SCH 3U - Periodic Trends

Summary of Trends to Date

Metals

• for metals, a low ionization energy (IE) results in a more reactive element 
• this is borne out by looking at the two trends on the periodic table above; when going down a metal column, the IE decreases (so, it gets easier for the atom to lose an electron), but the reactivity increases
• this suggests that metals react by losing electrons

Non-metals

• for non-metals, a high IE results in a more reactive element 
• this is borne out by looking at the two trends on the periodic table above; when going up a non-metal column, the IE increases (so, it gets more difficult for the atom to lose an electron) and the reactivity increases as well
  
• this suggests that non-metals react by hanging onto their electrons
  

Electron Affinity (EA)

• electron affinity (EA) is the energy released or gained when a neutral atom attracts an extra electron (in layman's terms, it's how much an atom likes electrons)
• note that high IE and high EA go hand-in-hand
• thus, metals with a low IE will also have a low EA (ie, metals don't really love electrons and happily give them away
  
• conversely, non-metals with a high IE, will also have a high EA (ie, non-metals really love electrons and hang onto the electrons that they have fiercely, as well as attempting to get more)
• take a look at Na and Cl below:
  

• Na, which is a metal, has a low IE and a low EA, so it easily loses its valence electron
• Cl, which is a non-metal, has a high IE and it will hang onto its existing valence electrons strongly; also it has a high EA, so it will accept an extra electron readily

Noble Gases

• above, we saw that the higher the IE, the more reactive the non-metal
• thus, by that logic, Noble gases, which have very high IE, should be very reactive
• however, Noble gases are the most unreactive elements on the Periodic Table
• so, what gives???
• taking a further look at the reaction of Na and Cl above, provides an answer
  
• when sodium loses an electron, it becomes isoelectronic with Ne
• when Cl gains an electron, it becomes isoelectronic with Ar
• thus, all noble gases are very stable just they way they are and the other elements gain or lose electrons to attain the same electronic structure (isoelectronic) of a noble gas.

Atomic Radius (AR)

• atomic radius is the distance from the centre of the nucleus to the valence shell 

• if the Noble Gases have a full valence shell, other properties of the atoms should reflect this.
• look at the atomic radii for three consecutive elements: F (72 pm), Ne (71 pm), Na (186 pm)
• notice that F and Ne are similar in size, but the Na is significantly larger
  
• looking at the location of the three elements on the Periodic Table provides an explanation
• F and Ne are in the same period and sit right next to each other and so have very little difference in size, since they both have a total of two shells around the nucleus
  
• however, with Na, the extra electron begins a whole period, resulting in the addition of a shell, causing a large increase in radius

Electrostatic Force

• electrostatic force is the force of attraction between (+, -) and the force of repulsion between (-, - or +,+) and this force determines an element’s chemical behaviour and many of its properties.
• the strength of the force varies inversely as the distance between objects squared (F ∝ 1/d², which reads as force is proportional to one over the distance squared)
  
• this relationship between F and d indicates that force of attraction between a proton and an electron increases rapidly as the distance between these two subatomic particles decreases (or rapidly decreases as the distance increases)
• likewise, the force of repulsion between two electrons is highest when the electrons are close to each other
  

Atomic Radius & Electrostatic Force

• look at Be and B
• boron's 5 protons in the nucleus will attract an electron more than 4 protons that beryllium has in its nucleus
  
• thus, boron has the smaller radius because it attracts its electrons more strongly
  

• thus, as one moves from left to right across a period, the AR decreases - each step to the right means the addition of a proton to the nucleus of the element, which attracts the valence electrons more strongly, drawing them closer to the nucleus and decreasing the size of the atom
  

Shielding Effect

• how do the core electrons affect the attraction of the protons for valence electrons?
• for most elements, the core electrons (electrons in core shells) shield the valence electrons from feeling the full positive charge in the nucleus
• also, the repulsion between electrons in a shell reduces the effect of the positive nucleus on any one electron
• thus, the more electrons in the valence shell and the closer together they are, the greater the electron-electron repulsion

➤ the above two opposing forces help determine the atomic radius

Explaining Ionization Energy

• as discussed above, as you move left to right across a period, the nuclear charge (# of p⁺) and number of valence e^- increases
• this results in the valence e^- being pulled in more closely, decreasing the atomic radius and increasing electron-nucleus attraction
• metals, which have a relatively large radius, coupled with a low atomic core charge, will therefore have a low first IE
• non-metals, which have a relatively small radius, coupled with a high atomic core charge, will therefore have a high first IE

Explaining Electron Affinity

• since non-metals have small radii, extra electrons can get very close to the positive nucleus
• they become trapped by the strong forces of attraction

TryIt!  Explain why sulfur has a higher electron affinity than aluminum.

Answer: 

Sulfur, has a small atomic radius and a larger core charge.  Thus, it not only hangs onto its original electrons, but it is able to attract extra electrons to itself.  

 

Aluminum, on the other hand, has a larger atomic radius and a smaller core charge, so not only does it not attract extra electrons but it doesn't hold on tightly to the ones it already has.

Homework

Chapter 1 Review, p. 45 # 1, 4, 6, 7, 9, 10, 11, 21-23, 28, 36, 39

Chapter 1 Self Assessment, p. 48 # 3, 4, 9, 15, 17-19, 22

Explain why the ionization energy of fluorine is so much larger than that of lithium.

 

 

Answers:

 

Success Criteria:

• nomenclature - be able to write name/formula for any given compound, using an appropriate naming system.
• be able to provide and use definitons/examples for all terminology (matter, state, changes of state, element, compound, solution, mechanical mixture, physical/chemical properties, quantitative/qualitative properties, physical/chemical changes, ion, ionization energy, etc).
• be familiar with the history of atomic theory, including the contributions of all discussed scientists and theories.
• be able to convert between standard atomic notation, Bohr diagrams and Lewis diagrams.
• be familiar with the history of Medeleev and Modern Periodic Table.
• know the names and locations of families, periods, etc on the Periodic Table.
• be able to provide the characteristics of metals, non-metals and metalloids
• be able to state and use the periodic trends (reactivity, IE, EA, AR) to make predictions
• be able to explain the trends