Pieces of the alkali metals Li, K and Cs are added to water. consider the genera
ID: 993265 • Letter: P
Question
Pieces of the alkali metals Li, K and Cs are added to water. consider the general reaction for an alkali metal (M) with water, which results in the ionization of the alkali metal (M^+ forms) and images of the reactions are shown below. 2M(s) + 2H_2O(I) rightarrow 2M^+(aq) + 2OH^-(aq) + H_2(g) Explain why the reactions become progressively more violent from lithium to cesium. The explanation must be based on atomic (i.e. numbers of protons and electrons) and electronic structure (i.e., electronic configuration) as well as relevant periodic trends. Consider the graph plotting first ionization energies for second period elements versus atomic numbers (Z) to answer the following questions. Provide a brief explanation for the general trend in ionization energy observed across the second period.Explanation / Answer
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[A] Reaction of alkali metals.
As the general reaction states, when water is added to an alkali metal, the metal loses an electron (oxidizes) and turns into a hydroxide reducing hydrogen in the process. Lithium, Potassium and Cesium are all members of the 1A (or IA) family in the periodic table. Those elements have in common that they all have 1 electron in an s orbital of the highest principal quantum number. Lithium’s last electron is a 2s1, Potassium’s is a 4s1 and Cesium’s last electron is a 6s1. Also, the transference of an electron requires energy. An energy that is necessary to pull that lonely electron from its orbital. In fact the speed of the reaction depends on that energy, because the fastest that the electron is removed, the fastest the speed. Now we see that in the experiment Cesium has the most violent reaction. This means that the energy to remove the electron in cesium should be the lowest. And that is correct. Cesium has the last electron in the level 6, which is farther away from the nucleus and its attraction force than for the other two elements. So the electron in cesium is easily pulled out and the reaction is very fast and violent. On the other hand potassium is a little slower and lithium even more slow. This also relates to the strength by which the nucleus holds the electrons. In potassium the electron is in a level 4, closer to the nucleus than in the case of cesium but farther apart than in lithium.
[B] Ionization energy is the minimum amount of energy required to remove an electron from a gaseous atom in its ground state. The atom turns into a cation (+) plus an electron. The ionization energy is normally expressed in electronvolts (eV) or in kilojoules per mole of atoms. Ionization energy is a measure of how difficult it is to remove an electron. Remember that electrons are hold to the atom by electrostatic forces established between the positive nucleus and the negative electrons. So, the more an electron is held to the nucleus, the higher the ionization energy will be. Picture something that you want to grab: you will need to apply more strength to pull something that is strongly held than to pull apart something that is weekly held.
In the graph we can see that it is a serial of the elements of the same period. When you go from lithium to neon you are not changing the principal level (n=2) but the atomic number is increasing. As the atomic number increases, the number of protons in the nucleus increases too. Since all the electrons are in the same level (but different subshell) they feel an increment in the effective nuclear charge which means they are more tightly held. As a consequence the ionization energy tends to increase.
We can also see that there are some “irregular values” in the graph. The first one between Be and B. When we pass from Be to B we pass from a 1s2 2s2 distribution to an 1s2 2s2 2p1 distribution. The single electron in the outermost p subshell is more easily removed because the two 2s electrons shield the nuclear charge. Therefore, less energy is needed to remove a single p electron than to remove an s electron from the same principal energy level.
The second case is between nitrogen and oxygen. In this case, oxygen has a slightly lower ionization energy than that of nitrogen because in a nitrogen atom each p-orbital has one electron (Hund’s rule). On the other hand, in oxygen the eighth electron is paired with an electron already occupying an orbital. The repulsion between the two electrons in the same orbital raises their energy and makes one of them easier to remove from the atom. This effect is not present in nitrogen
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