OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:
OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:
5th Edition
ISBN: 9781285460369
Author: STANITSKI
Publisher: Cengage Learning
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Chapter 20, Problem 72QRT

(a)

Interpretation Introduction

Interpretation:

Number of unpaired electrons in [Cr(H2O)6]2+ has to be given.

(a)

Expert Solution
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Explanation of Solution

Given complex ion is [Cr(H2O)6]2+.  The central metal ion is Cr2+Cr2+ ion is said to contain four 3d electrons.  For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less.  The splitting of orbitals depends upon the ligand.  If the ligand is weak field ligand, then the crystal field splitting energy (Δo) will be smaller resulting in unpaired electrons.  If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons.  The crystal field model for high-spin and low-spin complex of Cr2+ can be given as,

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 20, Problem 72QRT , additional homework tip  1

Number of unpaired electrons:

The ligand that is present in the complex is water.  Water is a weak field ligand.  Therefore, it forms high spin complex.  From the above drawn crystal field model, it is found that there are four unpaired electrons in high-spin complex.

(b)

Interpretation Introduction

Interpretation:

Number of unpaired electrons in [Mn(H2O)6]2+ has to be given.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given complex ion is [Mn(H2O)6]2+.  The central metal ion is Mn2+Mn2+ ion is said to contain five 3d electrons.  For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less.  The splitting of orbitals depends upon the ligand.  If the ligand is weak field ligand, then the crystal field splitting energy (Δo) will be smaller resulting in unpaired electrons.  If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons.  The crystal field model for high-spin and low-spin complex of Mn2+ can be given as,

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 20, Problem 72QRT , additional homework tip  2

Number of unpaired electrons:

The ligand that is present in the complex is water.  Water is a weak field ligand.  Therefore, it forms high spin complex.  From the above drawn crystal field model, it is found that there are five unpaired electrons in high-spin complex.

(c)

Interpretation Introduction

Interpretation:

Number of unpaired electrons in [FeF6]3 has to be given.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given complex ion is [FeF6]3.  The central metal ion is Fe3+Fe3+ ion is said to contain five 3d electrons.  For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less.  The splitting of orbitals depends upon the ligand.  If the ligand is weak field ligand, then the crystal field splitting energy (Δo) will be smaller resulting in unpaired electrons.  If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons.  The crystal field model for high-spin and low-spin complex of Fe3+ can be given as,

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 20, Problem 72QRT , additional homework tip  3

Number of unpaired electrons:

The ligand that is present in the complex is fluorine.  Fluorine is a weak field ligand.  Therefore, it forms high spin complex.  From the above drawn crystal field model, it is found that there are five unpaired electrons in high-spin complex.

(d)

Interpretation Introduction

Interpretation:

Number of unpaired electrons in [Cr(en)3]3+ has to be given.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given complex ion is [Cr(en)3]3+.  The central metal ion is Cr3+Cr3+ ion is said to contain three 3d electrons.  For high-spin complex, the number of unpaired electrons will be more and for low-spin complex, the number of unpaired electrons will be less.  The splitting of orbitals depends upon the ligand.  If the ligand is weak field ligand, then the crystal field splitting energy (Δo) will be smaller resulting in unpaired electrons.  If the ligand is a strong field ligand, then the crystal field splitting energy will be larger resulting in pairing of electrons.  In this ion, there are only three d electrons and the lower-energy orbitals has to be singly filled first.  Therefore, only one configuration is possible and it is shown as given below.

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 20, Problem 72QRT , additional homework tip  4

Number of unpaired electrons:

The ligand that is present in the complex is ethylenediamine.  From the above drawn crystal field model, it is found that there are three unpaired electrons in complex.

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Chapter 20 Solutions

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:

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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