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Which of the following is not a rule for calculating oxidation number? (a) For ions, oxidation number is equal to the charge on the ion. (b) The oxidation number of oxygen is \(-2\) in all of its compounds. (c) The oxidation number of fluorine is \(-1\) in all of its compounds. (d) Oxidation number of hydrogen is \(+1\) except in binary hydrides of alkali metals and alkaline earth metals where it is \(-1\).

Short Answer

Expert verified
The incorrect rule for calculating oxidation numbers is (b) The oxidation number of oxygen is always -2 in all of its compounds.

Step by step solution

01

Identify Incorrect Rule

Go through each statement and verify its accuracy based on the known rules for assigning oxidation numbers.
02

Evaluate Statement (a)

For any ion, the oxidation number is indeed equal to its charge. Hence, this rule is correct.
03

Evaluate Statement (b)

The oxidation number of oxygen is generally -2 in its compounds except in peroxides (where it is -1) and in some compounds with fluorine (where it can be positive). Thus, this statement is not universally true, making it the incorrect rule.
04

Evaluate Statement (c)

Fluorine is the most electronegative element and always has an oxidation number of -1 in its compounds. This rule is correct.
05

Evaluate Statement (d)

Hydrogen generally has an oxidation number of +1, but in metal hydrides such as those with alkali metals and alkaline earth metals, it is -1. This rule is also correct.

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Key Concepts

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

Assigning Oxidation Numbers
Understanding how to assign oxidation numbers is crucial for mastering redox reactions in chemistry. Oxidation numbers, often referred to as oxidation states, indicate the degree of oxidation or reduction of an atom in a chemical compound.

To assign these numbers properly, certain rules must be followed:
  • The oxidation number of any pure element is 0.
  • For monatomic ions, the oxidation number is the same as the ion's charge.
  • Hydrogen typically has an oxidation number of +1, but in metal hydrides with alkali and alkaline earth metals, it is -1.
  • Oxygen usually has an oxidation number of -2, except in peroxides where it is -1, and in compounds with fluorine where it can vary.
  • Fluorine always has an oxidation number of -1 in its compounds due to its high electronegativity.
  • The sum of oxidation numbers in a neutral compound must equal 0, while in a polyatomic ion, it should equal the ion's charge.

These rules provide a systematic approach to track how electrons are transferred between atoms in chemical reactions. When assigning oxidation numbers, always start with the atoms that have known values based on these rules, and then determine the remaining atoms' numbers so that the overall charge of the molecule or ion is consistent with the rule regarding sums.
Oxidation States in Chemistry
The concept of oxidation states is essential for understanding the electronic structure of atoms in molecules and ions. It's a theoretical charge that an atom would have if all bonds to atoms of different elements were completely ionic.

An atom's oxidation state is affected by its electronegativity, the tendency of an atom to attract electrons towards itself. In a bond between two atoms of different electronegativities, the more electronegative atom will have a negative oxidation state, while the less electronegative will have a positive one.

Common Misconceptions

It's important to note that oxygen usually has an oxidation state of -2, but this is not a fixed rule. In compounds such as peroxides, oxygen's oxidation state is -1 due to the different bonding environment. Similarly, hydrogen is often +1 but takes a -1 state when bonded with more electropositive elements, such as in metal hydrides.

The oxidation state can be an actual charge in ions, or it can be a formal charge in molecules where the actual charge distribution is more complex. Understanding oxidation states is fundamental to predicting the results of redox reactions and in balancing redox equations.
Redox Reactions
Redox reactions, which encompass all chemical reactions where the oxidation state of atoms is altered, are fundamental to numerous natural and industrial processes. They consist of two half-reactions: oxidation, where an atom or compound loses electrons, and reduction, where they gain electrons.

In any redox reaction, the transfer of electrons from the oxidized species to the reduced one is evident when tracking the changes in oxidation numbers. For example, in the combustion of hydrocarbons, oxygen is reduced (gains electrons) and the hydrocarbon is oxidized (loses electrons).

Balancing Redox Equations

When balancing redox reactions, it's essential to ensure that the number of electrons lost in the oxidation half-reaction equals the number gained in the reduction half-reaction. This provides an additional layer to the usual balancing of chemical equations, where mass balance (the same number of each type of atom on both sides of the reaction) must be accomplished.

Redox reactions are key for applications such as energy storage in batteries, biochemical processes within our bodies, and industrial oxidation of metals. Understanding and identifying the changes in oxidation numbers reveals the direction of electron flow, which is integral for harnessing chemical energy and predicting reaction outcomes.

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Most popular questions from this chapter

Which of the following is true about the given redox reaction? $$ \mathrm{SnCl}_{2}+2 \mathrm{FeCl}_{3} \rightarrow \mathrm{SnCl}_{4}+2 \mathrm{FeCl}_{2} $$ (a) \(\mathrm{SnCl}_{2}\) is oxidised and \(\mathrm{FeCl}_{3}\) acts as oxidising agent. (b) \(\mathrm{FeCl}_{3}\) is oxidised and acts as oxidising agent. (c) \(\mathrm{SnCl}_{2}\) is reduced and acts as oxidising agent. (d) \(\mathrm{FeCl}_{3}\) is oxidised and \(\mathrm{SnCl}_{2}\) acts as a oxidising agent.

Given \(E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}=+0.80 \mathrm{~V} ; \quad E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V}\) \(E_{\mathrm{Ft}^{3}+i \mathrm{Fe}^{2}}=+0.76 \mathrm{~V} ; E^{\circ} \mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}=+1.60 \mathrm{~V}\) Which of the following statements is not correct? (a) \(\mathrm{Fe}^{3+}\) does not oxidise \(\mathrm{Ce}^{3+}\). (b) Cu reduces \(\mathrm{Ag}^{+}\)to \(\mathrm{Ag}\). (c) Ag will reduce \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\). (d) \(\mathrm{Fe}^{3+}\) reduces \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\).

Arrange the oxides of nitrogen in increasing order of oxidation state of \(\mathrm{N}\) from \(+1\) to \(+5\). $$ \begin{aligned} &\mathrm{N}_{2} \mathrm{O}<\mathrm{N}_{2} \mathrm{O}_{3}<\mathrm{NO}_{2}<\mathrm{N}_{2} \mathrm{O}_{5}\\\ &\begin{aligned} &\text { (a) } \mathrm{N}_{2} \mathrm{O}<\mathrm{NO}<\mathrm{N}_{2} \mathrm{O}_{3}<\mathrm{NO}_{2}<\mathrm{N}_{2} \mathrm{O}_{5} \\ &\text { (b) } \mathrm{N}_{2} \mathrm{O}<\mathrm{NO}_{2}<\mathrm{N}_{2} \mathrm{O}_{3}<\mathrm{NO}<\mathrm{N}_{2} \mathrm{O} \\ &\text { (c) } \mathrm{N}_{2} \mathrm{O}_{5}<\mathrm{N}_{2} \mathrm{O}<\mathrm{NO}_{2}<\mathrm{N}_{2} \mathrm{O}_{3}<\mathrm{N}_{2} \mathrm{O}_{5} \end{aligned} \end{aligned} $$

Oxidation state of iron in \(\mathrm{Fe}(\mathrm{CO})_{4}\) is (a) \(+1\) (b) \(-1\) (c) \(+2\) (d) 0

Which of the following will act as cathode when connected to standard hydrogen electrode which has \(E^{\circ}\) value given as zero? (i) \(Z n^{2+} / Z n, E^{\circ}=-0.76 \mathrm{~V}\) (ii) \(\mathrm{Cu}^{2+} / \mathrm{Cu}, E^{\circ}=+0.34 \mathrm{~V}\) (iii) \(\mathrm{Al}^{3+} / \mathrm{Al}, E^{\circ}=-1.66 \mathrm{~V}\) (iv) \(\mathrm{Hg}^{2+} / \mathrm{Hg}, E^{\circ}=+0.885 \mathrm{~V}\) (a) (i) and (ii) (b) (ii) and (iv) (c) (i) and (iii) (d) (i), (ii), (iii) and (iv)

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