Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 23, Problem 23.107P

(a)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Iron (III) ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

(a)

Expert Solution
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Explanation of Solution

Given,

Iron (III) ion reacts with hypochlorite ion in basic solution to form ferrate ion

    (2FeO42), Cl, and H2O.

The balanced equation is

2Fe3+(aq)+6OCl(aq)+4OH(aq)2FeO4(aq)2+6Cl(aq)+2H2O(l)

Based on the chemical equation, the total number of iron oxidation is +6 each Iron has +3 oxidation state.

(b)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Manganese and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

(b)

Expert Solution
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Explanation of Solution

Given,

Potassium hexacyanomanganate (II) reacts with K metal to form K6[Mn(CN)6].

The balanced equation is

K4[Mn(CN)6](s)+2K(s)K6[Mn(CN)6](s)

Based on the chemical equation six potassium cation balanced by six cyanide ions.  Hence, the oxidation of Manganese is zero.

(c)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Cobalt ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

(c)

Expert Solution
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Explanation of Solution

Given,

Heating sodium superoxide (NaO2) with Co3O4 produces Na4CoO4 and O2 gas.

The balanced equation is

  12NaO2(s)+Co3O4(s)3Na4CoO4(s)+8O2(g)

Based on the chemical equation oxygen eight negative charge balance by four sodium and another four balance by Cobalt.  Therefor the Cobalt oxidation state is four.

(d)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Vanadium ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

(d)

Expert Solution
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Explanation of Solution

Vanadium (III) chloride reacts with Na metal under a CO atmosphere to produce

    Na[V(CO)6] and NaCl.

The balanced equation is

VCl3(s)+4Na(s)+6CO(g)Na[V(CO)6](s)+3NaCl(s)

Based on the chemical equation sodium has one positive charge that positive charge balanced by vanadium so the vanadium has -1 charge.

Therefore, the oxidation state of Vanadium is -1

(e)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Nickel ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

(e)

Expert Solution
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Explanation of Solution

Given,

Barium peroxide reacts with nickel (II) ions in basic solution to produce BaNiO3.

The balanced equation is

2Fe3+(aq)+6OCl(aq)+4OH(aq)2FeO4(aq)2+6Cl(aq)+2H2O(l)

  BaNiO3.+2+x+3(2)=0+2+x6=0x4=0x=+4

Hence, the oxidation of Nickel is +4

(f)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Cobalt ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

(f)

Expert Solution
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Explanation of Solution

Bubbling  CO through a basic solution of cobalt (II) ion Produces [Co(CO)4], CO32, and water.

The balanced equation is

11CO(g)+12OH(aq)+2Co(aq)2+2[Co(CO)4](aq)+3CO3(aq)2+6H2O(l)

[Co(CO)4](aq) here the ligand is neutral.  So the oxidation state of the Cobalt is -1.

(g)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Copper ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

(g)

Expert Solution
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Explanation of Solution

Heating cesium tetrafluorocuprate (II) with F2 gas under pressure gives Cs2CuF6.

The balanced equation is

Cs2[CuF4](s)+F2(g)Cs2CuF6(s)

  Cs2CuF62(+1)+x+6(1)=0+2+x6=0x4=0x=+4

Hence, the oxidation state of the copper is +4.

(h)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Tantalum ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

(h)

Expert Solution
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Explanation of Solution

Given,

Heating tantalum (V) chloride with Na metal produces NaCl and Ta6Cl15, in which half of the Ta is in the +2 state.

The balanced equation is

6TaCl5(s)+15Na(s)15NaCl(s)+Ta6Cl15(s)

Ta6Cl15(s) here the ligand is chloride, half of the Tandalum +2 oxidation state and another half of the tantalum is +3 oxidation state.

Hence, 3 Tantalum is +3 oxidation state and 3 Tantalum is +2 oxidation state.

(i)

Interpretation Introduction

Interpretation:

The oxidation state transition metal is Nickel ion and balanced equation has to be written.

Concept introduction:

Oxidation number: (Oxidation state) A number equal to the magnitude of the charge an atom would have if its shared electrons were transferred to the atom that attracts them more strongly.

A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

(i)

Expert Solution
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Explanation of Solution

Potassium tetracyanonickelate (II) reacts with hydrazine (N2H4) in basic solution to form

    K4[Ni2(CN)6] and N2 gas.

The balanced equation is

4K2[Ni(CN)4](s)+N2H4(aq)+4OH(aq)2K4[Ni2(CN)6](s)+N2(g)+4H2O(l)+4CN(aq)

K4[Ni2(CN)6](s)4(+1)+2x+6(1)=04+2x6=02x2=02x=+2x=+1

Hence, the oxidation state of the Nickel is +1.

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Chapter 23 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

Ch. 23.4 - Prob. 23.6AFPCh. 23.4 - Prob. 23.6BFPCh. 23.4 - Prob. 23.7AFPCh. 23.4 - Prob. 23.7BFPCh. 23.4 - Prob. B23.1PCh. 23.4 - Prob. B23.2PCh. 23 - Prob. 23.1PCh. 23 - Write the general electron configuration of a...Ch. 23 - Prob. 23.3PCh. 23 - Prob. 23.4PCh. 23 - Prob. 23.5PCh. 23 - Prob. 23.6PCh. 23 - (a) What is the range in electronegativity across...Ch. 23 - (a) Explain the major difference between the...Ch. 23 - (a) What behavior distinguishes paramagnetic and...Ch. 23 - Prob. 23.10PCh. 23 - Using the periodic table to locate each element,...Ch. 23 - Using the periodic table to locate each element,...Ch. 23 - Using the periodic table to locate each element,...Ch. 23 - Prob. 23.14PCh. 23 - Prob. 23.15PCh. 23 - What is the highest oxidation state for (a) Ta;...Ch. 23 - What is the highest oxidation state for (a) Nb;...Ch. 23 - Prob. 23.18PCh. 23 - Prob. 23.19PCh. 23 - Prob. 23.20PCh. 23 - Prob. 23.21PCh. 23 - Prob. 23.22PCh. 23 - Prob. 23.23PCh. 23 - Prob. 23.24PCh. 23 - Prob. 23.25PCh. 23 - Prob. 23.26PCh. 23 - What atomic property of the lanthanides leads to...Ch. 23 - Prob. 23.28PCh. 23 - Prob. 23.29PCh. 23 - Give the electron configuration of (a) La; (b)...Ch. 23 - Prob. 23.31PCh. 23 - Only a few lanthanides show an oxidation state...Ch. 23 - Prob. 23.33PCh. 23 - Prob. 23.34PCh. 23 - Describe the makeup of a complex ion, including...Ch. 23 - Prob. 23.36PCh. 23 - Prob. 23.37PCh. 23 - Prob. 23.38PCh. 23 - Prob. 23.39PCh. 23 - Prob. 23.40PCh. 23 - Prob. 23.41PCh. 23 - Prob. 23.42PCh. 23 - Prob. 23.43PCh. 23 - Prob. 23.44PCh. 23 - Prob. 23.45PCh. 23 - Prob. 23.46PCh. 23 - What are the charge and coordination number of the...Ch. 23 - What are the charge and coordination number of the...Ch. 23 - Prob. 23.49PCh. 23 - Give systematic names for the following...Ch. 23 - What are the charge and coordination number of the...Ch. 23 - What are the charge and coordination number of the...Ch. 23 - Prob. 23.53PCh. 23 - Prob. 23.54PCh. 23 - Prob. 23.55PCh. 23 - Prob. 23.56PCh. 23 - Prob. 23.57PCh. 23 - Prob. 23.58PCh. 23 - Prob. 23.59PCh. 23 - Prob. 23.60PCh. 23 - Prob. 23.61PCh. 23 - Prob. 23.62PCh. 23 - For any of the following that can exist as...Ch. 23 - Prob. 23.64PCh. 23 - Prob. 23.65PCh. 23 - Prob. 23.66PCh. 23 - Chromium(III), like cobalt(III), has a...Ch. 23 - When MCl4(NH3)2 is dissolved in water and treated...Ch. 23 - Prob. 23.69PCh. 23 - What is a coordinate covalent bond? Is such a...Ch. 23 - Prob. 23.71PCh. 23 - Prob. 23.72PCh. 23 - Prob. 23.73PCh. 23 - In terms of the theory of color absorption,...Ch. 23 - Prob. 23.75PCh. 23 - Prob. 23.76PCh. 23 - Prob. 23.77PCh. 23 - How do the relative magnitudes of Epairing and Δ...Ch. 23 - Prob. 23.79PCh. 23 - Give the number of d electrons (n of dn) for the...Ch. 23 - Give the number of d electrons (n of dn) for the...Ch. 23 - Prob. 23.82PCh. 23 - How many d electrons (n of dn) are in the central...Ch. 23 - Prob. 23.84PCh. 23 - Prob. 23.85PCh. 23 - Prob. 23.86PCh. 23 - Prob. 23.87PCh. 23 - Prob. 23.88PCh. 23 - Prob. 23.89PCh. 23 - Prob. 23.90PCh. 23 - Prob. 23.91PCh. 23 - Prob. 23.92PCh. 23 - Prob. 23.93PCh. 23 - Prob. 23.94PCh. 23 - Prob. 23.95PCh. 23 - Prob. 23.96PCh. 23 - Prob. 23.97PCh. 23 - Prob. 23.98PCh. 23 - When neptunium (Np) and plutonium (Pu) were...Ch. 23 - Prob. 23.100PCh. 23 - Prob. 23.101PCh. 23 - For the compound [Co(en)2Cl2]Cl, give: The...Ch. 23 - Prob. 23.103PCh. 23 - Prob. 23.104PCh. 23 - Prob. 23.105PCh. 23 - Prob. 23.106PCh. 23 - Prob. 23.107PCh. 23 - Prob. 23.108PCh. 23 - Prob. 23.109PCh. 23 - Prob. 23.110PCh. 23 - Prob. 23.111PCh. 23 - The actinides Pa, U, and Np form a series of...Ch. 23 - Prob. 23.113PCh. 23 - Prob. 23.114PCh. 23 - Prob. 23.115PCh. 23 - Prob. 23.116PCh. 23 - Prob. 23.117PCh. 23 - Prob. 23.118PCh. 23 - Prob. 23.119PCh. 23 - Prob. 23.120PCh. 23 - Prob. 23.121PCh. 23 - Prob. 23.122P
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