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Chlorine has several oxidation states. (a) Determine the oxidation state of chlorine in \(\mathrm{Cl}_{2}, \mathrm{HOCl}, \mathrm{ClO}_{2}, \mathrm{NH}_{4} \mathrm{Cl},\) and \(\mathrm{NaCl}\). (b) Chloride ion is the most stable form. Predict which of the other substances given in part (a) is an oxidizing or a reducing agent and rank them in decreasing order of their oxidizing strengths.

Short Answer

Expert verified
ClO鈧 is the strongest oxidizing agent, followed by Cl鈧 and HOCl.

Step by step solution

01

Understand Oxidation State Rules

To determine the oxidation state of chlorine in various compounds, we first need to understand the basic rules: 1) The oxidation state of a pure element is 0; 2) In compounds, hydrogen is usually +1 and oxygen is -2; 3) The sum of the oxidation states in a neutral compound is 0 and in an ion is equal to the ion charge.
02

Determine Oxidation State of Chlorine in Each Compound

Now we apply the rules to find the oxidation states: - In Cl鈧, as it is a pure element, oxidation state = 0. - In HOCl, the sum should be 0: (+1) + Cl + (-2) = 0. So, Cl = +1. - In ClO鈧: (-2 * 2) + Cl = 0. Thus, Cl = +4. - In NH鈧凜l, NH鈧勨伜 as a polyatomic ion has oxidation state: (+1 * 4) + N = +1, making Cl = -1. - In NaCl, Na is +1, Cl = -1 to balance.
03

Compare Oxidation States to Stability

Chloride ion (Cl鈦) is the most stable form with an oxidation state of -1. Compounds with high positive oxidation states are generally good oxidizing agents. Flatten the list: Cl鈧 (0), HOCl (+1), ClO鈧 (+4), NH鈧凜l (-1), NaCl (-1).
04

Identify Oxidizing/Reducing Abilities

A higher positive oxidation state can accept electrons easily and thus act as an oxidizing agent. Cl鈧 (0) and ClO鈧 (+4) are likely to accept electrons and are oxidizing agents. All Cl compounds with chlorine at -1 are stable and not oxidizing.
05

Rank Oxidizing Strength

Rank the oxidizing agents from strongest to weakest: ClO鈧 (+4), Cl鈧 (0). HOCl (+1) is also an oxidizing agent but less so than Cl鈧 and ClO鈧 due to lower positive oxidation state.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation Number Rules
Understanding oxidation number rules is essential for predicting the behavior of elements in chemical reactions. Let's break these rules down step-by-step to make them simple and user-friendly.
  • The oxidation state of a pure element is 0. This rule applies to elements in their natural, uncombined form, like \( \mathrm{Cl}_{2} \), which has an oxidation state of 0 because it is chlorine in its elemental form.

  • Hydrogen typically has an oxidation state of +1 and oxygen has -2 in compounds. However, there are exceptions, such as in metal hydrides (where hydrogen is -1) and peroxides (where each oxygen is -1).

  • The sum of oxidation states in a neutral compound is always 0. For example, in the molecule \( \mathrm{HOCl} \), the roles of hydrogen and oxygen help determine the oxidation state of chlorine. Here, chlorine helps satisfy the total sum of 0 by being +1.

  • In polyatomic ions, the sum of oxidation states equals the ion's charge. Taking \( \mathrm{NH}_{4}^{+} \) as an example, where hydrogen and nitrogen contribute to the charge, the chlorine in \( \mathrm{NH}_{4}\mathrm{Cl} \) is -1 to achieve an overall ion charge that is balanced.
Identifying Oxidizing Agents
Identifying oxidizing agents is crucial in understanding how certain substances drive chemical reactions. An oxidizing agent gains electrons and, in the process, oxidizes another substance. Let's look at how this works.Substances with high positive oxidation states are often eager to gather electrons, making them good oxidizing agents. As such, ClO鈧 (with chlorine at +4) and Cl鈧 (at 0) are strong oxidizing agents because they have room to accept electrons.Cl鈧, being elemental chlorine, has an oxidation state of 0, so it acts as a moderate oxidizer compared to ClO鈧, which has chlorine in a more positive state of +4, thus a stronger oxidizing ability.When organizing oxidizing strength:
  • Start with the highest oxidation state among the chlorine compounds in the list.
  • ClO鈧 comes first due to chlorine at an oxidation state of +4, making it the strongest oxidizer.
  • Next is Cl鈧, with its element-form of 0 oxidation state.
  • HOCl, in comparison, is weaker because its chlorine is only at +1.
  • Substances like \( \mathrm{NaCl} \) and \( \mathrm{NH}_{4}\mathrm{Cl} \) have chlorine in its most stable form (-1), so they don't act as oxidizers.
Stability of Chloride Ions
Chloride ions \( (\mathrm{Cl}^{-}) \) are incredibly stable, which explains why they are commonly found in nature. Let's understand what gives them this stability.
  • Negative Oxidation State: Chloride ions have an oxidation state of -1. This is a favored, low-energy state for chlorine, which makes it stable.

  • Completeness: The negative charge means it has gained an electron, thus completing its valence shell to resemble the nearest noble gas, Argon. This electronic configuration is highly stable.

  • Prevalence in Compounds: Chloride ions appear in various compounds, like \( \mathrm{NaCl} \) and \( \mathrm{NH}_{4}\mathrm{Cl} \), due to their stability and readiness to bond. This makes them essential in both biological and industrial processes.
Their very low energy and complete outer shell ensure chloride ions are less reactive compared to other forms with higher oxidation numbers. This stability makes them vital in maintaining the chemical equilibrium in various environments and reactions.

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Most popular questions from this chapter

You are given \(0.954 \mathrm{~g}\) of an unknown acid, \(\mathrm{H}_{2} \mathrm{~A},\) which reacts with \(\mathrm{NaOH}\) according to the balanced equation $$\mathrm{H}_{2} \mathrm{~A}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{Na}_{2} \mathrm{~A}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$ If \(36.04 \mathrm{~mL}\) of \(0.509-\mathrm{M} \mathrm{NaOH}\) is required to react with all of the acid, calculate the molar mass of the acid.

The ceramic silicon nitride, \(\mathrm{Si}_{3} \mathrm{~N}_{4}\), is made by heating silicon and nitrogen at an elevated temperature. $$3 \mathrm{Si}(\mathrm{s})+2 \mathrm{~N}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Si}_{3} \mathrm{~N}_{4}(\mathrm{~s})$$ Calculate the mass of silicon that must combine with excess \(\mathrm{N}_{2}\) to produce \(1.0 \mathrm{~kg} \mathrm{Si}_{3} \mathrm{~N}_{4}\) if this process is \(92 \%\) efficient.

Chlorofluorocarbons are involved in creating the hole in the ozone layer. Their manufacture begins with the preparation of HF from the mineral fluorspar, \(\mathrm{CaF}_{2},\) according to this unbalanced equation: $$\mathrm{CaF}_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{HF}(\mathrm{g})+\mathrm{CaSO}_{4}(\mathrm{~s})$$ HF is combined with, for example, \(\mathrm{CCl}_{4}\) in the presence of \(\mathrm{SbCl}_{5}\) to make \(\mathrm{CCl}_{2} \mathrm{~F}_{2},\) called dichlorodifluoromethane or CFC-12, and other chlorofluorocarbons. $$2 \mathrm{HF}(\mathrm{g})+\mathrm{CCl}_{4}(\ell) \longrightarrow \mathrm{CCl}_{2} \mathrm{~F}_{2}(\mathrm{~g})+2 \mathrm{HCl}(\mathrm{g})$$ (a) Balance the first equation and name each substance. (b) Is the first reaction best classified as an acid-base reaction, an oxidation-reduction reaction, or a precipitation reaction? (c) Give the names of the compounds \(\mathrm{CCl}_{4}, \mathrm{SbCl}_{5},\) and \(\mathrm{HCl}\). (d) Another chlorofluorocarbon produced in the reaction is composed of \(8.74 \% \mathrm{C}, 77.43 \% \mathrm{Cl}\), and \(13.83 \% \mathrm{~F}\). Determine the empirical formula of the compound.

In a reaction, \(1.2 \mathrm{~g}\) element A reacts with exactly \(3.2 \mathrm{~g}\) oxygen to form an oxide, \(\mathrm{AO}_{x} ; 2.4 \mathrm{~g}\) element A reacts with exactly \(3.2 \mathrm{~g}\) oxygen to form a second oxide, \(\mathrm{AO}_{y}\) (a) Determine the ratio \(x: y\). (b) If \(x=2\), determine what the identity of element A might be.

If 1.5 mol Cu reacts with a solution containing 4.0 mol \(\mathrm{AgNO}_{3},\) what ions will be present in the solution at the end of the reaction? $$\mathrm{Cu}(\mathrm{s})+2 \mathrm{AgNO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$$

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