2nd period, moving from alkali metals to noble gases. Explain why this is so. Is the trend periodic? In the second period the atomic radius decreases from alkali metal to noble gas, as the atomic number increases. This can be explained by the increase in nuclear charge as the atomic number, the number of protons, increases. Furthermore, the number of electrons in the atom increases within the valence level, meaning that repulsion between electron levels is not observed during this period. Therefore, the effect of increasing the proton and the resulting increase in nuclear charge is disproportionate to the effect of increasing the electrons, causing the electrons to be attracted towards the nucleus and the atom to contract. The trend is periodic and occurs for every period of the periodic table, however, there are some inconsistencies within the different periods that can be explained by the electronic configurations of those specific elements. The horizontal trend described above shows a constant decrease in atomic radius going from alkali metals to noble gases over a period. As seen in the graph, there are exceptions, trend peaks that occur in aluminum, gallium and indium. Looking at the periodic table, you can see What consistently happens to the size of atoms when you go from a noble gas of one period to an alkali metal of the next period? Explain why this is so? Is the trend periodic? Say no to plagiarism. Get a tailor-made essay on "Why Violent Video Games Shouldn't Be Banned"? Get an original essay The size of the atom increases dramatically when you go from a noble gas of one period to an alkali metal of the next period. This jump can be seen clearly in the graph, where the atomic radius gradually decreases through the period, and then increases to a value often greater than the atomic radius of the alkali metal of the previous period. This happens due to the addition of the valence electron shell in the atom. The noble gas atom has a completely filled valence electron shell, as well as a strong nuclear charge. All this changes in the atomic structure of the next element in the table, the alkali metal, where there is a single unpaired electron in its valence shell, the appearance of the new orbital, causing further shielding between the shells and an inflated size. The increase in electron shells causes an increase in size due to the repulsion between electrons. This trend occurs after every period. Considering just one group, that of the alkali metals, what happens to the size of the atoms ranging from lithium to rubidium? Explain why this is so. The size of atoms increases moving down the alkali metal group due to the constant increase in the level of an electron. Each member of the family gains a level of electrons, meaning there is increased shielding between levels and the repulsion results in an increase in size. From your graph, describe what happens to the first ionization energy for 2nd period elements, going from alkali metal to noble gas. Explain both the overall trend and the reasons for variations within the overall trend. Is the trend periodic? The ionization energy of the elements increases going from alkali metals to noble gases with the exception of boron and oxygen. Generally elements have an increase in ionization energy due to the increase in electrons within the same main level. This means that there is an increase in nuclear charge without an increase in shielding. The electron's valence shell consequently becomes more stable and less willing to become a less stable state ion. The variations observed in the trendcan be explained by examining their electronic configuration. In the second period, we can see that boron has an electronic configuration of [He] 2s22p1 and the electronic configuration of oxygen is [He] 2s22p4. In both of these cases the ionization energy is decreased compared to the previous elements because their configuration is less stable. Going from beryllium to boron, beryllium has an electronic configuration of [He]2s2, a filled valence orbital, making it more stable than a boron atom. Therefore, less energy is required to remove an electron from a boron atom since it would later reach the same configuration as beryllium. The same goes for oxygen. Oxygen is less stable than nitrogen due to its 2p4 configuration and has a lower ionization energy since ionization would allow the atom to have a half-filled valence p orbital, a more stable configuration (since the orbital shell ap with 4 electrons, as seen in oxygen, means that there are two unpaired electrons in 2 of the orbital p and 2 paired electrons in one orbital, the repulsion that occurs between the two paired electrons of opposite spin allows it to be more easily REMOVED ). This trend occurs in each period, moving from alkali metals to noble gas, and the graph differs slightly with the addition of block D elements in period 4, but still continues the same trend. What consistently happens to the first ionization energy when moving from a noble gas of one period to an alkali metal of the next period? Explain why this is so. Is the trend periodic? The first ionization energy drops dramatically when moving from the noble gas to the alkali metal of the next period and is actually lower than the ionization energy of the alkali metal of the previous period. This is due to the increase in electronic shells and therefore a dramatic increase in electronic shielding. Alkali metals have the lowest nuclear charge acting on their valence electrons due to their ratio of protons to internal electron shells, while noble gases have the highest nuclear charge acting on their valence electrons, giving them stability. Since there is very little attraction between the electrons and the nucleus, it is very easy to remove an electron from an atom of an alkali metal, making its first ionization energy much lower than that of a noble gas, which explains the observed sharp drop in the graph. This trend is periodic. Considering only one group, that of the noble gases, what happens to the first ionization energy that goes from helium to krypton? Explain why this is so. The ionization energy decreases as you descend into the noble gas group. This is due to the increase in the number of electron shells. Elements further down the group have a greater number of internal electron shells, meaning there is more shielding and repulsion between these shells, making it easier to remove the single external valence electron. Part C: Trends for Successive Ionization Energies Looking at the data for phosphorus only, describe the trend you see. Explain why this is so. The ionization energies of phosphorus increase slightly from the 1st IE up to the 5th IE, where there is a large increase in the energy required for the 6th ionization energy. The increase in energy for each subsequent electron is due to the fact that each time an electron is removed, the atom becomes more positive and there is greater attraction between the electrons and the nucleus. Looking at the electronic configuration of phosphorus, [Ne]3s23p3, we can see that it has five valence electrons. This explains the very gradual increase in energy visible in the graph; the first five electrons belonged to the same energy level, which is why the energy increases were slight. That is.