2nd period, ranging from alkali metal to noble gases. Explain why this is so. Is the trend periodic? The atomic radius decreases from the alkali metal to the noble gases in the second period, as the atomic number increases. This can be explained by the increase in nuclear charge as the atomic number, the number of protons, increases. Additionally, the number of electrons in the atom increases within the valence shell, which means that repulsion between electronic shells is not visible over the period. Thus, the effect of increasing the proton and the resulting increase in nuclear charge are disproportionate to the effect of increasing electrons, causing the electrons to be attracted to the nucleus and the nucleus to contract. the atom. The trend is periodic and occurs for each period of the periodic table. However, there are some inconsistencies within different periods that can be explained by the electronic configurations of these specific elements. The horizontal trend described above shows a steady decrease in atomic radius from alkali metals to noble gases over a given period. As the chart shows, there are exceptions, with trend peaks occurring in aluminum, gallium, and indium. By looking at the periodic table, you can see What is constantly happening to the size of atoms when you go from a noble gas in one period to an alkali metal in 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 Should Not Be Banned”? Get the original essay The size of the atom increases dramatically when going from a noble gas of one period to an alkali metal of the next period. This jump is clearly visible in the graph, where the atomic radius gradually decreases over the period, then increases to a value often greater than the atomic radius of the alkali metal from the previous period. This happens due to the addition of a layer of valence electrons in the atom. The noble gas atom has a completely filled valence electron shell, as well as a strong nuclear charge. All this changes 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 additional shielding between the shells and a swollen waist. Increasing the number of electronic shells results in an increase in size due to repulsion between electrons. This trend occurs after each period. Considering just one group, the alkali metals, what happens to the size of atoms ranging from lithium to rubidium? Explain why this is so. The size of the atoms increases moving down the alkali metal group due to the constant increase in an electron shell. Each element in the family gains an electron layer, meaning there is more shielding between layers and repulsion leads to an increase in size. From your graph, describe what happens to the first ionization energy for the 2nd period elements as they change from alkali metal to noble gases. Explain both the general trend and the reasons for variations within the general trend. Is the trend periodic? The ionization energy of the elements increases from the alkali metal to the rare gases with the exception of boron and oxygen. Generally, elements have an increase in ionization energy due to the increase in the number of electrons in the same major shell. This means that there is an increase in the nuclear chargewithout increasing shielding. The electronic valence shell therefore becomes more stable and less willing to become an ion in a less stable state. The observed variations in the trend can 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 these two cases, the ionization energy has decreased compared to its previous elements because their configuration is less stable. Moving from beryllium to boron, beryllium has an electronic configuration of [He]2s2, a filled valence orbital, making it more stable than a boron atom. Thus, less energy is required to remove an electron from a boron atom, because it would then obtain 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 (since an ap orbital shell with 4 electrons, as seen in oxygen, means that there are two unpaired electrons in 2 of the p orbital, and 2 paired electrons in one orbital, with repulsion occurring between the two paired electrons of opposite spin makes it easier to eliminate). This trend occurs in every period, going from alkali metal to noble gases, and the graph differs slightly with the addition of Block D elements in period 4, but still continues the same trend. What happens systematically to the first ionization energy when going 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 decreases sharply when moving from the noble gas to the alkali metal of the next period, and is actually lower than the alkali metal ionization energy of the previous period. This is due to the increase in the number of electronic layers and therefore a dramatic increase in electronic shielding. Alkali metals have the lowest nuclear charge acting on their valence electron due to their internal proton to electron shell ratio, while noble gases have the highest nuclear charge acting on their valence electrons, giving them their stability. Since there is very little attraction between the electrons and the nucleus, it is very easy to remove an electron from an alkali metal atom, making its first ionization energy much lower than that of an alkali metal atom. rare gas, which explains the sharp drop observed in the graph. This trend is periodic. Considering just one group, the noble gases, what happens to the first ionization energy going from helium to krypton? Explain why this is so. The ionization energy decreases moving down the noble gas group. This is due to the increase in the number of electronic layers. Lower elements in the group have a greater number of internal electronic layers, which means there is more shielding and repulsion between these layers, making it easier to eliminate the single outer valence electron. Part C: Trends in successive ionization energies Looking only at the phosphorus data, describe the trend you see. Explain why this is so. The ionization energies of phosphorus increase slightly from the 1st IE until the 5th IE where there is a large increase in the energy required for the 6th ionization energy. The increase in energy for each successive electron is because each time an electron is removed, the atom becomes more positive and there is more attraction between the electrons.