TOPIC 2 ATOMIC STRUCTURE 12.1 ELECTRONS IN ATOMS

ESSENTIAL IDEA The quantized nature of energy transitions is related to the energy states of electrons in atoms and molecules. NATURE OF SCIENCE (1.8) Experimental evidence to support theories – emission spectra provide evidence for the existence of energy levels.

INTERNATIONAL-MINDEDNESS In 2012 two separate international teams working at the Large Hadron Collider at CERN independently announced that they had discovered a particle with behavior consistent with the previously predicted “Higgs boson”.

THEORY OF KNOWLEDGE “What we observe is not nature itself, but nature exposed to our method of questioning.” – Werner Heisenberg. An electron can behave as a wave or a particle depending on the experimental conditions. Can sense perception give us objective knowledge about the world? The de Broglie equation shows that macroscopic particles have too short a wavelength for their wave properties to be observed. Is it meaningful to talk of properties which can never be observed from sense perception?

UNDERSTANDING/KEY IDEA 12.1.A In an emission spectrum, the limit of convergence at higher frequency corresponds to the first ionization energy.

Energy levels in the hydrogen atom converge at higher energy. We can calculate the ionization energy from the convergence limit at higher frequency. The transition from n=1 to n=∞ corresponds to ionization which is the removal of the electron from the 1s orbital.

APPLICATION/SKILLS Be able to solve problems using E=hv = hc/λ.

GUIDANCE The value of Planck’s constant (h) and E=hv are given in the data booklet in sections 1 and 2.

Example problem 1 Determine the energy in joules of a photon of red light, correct to 4 significant figures, given that the wavelength is nm. h=6.626x Js; c=2.998 x10 -8 m/s E=hv = hc/λ convert nm to m E=(6.626x Js)(2.998 x10 -8 m/s 6.500x m E=3.056x J

APPLICATION/SKILLS Be able to calculate the value of the first ionization energy from spectral data which gives the wavelength or frequency of the convergence limit.

Example problem 2 Calculate the first ionization energy in kJ/mol for hydrogen given that its shortest wavelength line in the Lyman series is nm. N A = 6.022x10 23 mol -1 h=6.626x Js; c=2.998 x10 -8 m/s E=hv = hc/λ convert nm to m E=(6.626x Js)(2.998 x10 -8 m/s 9.116x m E=2.179x J expressed in kJ/mol IE 1 = (2.179x J) x (6.022x10 23 mol -1 ) =1.312x10 6 J/mol = 1312 kJ/mol

The value of the first ionization energy for hydrogen is given in section 8 of the Data booklet.

UNDERSTANDING/KEY IDEA 12.1.B Trends in first ionization energy across periods account for the existence of main energy levels and sub-levels in atoms.

IONIZATION ENERGY Ionization energy is the energy needed to remove one mole of electrons from the ground state of one mole of the gaseous atom. 1 st Ionization energy – energy needed to remove the 1 st outermost electron 2 nd Ionization energy – energy needed to remove the 2 nd outermost electron 3 rd Ionization energy – energy needed to remove the 3 rd outermost electron. Etc, etc…

EVIDENCE FROM IONIZATION ENERGIES The graph from the previous slide shows evidence that each energy level contains a certain number of electrons before it becomes full.

UNDERSTANDING/KEY IDEA 12.1.C Successive ionization energy data for an element give information that shows relations to electron configurations.

KEY POINTS There is an increase in successive ionization energies. As you remove one electron, it takes more energy to remove the next electron because the effective nuclear charge is stronger upon the removal of each electron. There are jumps in energies needed to remove electrons depending upon how close you are to the nucleus. The valence electrons are the easiest to remove. You will have a significant jump in ionization energy as you move into the core electrons or as you enter a new sublevel.

SUCCESSIVE IONIZATION ENERGIES Upon studying successive ionization energies, scientists have additional evidence that orbitals do have sub levels. Aluminum 2,8,3 If one looks at the 8 electrons in the second energy level of aluminum (the 1 st three valence electrons are already pulled off), there is a jump in energy between the 6 electrons in the p-sublevel and the 2 in the s-sublevel. The 2 electrons in the s-sublevel are closer to the nucleus and harder to pull off.

IONIZATION ENERGIES AND ELECTRON CONFIGURATION Generally there is an increase in ionization energies as one moves from left to right on the periodic table. As you move from left to right, the nuclear charge increases therefore making it more difficult to remove an electron. As you move down the periodic table, there is a decrease in ionization energies because the outermost electrons are further away from the nucleus.

APPLICATION/SKILLS Be able to deduce the group of an element from its successive ionization energy data.

When the ionization energy makes a large jump, you can tell what group the element is from. 200 to 8000 would mean one electron was easily pulled off so you would be in Group ,400,600, would mean Group 3 because the large jump occurred after the 3 rd electron was pulled off.

APPLICATION/SKILLS Be able to explain the trends and discontinuities in first ionization energy across a period.

1 st IONIZATION ENERGIES FOR THE FIRST 20 ELEMENTS

TWO SIGNIFICANT POINTS Why did the energy drop from Be to B if they are the same row and why did the energy drop from Mg to Al? There is another drop from N to O and from P to S, why did this occur?

Decrease from Group 2 to Group 3 One would expect an increase from Be to B since they are in the same energy level, however, there is a decrease and this is due to electron configuration. The outer electrons being removed from Be and Mg in Group 2 are in the “s” orbitals. The outer electrons being removed from B and Al in Group 3 are in the “p” orbitals. The “p” orbital electrons take less energy to remove than the “s” orbital electrons which are closer to the nucleus.

Decrease from Group 5 to Group 6 One would expect an increase from Group 5 to Group 6, but this is not so. Look again at the e - configurations. Group 5 elements have the basic configuration of ns 2, np x 1, np y 1, np z 1 Group 6 elements have the basic configuration of ns 2, np x 2, np y 1, np z 1 An electron is easier to remove from a doubly occupied orbital because it is being repelled by its partner electron and takes less energy to pull off than electrons in singly occupied orbitals.

Citations International Baccalaureate Organization. Chemistry Guide, First assessment Updated Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, Print. Most of the information found in this power point comes directly from this textbook. The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.