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Give the oxidation numbers for the underlined atoms in the following molecules and ions: (a) \(\mathrm{ClF},(\mathrm{b}) \mathrm{IF}_{7}\) (c) \(\underline{\mathrm{C}} \mathrm{H}_{4}\) (d) \(\underline{\mathrm{C}}_{2} \mathrm{H}_{2}\) (e) \(\underline{\mathrm{C}}_{2} \mathrm{H}_{4}\) (f) \(\mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4},(\mathrm{~g}) \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (h) \(\mathrm{KMnO}_{4}\), (i) \(\mathrm{NaHCO}_{3},(\mathrm{j}) \mathrm{Li}_{2},(\mathrm{k}) \mathrm{NaIO}_{3},\) (I) \(\mathrm{KO}_{2}\), \((\mathrm{m}) \mathrm{PF}_{6}^{-},(\mathrm{n}) \mathrm{K} \mathrm{AuCl}_{4}\)

Short Answer

Expert verified
(a) +1, (b) +7, (c) -4, (d) -1, (e) -2, (f) +6, (g) +6, (h) +7, (i) +4, (j) 0, (k) +5, (l) -1/2, (m) +5, (n) +3.

Step by step solution

01

Determine the oxidation number for compound (a) Chlorine in ClF

In \( \text{ClF} \), the compound involves chlorine (Cl) and fluorine (F). Since fluorine is the most electronegative element, it always has an oxidation number of -1. Thus, chlorine must have an oxidation number of \(+1\) to balance the -1 from fluorine, since the molecule is neutral.
02

Determine the oxidation number for compound (b) Iodine in IF7

In \( \text{IF}_7 \), iodine (I) is less electronegative than fluorine (F). Each fluorine atom has an oxidation number of -1. Therefore, with seven fluorine atoms, the Iodine must have an oxidation number of \(+7\) to balance the total of -7 from fluorine.
03

Determine the oxidation number for compound (c) Carbon in CH4

In \( \text{CH}_4 \), hydrogen has an oxidation number of +1. Since there are four hydrogen atoms, their total oxidation is +4. To balance the molecule to zero (neutral), carbon must have an oxidation number of \(-4\).
04

Determine the oxidation number for compound (d) Carbon in C2H2

In \( \text{C}_2\text{H}_2 \), each hydrogen has an oxidation number of +1, leading to a total of +2 for two hydrogen atoms. Therefore, the two carbons combined must have an oxidation number of -2, making each carbon \(-1\).
05

Determine the oxidation number for compound (e) Carbon in C2H4

In \( \text{C}_2\text{H}_4 \), each hydrogen has +1, summing to +4 for four hydrogens. The two carbon atoms need a total oxidation of -4 combined, so each carbon has an oxidation state of \(-2\).
06

Determine the oxidation number for compound (f) Chromium in K2CrO4

In \( \text{K}_2\text{CrO}_4 \), potassium has +1 each and oxygen has -2 each. Total potassium is +2, total oxygen is -8. Chromium must be +6 to balance to zero overall.
07

Determine the oxidation number for compound (g) Chromium in K2Cr2O7

In \( \text{K}_2\text{Cr}_2\text{O}_7 \), each potassium is +1, and each oxygen is -2 (total -14). Two potassiums = +2, so two chromiums must = +12 (each \(+6\)) to balance.
08

Determine the oxidation number for compound (h) Manganese in KMnO4

In \( \text{KMnO}_4 \), potassium is +1, oxygen is -2 each (total -8). For the compound to be neutral, manganese must have an oxidation of \(+7\).
09

Determine the oxidation number for compound (i) Carbon in NaHCO3

In \( \text{NaHCO}_3 \), sodium is +1, hydrogen is +1, each oxygen is -2 (total -6). Balancing gives carbon an oxidation state of \(+4\).
10

Determine the oxidation number for compound (j) Lithium in Li2

In \( \text{Li}_2 \), both lithium atoms are in a neutral elemental state, so each has an oxidation number of \(0\).
11

Determine the oxidation number for compound (k) Iodine in NaIO3

In \( \text{NaIO}_3 \), sodium is +1, each oxygen is -2 (total -6). The iodine must be +5 to balance the compound to neutrality.
12

Determine the oxidation number for compound (l) Oxygen in KO2

In \( \text{KO}_2 \), potassium is +1. This compound is a superoxide, where each oxygen is typically \(-\frac{1}{2}\). Therefore, the total for two oxygens is -1.
13

Determine the oxidation number for compound (m) Phosphorus in PF6^-

In \( \text{PF}_6^- \), each fluorine is -1 (total -6). Overall charge is -1, so phosphorus is +5.
14

Determine the oxidation number for compound (n) Gold in KAuCl4

In \( \text{KAuCl}_4 \), potassium is +1, each chlorine is -1 (total for four is -4). Gold must balance with +3.

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Key Concepts

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

Oxidation States
Oxidation states, also known as oxidation numbers, are essential in understanding how electrons interact within a chemical compound. They allow us to determine the electron distribution among atoms, indicating which atoms might lose or gain electrons during a chemical reaction.
The oxidation state of an element in a compound is a hypothetical charge, assuming that all bonds are ionic. For instance, in the molecule ClF, fluorine, being the most electronegative element, is assigned an oxidation state of -1. Consequently, chlorine's oxidation state becomes +1, so the compound remains neutral. Usually, group 1 elements have an oxidation state of +1, while group 2 elements have +2.
In molecules,
  • The oxidation state of oxygen is typically -2, except in peroxides or superoxides where it is -1 or -1/2, respectively.
  • Hydrogen has an oxidation state of +1 when paired with more electronegative elements, like carbon or oxygen, but behaves differently in metal hydrides.
Understanding these rules helps predict the outcome of chemical reactions and balance chemical equations.
Chemical Compounds
Chemical compounds are formed when two or more atoms are chemically bonded together, resulting in a substance with unique properties differing from the individual elements involved.
Compounds can be characterized by their oxidation states, which reveal how electrons are shared or transferred among atoms within a compound. For instance, sodium chloride (NaCl) consists of sodium (Na) with an oxidation state of +1 and chlorine (Cl), having -1.
Each element in a compound influences the overall charge. For example, in sodium iodate (NaIO extsubscript{3}), the oxidation state of sodium is +1, iodine is +5, and each oxygen is -2. Together, they balance out to form a stable compound with a neutral charge.
The understanding of these compounds offers vital insight into chemical behavior, reactions, and properties, which is crucial for tasks such as synthesis and chemical analysis. This formalizes the connection between chemical composition and function, making it impactful in areas ranging from industry to biology.
Electronegativity
Electronegativity refers to the ability of an atom in a molecule to attract electrons towards itself. This concept is pivotal in determining how atoms will interact, especially in deciding oxidation states.
A highly electronegative atom, such as fluorine, will attract electrons strongly towards itself. This means that in compounds like IF extsubscript{7}, fluorine takes electrons away from iodine, assigning itself an oxidation state of -1, leaving iodine with +7.
Generally, electronegativity increases across a period from left to right and decreases down a group in the periodic table. Elements with high electronegativity often pull electrons towards them, thus commonly exhibiting negative oxidation states in compounds.
  • Fluorine is the most electronegative element and always has an oxidation state of -1 in its compounds.
  • Oxygen is also highly electronegative, usually exhibiting an oxidation state of -2.
  • In contrast, hydrogen, which is less electronegative, tends to have an oxidation state of +1.
This property is critical for predicting how elements will combine and the type of bonds they will form.
Neutral Molecules
Neutral molecules are compounds with no net charge. This is achieved when all the positive and negative charges within a molecule are balanced.
In the context of oxidation states, the algebraic sum of all the oxidation numbers in a neutral molecule must be zero. For instance, in methane (CH extsubscript{4}), carbon has an oxidation state of -4, while each hydrogen is +1, leading to a sum of -4 + 4 = 0, which confirms its neutrality.
Neutrality is observed in many molecular compounds, such as PF extsubscript{6} extsuperscript{-}. Here, the overall charge of -1 (due to the extra electron) is balanced by the phosphorus atom's +5 charge and the -1 from each of the six fluorine atoms.
  • In such compounds, ensuring charge balance is essential to predict the chemical behavior and reactivity.
  • Neutrality also affects solubility and conductivity, pivotal for understanding their applications in real-world scenarios.
  • Particular attention must be paid to ensure the correct electron count, especially in complex ions and molecules.
Understanding neutral molecules and their charge distribution allows chemists to effectively manipulate and utilize these compounds in various chemical contexts.

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Most popular questions from this chapter

A student carried out two titrations using an \(\mathrm{NaOH}\) solution of unknown concentration in the burette. In one titration, she weighed out \(0.2458 \mathrm{~g}\) of KHP ([see page \(166 .]\) ) and transferred it to an Erlenmeyer flask. She then added \(20.00 \mathrm{~mL}\) of distilled water to dissolve the acid. In the other titration, she weighed out \(0.2507 \mathrm{~g}\) of KHP but added \(40.00 \mathrm{~mL}\) of distilled water to dissolve the acid. Assuming no experimental error, would she obtain the same result for the concentration of the \(\mathrm{NaOH}\) solution?

What volume of \(0.112 M\) ammonium sulfate contains \(5.75 \mathrm{~g}\) of ammonium ion?

Acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right)\) is a monoprotic acid commonly known as "aspirin." A typical aspirin tablet, however, contains only a small amount of the acid. In an experiment to determine its composition, an aspirin tablet was crushed and dissolved in water. It took \(12.25 \mathrm{~mL}\) of \(0.1466 \mathrm{M} \mathrm{NaOH}\) to neutralize the solution. Calculate the number of grains of aspirin in the tablet (one grain \(=0.0648 \mathrm{~g}\) ).

Ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) is one of the most important nitrogen-containing fertilizers. Its purity can be analyzed by titrating a solution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) with a standard \(\mathrm{NaOH}\) solution. In one experiment a \(0.2041-\mathrm{g}\) sample of industrially prepared \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) required \(24.42 \mathrm{~mL}\) of \(0.1023 \mathrm{M} \mathrm{NaOH}\) for neutralization. (a) Write a net ionic equation for the reaction. (b) What is the percent purity of the sample?

Describe the basic steps involved in an acid-base titration. Why is this technique of great practical value?

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