Chapter 6 Chemical & Physical Properties of the Elements and the Periodic Table

Review Quiz Chapter 6 Heats of (kJ/mol) conversion. ∆H summation formula.

Valence Electrons The valence electrons are the electrons in the outer energy level (valence shell). All other electrons are termed core electrons (electrons not in the outer energy shell).

Alkali Metals

Alkaline Earth Metals

Transition Elements (Metals)

Halogens

Noble Gases

Trends in the Periodic Table The periodic table can be used to predict: –Covalent radii (atomic size) –Ionic radii (ionic size) –First Ionization energy

Covalent radius Covalent radius is essentially the size of an atom.

Covalent Radii (atomic radii) Atomic Radius

Ionic Radius Ionic Radius is the size of an ion.

Isoelectronic Series Substances are isoelectronic if they have the same electron configuration. Name two isoelectronic species.

Ionization Energy Ionization energy is the energy needed to remove an electron from an atom or ion.

First Ionization Energy First Ionization energy is the energy needed to remove the first electron from an atom.

Multiple Ionization Energies Second Ionization energy is the energy needed to remove the second electron from an atom. Third Ionization energy is the energy needed to remove the third electron from an atom. Etc.

Ionization Energies in kJ/mol H1312 He Li Be B C N Write the equation representing the first ionization energy of hydrogen.

First Ionization Energy of H H kJ → H + + e -

Effective Nuclear Charge (Z eff ) You will find many of the notes for effective nuclear charge on a sheet in your notebook titled “Effective Nuclear Charge”. The effective nuclear charge (Z eff ) of an atom is basically how well it is able to hold on to its most loosely held electron. Effective nuclear charge is a direct result of Coulomb’s Law.

Coulomb's law helps describe the forces that bind electrons to an atomic nucleus. Based on Coulomb’s Law, the force between two charged particles is proportional to the magnitude of each of the two charges and inversely proportional to the square of the distance (radius) between them.

Effective Nuclear Charge and Coulomb's law There are certain properties that depend upon how well the nucleus is holding on to an electron(s). These properties include: –Ionization energy –Atomic and ionic radii –Electronegativity

Effective Nuclear Charge and Coulomb's law By applying Coulomb’s law we can better understand the force of attraction between the nucleus and an electron which is essentially the effective nuclear charge.

Effective Nuclear Charge (Z eff ) We can estimate the effective nuclear charge of an atom by using the following: 1.The nuclear charge (Z) 2.The shielding effect 3.Electron repulsions

The Nuclear Charge (Z) Based on the number of protons in the nucleus. –Example: Carbon vs. Nitrogen

The Nuclear Charge (Z)

The greater the number of protons in the nucleus the greater the effective nuclear charge.

Nuclear Charge and Z eff

Shielding Effect. Core electrons are generally closer to the nucleus than valence electrons, and they are considered to shield the valence electrons from the full electrostatic attraction of the nucleus. This shielding effect can be used in conjunction with coulomb’s law to explain relative ionization energies.

Shielding Effect. Shielding can be understood by examining the electron configuration for an atom or ion.

Shielding Effect Energy Levels vs. Sublevels Energy levels have the greatest effect on shielding. Sublevels increase shielding but to a far lesser extent.

Ionization Energies in kJ/mol H1312 He Li Be B C N

Z eff can help us explain the ionization energies.

Explain the first ionization energies of Be and B A

Explain the first ionization energies of Be and Mg

Effective Nuclear Charge can be used to help explain atomic radius. Atomic Radius

Explain the difference in atomic radii for Li and Be. Which are 1.52 and 1.11 angstroms respectively.

Explain the difference in atomic radii for Li and Na. Which are 1.52 and 1.86 angstroms respectively.

Effective Nuclear Charge can be used to help explain atomic radius. Based on nuclear charge and shielding.

Nitrogen vs. Oxygen First Ionization Energy

Electron Repulsions: Paired vs. Unpaired Electrons Differences in electron – electron repulsion result from the pairing of electrons within the orbitals of a particular subshell. This pairing of electrons is responsible for the differences in ionization energy for electrons within the same subshell.

Electron Repulsions: Paired vs. Unpaired Electrons A paired electron has increased electron – electron repulsion acting upon it which acts to lessen the hold of the nucleus on a paired electron lowering the effective nuclear charge. Therefore it is easier (takes less energy) to remove a paired electron than it does to remove an unpaired electron. We check the pairing of electrons in the outer sublevel by writing an orbital filling diagram.

Nitrogen vs. Oxygen First Ionization Energy

It is much harder to remove an electron from helium than it is Li. This is Illustrated by their respective ionization energies given below. Explain. He = 2370 kJ/mol Li = 520 kJ/mol Stability Schmability

Penetration Effect Electrons in a higher energy level can often penetrate (dive) through lower energy levels because of the attraction that the nucleus has on them. Smaller sublevels can penetrate closer to the nucleus than larger sublevels.

Explain the relative energies of the sublevels within the fourth energy level. The s sublevel penetrates closer to the nucleus followed by the p, d and the f has the least penetration. The closer to the nucleus the lower the energy and therefore the relative energies of the sublevels in the fourth energy level is: 4s < 4p < 4d < 4f.

Explain why a 4s sublevel has a lower energy than 3d. A 4s sublevel penetrates closer towards the nucleus than does a 3d so even though the 3d is part of the third energy level the 4s on average is closer to the nucleus and is therefore lower in energy than the 3d.

Reactivity of Metals Which alkali metal would you expect ot be the most reactive? Explain the trend in the reactivity of the alkali metals?

Alkali Metals in Water Accurate

Lab - Spectrophotometry of Cobalt(II)

Lab - Spectrophotometry of Cobalt(II) The Beer – Lambert Equation Beer’s Law

Beer – Lambert Law The amount of light absorbed by a solution can be used to measure the concentration of the absorbing molecule in that solution by using the Beer – Lambert Law.

Beer – Lambert Law A = Ɛ Cl where A is the absorbance, Ɛ is the molar absorption coefficient, C is the molar concentration (molarity), and 1 is the sample length.

In this lab you will prepare solutions of CoCl 2 and use Beers Law to determine [Co 2+ ]

How can we use the slope of the line to determine Ɛ, the molar absorption coefficient?

A = Ɛ Cl

Transmittance A = -logT The transmittance is the percentage of the light in the original light beam that passes through the sample and reaches the detector.

Why do we use absorbance instead of transmittance?

Homework Write up the Lab Summary. Complete the pre-lab assignment on a separate sheet of paper. –You will need a sheet of graph paper for the pre-lab assignment. Finish your homework for Chapter 6.