7.5. Exercises

7.1 Assign oxidation number to the underlined elements in each of the following species: (a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4 (e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2O

In this exercise, you are tasked with determining the oxidation numbers of specified elements within various chemical compounds. Oxidation numbers indicate the degree of oxidation (loss of electrons) of an atom in a chemical compound. To do this, recall the rules set forth for assigning oxidation numbers, such as the oxidation number of free elements being zero, the oxidation number of monoatomic ions being equal to their charge, and certain common oxidation states for particular elements (e.g., oxygen usually as -2, hydrogen as +1).

Think of oxidation numbers like measuring the charge of a battery. Just like a fully charged battery has a certain energy level, each element in a compound has a specific oxidation number that reflects its 'charge' relative to its neutral state. For example, in water (H2O), hydrogen has a +1 oxidation number, indicating it is more 'positive' than when it is in its elemental form.

7.2 What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results? (a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH

This exercise focuses on not just assigning oxidation numbers but also explaining how you arrived at your answers. This logical reasoning is key in chemistry. For instance, in KI3, potassium (K), a Group 1 metal, has a +1 oxidation state, while Iodine typically varies between -1 and +5 depending on its chemical environment. Rationalizing results requires an understanding of how different atoms interact based on their electronegativities and common oxidation states.

Imagine a game where each player (element) has a score (oxidation number) that changes depending on the team they are playing on (chemical environment). A player like potassium always maintains a high score regardless of who they are playing with, while iodine’s score can change drastically based on the rules of the game (the environment created by its surrounding atoms).

7.3 Justify that the following reactions are redox reactions: (a) CuO(s) + H2(g) → Cu(s) + H2O(g) (b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) (c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3(s) (d) 2K(s) + F2(g) → 2K+F–(s) (e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)

In this exercise, students need to recognize and justify that the provided reactions involve electron transfer, which is the hallmark of redox processes. For example, in reaction (a), CuO is reduced to Cu while H2 is oxidized to H2O, demonstrating the simultaneous occurrence of oxidation and reduction. This concept is central to understanding how energy changes in chemical reactions and the flow of electrons.

Think of redox reactions like a relay race where electrons are passed between runners; as one runner (an atom) sprints ahead to donate their baton (electrons), another runner is waiting to receive it, thus keeping the flow going. Just as the race cannot continue without passing the baton, a redox reaction cannot occur without the exchange of electrons.

7.4 Fluorine reacts with ice and results in the change: H2O(s) + F2(g) → HF(g) + HOF(g) Justify that this reaction is a redox reaction.

In this exercise, students need to analyze how fluorine reacts with ice and determine the changes in oxidation states. In this case, fluorine is reduced as it gains electrons and achieves a more stable electronic configuration. The compound H2O loses some hydrogen atoms, demonstrating its role as a reducing agent. The recognition of this transfer is fundamental to classifying the reaction as redox.

Consider this reaction as a dramatic transformation in a movie where the hero (fluorine) breaks into a scene (ice), changes the narrative (reaction), and while doing so, transforms some characters (hydrogens) into different versions of themselves (HF and HOF), illustrating the classic tale of oxidation and reduction.

7.5 Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72– and NO3–. Suggest structure of these compounds. Count for the fallacy.

In this exercise, students will calculate the oxidation states of sulphur, chromium, and nitrogen in specified compounds. It requires an application of the oxidation rules while also emphasizing the need to visualize their molecular structures. Additionally, students should reflect on any discrepancies or inconsistencies in their calculations – a practice that enhances critical thinking in chemistry.

Imagine trying to figure out how many cookies each friend brought to a party while some of them may have estimated incorrectly. Similarly, while calculating oxidation numbers, we must keep track of all elements to ensure no one is overestimated or underestimated, leading to hidden complexities that may arise from incorrect assumptions.

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