This has two effects. The explanation for the drop between magnesium and aluminum is the same, except that everything is happening at the 3-level rather than the 2-level. The 3p electron in aluminum is slightly more distant from the nucleus than the 3s, and partially screened by the 3s 2 electrons as well as the inner electrons. Both of these factors offset the effect of the extra proton.
Once again, you might expect the ionization energy of the group 6 element to be higher than that of group 5 because of the extra proton. What is offsetting it this time? The screening is identical from the 1s 2 and, to some extent, from the 2s 2 electrons , and the electron is being removed from an identical orbital.
The difference is that in the oxygen case the electron being removed is one of the 2p x 2 pair. The repulsion between the two electrons in the same orbital means that the electron is easier to remove than it would otherwise be. The drop in ionization energy at sulfur is accounted for in the same way. As you go down a group in the Periodic Table ionization energies generally fall. You have already seen evidence of this in the fact that the ionization energies in period 3 are all less than those in period 2.
Taking Group 1 as a typical example:. Why is the sodium value less than that of lithium? There are 11 protons in a sodium atom but only 3 in a lithium atom, so the nuclear charge is much greater.
You might have expected a much larger ionization energy in sodium, but offsetting the nuclear charge is a greater distance from the nucleus and more screening. Lithium's outer electron is in the second level, and only has the 1s 2 electrons to screen it.
The sodium's outer electron is in the third level, and is screened from the 11 protons in the nucleus by a total of 10 inner electrons. In other words, the effect of the extra protons is compensated for by the effect of the extra screening electrons.
The only factor left is the extra distance between the outer electron and the nucleus in sodium's case. That lowers the ionization energy. Similar explanations hold as you go down the rest of this group - or, indeed, any other group. Apart from zinc at the end, the other ionization energies are all much the same. All of these elements have an electronic structure [Ar]3d n 4s 2 or 4s 1 in the cases of chromium and copper. The electron being lost always comes from the 4s orbital.
As you go from one atom to the next in the series, the number of protons in the nucleus increases, but so also does the number of 3d electrons. The 3d electrons have some screening effect, and the extra proton and the extra 3d electron more or less cancel each other out as far as attraction from the center of the atom is concerned.
In each case, the electron is coming from the same orbital, with identical screening, but the zinc has one extra proton in the nucleus and so the attraction is greater. There will be a degree of repulsion between the paired up electrons in the 4s orbital, but in this case it obviously isn't enough to outweigh the effect of the extra proton.
The lower the ionization energy, the more easily this change happens:. You can explain the increase in reactivity of the Group 1 metals Li, Na, K, Rb, Cs as you go down the group in terms of the fall in ionization energy.
Whatever these metals react with, they have to form positive ions in the process, and so the lower the ionization energy, the more easily those ions will form. The danger with this approach is that the formation of the positive ion is only one stage in a multi-step process. For example, you would not be starting with gaseous atoms; nor would you end up with gaseous positive ions - you would end up with ions in a solid or in solution.
The energy changes in these processes also vary from element to element. Ideally you need to consider the whole picture and not just one small part of it. Write a chemical equation representing the second ionization energy for lithium. It is possible to determine the ionization energy for hydrogen using the Bohr equation. Which two elements have the same number of energy levels with electrons in them? Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength Energy of a photon of this light is Write an equation that shows the process corresponding to.
Choose the element with the highest ionization energy element: Na, Mg, Al, P, S In my book: ionization energy increase from left to right and bottom to top. Which value represents the first ionization energy of a nonmetal? I think the answer is Am I correct? Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Db Sg Bh Hs Mt Uuq n. Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
0コメント