Chemistry: The Molecular Science, Hybrid Edition (with OWLv2 24-Months Printed Access Card)
Chemistry: The Molecular Science, Hybrid Edition (with OWLv2 24-Months Printed Access Card)
5th Edition
ISBN: 9781285461847
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Textbook Question
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Chapter 20, Problem 16QRT

Which Period 4 transition-metal ions are isoelectronic with

  1. (a) Zn2+
  2. (b) Mn2+
  3. (c) Cr3+
  4. (d) Fe3+

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Transition metal ion from period 4 that is isoelectronic with Zn2+ has to be given.

Concept Introduction:

Isoelectronic means same electronic configuration.  An atom and an ion can be isoelectronic if they have same number of electrons in it.  This is because the electronic configuration will be same.  There will not be any change in atomic number or identity of the element or ion.

Explanation of Solution

Given ion is Zn2+.  Zinc has an atomic number of 30.  The electronic configuration of zinc atom can be given as shown below.

    Zn: 1s2 2s2 2p6 3s2 3p6 4s2 3d10

Zinc atom loses two electrons to form Zn2+ ion.  This can be represented as shown below.

    ZnZn2++2e

The electronic configuration of Zn2+ is given below.

    Zn2+1s2 2s2 2p6 3s2 3p6 3d10

In period 4, the transition metal ion that is isoelectronic with Zn2+ ion must also contain 28 electrons.  Copper is an element that has 29 electrons in its electronic configuration.  Copper loses one electron to form Cu+ ion that has 28 electrons.  Electronic configuration of copper can be given as shown below.

    Cu: 1s2 2s2 2p6 3s2 3p6 4s1 3d10

Copper atom loses one electron to form Cu+ ion.  This can be represented as shown below.

    CuCu++1e

The electronic configuration of Cu+ is given below.

    Cu+1s2 2s2 2p6 3s2 3p6 3d10

As the electronic configuration of Zn2+ ion and Cu+ ion are same, they both are said to be isoelectronic.  Thus Cu+ is isoelectronic with Zn2+.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Transition metal ion from period 4 that is isoelectronic with Mn2+ has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is Mn2+.  Manganese has an atomic number of 25.  The electronic configuration of manganese atom can be given as shown below.

    Mn: 1s2 2s2 2p6 3s2 3p6 4s2 3d5

Manganese atom loses two electrons to form Mn2+ ion.  This can be represented as shown below.

    MnMn2++2e

The electronic configuration of Mn2+ is given below.

    Mn2+1s2 2s2 2p6 3s2 3p6 3d5

In period 4, the transition metal ion that is isoelectronic with Mn2+ ion must also contain 23 electrons.  Iron is an element that has 26 electrons in its electronic configuration.  Iron loses three electrons to form Fe3+ ion that has 23 electrons.  Electronic configuration of iron can be given as shown below.

    Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6

Iron atom loses three electrons to form Fe3+ ion.  This can be represented as shown below.

    FeFe3++3e

The electronic configuration of Fe3+ is given below.

    Fe3+1s2 2s2 2p6 3s2 3p6 3d5

As the electronic configuration of Mn2+ ion and Fe3+ ion are same, they both are said to be isoelectronic.  Thus Fe3+ is isoelectronic with Mn2+.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Transition metal ion from period 4 that is isoelectronic with Cr3+ has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is Cr3+.  Chromium has an atomic number of 24.  The electronic configuration of chromium atom can be given as shown below.

    Cr: 1s2 2s2 2p6 3s2 3p6 4s1 3d5

Chromium atom loses three electrons to form Cr3+ ion.  This can be represented as shown below.

    CrCr3++3e

The electronic configuration of Cr3+ is given below.

    Cr3+1s2 2s2 2p6 3s2 3p6 3d3

In period 4, the transition metal ion that is isoelectronic with Cr3+ ion must also contain 21 electrons.  Vanadium is an element that has 23 electrons in its electronic configuration.  Vanadium loses two electrons to form V2+ ion that has 21 electrons.  Electronic configuration of vanadium can be given as shown below.

    V: 1s2 2s2 2p6 3s2 3p6 4s2 3d3

Vanadium atom loses two electrons to form V2+ ion.  This can be represented as shown below.

    VV2++2e

The electronic configuration of V2+ is given below.

    V2+1s2 2s2 2p6 3s2 3p6 3d3

As the electronic configuration of Cr3+ ion and V2+ ion are same, they both are said to be isoelectronic.  Thus V2+ is isoelectronic with Cr3+.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Transition metal ion from period 4 that is isoelectronic with Fe3+ has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is Fe3+.  Iron has an atomic number of 26.  The electronic configuration of iron atom can be given as shown below.

    Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6

Iron atom loses three electrons to form Fe3+ ion.  This can be represented as shown below.

    FeFe3++3e

The electronic configuration of Fe3+ is given below.

    Fe3+1s2 2s2 2p6 3s2 3p6 3d5

In period 4, the transition metal ion that is isoelectronic with Fe3+ ion must also contain 23 electrons.  Manganese is an element that has 25 electrons in its electronic configuration.  Manganese loses two electrons to form Mn2+ ion that has 23 electrons.  Electronic configuration of manganese can be given as shown below.

    Mn: 1s2 2s2 2p6 3s2 3p6 4s2 3d5

Manganese atom loses two electrons to form Mn2+ ion.  This can be represented as shown below.

    MnMn2++2e

The electronic configuration of Mn2+ is given below.

    Mn2+1s2 2s2 2p6 3s2 3p6 3d5

As the electronic configuration of Fe3+ ion and Mn2+ ion are same, they both are said to be isoelectronic.  Thus Mn2+ is isoelectronic with Fe3+.

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Chapter 20 Solutions

Chemistry: The Molecular Science, Hybrid Edition (with OWLv2 24-Months Printed Access Card)

Ch. 20.5 - Use Le Chatelier’s principle to explain how the...Ch. 20.5 - At what pH does Ecell = 0.00 V for the reduction...Ch. 20.6 - Prob. 20.6PSPCh. 20.6 - Prob. 20.8CECh. 20.6 - (a) Name this coordination compound:...Ch. 20.6 - Prob. 20.9CECh. 20.6 - Prob. 20.8PSPCh. 20.6 - Prob. 20.10CECh. 20.6 - Prob. 20.11CECh. 20.6 - Prob. 20.9PSPCh. 20.6 - Prob. 20.12ECh. 20.7 - Prob. 20.10PSPCh. 20.7 - Prob. 20.13CECh. 20.7 - Prob. 20.14CECh. 20 - Prob. 1QRTCh. 20 - Prob. 2QRTCh. 20 - Prob. 3QRTCh. 20 - Prob. 4QRTCh. 20 - Prob. 5QRTCh. 20 - Prob. 6QRTCh. 20 - Prob. 7QRTCh. 20 - Prob. 8QRTCh. 20 - Prob. 9QRTCh. 20 - Prob. 10QRTCh. 20 - Prob. 11QRTCh. 20 - Prob. 12QRTCh. 20 - Prob. 13QRTCh. 20 - Prob. 14QRTCh. 20 - Prob. 15QRTCh. 20 - Which Period 4 transition-metal ions are...Ch. 20 - Prob. 17QRTCh. 20 - Prob. 18QRTCh. 20 - Prob. 19QRTCh. 20 - Prob. 20QRTCh. 20 - Prob. 21QRTCh. 20 - Prob. 22QRTCh. 20 - Prob. 23QRTCh. 20 - Prob. 24QRTCh. 20 - Prob. 25QRTCh. 20 - Prob. 26QRTCh. 20 - Prob. 27QRTCh. 20 - Prob. 28QRTCh. 20 - Prob. 29QRTCh. 20 - Prob. 30QRTCh. 20 - Prob. 31QRTCh. 20 - Prob. 32QRTCh. 20 - Prob. 33QRTCh. 20 - Prob. 34QRTCh. 20 - Prob. 35QRTCh. 20 - Prob. 36QRTCh. 20 - Prob. 37QRTCh. 20 - Prob. 38QRTCh. 20 - Prob. 39QRTCh. 20 - Prob. 40QRTCh. 20 - Prob. 41QRTCh. 20 - Prob. 42QRTCh. 20 - Prob. 43QRTCh. 20 - Prob. 44QRTCh. 20 - Prob. 45QRTCh. 20 - Prob. 46QRTCh. 20 - Prob. 47QRTCh. 20 - Prob. 48QRTCh. 20 - Prob. 49QRTCh. 20 - Prob. 50QRTCh. 20 - Prob. 51QRTCh. 20 - Prob. 52QRTCh. 20 - Give the charge on the central metal ion in each...Ch. 20 - Prob. 54QRTCh. 20 - Prob. 55QRTCh. 20 - Classify each ligand as monodentate, bidentate,...Ch. 20 - Prob. 57QRTCh. 20 - Prob. 58QRTCh. 20 - Prob. 59QRTCh. 20 - Prob. 60QRTCh. 20 - Prob. 61QRTCh. 20 - Prob. 62QRTCh. 20 - Prob. 63QRTCh. 20 - Prob. 64QRTCh. 20 - Prob. 65QRTCh. 20 - Prob. 66QRTCh. 20 - Prob. 67QRTCh. 20 - Prob. 68QRTCh. 20 - Prob. 69QRTCh. 20 - Prob. 70QRTCh. 20 - Prob. 71QRTCh. 20 - Prob. 72QRTCh. 20 - Prob. 73QRTCh. 20 - Prob. 74QRTCh. 20 - How many unpaired electrons are in the high-spin...Ch. 20 - Prob. 76QRTCh. 20 - Prob. 77QRTCh. 20 - Prob. 78QRTCh. 20 - An aqueous solution of [Rh(C2O4)3]3− is yellow....Ch. 20 - Prob. 80QRTCh. 20 - Prob. 81QRTCh. 20 - Prob. 82QRTCh. 20 - Prob. 83QRTCh. 20 - Prob. 84QRTCh. 20 - Give the electron configuration of (a) Ti3+. (b)...Ch. 20 - Prob. 86QRTCh. 20 - Prob. 87QRTCh. 20 - Prob. 88QRTCh. 20 - Prob. 89QRTCh. 20 - Prob. 90QRTCh. 20 - Prob. 91QRTCh. 20 - Prob. 92QRTCh. 20 - Prob. 93QRTCh. 20 - Prob. 94QRTCh. 20 - Prob. 95QRTCh. 20 - Prob. 96QRTCh. 20 - Prob. 97QRTCh. 20 - Prob. 98QRTCh. 20 - Prob. 99QRTCh. 20 - Prob. 100QRTCh. 20 - Prob. 101QRTCh. 20 - Prob. 103QRTCh. 20 - Prob. 104QRTCh. 20 - Prob. 105QRTCh. 20 - Prob. 106QRTCh. 20 - Repeat the directions for Question 106 using a...Ch. 20 - Prob. 113QRTCh. 20 - Prob. 114QRTCh. 20 - Prob. 115QRTCh. 20 - Prob. 116QRTCh. 20 - Prob. 117QRTCh. 20 - Prob. 118QRTCh. 20 - Prob. 119QRTCh. 20 - Prob. 120QRTCh. 20 - The glycinate ion (gly) is H2NCH2CO2. It can act...Ch. 20 - Five-coordinate coordination complexes are known,...Ch. 20 - Prob. 123QRTCh. 20 - Prob. 124QRTCh. 20 - Two different compounds are known with the formula...Ch. 20 - Prob. 126QRT
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