Consider the two waves shown in the diagram below.

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2 The way matter interacts with electromagnetic radiation gives a great deal of information about the electronic structure of atoms. In this section we will look at the electromagnetic spectrum and then go on to look at the structure of the atom. Electromagnetic radiation is a form of energy. Light, X-rays, radio signals and microwaves are all forms of electromagnetic radiation. Visible light is only a small part of the range of the electromagnetic spectrum. Electromagnetic radiation may be described in terms of waves of varying length between m and 10 4 m that travel at a speed of 3 x 10 8 m s -1 (the speed of light given the symbol, c) Consider the two waves shown in the diagram below. Wavelength has the symbol,, (lambda) and it is the distance between two wave crests. Wavelength is usually measured in metres but NANOMETRES (nm) are often used. One nanometre = 10-9 m. In the diagram above wave A has twice the wavelength of wave B. Waves can also be described by their FREQUENCY, that is the number of waves that travel past some point in a second. Frequency is given the symbol, f and measured in hertz (Hz) the unit is s -1

3 The relationship between wavelength, frequency and the speed of light is given by the expression, c = f x Some points to remember All electromagnetic radiation travels at the speed of light this is always constant 3 x 10 8 metres per second. The frequency of radiation is directly proportional to energy. High frequency means high energy. Frequency is inversely proportional to wavelength as frequency increases wavelength decreases. Wavelength is inversely proportional to energy. Large wavelengths wave low energy. The visible part of the spectrum runs from around 400 nm (violet light) to around 700 nm (red light). Violet light has a shorter wavelength and a higher frequency than red light and so violet light has more energy than red light. Another unit that you will come across, especially when dealing with infra-red spectra, is the WAVENUMBER. Wavenumber is the reciprocal of the wavelength { 1/ } 1. Calculate the wavelength of radio waves corresponding to 98 MHz. Use c f The frequency is 98 megahertz ( 98 X 10 6 Hz) and the speed of light is 3 X 10 8 m s -1. Putting these values into the formula = 3 X 10 8 / 98 X 10 6 = 3.1 metres

4 2. O H bonds in molecules can absorb infra-red radiation with a wave number of 3600cm -1. Calculate the frequency of this radiation. Remember that wave number is the reciprocal of wavelength. So 1/3600 = 2.8 x 10-4 cm = 2.8 x 10-6 m (wavelength must be in metres) Now use Frequency = Speed / Wavelength = 3 x 10 8 / 2.8 x 10-6 = 1.07 x Hz. 1. A typical microwave oven operates at a frequency of 2.5 x 10 9 Hz. Calculate the wavelength of this radiation. Give your answer in nanometres. 2. Data book information records the flame colour of potassium as lilac and with a wavelength of 405 nm. Calculate the frequency of this radiation. 3. Use the data book to find the wavelength of light emitted by a sample of copper in a flame and thus calculate its frequency. 4. Arrange these types of radiation in order of increasing frequency. Ultra violet, X rays, radio, visible, gamma 5. A certain colour of light has a wavelength of 580 nm. Calculate the frequency of this light and suggest its colour. 6. The absorption spectrum for an alkanal molecule contains a line at wavenumber 1700 cm -1. Calculate frequency of this radiation.

5 When white light is passed through a prism or a diffraction grating, the light is split into the colours of the rainbow. This is called a continuous spectrum there are no gaps or breaks in the colour pattern. However, if sufficient energy is supplied to a chemical ( i.e. heating sodium chloride in a bunsen or by passing electricity through a gas) and this results in the production of light the spectrum produced is NOT CONTINUOUS but rather a SERIES OF LINES of different wavelengths and thus of different colours. Spectra that show energy being given out by an atom or ion are called atomic emission spectra. The pattern of lines in such a spectrum is characteristic of each element and, like a fingerprint, can be used to identify the element. The spectrum on the right shows the lines formed when hydrogen gas is subjected to high voltage. The pattern of lines are at specific wavelengths and no other element will produce the same pattern. As part of the Advanced Higher course you need to be able to explain how these lines arise and be able to calculate the energy associated with the lines.

6 Atomic emission spectra provides evidence for the structure of the atom. In particular it explains the electron energy levels within the atom. An electron is normally in the lowest energy level. This is often called the ground state. If the electron absorb energy (from a flame or by electrical discharge) it can be promoted to a higher energy level. This is often called an excited state. When the electron drops back down energy is given out. Energy in Energy out A B In A, the electron is in its ground state (E 1 ). Energy is absorbed and the electron is promoted to a higher energy level. In B, the electron is in an excited state (E 2 ). When the electron falls back to a lower energy level, energy is released. If hydrogen gas is subjected to a high voltage it will emit light and produce an emission spectrum as a series of lines. Each line represents the DIFFERENCE in energy between the energy levels. Every line has its own separate FREQUENCY and WAVELENGTH This picture shows the light emitted by a hydrogen filled discharge lamp. The colour we observe (pinky purple) is a result of our eyes merging together the individual colours of the line spectrum.

7 This diagram shows the lines produced (visible region) and the electron transitions which cause them. Notice two things about these transitions. 1. They are all caused by electron falling from higher energy levels to the SECOND energy level. 2. The larger the difference in energy cause lines of higher frequency. E.G. The red line is caused by the smallest energy transition (3 2) This series of lines is known as the BALMER series after the Swiss mathematician Johann Balmer who derived a formula for the lines. The Balmer series shows only the lines in the visible region of the spectrum the entire hydrogen emission spectrum is a little more complicated Points Each electron transition causes a separate line on the spectrum. The highest energy lines (those in the UV region) are caused by the largest transitions. As the energy levels get further from the nucleus, they get closer together until they are so close that they form a CONTINUUM. If an electron is promoted from n = 1 to the continuum it is considered to have been removed from the atom altogether. The energy associated with this transition is called the IONISATION energy.

8 Spectra called atomic absorption spectra also exist. These look like the continuous spectrum with the emission lines missing. The diagram on the right shows both types of spectra for hydrogen. If you were to superimpose these on top of each other it would produce the continuous visible spectrum. As mentioned before each element on the Periodic table has its own emission spectrum. Some of these are shown below. Finally It is clear that the unknown sample is strontium. 1. When some elements/compounds are placed in a Bunsen or subjected to a high voltage produce no colours at all. The reasons for this could be that the energy supplied is not sufficient to promote electrons to an excited state or the light emitted is not in the visible region of the electromagnetic spectrum. 2. Different elements produce different spectra because they have different electron configurations and these electrons are attracted to the nucleus of the atoms by differing numbers of protons. 3. All element spectra are shown here:

9 1. The diagram below shows some of the electron transitions when hydrogen gas is subjected to high voltage at low pressure. The line spectrum produced when the light produced is passed through a prism is also shown. a. Explain why the hydrogen emission spectrum shows up as a series of lines. b. What term is given to the energy needed to promote an electron from energy level n=1 to infinity? c. Why do the lines formed when electrons fall back to energy level one NOT appear on spectrum shown above? d. Which of the observed lines has the lowest frequency? 2. A student observed three different spectra and drew diagrams of each. a. What do the numbers 0.4 and 0.7 represent? b. Why are there gaps with no colour in the spectra of helium and neon? c. Why are the patterns produced for helium and neon different? d. Which of the lines in the helium spectrum represents the highest energy?

10 3. Use the spectra in the box to help answer this question. a. Samples A and B were taken from two different contaminated soils. (i) Which element responsible for the triplet of lines at 650 nm? (ii) Which other element is present in both samples? (iii) Thallium salts are used as rat poison. Is there any evidence of thallium in any of the samples? (iv) Is there any evidence of mercury in sample B? b. What would the absorption spectrum of sodium look like? c. Which element has the line created by the lowest energy transition? d. Which of the spectra, if any, would produce the same spectrum as potassium? e. What colour will the line in the spectrum of thallium be? f. Calcium compounds produce a brick red colour when they are held in a Bunsen flame. How can you tell this form the calcium spectrum?

11 Max Planck developed the theory that under certain circumstances electromagnetic radiation may be regarded as a stream of particles. These particles are called photons. The energy carried by a photon is related to its frequency by the equation: E = h x f Where h = Planck s constant = 6.63 x Js and f is the frequency in s -1. For one mole of photons Avogadro s constant (L) is added to the equation and in chemistry we usually use wavelength of light not frequency. Answers are quoted in kilojoules (kj). This gives the equation It is very important to ensure the wavelength of the radiation is given in METRES as the speed of light (c) is quoted in metres per second (m s -1 ) 1. Calculate the energy, in kjmol -1, of blue light with a wavelength of 450 nm. E = L x h x c/1000 E = (6.02 x ) x (6.63 x ) x (3 x 10 8 ) = 270 kj mol x 450 x 10-9 This changes nm to metres 2. The ionisation energy of hydrogen is 1310 kj mol -1. Calculate the wavelength of light, in nanometres, associated with this ionisation energy. = L x h x c/1000 E = (6.02 x ) x (6.63 x ) x (3 x 10 8 ) 1000 x 1310 = metres = 91 nm

12 1. The diagram shows ozone molecules being destroyed by ultra violet radiation Calculate the energy, in kj mol -1, associated with the ultra violet radiation. 2. The red line in the hydrogen spectrum has a wavelength of 656 nm. Calculate the energy value, in kj mol -1, for one mole of photons at this wavelength. 3. An advertising sign gives off red light and green light. a. Which colour of light has the smallest wavelength? b. The green light has a wavelength of 510 nm. Calculate the energy, in kj mol -1, of this green light. 4. The diagram shows the helium emission spectrum (visible range). a. What causes helium to produce the lines seen on the spectrum? b. Which of the lines has the highest energy value? c. Which of the lines is most likely to be violet in colour? d. Calculate the energy, in kj mol -1, of the line with a wavelength of nm

13 Till now our picture of atomic structure has been a very simple one. It consists of the atom comprising a nucleus containing protons and neutrons. Electrons orbit the nucleus in areas we have called energy levels. As the energy levels get further from the nucleus they get larger and can hold more electrons. Emission spectroscopy gives more information about atomic structure The line spectrum of hydrogen can be adequately explained using the theory already discussed. However, emission spectra of elements with more than one electron show evidence of sub levels or atomic orbitals within each principal level above the first. To explain these sub levels, the theory of Quantum Mechanics was put forward. This theory relies heavily on complex mathematics beyond the scope of Advanced Higher Chemistry. Luckily, we only have to be able to use the theory to explain the electron arrangements of elements. The basis of quantum theory states that the energy an electron has can be defined by a set of FOUR QUANTUM NUMBERS. Each quantum number tells us something about an electron in terms of its energy and its position in the electronic structure of the atom - in other words - the atomic orbital it is found in. Main energy level. As the value of n increases the energy and size of the level increases. These levels are sometimes called shells Subshell. Indicates the number and the shape of each sub-shell within each main energy level. Direction. Indicates the orientation in space of each atomic orbital within a subshell. Spin. Indicates if the electron is spinning clockwise or anti-clockwise.

14 Each quantum number has a set of values which provide information about each atomic orbital. Principal n Values 1,2,3,4 etc. The numbers determine the size and energy of the main energy level. Angular momentum l Values of zero to (n-1). The numbers determine the shape of the atomic orbitals within a sub-shell. Each values of l is assigned a letter which indicates each sub shell. If l = 0 the subshell is labelled If l = 1 the subshell is labelled If l = 2 the subshell is labelled If l = 3 the subshell is labelled The table shows the possible values of l for the first four energy levels Principal QN - n Angular QN l {0 to (n-1)} Subshell type 1 0 s s p s p d s p d f

15 In simple terms this means energy level one has one type of subshell called an s subshell, energy level two has two types of subshell called s and p subshells, energy level three has three types of subshell called s,p and d subshells, energy level four has four types of subshell called s,p,d and f subshells. Magnetic m For example: Values of l to +l. The value indicates the orientation in space of the atomc orbital. If l = 2 this would indicate a d subshell. The magnetic quantum number- m would have the values This indicates that there are five d orbitals each one with a different orientation in space. The table shows the possible values of m for the first three energy levels Subshell Value(s) of l Value(s) of m (-l to +l) 1s 0 0 2s 2p 3s 3p 3d , 0, , 0, , -1, 0, +1, +2 In simple terms this means there is only one type of orbital, there are three types of orbital, there are five types of orbital.

16 The shapes of the orbitals are shown below. You must be able to identify the type of orbital from a diagram of its shape. orbitals are spherical. orbitals are roughly dumbbell shaped. All energy levels above level one have a p subshell. Within each p subshell there are three individual p orbitals one along each of the axis x, y and z. orbitals have more complex shapes All energy levels above level two have a d subshell. Within each d subshell there are five individual d orbitals. Each d orbital has a different orientation

17 Spin s The final quantum number indicates the direction in which an electron is spinning. This can be either clockwise or anti-clockwise. It has two values +½ and ½ An atomic orbital is a region of space where an electron is likely to be found. An atomic orbital can hold a maximum of TWO electrons. Atomic orbitals increase in size and energy as the value of n increases. Atomic orbitals in the same subshell have the same energy. Orbitals which have the same energy are said to be DEGENERATE. The three 2p orbitals are degenerate as are the three 3p orbitals. However an individual 3p orbital is larger and has more energy than an individual 2p orbital. The first energy level has one s subshell which holds a maximum of two electrons. The second energy level has two subshells s and p. This means there are four individual orbitals in the second energy level one s and 3p orbitals. These four orbitals hold a maximum of 4 x 2 = 8 electrons. The total number of orbitals in an energy level = n 2. For example; the third energy level has 3 2 = 9 orbitals in total. The total number of electrons an energy level can hold = 3n 2. For example; the third energy level can hold 3 x 2 2 = 18 electrons. If two electrons occupy the same atomic orbital they MUST have opposite spins.

18 1. There are four quantum numbers which give information about the energy levels in atoms. They have the symbols n,l,m and s. a. What information does the quantum number n give about an atomic orbital? b. What are the values for the quantum number m if the value of l is 3? c. What are the values for the quantum number l if the value of n is 4? 2. The diagram shows the atomic orbitals for principal quantum number three. a. How many subshells exist in the third energy level? b. How many orbitals exist in the third subshell? c. What is the maximum number of electrons the 3p subshell can hold? d. What is the maximum number of electrons the 3d subshell can hold? e. What is the maximum number of electrons the third energy level can hold? f. Explain why the 3p and the 3d orbitals are not degenerate? g. State one way the 2s and 3s subshells are similar. h. State one way the 2s and 3s subshells are different. i. What is the value of the quantum number l in the 3s subshell? j. What values does the quantum number m have in the 3d subshell?

19 We already know how to write simplified electronic configurations for atoms and ions. Both at Standard Grade and Higher this was simply a matter of looking up the data book and either copying the arrangement for an atom or adding/subtracting electrons to show the arrangement for an ion. If you are not confident about how to do this revision is necessary NOW. We now know the diagram on the right is a very oversimplified picture of the electron shells for an atom of sodium. Previously we would write the electronic configuration as - 2, 8, 1 While this is still true, in order to show atomic orbitals, you must now be able to write this electronic arrangement in two new ways. Spectroscopic notation uses numbers to designate the principal energy level and letters to identify a subshell. A superscripted number indicates the number of electrons in a subshell. An orbital box diagram uses boxes to represent the individual atomic orbitals within each subshell and arrows to represent electrons. Spectroscopic notations are sometimes made shorter by labelling the core of filled inner shells with the configuration of the preceding noble gas. This form is rarely asked for in exams. In the following pages we will look at the rules which govern how to write the electronic configurations for the elements from hydrogen to krypton. In addition, we will examine how to write the electronic configurations for ions and study ionisation energy values. We will then relate the electron arrangements to the position of the elements in the periodic table.

20 In order to write electronic configurations of atoms/ions several rules must be followed. You are required to be able to state these rules and identify if an electron if configuration is written correctly. Aufbau Principle When electrons are placed into orbitals the energy levels are filled up in order of increasing energy. Generally as the energy level gets further from the nucleus the more energy it will possess. There are exceptions You will notice that the diagram shows that the 4s subshell has been placed below the 3d subshell. This is Not a mistake the 4s subshell is larger and further from the nucleus than the 3d subshell. However, the 3d subshell has more energy. The 4s subshell fills before the 3d subshell. You need to remember the filling order up to atomic number 36 krypton. The order is 1s 2s 2p 3s 3p 4s 3d 4p increasing energy Pauli Exclusion Principle This states that an orbital cannot contain more than two electrons and they must have opposite spins.

21 Hund s Rule This states that in degenerate orbitals (eg, 2p or 3d) electrons will fill each orbital singly with spins parallel before pairing occurs. 1. The orbital box and spectroscopic electronic configurations for the first twelve elements are shown below. Your task is to fill in any blank spaces and correct mistakes. Write AP (Aufbau principle), PEP (Pauli exclusion principle) or HR (Hund s rule), to identify which rule has been broken, next to the name of the element.

22 2. Write the electronic configuration for the elements with atomic numbers 13 to 20 in terms of s and p orbitals. 3. Explain why the electronic configuration, in spectroscopic notation, for a nitrogen atom shown below is incorrect 4. The values of the four quantum numbers for a 1s electron in a hydrogen atom are n l m s ½ State the values of the four quantum numbers for a. the 4s electron in potassium b the four electrons in the 2p subshell in oxygen. 5. Phosphorus has the atomic number 15. A student wrote the following electron configuration for phosphorus. 1s 2s 2p x 2p y 2p z 3s 4p x 4p y 4p z Explain the two mistakes in the student s answer. 6. The electron configuration of an atom of element Y in the ground state is shown below Energy 4s 3p 3d 3s 2p 2s 1s Identify element Y.

23 7. (i) Which elements have the following electron configurations a. 1S 2 b. 1S 2 2s 2 2p 4 c. 1s 2 2s 2 2p 6 3s 2 3p 6 (ii) Write the orbital box notation for element b. 8. Which element has the electron configuration [Ar]3d 10 4s 2 4p 5? 9. The element carbon has atoms which contain six electrons. a. Write the orbital box notation for the electronic configuration of a carbon atom b. explain why carbon atoms have two unpaired electrons c. How many unpaired electrons do the following elements have (i) fluorine (ii) sulphur (iii) beryllium 10 The diagram shows the shapes of the 5 d orbitals It is often more convenient to show these orbitals as boxes. a. Which quantum number dictates the shape of these orbitals? b. Use your knowledge of quantum theory to explain why there are five d orbitals c. What is the maximum number of electrons each orbital can hold? d. Why can these orbitals be described as degenerate? e. Name the element which is the first to have an electron in a d subshell. f. Write the electronic configuration, in terms of s, p and d orbitals for this element.

24 In general metal atoms will lose electrons and form positive ions while non metal atoms gain electrons and become negative ions. The number of electrons involved is usually equal to the valency number of the element. When writing the spectroscopic or orbital box notation for ions, the appropriate number of electrons are removed (+ ion) or added (- ion) to the outermost atomic orbital of the atom. Positive ion Negative ion The Periodic Table can be divided into blocks which show the atomic orbitals being filled as the atomic number increases. The blocks are named after the type of outermost orbital being filled s, p,d or f. The noble gases are all classed as p- block elements while the alkali metals are all classed as s block elements.

25 At this point it is worth mentioning some issues with the electron arrangements of the 3d transition metals. The filling of the d orbitals follows the Aufbau principle with the exceptions of chromium and copper. These exceptions are due to a special stability associated with all the d orbitals being half filled or completely filled. Transitions metal atoms generally form positive ions. In doing so you will ALWAYS remove the electrons in the 4s subshell BEFORE any of the electrons in the 3d subshell. For example Vanadium atom 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2 Vanadium(III) ion (V 3+ ) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 1. Write down the electronic configurations, in terms of s and p orbitals for the following ions. a. Ca 2+ b. N 3- c. O 2- d. K + e. Al Classify the following as s,p,d or f block elements. a. Aluminium b. Scandium c. 1s 2 2s 2 2p 6

26 3. Complete the diagram by adding the appropriate arrows in the boxes 4. Identify the elements with the following electronic configurations a. [Xe] 6s 2 b. [Ar] 4s 2 4p 4 c. [Ne] 3s 2 5. A neutral atom has two electrons with n = 1, eight electrons with n = 2, eight electrons with n = 3 and two electrons with n = 4 Assuming this atom is in its ground state supply the following information about the atom. a. Atomic number. b. Total number of s electrons. c. Total number of p electrons d. Total number of d electrons. e. Is the atom metallic or non metallic? 6. How many unpaired electrons are there in the following species? a. chromium atom b. copper(i) ion c. oxide ion d. iron(iii) ion e. fluorine atom f. manganese(iv) ion

27 Ionisation energy is defined as the energy required to remove one mole of electrons from one mole of gaseous atoms or ions Equations can be written to describe ionisation energies (these are given in the data book.) First ionisation energy E(g) E + (g) + e - Second ionisation energy E + (g) E 2+ (g) + e - The Periodic Table above shows two obvious patterns in ionisation. Ionisation energy INCREASES across a period. Ionisation energy DECREASES down a group. These patterns are explained by the following facts 1. Atomic size - electrons are further from the influence of the nucleus in larger atoms-ionisation energy will decrease with atom size. 2. Nuclear charge - more protons mean more nuclear pull on electrons this in turn increases ionisation energy. 3. Electron screening - inner electrons shield outer electrons from the full force of the nucleus this means atoms with more electron shells will have lower ionisation energies. Examples a. The first ionisation energy of sodium, 2,8,1, is less than that of lithium, 2,1 because lithium atoms are smaller. This means the outer electron is closer to the attractive force of the nucleus. The extra energy level in sodium also provides a screening effect which shields the outer electron from the nucleus. b. The first ionisation energy of fluorine is larger than that of oxygen because fluorine atoms are smaller and have a more positively charged nucleus. Both these properties bind the electrons more tightly to the nucleus.

28 Analysing this graph shows that, in general, ionisation energies increase across period 3. However it can also be seen that some atoms do not follow the general pattern. These apparent anomalies can be explained by the relative stabilities of the subshells the electrons are in. Mg 1s 2 2s 2 2p 6 3s 2 Al 1s 2 2s 2 2p 6 3s 2 3p 1 The outer electron in aluminium is in the 3 p sub-level. This is higher in energy than the outer electron in magnesium, which is in an s sublevel. In addition Mg a full outer subshell so less energy is needed to remove it. The 3p electrons in phosphorus are all unpaired. In sulphur, two of the 3p electrons are paired. There is some repulsion between paired electrons in the same sub-level. This reduces the force of their attraction to the nucleus, so less energy is needed to remove one of these paired electrons than is needed to remove an unpaired electron from phosphorus. Although not always the case, half filled and filled subshells confer some extra stability to the electrons in atoms and so can explain the trends in ionisation energies.

29 1. Which group in the periodic table contains the elements with the lowest 1 st ionisation energy. 2. Why do the 1 st ionisation energy values decrease from helium to krypton? 3. Write the equation for the 1 st ionisation energy,i.e, of chlorine and use the data book to find the value for this change. 4. From their positions in the Periodic Table it would be expected that boron should have a higher 1 st I.E. than beryllium. Why is the 1 st I.E of boron smaller than the 1 st I.E. of beryllium? 5. What is unusual about the trend in 1 st I.E. for the first transition series from scandium to zinc compared to the period from Li to Ne and from Na to Kr? 6. The first five ionisation energies, in kj mol -1, of two elements Q and R are shown below: First Second Third Fourth Fifth Q R a. In which groups of the Periodic Table would you expect to find elements Q and R? Explain your reasoning for element Q. b. Write an equation which represents the 2 nd ionisation energy of element R 7. The graph below shows the first ionisation energies of the elements from hydrogen to sodium. a. Explain why there is an overall rise in first ionisation energy from lithium to neon. b. Explain why there is a fall in the first ionisation energy from nitrogen to oxygen. c. Explain why the first ionisation energy falls markedly from helium to lithium and again from neon to sodium.

30 A list of learning outcomes for the topic is shown below. When the topic is complete you should review each learning outcome. Your teacher will collect your completed notes, mark them, and then decide if any revision work is necessary. Need Help Revise Understand State that visible light is part of the electromagnetic spectrum with wavelengths between 400nm to 700 nm State that blue light has higher energy than red light. State that electromagnetic radiation travels in waves with a speed of 3x10 8 m s -1 Be able to use the relationship c = f where f is the frequency in Hertz and is is the wavelength in metres. State that an atomic emission spectrum is not continuous like the visible spectrum but made of individual lines of a specific frequency. Be able to explain the origin of lines in an emission spectrum Be able to use the formula E = L h c/1000 to calculate the energy, in kj mol -1, of a specified wavelength or frequency of light. State the connection between the four quantum number n,l m and s and what they indicate about atomic orbitals. Be able to recognise s, p and d atomic orbitals from diagrams. State the meaning of the term degenerate in relation to atomic orbitals. Be able to write the electron configurations for atoms and ions of the first 36 elements in both spectroscopic and orbital box notation. Be able to state the Aufbau principle, the Pauli exclusion principle and Hund s rule and to use the rules correctly to place electrons in atomic orbitals. Be able to define and write appropriate equations for the first and second ionisation energy.

31 Be able to explain ionisation energies in terms of the relative stablilities of atomic orbitals. I have discussed the learning outcomes with my teacher. My work has been marked by my teacher. Teacher Comments. Date. Pupil Signature. Teacher Signature.