ANALYSIS OF THE IONIZATION ENERGY & ATOMIC RADIUS GRAPHS: ANSWER KEY Part A: Determining the trends 1. On the graphs, beside the appropriate data points, write the symbols of all the Group I elements (H, Li, etc…) and all the Noble Gases. 2. Label on both graphs the first period, the second period, etc…. 3. What is the trend evident in atomic radius, as you move across each period from left to right? decreases 4. What is the trend evident in Ionization energy as you move across each period from left to right? increases 5. Look at the Noble gases (He, Ne, Ar, Kr) are in the same group. a) What is the trend in atomic size from He to Kr? increases b) What is the trend in ionization energy from He to Kr decreases 6. Look at the Alkali metals (Li, Na, K). a) What is the trend in atomic size from Li to K? increases b) What is the trend in ionization energy from Li to K decreases (i.e. easiest to remove e- from K)

7. What is the relationship between the peaks and troughs in the Atomic Radius graph and the peaks and troughs in the Ionization Energy graph? The peaks in AR graph become the troughs in the IE graph; the troughs in AR graph become the peaks in the IE graph 8. What do you estimate to be the missing value for atomic radius of

a) 25Mn? 0.117nm

b) 34Se? 0.116 nm

9. What do you estimate to be the missing value for the first ionization energy of

a) 25Mn? 700 kJ/mol

b) 34Se? 1050 kJ/mol

Part B: Explaining the Trends

10. Based on the patterns that you observe, would you expect the atomic radius of 37Rb to be larger or smaller than element 36Kr? Explain. Rb is larger because:  Rb is at the start of a new period which means there is one more added shell.  Though Rb has more protons in the nucleus which makes e-nucleus attraction stronger, there are more shielding of the core e-.

 Note: if you use the following approach to answer question 10-13, you will not get full or any mark Bad answer example: From L to R, atomic radii decreases therefore Rb is larger Down a group atomic radii increases therefore Rb is larger

11. Based on the patterns that you observe, would you expect the ionization energy of 37Rb to be larger or smaller than element 36Kr? Explain. IE for Rb is smaller because:  Rb valence e- of RB is loosely held due to more shielding of core e-. It is therefore easier; thus, requires less energy to remove its valence e-  Kr’s e-nucleus attraction is highest within its period while shielding of core e- is constant; thus, it requires lots of energy to remove a valence e- from Kr 12. Explain with the use of condensed Bohr diagrams the underlying factors that explain the difference in atomic radius between 14Si and 17Cl. Cl with 3 e- more than Si is smaller than Si for the following reasons:

 Those 3 additional e- in Cl are put in the same valence shell; they are not put in the core shell to increase shielding.

 The nuclear-electron attraction is higher in Cl than in Si due to the increase # of protons in Cl. Thus, valence e- are pull closer to the nucleus in Cl

13. Explain with the use of condensed Bohr diagrams the underlying factors that explain the dip in first ionization energy between 18Ar and 19 K

IE K< IE Ar because - Shielding toward the valence shell e- increases in K compared to Ar due to an additional layer of core e- - While nuclear charge increases in K, increased shielding causes the last shell to be further from the nucleus making it easier (i.e. lower energy) to ionize K