𝒇 𝒏𝟐𝒊 𝟏 𝟏 ∆𝜠 = (−)𝟐. 𝟏𝟕𝟗 𝐱 𝟏𝟎−𝟏𝟖 𝐉 − ; 𝒏𝟐𝒇 𝒏𝟐𝒊 for IE (ionization energy) of an atom, 𝒏𝒊 = ∞. Note: Based on a transition, this formula can reflect either emission (- ve value of ∆𝜠) or absorption (+ve value of ∆𝜠). 29 Cont’d Energy levels, spectrum, & ionization energy of the hydrogen atom-Based on line spectra: Fig. 1: The emission spectrum of atomic hydrogen: some transitions for Lyman & Balmer series. 30 Cont’d Energy levels, spectrum, & ionization energy of the hydrogen atom-Based on line spectra: ▪ The series with 𝒏𝒇 = 𝟏 is called the Lyman series & lies in the UV region. ▪ The series with 𝒏𝒇 = 𝟐 is called the Balmer series & lies in the visible region. ▪ The infrared series include the Paschen series (𝒏𝒇 = 𝟑), Brackett series (𝒏𝒇 = 𝟒), & the Pfund series (𝒏𝒇 = 𝟓). ▪ All spectral lines in all the series obey the general Balmer-Rydberg equation. 31 Cont’d Energy levels, spectrum, & ionization energy of the hydrogen atom-Based on line spectra: ▪ Unfortunately, the equation could not explain why the hydrogen atom produced line spectra as opposed to a continuous spectrum & even why this equation worked or failed to predict the location of atomic line spectra for many- electron elements. Fig. 2: The spectrum of atomic hydrogen & its analysis 32 into series. Cont’d The Bohr Model of the Hydrogen Atom: 33 Cont’d Example: Computing the wavelength of a line in the hydrogen spectrum Calculate the wavelength of the line in the Balmer series of hydrogen corresponding to the transition from 𝑛 = 5 to 𝑛 = 2. Analyse: The transition is from a higher to a lower energy level, so energy (a photon) is emitted by the atom. Two approaches: (a) The wavelength, 𝜆, can be calculated directly using Balmer-Rydberg equation. ℎ𝑐 (b) The energy of the emitted photon 𝐸photon = ℎ𝜈 = is equal to Δ𝐸 , the magnitude of the energy difference between the two levels involved, & 𝜆 can 𝜆 finally be solved. 34 Cont’d Solve: (a) 1 1 1 = 𝑅𝐻 2 − 2 𝜆 𝑛𝑓 𝑛𝑖 1 1 1 = 1.097 𝑥 107 m−1 2 − 2 = 1.097 𝑥 107 m−1 0.250 − 0.040 = 2.303700 𝑥 106 m−1 𝜆 2 5 1 𝜆= 6 −1 = 𝟒𝟑𝟒. 𝟏 𝒙 𝟏𝟎−𝟗 𝐦 = 𝟒𝟑𝟒. 𝟏 𝐧𝐦 (2.303700 𝑥 10 m ) 35 Cont’d Solve: (b) 1 1 1 1 ∆𝛦 = −2.179 𝑥 10−18 J 2 − 2 = −2.179 𝑥 10 −18 𝐽 2− 2 𝑛𝑓 𝑛𝑖 2 5 = −2.179 𝑥 10−18 J 0.250 − 0.040 = −𝟒. 𝟓𝟕𝟔 𝒙 𝟏𝟎−𝟏𝟗 𝐉 1 1 Note: A one may use ∆𝐸 = 𝑅𝐻 ( 2 − 2) to directly obtain a positive value without 𝑛𝑖 𝑛𝑓 bothering if the energy is an emission or an absorption. ℎ𝑐 We can compute the wavelength using the equation: 𝜆 =. |Δ𝐸| (6.626 𝑥 10−34 J s photon−1 )(2.998 𝑥 108 m s −1 ) −𝟕 𝐦 = 𝟒𝟑𝟒. 𝟏 𝐧𝐦 𝜆= = 𝟒. 𝟑𝟒𝟏 𝒙 𝟏𝟎 | − 4.576 𝑥 10−19 J photon−1 | 36 Two Ideas Leading to Quantum Mechanics Wave-particle Duality of light & an electron (Remember the Photoelectric effect): ▪ Wave-particle duality was coined by: ▪ Louis de Broglie phase wave equation, ▪ & Neils Bohr and Werner Heisenberg uncertainty principle. Electrons & Light can behave as particle or as wave. 37 Cont’d The wavelike properties of matter: ▪ The wavelength of a particle of matter can be derived by combining the equation for the energy of a photon ▪ Remember that light has & Einstein’s equation for the properties of waves & of relationship of mass & energy; particles. ▪ substituting the speed of light (c) with the speed of a particle (𝝊): ▪ Louis de Broglie suggested that if light can have particle-like 𝜠 = 𝒉𝝂 = 𝒉𝒄/𝝀, 𝜠 = 𝒎𝒄𝟐 properties, The de Broglie phase wave relation: ▪ then matter can have wave-like 𝒉 properties due to phenomena 𝝀= observed from the interaction of 𝒎𝝊 matter & radiant energy, i.e., The photoelectric effect. 38 Cont’d ▪ Both matter & light are both wave- like & particle-like. ▪ Thus, to model how electrons are arranged inside atoms, both their particle properties & their wave ▪ Based on the examples above; a matter with large mass (i.e., football, properties must be considered. car, or person), ▪ Note: electrons are particles & have ▪ the de Broglie wavelength is so small that wave-like properties are minute, & a mass; the objects behave entirely like particles. ▪ they can diffract just like light does & ▪ & matter with small masses (i.e., produce interference patterns (double electrons or photons) the de Broglie slit experiment), wavelength is large enough to be measurable, ▪ which means electrons possess wave properties. ▪ & the wave properties of these particles become important. 39 Example: Calculating the wavelength associated with a beam of particles Calculate the wavelength associated with electrons traveling at one-tenth the speed of light. Analyse: 𝒉 Use de Broglie phase wave equation, 𝝀 =. We need electron mass & Planck’s constant. 𝒎𝝊 Solve: 1 The electron velocity is: 𝑣 = 𝑥 𝑐 = 0.100 𝑥 2.998 𝑥 108 m s −1 = 2.998 𝑥 107 m s −1 10 Planck’s constant, ℎ = 6.626 𝑥 10−34 J s = 6.626 𝑥 10−34 kg m2 s −1 The electron mass, 𝑚𝑒 = 9.109 𝑥 10−31 ℎ 6.626 𝑥 10−34 kg m2 s −1 −𝟏𝟏 𝐦 = 𝟐𝟒. 𝟐 𝐩𝐦 𝜆= = = 𝟐. 𝟒𝟐 𝒙 𝟏𝟎 𝑚𝜐 (9.109 𝑥 10−31 kg)(2.998 𝑥 107 m s −1 ) 40 Cont’d The Heisenberg Uncertainty Principle: ▪ According to Louis de Broglie, electromagnetic (EM) radiation could be considered to consist of particles called photons yet at the same time exhibit wave-like properties (i.e., diffraction, interference), ▪ then same might be true of electrons. ▪ This dual nature is referred as wave-particle duality. ▪ The immediate consequence of duality is that it is impossible to know the linear momentum (𝑝 = 𝑚𝜐) & the location of an electron (& any particle) as a wave simultaneously. 41 Cont’d ▪ This restriction for an electron is called the Heisenberg uncertainty principle. ▪ The Heisenberg uncertainty principle can be stated as follows: ▪ that the product of the uncertainty in momentum & the uncertainty in position cannot be less than a quantity of the order of Planck’s constant: 𝒉 𝟏 ∆𝒙 𝒎∆𝝊 ≥ or ∆𝒙 𝒎∆𝝊 ≥ 𝟒𝝅 𝟐ℏ ▪ stated qualitatively: that it is impossible to know simultaneously the exact position & momentum of an electron-wave beyond a certain level of precision. ▪ Position & velocity are complementary properties such that you know more about one, the less you know about the other. 42 Quantum Theory of the Hydrogen Atom: The Schrodinger Equation The Quantum Mechanics (QM) model of the Atom: Wave-nature of electrons ▪ The Bohr model was superseded by the Schrodinger equation due to its limitations. ▪ QM is the discipline of physics & chemistry which describes the properties of light & matter at the atomic scale. ▪ QM unifies the wave & particle models into a single description of reality. ▪ QM concentrate more on the electron’s wave-like properties than its particle properties; ▪ (classical quantum theory-electron is treated as a particle) where is a moving particle around the nucleus in a defined 43 path. Cont’d ▪ Quantum mechanics (QM) is also known as Wave mechanics. ▪ Note: Erwin Schrödinger, Louis de Broglie, & Werner Heisenberg independently combined the wave-particle duality of matter & energy in the QM model of the atom. ▪ The electron bound to the nucleus was perceived as a standing wave. ▪ Schrodinger produced a mathematical treatment to fit this wave behaviour of an electron, but its too complicated to be detailed here. ▪ The form of Schrodinger equation is ෡ = 𝐸𝜓 𝐻𝜓 44 Cont’d Standing waves in atoms: ▪ So, only standing wave with a whole number of wavelengths (or An electron is a circular standing wave integer numbers of wavelengths); ▪ fit-well the size of circular orbits of an electron in a hydrogen atom; ▪ And all other fractional number of half-waves in orbits generate destructive interference & are not allowed. 45 Cont’d ෡ = 𝐸𝜓 𝐻𝜓 ▪ The wavefunction (𝝍) is a mathematical function of the position coordinates (𝑥, 𝑦, 𝑧)-3D space, which describes the behaviour of an electron. ▪𝑯෡ represents a set of mathematical instructions called an operator. ▪ 𝑬 represents the total energy of the atom (𝐾𝐸 plus potential energy (𝑉)). ▪ The solutions from the Schrodinger wave equation consists of a wavefunction (𝜓), that is characterized by certain values of 𝑬. ▪ Therefore, the quantisation of energy, value of 𝒏, & thus an electron can possess only certain discrete energies in an atom. 46 Cont’d ▪ When the Schrodinger wave equation, a solution to a specific wavefunction (𝜓), for an electron in an atom is called an atomic orbital (which describes behaviour of an electron in a region of space). ▪ The probability of finding an electron at a given location/region of space is proportional to the square of the wavefunction at that point, 𝝍𝟐. ▪ Therefore, there is a high probability of finding an electron where 𝝍𝟐 is large, & the electron will not be found where 𝝍𝟐 is zero. ▪ The quantity 𝝍𝟐 is called the probability density of the electron. 47 Cont’d ▪ The Schrödinger Wave Equation can be solved exactly for several simple systems-only one electron (such as 1H atom; 42He+ ) ▪ And can be solved approximately for many more complex systems (such as many-electron atoms & molecules). 48 Cont’d Atomic Orbitals Key points: ▪ The energy of the bound electron is determined by 𝒏, the principal quantum number (specifies energy & size of the orbital). ▪ The 𝒍 specifies the magnitude of the orbital angular momentum quantum number (formerly ‘azimuthal or secondary quantum number’) or angular shape of the orbital. ▪ The 𝒎𝒍 , magnetic quantum number, specifies the orientation of the orbital angular momentum (shape) or its orientation in space. ▪ The 𝒎𝒔 specifies the orientation of spin axis or how many electrons can occupy that orbital. 49 Cont’d Atomic Orbitals Shapes + Wavefunction & interference: + - + - + ▪ Wavefunctions (atomic orbitals) generally have - regions of +ve & -ve amplitude, or signs (+ or -); normally shown as colour shades. + - + + - - - + - ▪ The signs are important when the wavefunctions + + spread into the same region of space & interact. - + - + ▪ One orbital can hold no more than two + - + - electrons. 50 Cont’d Quantum numbers: ▪ Remember: ▪ Each wavefunction for an electron in an atom, obtained by solving Schrödinger wave equation for a hydrogenic atom is called atomic orbital; & is uniquely labelled by a set of quantum numbers: 𝒏, 𝒍, 𝒎𝒍 ,𝒎𝒔. ▪ 𝒏 (principal quantum number) is an integer ranging from 1 to infinity, 𝒏 = 1, 2, 3,..., ∞. ▪ A larger 𝒏 means higher-energy orbitals & more diffuse than low 𝒏 (low-energy orbital)- where orbitals are more compact, tightly bound. ▪ The allowed energies are specified by 𝒏, as 𝒏 increases, the number of allowed orbitals increases as well as their size increases; ▪ allowing an electron to be farther from the nucleus & higher in energy. 51 Cont’d ▪ The total number of orbitals for a given 𝒏 value is 𝒏𝟐. ▪ Shells with 𝒏 = 1, 2, 3,... are sometimes referred to as K, L, M,... (esp. when electronic transitions between these shells are referred to in X-ray spectroscopy). 52 Cont’d Angular Momentum Quantum Number, 𝒍: ▪ The orbitals belonging to each shell (𝒏) are classified into subshells distinguished by a quantum number 𝒍. ▪ 𝒍 divides the shells into smaller groups of orbitals called subshells (sublevels). ▪ 𝒍 specifies the shape of an orbital with a particular value of 𝒏. Determines the angular momentum. ▪ 𝒍, is an integer ranging from 0 to 𝒏 − 𝟏 for a given value of 𝒏: 𝒍 = 𝟎, … , 𝒏 − 𝟏, giving 𝒏 different values in ALL. ▪ Normally, a subshell, 𝒍, is referred to by letter to avoid confusion with 𝒏: Value of 𝒍 0 1 2 3 4 Subshell designation 𝒔 𝒑 𝒅 𝒇 𝒈 53 Cont’d ▪ Also, the value of 𝒍 has a slight effect on the energy of the subshell; ▪ the energy of the subshell increases with 𝒍 (𝒔 < 𝒑 < 𝒅 < 𝒇). 54 Cont’d Magnetic Quantum Number, 𝑚𝑙 : ▪ 𝒎𝒍 specifies the orientation of the momentum or orientation in space of an orbital of a given energy (𝒏) & shape (𝒍), i.e., whether the circulation is clockwise or anticlockwise. Also determines the behavior of particle (electron) in magnetic field. ▪ 𝒎𝒍 divides the subshell (𝒍) into individual𝟐 orbitals which hold the electrons. 𝒍 consists of 𝟐𝒍 + 𝟏 individual orbitals, & 𝒏 orbitals in each shell. ▪ These orbitals are distinguished by 𝒎𝒍 which can have the integer values from − 𝒍 down to +𝒍. ▪ Thus, the 𝒑 subshell (𝒍 = 𝟏), consists of 3 individual atomic orbitals such that 𝒎𝒍 = +1, 0, -1. Write the sign of even when it is +ve, i.e., 𝒎𝒍 = +3, NOT 𝒎𝒍 = 3. 55 Cont’d The Pauli Exclusion Principle: ▪ States that NO two electrons in an atom can have the same values for all four of the quantum numbers; ▪ This means that if two electrons have the same orbital with the same quantum numbers (𝒏, 𝒍, 𝒎𝒍 ), then they must have opposite spins No two electrons in the same atom can have (or spin-paired, 𝒎𝒔 ), & thus, an the same four quantum numbers. orbital can hold at most two electrons. 56 A Fourth Quantum Number: Electron Spin Spin magnetic quantum number, 𝑚𝑠 : ▪ The intrinsic spin angular momentum of an electron is defined by two quantum numbers 𝒔 & 𝒎𝒔. ▪ We discussed 3 quantum numbers (𝒏, 𝒍, 𝒎𝒍 ), which relate to the intrinsic angular momentum of an electron, its spin. ▪ The four quantum numbers (𝒏, 𝒍, 𝒎𝒍 , 𝒎𝒔 ) are needed to define the state of an electron in an atom. ▪ 𝒎𝒔 tells us how many electrons can occupy a particular orbital & the orientation of the spin axis of an electron. 57 Cont’d ▪ An electron can be regarded as having an angular momentum arising from a spinning motion, rather like the daily rotation of the earth (𝒔) & travels in its orbit around the Sun (𝒍). Spin is described by two quantum numbers, 𝒔 & 𝒎𝒔 : ▪ 𝒔 is the analogue of 𝒍 orbital motion but restricted to the single, unchangeable value 𝒔 = 𝟏/𝟐. 𝟏 ▪ The 2nd quantum number (QN), 𝒎𝒔 ,may take two values: + (anticlockwise 𝟐 𝟏 spin) & − (clockwise spin). 𝟐 58 Cont’d ▪ An electron spins either clockwise or anti-clockwise to create a tiny magnetic field, which can be ▪ The two 𝒎𝒔 states are often oriented up or down: this quantum represented by the two arrows phenomena is exploited in NMR or 𝟏 (‘spin-up’, 𝒎𝒔 = + ) & (‘spin- MRI instruments. 𝟐 𝟏 down’, 𝒎𝒔 = − ) or by the 𝟐 Greek letters 𝛼 & 𝛽, respectively. 59 Cont’d Example: Applying relationships among quantum numbers (a) Identify a set of orbitals defined by 𝑛 = 4 & 𝑙 = 1, & state how many orbitals are there in this set. (b) List the possible types of atomic orbital for 𝑛 = 4. Analyse: Determine whether the given set of quantum numbers is allowed by the rules expressed above for each quantum number. Solve: The 𝑙 quantum number cannot be greater than 𝑛 − 1, & 𝑚𝑙 can only be equal to +𝑙,…,0,…, −𝑙. (a) Thus, if 𝑛 = 4 & 𝑙 = 1 (a 𝑝 orbital), them 𝑚𝑙 = +1,0, −1 (which means there are 3 𝑝 orbitals). (b) Thus, if 𝑛 = 4, 𝑙 maximum = 𝑛 − 1 = 4 − 1 = 3, so that 𝑙 = 0, 1, 2, 3 , which are s, 𝑝, 𝑑, 𝑓 orbitals, respectively. Orbital designations are based on 𝒏𝒍, the list is 𝟒𝒔, 𝟒𝒑, 𝟒𝒅, 𝟒𝒇. 60 Cont’d Example: Relating orbital designations & quantum numbers Write an orbital designation corresponding to the quantum numbers 𝑛 = 4, 𝑙 = 2, 𝑚𝑙 = 0. Analyse: To write orbital designations, you must recall the rules & conventions associated with the quantum numbers 𝑛 and 𝑙. The magnetic quantum number, 𝑚𝑙 , is not reflected in the orbital designation, but only 𝑛 & 𝑙 is reflected as 𝑛𝑙. Solve: The type of orbital is determined by the 𝑙 quantum number, becoz 𝑙 = 2, the orbital is of the 𝑑 type. Since 𝑛 = 4, the orbital designation is 4𝑑. 61 Cont’d Example: Choosing an appropriate combination of the four quantum numbers: 𝒏 ,𝒍 ,𝒎𝒍 ,𝒎𝒔. From the following sets of quantum numbers (𝑛, 𝑙, 𝑚𝑙 , 𝑚𝑠 ), identify the set that is correct, and state the orbital designation for those quantum numbers. (2,1,0,0) (2, 0,1,1/2) (2,2,0,1/2) (2,-1,0,1/2) (2,1,0,-1/2). Analyse: If 𝑛 = 2, 𝑙 has two possible values: 0 or 1. The range of values for 𝑚𝑙 : 𝑚𝑙 1 = +𝑙,…,0,…, −𝑙 (𝑚𝑙 can never be greater than 𝑙), and 𝑚𝑠 = ±. From the 2 information provided, you can judge which combination is correct. 62 Cont’d Solve: (𝒏, 𝒍, 𝒎𝒍 , 𝒎𝒔 ) Comment (2, 1, 0, 0) The value of 𝒎𝒔 is incorrect. (2, 0, 1, 1/2) The value of 𝒎𝒍 is incorrect. (2, 2, 0, 1/2) The value of 𝒍 is incorrect. (2, -1, 0, 1/2) The value of 𝒍 is incorrect. (2, 1, 0, -1/2) All the quantum numbers are correct. The correct combination is (2,1,0,-1/2), which corresponds to a 2𝑝 orbital. 63 Multielectron Atoms Multielectron atom: ▪ An atom with more than one electron. ▪ The exact solution of the Schrödinger equation for an atom with 𝑵 electrons would be a function of the 𝟑𝑵 coordinates of ALL the electrons, therefore, it will be extremely challenging to find exact formulas for such complicated systems. ▪ Orbital approximation in which each electron occupies an atomic orbital that resembles those found in hydrogenic atoms are relied upon. ▪ Multi-electron orbitals are similar in shape & have the same nodal structure as corresponding hydrogen one-electron orbitals. ▪ So, an electron occupying an atomic orbital is described by the corresponding wavefunction & set of quantum numbers. 64 Electron Configurations & The Periodic Table of Elements The building-up principle/The Aufbau Principle: ▪ Is a procedure that leads to plausible ground-state electron configurations; ▪ a set off rules that guide the order in which orbitals are filled; ▪ provides a theoretical framework for understanding the structure & implications of the periodic table; ▪ orbitals of neutral atoms are being occupied in the order determined by 𝒏 & in part by in 𝒍 in multi-electron atoms. 65 Cont’d Energy levels of one-electron atoms (H, For many-electron atoms, the He+): the energy of the orbitals depends only orbital energy depends on 𝒏 & 𝒍 on 𝒏. (𝒔 < 𝒑 < 𝒅 < 𝒇). 1-electron energies Multi-electron energies 66 Cont’d Ground-state electron configuration: ▪ Each successive electron added to an atom occupies the lowest-energy orbital available before higher-energy orbitals. Order of occupation of orbitals: Note: The total number of electrons dictate the total number of orbitals that will be occupied. 67 Cont’d ▪ An orbital can accommodate only two electrons, which must be spin- paired (Pauli exclusion principle). ▪ For instance, the 3 orbitals in 𝒑 subshell hold a total of 6 electrons. ▪ When two or more degenerate orbitals are available for occupation, we adopt Hund’s rule: ▪ When more than one orbital has the same energy, electrons occupy separate orbitals & do so with parallel spins (↑↑) until all are half-full; only then does a 2nd electron occupy the orbital to be spin-paired; ▪ This is a consequence of the mutual repulsion between like charged- electrons to minimise the energy. ▪ Simply stated: degenerate orbitals are occupied singly before being doubly occupied. 68 Cont’d Note: The Aufbau principle is used in conjunction with Pauli exclusion principle & the Hund’s rules to determine ground-state electron configurations. 69 Cont’d Representing Electron Configuration: ▪ 𝑠𝑝𝑑𝑓 notation (condensed/complete): carbon, C 1𝑠 2 2𝑠 2 2𝑝2. ▪ Orbital level diagram: 1𝑠 2𝑠 2𝑝 ▪ Noble gas abbreviated electron configuration: [He]2𝑠 2 2𝑝2 or [He] 2𝑠 2𝑝 Note: Use may use or for orbital diagrams. 70 Cont’d An easy way to deduce the ground state electron configuration for the elements (atoms or ions): Give an example using Cu atom & Cu+ ion. Use abbreviated form. 71 Cont’d The ground state electron configuration for the elements (atoms): Note: The elements in the same group have the same valence electron configuration as shown in the periodic table. 72 Cont’d ▪ Core electrons occupy lower energy quantum levels than valence electrons & don’t participate in bonding. ▪ The valence electrons of an element determine its chemical & physical properties. ▪ All the elements in the same group on the periodic table have similar electron config. for their valence shells, & therefore, have similar chemical properties. ▪ This is the power of the quantum mechanical model. ▪ Generally, only the valence electrons participate in chemical reactions. 73 Cont’d Electron configurations & the Periodic Table: The periodic table of the elements can be related the Aufbau process by dividing the table into the following four blocks according to the subshells being filled last: ▪ 𝑠-block: The 𝑠 orbital of highest principal quantum number (𝑛) is last to be filled. Consists of Group 1 & 2 (plus He in Group 18). ▪ 𝑝-block: The 𝑝 orbitals of highest principal quantum number (𝑛) is last to be filled. Consists of Group 13 to 18 (except He). ▪ 𝑑-block: The 𝑑 orbitals of the electronic shell (𝑛 − 1) is last to be filled. Consists of Group 3 to 12. ▪ 𝑓-block: The 𝑓 orbitals of the electronic shell (𝑛 − 2) is last to be filled. Includes only the lanthanoids & the actinoids. 74 Cont’d The periodic table showing the blocks: 75 Cont’d Example: Representing electron configurations: Using 𝑠𝑝𝑑𝑓 notation & orbital diagram (a) Write the noble gas abbreviated ground state electron configuration of arsenic using 𝑠𝑝𝑑𝑓 notation. (b) Write a valence electron orbital level diagram for the electron configuration of tin and indicate the number of valence electrons and unpaired electrons in the atom. (c) Show the abbreviated form of the electron configuration of Sn2+ using the 𝑠𝑝𝑑𝑓 notation. Analyse: The total number of electrons in a neutral atomic species is equal to the atomic number of the elements. All electrons must be accounted for in an electron configuration. For ions, charge = number of protons – number of electrons, & therefore, the number of remaining electrons can be accounted for. 76 Cont’d Solve: a) Arsenic, As (𝑍 = 33) is in period 4 & group 15. The noble gas that closes the third period is Ar (𝑍 = 18, equivalent to core electrons), & the valence subshells that fill in the fourth period are 4𝑠, 3𝑑, & 4𝑝, in that order according to Aufbau principle (filled with remaining 33 − 18 = 15 electrons-5 which are valence electrons, since this is group 15), for ground-state electron configuration, the valence subshells should be arranged in order of increasing value of 𝑛 and 𝑙. Note that we must account for 33 electrons in the configuration: [Ar]3𝑑10 4𝑠 2 4𝑝3 77 Cont’d Solve: (b) Tin, Sn (𝑍 = 50) is in period 5 & group 14. The noble gas that closes the fourth period is Kr (𝑍 = 36, equivalent to core electrons), & the valence subshells that fill in the fifth period are 5𝑠, 4𝑑, & 5𝑝, in that order according to Aufbau principle (filled with remaining 50 − 36 = 14 electrons-4 which are valence electrons, since this is group 14), for ground-state electron configuration, the valence subshells should be arranged in order of increasing value of 𝑛 and 𝑙. Note that we must account for 50 electrons in the configuration: [Kr] 5𝑠 4𝑑 5𝑝 Therefore, the are four valence electrons, i.e., 5𝑠 2 5𝑝2 , & two unpaired electrons with parallel spins in785𝑝. Cont’d Solve: (c) Note: To deduce the electron configuration of an ion (cationic or anionic) always start with the corresponding ground state electron configuration for the neutral atom, they you may subtract or add the corresponding number of electrons to obtain the correct charge of the ion. So, Tin, Sn (𝑍 = 50) is in period 5 & group 14. The noble gas that closes the fourth period is Kr (𝑍 = 36, equivalent to core electrons), & the valence subshells that fill in the fifth period are 5𝑠, 4𝑑, & 5𝑝, in that order according to Aufbau principle (filled with remaining 50 − 36 = 14 electrons-4 which are valence electrons, since this is group 14), for ground-state electron configuration, the valence subshells should be arranged in order of increasing value of 𝑛 and 𝑙. Note that we must account for 50 electrons in the configuration: Since we must loss/subtract 2 electrons to obtain a charge 𝐒𝐧: [Kr]4𝑑10 5𝑠 2 5𝑝2 of 2+, we have to loss all the 2 electrons from the outer valence orbital, 5𝑝2 , & drop it from the electron 𝐒𝐧𝟐+ : [Kr]4𝑑10 5𝑠 2 configuration becoz it will be empty. 79 Cont’d Note: Examples based on recognizing the correct & incorrect ground state & excited state atomic orbital diagrams or 𝑠𝑝𝑑𝑓 notation shall be provided through a tutorial. 80 EnD of Slides 81

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