“Test Review” Go through these to practice for your test on Tuesday (answers are on following slides):

Review questions for electron configuration: I can …. Write the longhand electron configurations for Rb Br Ni 2. Write the shorthand electron configuration for the above 3 elements.

I can… 3. Determine how many electrons will either be lost or gained for these elements to become stable & look like a noble gas electron configuration. Rb b. Br c. Ni 4. Tell the identity of the following elements: 1s22s22p63s1 1s22s22p63s23p4 [Ar]4s23d5

Review questions for periodic trends. I can… 5. Explain how the atomic radius changes as you go across a period on the periodic table and how it changes going down a family. 6. Arrange the following from smallest to largest atomic radius: Ca, Zn, K, Br 7. Arrange the following from least to greatest electronegativity: 8. Explain how an element becomes like a noble gas to have a full outer energy level?

I can… 9. Name the element that has the following last e- configuration: 4s2 3p2 3d5 6s2 [Ar]4s23d104p2 10. Why is it possible to find the element’s identity without using the entire e- configuration?

Check your work: 1. Rb 1s22s22p63s23p64s23d104p65s1 Br 1s22s22p63s23p64s23d104p5 Ni 1s22s22p63s23p64s23d8 2. [Kr] 5s1 [Ar] 4s23d104p5 [Ar] 4s23d8 3. Rb gives up 1 e-, Rb+1 ion Br gains 1 e-, Br-1 Ni, gives up 2 e-, Ni+2

Check your work 4. a. Na b. S, c. Mn 5. (draw PT & label with arrows how atomic radius change & include ionization energy & electronegativity 6. Br, Zn, Ca, L 7. K, Ca, Zn, Br 8. Elements transfer e- by gaining enough to reach 8 valence e- or giving up outer orbital e- to fall back on a previous full orbital. 9. a. Ca b. Si c. Mn d. Ba e. Ge 10. Aufbau principle because each sublevel should be filled up

I can pass the test on Tuesday because I know … the periodic trends and how to use them to compare elements to determine the differences in atomic radius, ionization energy, & electronegativity. Use the periodic table to write e- configurations, identify elements from e- configurations & use the e- configurations to determine the number of valence e- that will need to be gained or lost to become stable, like a noble gas configuration.

Part II: Periodic trends & e- configuration review I can… What is electronegativity? What is ionization energy? How is ionization energy & electronegativity related to the atomic radius of an element? Draw the Bohr diagram for lithium and chlorine. Then show how the electrons are transferred for the two to combine in making an ionic compound, LiCl. How is the atomic radius impacted when valence electrons are lost or gained by an element as it becomes an ion?

I can … 6. Explain how does the atomic radius of O-2 ion become smaller than when it was just O? (what is this called?)_________________ Ne Mg O 8. How can 2 or more elements have the same e- configuration? (look at 7 to explain) Do e- configuration Now make these an ion

I can… Which group on the periodic table has no electronegativity or ionization energy? ____________ Explain why that group would have not electronegativity or ionization energy?

Check your work: Electronegativity is … Ionization is … The atomic radius determines the ionization & electronegativity values because as the atomic radius decreases it causes an increase in the ionization energy & electronegativity because the outer valence e- are being pulled in closer to the nucleus because of the nuclear charge, so the valence e- are being held on tighter (requiring a higher ionization value) and are more likely to attract electrons more (making for greater electronegativity value).

Draw Bohr diagram of Li and Cl Draw Bohr diagram of Li and Cl. The Li should have 2 orbital rings around the nucleus with 1 valence e- on the outer orbital and Cl should have 3 orbitals with 7 valence e- on the outer orbital. Li will give Cl the 1 valence e- it has, no lose the outer orbital around Li to become Li+1 ion, while Cl will gain the 1 valence e- that will add to the 7 valence e- making Cl stable with 8 valence e- now and Cl will become the ion Cl-1. Both ions (Li+1 and Cl-1) will have a smaller atomic radius than they did before the transfer of the e-. They both become smaller because Li+1 lost the outer orbital to get smaller atomic radius while the nuclear charge pulled in the outer orbital even more on Cl-1 after it gained another e- due to the “shielding effect”.

7. Ne Mg O As you can see more than 2 elements can have the same e- configuration once they become an ion, the e- configuration of the ion will ALWAYS look like that of a noble gas. 9.&10. Noble gases have a 0 electronegativity & ionization energy because they are stable, they do not need to attract a pair of e- nor do they need to gain any to have an ionization energy. Do e- configuration Now make these an ion 1s22s22p6 1s22s22p6 (stable not going to change to an ion) 1s22s22p63s2 Mg+2 1s22s22p6 (will give up 3s2) 1s22s22p4 O-2 1s22s22p6 (will gain 2 e-)

Do I really get stuff? What happens with electronegativity and metallic properties as the elements in Period 4 are considered in order of increasing atomic number? What happens with electronegativity and ionization as the elements in Group 2 are considered in order of increasing atomic number? In 1s2 2s2 2p4 how many more electrons does this element need to satisfy the octet rule? How many elements could have the electron configuration 1s22s22p63s23p6? (at least 10) WHY? What is the correct ionization energy order from low to highest for these I, Br, Cl, F ?