Inorganic chemistry (3)
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1.
The reactivity of the Group 17 elements decreases down the group. Explain this trend in terms of atomic structure and effective nuclear charge. Include a discussion of the factors affecting bond strength.
The reactivity of Group 17 elements decreases down the group due to a combination of factors related to atomic structure and effective nuclear charge. As you move down the group, the atomic radius increases. This means the valence electrons are further from the nucleus.
Effective Nuclear Charge (Zeff) is the net positive charge experienced by the valence electrons. While the number of protons increases down the group, the shielding effect of inner electrons also increases. However, the increase in nuclear charge is greater than the shielding effect, leading to a smaller increase in Zeff down the group. A smaller Zeff means the valence electrons are less strongly attracted to the nucleus.
Bond Strength is directly related to the strength of the attraction between the nucleus and the valence electrons. Weaker attraction results in weaker bonds. As the Zeff decreases down the group, the attraction between the nucleus and the valence electrons weakens, leading to weaker bonds. Therefore, the energy required to break these bonds (i.e., the activation energy for a reaction) is lower, resulting in lower reactivity.
In summary, the combination of increased atomic radius, decreasing Zeff, and weaker bonds explains the decreasing reactivity trend down Group 17.
2.
Sodium (Na) is a highly reactive metal. Describe the trends in the following properties of the alkali metals (Li, Na, K, Rb, Cs) from top to bottom in Group 1, explaining your reasoning in terms of atomic structure.
The following trends are observed in the alkali metals (Li, Na, K, Rb, Cs) from top to bottom in Group 1:
- Atomic Radius: Increases. As you move down the group, the number of electron shells increases. This leads to greater shielding of the valence electrons from the nucleus, resulting in a larger atomic radius.
- Ionisation Energy: Decreases. The ionization energy is the energy required to remove an electron. As atomic radius increases, the valence electrons are further from the nucleus and experience less effective nuclear charge. This makes it easier to remove an electron, hence the ionization energy decreases.
- Metallic Character: Increases. Metallic character refers to the ability of an element to lose electrons and form positive ions. As ionization energy decreases, it becomes easier for the alkali metals to lose their valence electron, increasing their metallic character.
- Reactivity with Water: Increases. The reactivity of alkali metals with water is related to their tendency to lose electrons. As ionization energy decreases, the alkali metals are more likely to react with water, producing hydrogen gas and a metal hydroxide. The reaction becomes more vigorous down the group.
Reasoning in terms of atomic structure: The trends are directly related to the increasing number of electron shells and the decreasing effective nuclear charge as you move down the group. The increased shielding effect reduces the attraction between the nucleus and the valence electrons, making it easier to remove electrons and increasing the atomic radius, ionization energy, metallic character, and reactivity with water.
3.
Question 2
Describe the reactions of sulfur with both sodium hydroxide and concentrated nitric acid. For each reaction, include the balanced chemical equation and state the products formed. Explain why the reaction of sulfur with sodium hydroxide is different from the reaction of sulfur with nitric acid.
Reaction with Sodium Hydroxide (NaOH): Sulfur reacts with sodium hydroxide to produce sodium thiosulfate and hydrogen sulfide. The balanced equation is:
S(s) + 2NaOH(aq) → Na2S(aq) + Na2S2O3(aq) + H2(g) |
This reaction is a redox reaction where sulfur is reduced to sulfide ions (S2-) and sodium hydroxide is oxidized to hydrogen gas. The sulfide ions react with sodium ions to form sodium sulfide. The hydrogen sulfide is a gas that is released.
Reaction with Concentrated Nitric Acid (HNO3): Sulfur reacts with concentrated nitric acid to produce sulfur dioxide and nitrogen dioxide. The balanced equation is:
S(s) + 6HNO3(aq) → H2S(g) + 6NO2(g) + 2H2O(l) |
This reaction is also a redox reaction. Sulfur is oxidized to sulfur dioxide (SO2), and nitric acid is reduced to nitrogen dioxide (NO2). The nitric acid acts as both an oxidizing and a reducing agent. The presence of concentrated nitric acid is crucial for the oxidation of nitrate ions to nitrogen dioxide.
The reaction of sulfur with sodium hydroxide is different from the reaction with nitric acid because the products are fundamentally different. With NaOH, the reaction is a simple displacement of sulfur by sulfide ions. With nitric acid, the reaction involves oxidation of sulfur and reduction of nitrate ions, resulting in the formation of oxides of nitrogen and sulfur.