General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
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Chapter 26, Problem 26.47P

(a)

Interpretation Introduction

Interpretation:

Electronic configuration of d-orbital has to be written for Nb(III) complex.

(a)

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Explanation of Solution

Given complex contains Nb(III) as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of niobium is 41.  Therefore, Nb(III) will have 38 electrons.

In the octahedral complex, the d orbitals splits into two energy levels and they are t2g and eg.  In this t2g is the lower energy degenerate orbital and eg is the higher energy degenerate orbital.

Electronic configuration of niobium is given as shown below;

  Nb:[Kr]5s2 4d3

Niobium loses three electrons to form Nb(III) ion.  The electronic configuration of Nb(III) is given as shown below.

    Nb:[Kr]5s2 4d3Nb(III):[Kr]5s0 4d2+3e

Therefore, there are two electrons in d-orbital.  The d-orbital electronic configuration of Nb(III) is t2g2eg0.

(b)

Interpretation Introduction

Interpretation:

Electronic configuration of d-orbital has to be written for Mo(II) complex that has Δ0 greater than electron-pairing energy..

(b)

Expert Solution
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Explanation of Solution

Given complex contains Mo(II) as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of molybdenum is 42.  Therefore, Mo(II) will have 40 electrons.

In the octahedral complex, the d orbitals splits into two energy levels and they are t2g and eg.  In this t2g is the lower energy degenerate orbital and eg is the higher energy degenerate orbital.

Electronic configuration of molybdenum is given as shown below;

  Mo:[Kr]5s2 4d4

Molybdenum loses two electrons to form Mo(II) ion.  The electronic configuration of Mo(II) is given as shown below.

    Mo:[Kr]5s2 4d4Mo(II):[Kr]5s0 4d4+2e

Therefore, there are four electrons in d-orbital.  As in the problem it is said that the Δ0 is greater than the electron-pairing energy, the electrons will get paired in t2g orbital itself.  Therefore, the d-orbital electronic configuration of Mo(II) is t2g4eg0.

(c)

Interpretation Introduction

Interpretation:

Electronic configuration of d-orbital has to be written for Mn(II) complex that has Δ0 lesser than electron-pairing energy..

(c)

Expert Solution
Check Mark

Explanation of Solution

Given complex contains Mn(II) as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of manganese is 25.  Therefore, Mn(II) will have 23 electrons.

In the octahedral complex, the d orbitals splits into two energy levels and they are t2g and eg.  In this t2g is the lower energy degenerate orbital and eg is the higher energy degenerate orbital.

Electronic configuration of manganese is given as shown below;

  Mn:[Ar]4s2 3d5

Manganese loses two electrons to form Mn(II) ion.  The electronic configuration of Mn(II) is given as shown below.

    Mn:[Ar]4s2 3d5Mn(II):[Ar]4s0 3d5+2e

Therefore, there are five electrons in d-orbital.  As in the problem it is said that the Δ0 is lesser than the electron-pairing energy, the electrons will get be filled in all the degenerate orbital before getting paired.  Therefore, the d-orbital electronic configuration of Mn(II) is t2g3eg2.

(d)

Interpretation Introduction

Interpretation:

Electronic configuration of d-orbital has to be written for Au(I) complex.

(d)

Expert Solution
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Explanation of Solution

Given complex contains Au(I) as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of gold is 79.  Therefore, Au(I) will have 78 electrons.

In the octahedral complex, the d orbitals splits into two energy levels and they are t2g and eg.  In this t2g is the lower energy degenerate orbital and eg is the higher energy degenerate orbital.

Electronic configuration of gold is given as shown below;

  Au:[Xe]6s1 4f14 5d10

Gold loses one electron to form Au(I) ion.  The electronic configuration of Au(I) is given as shown below.

    Au:[Xe]6s1 4f14 5d10Au(I):[Xe]6s0 4f14 5d10+1e

Therefore, there are ten electrons in d-orbital.  The d-orbital electronic configuration of Au(I) is t2g6eg4.

(e)

Interpretation Introduction

Interpretation:

Electronic configuration of d-orbital has to be written for Ir(III) complex that has Δ0 greater than electron-pairing energy..

(e)

Expert Solution
Check Mark

Explanation of Solution

Given complex contains Ir(III) as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of iridium is 77.  Therefore, Ir(III) will have 74 electrons.

In the octahedral complex, the d orbitals splits into two energy levels and they are t2g and eg.  In this t2g is the lower energy degenerate orbital and eg is the higher energy degenerate orbital.

Electronic configuration of iridium is given as shown below;

  Ir:[Xe]6s2 4f14 5d7

Iridium loses three electrons to form Ir(III) ion.  The electronic configuration of Ir(III) is given as shown below.

    Ir:[Xe]6s2 4f14 5d7Ir(III):[Xe]6s0 4f14 5d6+3e

Therefore, there are six electrons in d-orbital.  As in the problem it is said that the Δ0 is greater than the electron-pairing energy, the electrons will get paired in t2g orbital itself before entering eg orbital.  Therefore, the d-orbital electronic configuration of Ir(III) is t2g6eg0.

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Chapter 26 Solutions

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