We would like to begin by stating our sincere gratitude towards you and your company for giving us the opportunity to help analyze your data from the Photoelectron Spectroscopy (PES) instrument. Our job was to complete five main tasks. We fixed your corrupted data of chlorine with a new one of our own. We expanded your portfolio so you have the PES of the elements Titanium and Vanadium, while also providing a full analysis of all of the elements that were tested. We discovered a pattern in which the elements relate back to the periodic table. Our lab also had extra time to provide you with data on how the atomic size corresponds with the PES spectra. For your first request, you asked us to provide the PES for chlorine because yours was
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On page 24 you will find a list of the electron configurations for each of the elements. From this data you can find the total number of electrons, the total number of shells and subshells, as well as how many electrons are in each shell. The large number (the one in 1s2 ) will tell you the shell that the electrons are in. For example, helium (He) has one shell, so its electron configuration is 1s2. Furthermore, the electron configuration tells you what subshell each of the electrons occupy. The subshell is the letter next to the larger number. You will see subshells S, P, and D on the chart (24). The subshell that the valence electron occupies dictates the position of the element on the periodic table. The S subshell occupies groups 1 and 2, the D subshell is all of the transition metals, and the P subshell occupies groups 13-18. The D subshell is irregular in that it will always have a three in front of it, but the numbers in front of the other subshells will change based on what shell they are in. The three is there to show that the electrons in the D subshell are harder to remove than any electron in the fourth subshell. The peak for the electrons in the 3d shell also comes before the peaks for the electrons in the fourth shell on a photoelectron spectroscopy. Moreover, the exponent after the subshell is …show more content…
We have concluded that as the ionization energies decrease, the atomic sizes increase, and the same way vice versa. This is because the lower the ionization energy is, the lower the core charge is. When the core charge is low the nucleus has a weaker attraction to the electrons. And when the attraction with the electrons is weak, the electron shells are more separated, making the element larger. However, this is only the case between comparing elements with the same number of total electron shells. Because even though hydrogen has a core charge of +1, it is still smaller than the element Beryllium, regardless of the fact that hydrogen has a lower core charge. This is because hydrogen has a total of one shell whereas Beryllium has a total of two shells, making Beryllium
On page 24 you will find a list of the electron configurations for each of the elements. From this data you can find the total number of electrons, the total number of shells and subshells, as well as how many electrons are in each shell. The large number (the one in 1s2 ) will tell you the shell that the electrons are in. For example, helium (He) has one shell, so its electron configuration is 1s2. Furthermore, the electron configuration tells you what subshell each of the electrons occupy. The subshell is the letter next to the larger number. You will see subshells S, P, and D on the chart (24). The subshell that the valence electron occupies dictates the position of the element on the periodic table. The S subshell occupies groups 1 and 2, the D subshell is all of the transition metals, and the P subshell occupies groups 13-18. The D subshell is irregular in that it will always have a three in front of it, but the numbers in front of the other subshells will change based on what shell they are in. The three is there to show that the electrons in the D subshell are harder to remove than any electron in the fourth subshell. The peak for the electrons in the 3d shell also comes before the peaks for the electrons in the fourth shell on a photoelectron spectroscopy. Moreover, the exponent after the subshell is …show more content…
We have concluded that as the ionization energies decrease, the atomic sizes increase, and the same way vice versa. This is because the lower the ionization energy is, the lower the core charge is. When the core charge is low the nucleus has a weaker attraction to the electrons. And when the attraction with the electrons is weak, the electron shells are more separated, making the element larger. However, this is only the case between comparing elements with the same number of total electron shells. Because even though hydrogen has a core charge of +1, it is still smaller than the element Beryllium, regardless of the fact that hydrogen has a lower core charge. This is because hydrogen has a total of one shell whereas Beryllium has a total of two shells, making Beryllium