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Determine the oxidation number for the indicated element (b) \(\mathrm{Al}\) in each of the following compounds: (a) \(\mathrm{Co}\) in \(\mathrm{LiCoO}_{2}\), in \(\mathrm{NaAlH}_{4}\), (c) \(\mathrm{C}\) in \(\mathrm{CH}_{3} \mathrm{OH}\) (methanol), \((\mathrm{d}) \mathrm{N}\) in \(\mathrm{GaN},\) (e) Cl in \(\mathrm{HClO}_{2},\) (f) \(\mathrm{Cr}\) in \(\mathrm{BaCrO}_{4}\)

Short Answer

Expert verified
The oxidation numbers for the indicated elements in the given compounds are: (a) Co in LiCoO鈧: +3 (b) Al in NaAlH鈧: +3 (c) C in CH鈧僌H (methanol): -2 (d) N in GaN: -3 (e) Cl in HClO鈧: +3 (f) Cr in BaCrO鈧: +6

Step by step solution

01

Assign known oxidation numbers

Assign the oxidation number to lithium (Li) and oxygen (O) in LiCoO鈧: Li is a monoatomic ion, so its oxidation number is equal to its charge: Li鈦, oxidation number is +1. Oxygen (O) has an oxidation number of -2 in this compound (no indication of a peroxide).
02

Calculate the oxidation number of Co.

Since the compound is neutral, the sum of the oxidation numbers equals 0: (+1) + Co + [2 脳 (-2)] = 0 Now, solve for Co's oxidation number: Co - 3 = 0 Co = +3 The oxidation number of Co in LiCoO鈧 is +3. #b) Al in NaAlH4#
03

Assign known oxidation numbers

Assign the oxidation number to sodium (Na) and hydrogen (H) in NaAlH鈧: Na is a monoatomic ion, so its oxidation number is equal to its charge: Na鈦, oxidation number is +1. Since H is part of a metal hydride (AlH鈧勨伝), its oxidation number is -1.
04

Calculate the oxidation number of Al

Since NaAlH鈧 is neutral, the sum of the oxidation numbers equals 0: (+1) + Al + [4 脳 (-1)] = 0 Now, solve for Al's oxidation number: Al - 3 = 0 Al = +3 The oxidation number of Al in NaAlH鈧 is +3. #c) C in CH3OH (methanol)#
05

Assign known oxidation numbers

Assign the oxidation number to hydrogen (H) and oxygen (O) in CH鈧僌H: H in organic compounds usually has an oxidation number of +1. Oxygen (O) has an oxidation number of -2.
06

Calculate the oxidation number of C

Since CH鈧僌H is neutral, the sum of the oxidation numbers equals 0: C + [(3 脳 +1) + (-2) + (+1)] = 0 Now, solve for C's oxidation number: C + 2 = 0 C = -2 The oxidation number of C in CH鈧僌H is -2. #d) N in GaN#
07

Identify the charges of the ions

GaN is an ionic compound formed by Ga鲁鈦 and N鲁鈦 ions.
08

Determine the oxidation number

The oxidation numbers are equal to the charges of the ions: For Ga, the oxidation number is +3. For N, the oxidation number is -3. #e) Cl in HClO2#
09

Assign known oxidation numbers

Assign the oxidation number to hydrogen (H) and oxygen (O) in HClO鈧: H in compounds usually has an oxidation number of +1. Oxygen (O) has an oxidation number of -2.
10

Calculate the oxidation number of Cl

Since HClO鈧 is neutral, the sum of the oxidation numbers equals 0: (+1) + Cl + [2 脳 (-2)] = 0 Now, solve for Cl's oxidation number: Cl - 3 = 0 Cl = +3 The oxidation number of Cl in HClO鈧 is +3. #f) Cr in BaCrO4#
11

Assign known oxidation numbers

Assign the oxidation number to barium (Ba) and oxygen (O) in BaCrO鈧: Ba is a monoatomic ion, so its oxidation number is equal to its charge: Ba虏鈦, oxidation number is +2. Oxygen (O) has an oxidation number of -2.
12

Calculate the oxidation number of Cr

Since BaCrO鈧 is neutral, the sum of the oxidation numbers equals 0: (+2) + Cr + [4 脳 (-2)] = 0 Now, solve for Cr's oxidation number: Cr - 6 = 0 Cr = +6 The oxidation number of Cr in BaCrO鈧 is +6.

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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 vital for understanding chemical reactions. An oxidation state indicates the degree of oxidation of an element in a chemical compound. Each element can have different oxidation states depending on the chemical environment. This concept helps us determine how electrons are distributed in molecules and compounds, which is crucial for predicting molecular interactions.

To assign oxidation states, remember a few key guidelines:
  • Elements in their natural state have an oxidation number of zero. For instance, \( ext{O}_2\) or \( ext{N}_2\) are both zero.
  • Monoatomic ions have oxidation numbers equal to their charge. For example, \(Li^+\) is +1, while \(O^{2-}\) is -2.
  • In most compounds, oxygen is assigned an oxidation number of -2, while hydrogen is usually +1, except when it's bonded with metals in hydrides.
Neutral Compounds
Neutral compounds play an important role in chemistry. These are compounds where the total charge is zero, meaning the sum of all oxidation numbers in the compound equals zero. Understanding neutrality helps in the calculation of oxidation numbers.

When calculating the oxidation state for an element within a compound, the principle that the sum must equal zero guides us. For example, in \(\text{NaAlH}_4\), we know \(Na\) is +1 and each hydrogen atom contributes -1. To ensure the compound remains neutral, we solve for the oxidation number of aluminium. This calculation ensures that chemical compounds maintain electrical neutrality, meaning there's a balanced distribution of electrons. The neutrality of compounds is often confused with neutral pH, but here, neutrality purely indicates charge balance.
Chemical Compounds
Chemical compounds are substances formed by two or more elements chemically bonded together. Each compound has a specific chemical formula that provides information on the types and number of atoms in a molecule. Properties of compounds arise from the nature of bonds between these atoms, affecting their strength, reactivity, and behavior.

There are several types of chemical compounds, including molecular (covalent bonds) and ionic compounds. For instance, water, \(\text{H}_2\text{O}\), is molecular, while sodium chloride, \(\text{NaCl}\), is ionic. Understanding these compounds requires comprehending how electrons are shared or transferred, forming stable chemical structures. This electron movement during bonding can change the oxidation states of elements, affecting their functionalities and reactions.
Charge Balance Equations
Charge balance equations are fundamental for ensuring reactions remain neutral overall. These equations require that the sum of positive charges equals the sum of negative charges in a reaction or compound. This principle dictates that total net charge remains consistent before and after a chemical reaction.

For example, assigning oxidation numbers is part of using charge balance. In ionic compounds, like \(BaCrO_4\), you will add the charges of each element using their oxidation states to predict and confirm the compound's neutral status. Charge balance ensures that ions combine in proportions that equalize the charge difference, reinforcing the structure's stability.
  • This balance law is pivotal in redox (reduction-oxidation) reactions, where one species is oxidized, and another is reduced, balancing overall electron transfer.
  • Understanding charge balance allows chemists to write, balance, and predict chemical equations accurately.

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Most popular questions from this chapter

(a) Is the concentration of a solution an intensive or an extensive property? (b) What is the difference between \(0.50 \mathrm{~mol}\) \(\mathrm{HCl}\) and \(0.50 \mathrm{M} \mathrm{HCl} ?\)

An aqueous solution of an unknown solute is tested with litmus paper and found to be acidic. The solution is weakly conducting compared with a solution of \(\mathrm{NaCl}\) of the same concentration.Which of the following substances could the unknown be: \(\mathrm{KOH}, \mathrm{NH}_{3}, \mathrm{HNO}_{3}, \mathrm{KClO}_{2}, \mathrm{H}_{3} \mathrm{PO}_{3}, \mathrm{CH}_{3} \mathrm{COCH}_{3}\) (acetone)?

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) (b) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{KOH}(a q) \longrightarrow\).

Calculate (a) the number of grams of solute in \(0.250 \mathrm{~L}\) of \(0.175 \mathrm{M} \mathrm{KBr},\) (b) the molar concentration of a solution containing \(14.75 \mathrm{~g}\) of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(1.375 \mathrm{~L},\) (c) the volume of \(1.50 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) in milliliters that contains \(2.50 \mathrm{~g}\) of solute.

You choose to investigate some of the solubility guidelines for two ions not listed in Table \(4.1,\) the chromate ion \(\left(\mathrm{CrO}_{4}^{2-}\right)\) and the oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right) .\) You are given \(0.01 \mathrm{M}\) solutions (A, B, C, D) of four water-soluble salts: $$ \begin{array}{lll} \hline \text { Solution } & \text { Solute } & \text { Color of Solution } \\ \hline \text { A } & \mathrm{Na}_{2} \mathrm{CrO}_{4} & \text { Yellow } \\ \mathrm{B} & \left(\mathrm{NH}_{4}\right)_{2} \mathrm{C}_{2} \mathrm{O}_{4} & \text { Colorless } \\ \mathrm{C} & \mathrm{AgNO}_{3} & \text { Colorless } \\ \mathrm{D} & \mathrm{CaCl}_{2} & \text { Colorless } \\ \hline \end{array} $$ When these solutions are mixed, the following observations are made: $$ \begin{array}{lll} \hline \text { Expt } & \text { Solutions } & \\ \text { Number } & \text { Mixed } & \text { Result } \\ \hline 1 & \mathrm{~A}+\mathrm{B} & \text { No precipitate, yellow solution } \\\ 2 & \mathrm{~A}+\mathrm{C} & \text { Red precipitate forms } \\ 3 & \mathrm{~A}+\mathrm{D} & \text { Yellow precipitate forms } \\ 4 & \mathrm{~B}+\mathrm{C} & \text { White precipitate forms } \\ 5 & \mathrm{~B}+\mathrm{D} & \text { White precipitate forms } \\ 6 & \mathrm{C}+\mathrm{D} & \text { White precipitate forms } \end{array} $$ (a) Write a net ionic equation for the reaction that occurs in each of the experiments. (b) Identify the precipitate formed, if any, in each of the experiments.

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