POGIL 5 KEY Periodic Table Trends (Part 1)

Transcription

1 Honors Chem Block Name POGIL 5 KEY Periodic Table Trends (Part 1) The periodic table is often considered to be the best friend of chemists and chemistry students alike. It includes information about atomic masses and element symbols, but it can also be used to make predictions about atomic size, electronegativity, ionization energies, bonding, solubility, and reactivity. In this activity you will look at a few periodic trends that can help you make those predictions. Like most trends, they are not perfect, but useful just the same. 1. Let us learn about the first trend on the periodic table Atomic Radius Trend. Try answering the following questions (your instructor will help you): a. How would you define the atomic radius? Atomic radius is the distance between the nucleus and the place of highest probability of the furthest electron location b. What are the units for the atomic radius? The units are Angstroms (A o ) 1nm=10A o 1m=10 10 A o c. What is the group trend for atomic radius? Use the Atomic Radius graph on the back of this page to answer this question. (It is easier to see elements that belong to the same group by looking at the highest points on the graph or the lowest points on the graph). The group trend for atomic radius goes down the group. d. Using your knowledge of Coulombic attraction and the structure of the atom, explain the trend in atomic radius that you identified in Question c. Hint: You should discuss either a change in distance between the nucleus and outer shell of electrons or a change in the number of protons in the nucleus. As you go down the group the number of levels increases, so the distance between the nucleus and valence electrons becomes bigger, so does the atomic radius e. What is the trend in atomic radius as you go across a period? Use the graph to answer this question. As you go from right to left in a period, the atomic radius becomes bigger. f. Using your knowledge of Coulombic attraction and the structure of the atom, explain the trend in atomic radius that you identified in Question e. As you go from right to left in a period, the number of protons decreases, the force of attraction of the valence electrons becomes weaker, so the radius increases g. Which one influences the force of attraction more, the number of protons in the nucleus or the distance between the nucleus and the valence electrons that make the atomic radius smaller or bigger? It looks like the distance between a nucleus and valence electrons influences the force of attraction more. In a group as you go down, the nucleus becomes much more powerful, but still the force becomes weaker with the increase of # of levels. 1

2 Atomic Radius Trend K Atomic radius (pm) Li Na 50 H Ar Kr 0 He Ne Atomic number 2

3 2. Rank the following atoms in order from smaller to larger: Ge, Sr, S, Si, Rb, F, Ca, P, Ga F<S<P<Si<Ge<Ga<Ca<Sr<Rb 3. If we analyze atomic radius group trend further, there is a question we still need to answer. Why does an addition of one energy level or shell makes the atomic radius increase so much. Even if the force of attraction is decreased by the square of the distance, would it be at least partially compensated by the attraction coming from a much more positive nucleus? This phenomenon can be explained if we take the shielding of the valence electrons into account. Once a level of electrons appears between the valence shell and the nucleus, the nuclear attraction gets weakened. The electrons of the shell in the middle would shield the valence electrons from the nucleus. So, no matter how big the nuclear charge is (number of protons in the nucleus), the valence electrons cannot experience that attraction fully. In reality it is not the nucleus that attracts the valence electrons, it is the core of the atom or effective nuclear charge. The examples below show the core of the atom that is positively charged, the valence electrons and the calculation of the core charge or the effective nuclear charge Z eff 3

4 4. What is another way of figuring out the effective nuclear charge? (Hint: valence electrons). Effective nuclear charge (Z eff ) = number of valence electrons 5. What is the trend for Z eff on the periodic table? In a period (from left to right) Z eff goes up In a group Z eff stays the same 6. What is Z eff in: K atom +1 ; C atom _+4 ; As atom +5 Br atom _+7 7. Now we can answer the question why the increase of number of protons down the group cannot compete with the increase of the distance between the nucleus and the valence electrons. Protons cannot attract valence electrons because of shielding by inner shells. Since Zeff is not changing down the group, the distance is the only one that affects the force From now on, when describing any trend in a period, you will use Z eff only as an explanation and, when describing any trend in a group, you will use the number of energy levels only as an explanation of the trend. Ionic Radius Try to do this on your own: 8. What do the following atoms/ions have in common: Ne, F -, O 2-, Mg 2+, Na + These species are isoelectronic. 9. Arrange the above atoms/ions in order from smallest to largest. (Hint: consider the nucleus) Mg 2+, Na +, Ne, F, O In the above example, the number of electrons for each atom was the same. Let s consider examples where the numbers of electrons are different: Which is bigger: F or F -? Why? F >F. They have the same nuclear power. F has one extra electron, which brings more repulsion, and atoms become bigger. Which is bigger: Na or Na +? Why? Na>Na +. When an atom becomes a positive ion, it loses the entire valence shell. Na has 3 shells, Na + has 2 shells. 4

5 11. Atom/Ion Size Sample Problems Arrange each of the following sets of atoms/ions in order from smallest to largest radius: 1) Rn, Ra 2+, At -, Fr + They are isoelectronic to Rn. The nuclear charge decides the size. Ra 2+ <Fr + <Rn<At 2) Cl, Ar, Li +, Se, S, Se 2-, He Li +, He, Ar, Cl, S, Se, Se 2 3) Fe, Fe 2+, Fe 3+ Fe 3+, Fe 2+, Fe Ionization Energy The ionization energy is the amount of energy needed to remove an electron from an atom. 12. Which takes more energy, removing an electron from an atom where the nucleus has a tight hold on its electrons, or a weak hold on its electrons? Explain. It takes more energy to remove an electrons that is tightly held by the nucleus 13. Using your knowledge about the Z eff and the number of energy levels predict the group trend for ionization energy and explain your thinking. The group trend for IE goes up the group. Elements in the same group have the same Z eff. As you go up the group, the # of shells decreases, the force of attraction increases, harder to remove an electron, more energy needed. 14. Predict the period trend for the ionization energy and explain your thinking. The period trend will go right. As you go right in a period, Z eff increases, force of attraction increases, holding electrons tighter, it is harder to remove them, so more energy is needed. 15. What does 1 st Ionization Energy refer to as opposed to 2 nd Ionization Energy? First IE is needed to remove the first electron. Second IE is spent to remove the second electron. 5

6 16. Examine the following ionization energies. Explain the break in the sequence. Si kj/mol S has an IE drop compared to P. P S 1012 kj/mol kj/mol P: S: Cl 1255 kj/mol It is easier to remove an electron from sulfur than from phosphorus. The phosphorus configuration is more stable Ar 1520 kj/mol (p-sublevel is half filled). 17. Follow the instruction, given to you by your teacher to build the graph for ionization energy vs. atomic number, and check your predictions (this time you need to use the first and the third column from the data). 18. Look at your graph. What is happening to ionization energy as the atomic number increases? As atomic number increases, the IE goes up and down periodically in the same intervals. 19. How can you explain the interruptions in the graph? The interruptions are the IE drops. They are in groups 3A and 6A. In 6A, for the same reason as sulfur. In 3A the elements have only one electron in the p-subshell. It is easier to remove it than removing an electron from completely filled s-subshell. 20. Sample Problems 1. The ionization energies of all noble gases are extremely high. Explain. Noble gases have the most stable electron configuration 8 valence electrons (complete valence shell). They also have the highest Z eff, so it is very hard to remove electrons. 2. Arrange the following atoms in order of increasing ionization energy: In, Ar, Rb, Ga, Kr, Cs Cs<Rb<In<Ga<Kr<Ar 6