Periodic Trends in The World

2nd period, going from alkali metal to noble gas. Explain why this is so. Is the trend periodic? The atomic radius decreases going from alkali metal to noble gas 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, meaning the repulsion between electron shells is not seen within the period. Thus, the effect of the increased proton and resulting increase in nuclear charge is disproportionate to the effect of increased electrons, causing the electrons to be attracted to the nucleus and contracting the atom. The trend is periodic, occurring for every period of the periodic table, however, there are a few inconsistencies within different periods which can be explained by the electron configurations of those specific elements. The horizontal trend as described above shows a constant decrease in atomic radius going from alkali metals to noble gas within a period. As seen in the graph, there are exceptions, spikes in the trend occurring in aluminium, gallium, and indium. Looking at the periodic table, you can see What happens constantly to the size of atoms 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 size of the atom dramatically increases when going 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 across the period, 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 atom of the noble gas has a completely filled valence electron shell, as well as a strong nuclear charge. All of this changes in the atomic structure of following 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 shells and an inflated size. The increase in electron shells cause an increase in size due to repulsion between electrons. This trend occurs after every period. Looking at just one group, the alkali metals, what happens to the size of the atoms going from lithium to rubidium? Explain why this is so.

The size of the atoms increase going down the group of alkali metals because of the constant increase of one electron shell. Each element in the family gains one electron shell, meaning there is more shielding occurring between shells, and the repulsion results in an increased size.

From your graph, describe what happens to the first ionization energy for the elements of the 2nd period, going from alkali metal to noble gas. 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 going from alkali metal to noble gas with the exception of boron and oxygen. Generally, elements have an increase in ionization energy because of the increase in electrons within the same principal shell. This means there is an increase in nuclear charge without an increase in shielding. The electron valence shell consequently becomes more stable and less willing to become an ion of less stable state. The variations seen in the trend can be explained by examining their electron configuration. In the second period, we can see that boron has an electron configuration of [He] 2s22p1 and the electron configuration of oxygen is [He] 2s22p4. In both these cases, the ionization energy has decreased from its preceding elements as their configuration is less stable. Going from beryllium to boron, beryllium has an electron configuration of [He]2s2, a filled valence orbital, making it more stable than a boron atom. Thus, less energy is required to remove one electron from a boron atom as it would achieve the same configuration as beryllium afterwards. The same is true 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 a p orbital shell with 4 electrons, as seen in oxygen, means there are two unpaired electrons in 2 of the p orbital, and 2 paired electrons in one orbital, the repulsion occurring between the two paired electrons of opposite spin allows it to be more easily removed). This trend occurs every period, going from alkali metal to noble gas, and the graph differs slightly with the addition of the d-block elements in period 4, but still continues the same trend. What happens consistently 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 going from noble gas to alkali metal of the next period, and is actually smaller than the ionization energy of the alkali metal from the previous period. This is because of the increase in electron shells and thus a dramatic increase in electron shielding. Alkali metals have the lowest nuclear charge acting on their valence electron due to their ratio of protons to inner electron shells whereas noble gases have the most 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 atom of an alkali metal, making its first ionization energy much lower than that of a noble gas, which explains the steep drop seen in the graph. This trend is periodic. Looking at 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 going down the group of noble gases. This is because of the increase in number of electron shells. Elements lower in the group have a greater number of inner electron shells, meaning there is more shielding, repulsion between these shells, making it easier for the single outer valence electron to be removed. Part C: Trends for Successive Ionization Energies Looking at the data for phosphorus only, describe the trend that you see. Explain why this is soThe ionization energies for phosphorus increase slightly from the 1st I.E. until the 5th I.E. where there is a sharp increase in energy required for the 6th ionization energy. The increase in energy for each successive electron is because every time an electron is removed, the atom becomes more positive and there is more attraction between the electrons and the nucleus. Looking at the electron configuration of phosphorus, [Ne]3s23p3, we can see that it has five valence electrons. This explains the very gradual increase in energy seen in the graph; the first five electrons belonged to the same energy level, which is why the increases in energy were slight.

This also explains the dramatic increase in energy required to remove the sixth electron. The sixth electron is part of the inner electron shells, and experiences less shielding than the outer five electrons, thus more energy is required to remove it. We can also see that without the five outer electrons, phosphorus has the very stable noble gas configuration of neon, which helps illustrate why it so much more energy is required to remove the 6th electron. Describe the trend that you see relating the elements from group 1, 2, 13, 14, 15.

Going from left to right on the graph, that is, from 1st electron removed to 4th electron removed, the ionization energy increases gradually. The gradual increase is because the with every successive electron that is removed, the ion becomes more positive, meaning there is more attraction between the electrons and the nucleus and it is more difficult to remove successive electrons. For all elements, the ionization energy jumps steeply when the electron is removed when they have reached a noble gas configuration. The group 1 element has the jump for the 2nd electron, the group 2 element for the 3rd electron, the group 13 element for the 4th electron and so on. This trend points to when all outer electrons have already been removed (leaving the element with a noble gas configuration) and the electron has been removed from an inner electron shell.

To which group of the periodic table does this element likely belong? Explain why this is so. The element likely belongs to group 13 of the periodic table because the data shows the largest increase in ionization energy going from the 3rd I.E. to the 4th I.E., going from 3664 kJ/mol to 25060 kJ/mol, a roughly 580% increase. This means there was a significant change in energy level for the 4th electron, which corresponds with the electron configuration of a group 13 element. Elements in group 13 are p-block elements, and have 3 electrons in the p orbital shell. Once those electrons are removed, more energy is required to remove the 4th successive electron, which is found in the inner s orbital, which has much more nuclear charge acting on it. The data also points to group 13 element based on the second significant increase in energy going from 1st I.E. to 2nd I.E.. This points to a group 13 element as the increase corresponds with the increase in energy required to detach the second electron found in a filled, stable s orbital (2nd I.E.) as opposed to the lesser energy required to remove the single outermost electron from the p orbital. The element however must be a group 13 element from the second or third period for this explanation to apply, as explained in the following answer.

To which period of the periodic table does this element likely belong? Explain why this is so. This element likely belongs to the second period of the periodic table because of the relatively high I.E. values as well as the very large increase between 3rd I.E. and 4th I.E.. All elements in group 13 and group 3 have this same increase due to the position of the 4th electron in an inner orbital, however, it is known the ionization energy decreases going down a group due to the effects of shielding from increased number of electron shells. For this reason, it is likely that the element belongs to a period higher up on the periodic table, the second period. Elements of lower periods such as aluminum of gallium would not have as significant an increase going from 3rd I.E. to 4th I.E. as the 4th electron would be in a shell that experiences some nuclear shielding from additional interior shells, while a period 2 element such as boron would have the greatest increase as the 1p orbital from which the 4th electron would be taken experiences the most nuclear charge.