EXERCISES in Chem Report Anwers
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EXERCISES: 1. Identify the following elements: a) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 ANS: The element is iron, Fe b) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 7 ANS: The element is Rhodium, Rh c) 1s 2 2s 2 2p 6 3s 2 3p 4 ANS: The element is Sulfur, S d) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 4 ANS: The element is Polonium, Po 2. Write the electronic configuration of each element: a. As (Arsenic) ANS: What if we had an atom of arsenic, with atomic number 33? We would fill up all the orbital sublevels in the electron configuration pattern until we got to the 4p orbital sublevel. Then we would have only 3 electrons left. 2 + 2 + 6 + 2 + 6 + 2 + 10 + 3 = 33. The electron configuration for arsenic, then, is as follows: B, Li (Lithium)
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Transcript of EXERCISES in Chem Report Anwers
EXERCISES:
1. Identify the following elements:a) 1s2 2s2 2p6 3s2 3p6 4s2 3d6
ANS: The element is iron, Feb) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d7
ANS: The element is Rhodium, Rh
c) 1s2 2s2 2p6 3s2 3p4
ANS: The element is Sulfur, S
d) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p4
ANS: The element is Polonium, Po
2. Write the electronic configuration of each element:a. As (Arsenic)
ANS:
What if we had an atom of arsenic, with atomic number 33? We would fill up all the orbital sublevels in the electron configuration pattern until we got to the 4p orbital sublevel. Then we would have only 3 electrons left. 2 + 2 + 6 + 2 + 6 + 2 + 10 + 3 = 33. The electron configuration for arsenic, then, is as follows:
B, Li (Lithium)
3 positive charges (from the protons) and 2 negative charges (from the electrons) add up to a total of +1. So we have only 2 electrons, and the electron configuration for the lithium ION is:
C. F (Fluorine)
9 positive charges (from the protons) and 10 negative charges (from the electrons) add up to a total of -1. So we have 10 electrons and the electron configuration for the fluorine ION
