General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN: 9781305580343
Author: Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher: Cengage Learning
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Chapter 22, Problem 22.60QP

(a)

Interpretation Introduction

Interpretation:

Distribution of d-electrons and the geometry of the complex ion [Pt(NH3)2(NO2)2]2+ has to be determined.

Concept Introduction:

Complex compounds exist in following geometries - tetrahedral, square planar, octahedral etc.

When ligands approach the metal ion the degeneracy in d-orbitals of the metal ion is destroyed and they split into two different energy levels. In case of tetrahedral complex, the dx2y2and dz2 orbitals occupy lower energy level and the dxy,dxz,dyz orbitals occupy higher energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

In case of square planar complex, the 5 d-orbitals split into following pattern and it is given below with increasing order of energy.

dxy= dyz<dz2<dxydx2y2

(a)

Expert Solution
Check Mark

Answer to Problem 22.60QP

The d-electrons are distributed in the complex ion [Pt(NH3)2(NO2)2]2+ as follows –

General Chemistry - Standalone book (MindTap Course List), Chapter 22, Problem 22.60QP , additional homework tip  1

The complex has square planar geometry.

Explanation of Solution

In the complex ion [Pt(NH3)2(NO2)2]2+ the central metal ion Platinum is in +4 oxidation state as follows –

oxidation state of Pt = charge on complex - charge of ligands = +2-[2(0)+2(1)]  = +2-(2)=+4

Atomic number of Platinum is 78.  The electrons are filled on the basis of building-up rule. The electronic configuration of Platinum could be 1s22s22p63s23p63d104s24p64d105s25p64f145d96s1 which can be simplified as [Xe]4f145d96s1.   The electronic configuration of Pt+4 could be [Xe]4f145d66s0.   Thus there are six electrons in d-orbital of Pt+4 and it has two unpaired electrons which are distributed as follows –

General Chemistry - Standalone book (MindTap Course List), Chapter 22, Problem 22.60QP , additional homework tip  2

The distribution of d-electrons as shown above correlates to that of square planar geometry.  Hence the complex ion [Pt(NH3)2(NO2)2]2+ exists in square planar geometry.

Conclusion

Splitting of d-orbitals determines the geometry of the complex.

(b)

Interpretation Introduction

Interpretation:

Distribution of d-electrons and the geometry of the complex ion [MnCl4]2 has to be determined.

Concept Introduction:

Complex compounds exist in following geometries - tetrahedral, square planar, octahedral etc.

When ligands approach the metal ion the degeneracy in d-orbitals of the metal ion is destroyed and they split into two different energy levels. In case of tetrahedral complex, the dx2y2and dz2 orbitals occupy lower energy level and the dxy,dxz,dyz orbitals occupy higher energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

In case of square planar complex, the 5 d-orbitals split into following pattern and it is given below with increasing order of energy.

dxy= dyz<dz2<dxydx2y2

(b)

Expert Solution
Check Mark

Answer to Problem 22.60QP

The d-electrons are distributed in the complex ion [MnCl4]2 as follows –

General Chemistry - Standalone book (MindTap Course List), Chapter 22, Problem 22.60QP , additional homework tip  3

The complex has tetrahedral geometry.

Explanation of Solution

In the complex ion [MnCl4]2 the central metal ion Manganese is in +2 oxidation state as follows –

oxidation state of Mn = charge on complex - charge of ligands = -2-[4(1)]  = -2+4=+2

Atomic number of Manganese is 25.  The electrons are filled on the basis of building-up rule. The electronic configuration of Manganese could be 1s22s22p63s23p63d54s2 which can be simplified as [Ar]3d54s2.   The electronic configuration of Mn+2 could be [Ar]3d5.   Thus there are five electrons in d-orbital of Mn+2 and all of them are unpaired which are distributed as follows –

General Chemistry - Standalone book (MindTap Course List), Chapter 22, Problem 22.60QP , additional homework tip  4

The distribution of d-electrons as shown above correlates to that of tetrahedral geometry.  Hence the complex ion [MnCl4]2 exists in tetrahedral geometry.

Conclusion

Splitting of d-orbitals determines the geometry of the complex.

(c)

Interpretation Introduction

Interpretation:

Distribution of d-electrons and the geometry of the complex ion [NiCl4]2 has to be determined.

Concept Introduction:

Complex compounds exist in following geometries - tetrahedral, square planar, octahedral etc.

When ligands approach the metal ion the degeneracy in d-orbitals of the metal ion is destroyed and they split into two different energy levels. In case of tetrahedral complex, the dx2y2and dz2 orbitals occupy lower energy level and the dxy,dxz,dyz orbitals occupy higher energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

In case of square planar complex, the 5 d-orbitals split into following pattern and it is given below with increasing order of energy.

dxy= dyz<dz2<dxydx2y2

(c)

Expert Solution
Check Mark

Answer to Problem 22.60QP

The d-electrons are distributed in the complex ion [NiCl4]2 as follows –

General Chemistry - Standalone book (MindTap Course List), Chapter 22, Problem 22.60QP , additional homework tip  5

The complex has tetrahedral geometry.

Explanation of Solution

In the complex ion [NiCl4]2 the central metal ion Nickel is in +2 oxidation state as follows –

oxidation state of Ni = charge on complex - charge of ligands = -2-[4(1)]  = -2+4=+2

Atomic number of Nickel is 28.  The electrons are filled on the basis of building-up rule. The electronic configuration of Nickel could be 1s22s22p63s23p63d84s2 which can be simplified as [Ar]3d84s2.   The electronic configuration of Ni+2 could be [Ar]3d8.   Thus there are five electrons in d-orbital of Ni+2 and given that there are two electrons unpaired which are distributed as follows –

The distribution of d-electrons as shown above correlates to that of tetrahedral geometry.  Hence the complex ion [NiCl4]2 exists in tetrahedral geometry.

Conclusion

Splitting of d-orbitals determines the geometry of the complex.

(d)

Interpretation Introduction

Interpretation:

Distribution of d-electrons and the geometry of the complex ion [AuF4] has to be determined.

Concept Introduction:

Complex compounds exist in following geometries - tetrahedral, square planar, octahedral etc.

When ligands approach the metal ion the degeneracy in d-orbitals of the metal ion is destroyed and they split into two different energy levels. In case of tetrahedral complex, the dx2y2and dz2 orbitals occupy lower energy level and the dxy,dxz,dyz orbitals occupy higher energy level.  The energy gap between these two set of orbitals is termed as crystal field splitting energy.

In case of square planar complex, the 5 d-orbitals split into following pattern and it is given below with increasing order of energy.

dxy= dyz<dz2<dxydx2y2

(d)

Expert Solution
Check Mark

Answer to Problem 22.60QP

The d-electrons are distributed in the complex ion [AuF4] as follows –

General Chemistry - Standalone book (MindTap Course List), Chapter 22, Problem 22.60QP , additional homework tip  6

The complex has square planar geometry.

Explanation of Solution

In the complex ion [AuF4] the central metal ion Gold is in +3 oxidation state as follows –

oxidation state of Au = charge on complex - charge of ligands = -1-[4(1)]  = -1+4=+3

Atomic number of Gold is 79. The electrons are filled on the basis of building-up rule. The electronic configuration of Gold could be 1s22s22p63s23p63d104s24p64d105s25p64f145d106s1 which can be simplified as [Xe]4f145d106s1.   The electronic configuration of Au+3 could be [Xe]4f145d8.   Thus there are eight electrons in d-orbital of Au+3 and given that there are no electrons unpaired which are distributed as follows –

General Chemistry - Standalone book (MindTap Course List), Chapter 22, Problem 22.60QP , additional homework tip  7

The distribution of d-electrons as shown above correlates to that of square planar geometry.  Hence the complex ion [AuF4] exists in square planar geometry.

Conclusion

Splitting of d-orbitals determines the geometry of the complex.

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Chapter 22 Solutions

General Chemistry - Standalone book (MindTap Course List)

Ch. 22 - What characteristics of the transition elements...Ch. 22 - Prob. 22.2QPCh. 22 - Prob. 22.3QPCh. 22 - Prob. 22.4QPCh. 22 - Prob. 22.5QPCh. 22 - Prob. 22.6QPCh. 22 - Prob. 22.7QPCh. 22 - Prob. 22.8QPCh. 22 - Silver(I) ion in basic solution is reduced by...Ch. 22 - What evidence did Werner obtain to show that the...Ch. 22 - Define the terms complex ion, ligand, and...Ch. 22 - Prob. 22.12QPCh. 22 - Prob. 22.13QPCh. 22 - Prob. 22.14QPCh. 22 - Prob. 22.15QPCh. 22 - Prob. 22.16QPCh. 22 - Explain the difference in behavior of d and l...Ch. 22 - Prob. 22.18QPCh. 22 - Prob. 22.19QPCh. 22 - a Describe the steps in the formation of a...Ch. 22 - Prob. 22.21QPCh. 22 - Prob. 22.22QPCh. 22 - Prob. 22.23QPCh. 22 - Prob. 22.24QPCh. 22 - Prob. 22.25QPCh. 22 - Prob. 22.26QPCh. 22 - Prob. 22.27QPCh. 22 - Prob. 22.28QPCh. 22 - What is the correct name for the coordination...Ch. 22 - Prob. 22.30QPCh. 22 - Prob. 22.31QPCh. 22 - Prob. 22.32QPCh. 22 - Prob. 22.33QPCh. 22 - Prob. 22.34QPCh. 22 - Prob. 22.35QPCh. 22 - Prob. 22.36QPCh. 22 - Prob. 22.37QPCh. 22 - Prob. 22.38QPCh. 22 - Prob. 22.39QPCh. 22 - Prob. 22.40QPCh. 22 - Prob. 22.41QPCh. 22 - Prob. 22.42QPCh. 22 - Prob. 22.43QPCh. 22 - Prob. 22.44QPCh. 22 - Consider the complex ion [CoCl(en)2(NO2)]+. a What...Ch. 22 - Prob. 22.46QPCh. 22 - Prob. 22.47QPCh. 22 - Name the following complexes, using IUPAC rules. a...Ch. 22 - Prob. 22.49QPCh. 22 - Prob. 22.50QPCh. 22 - Prob. 22.51QPCh. 22 - Give the structural formula for each of the...Ch. 22 - Prob. 22.53QPCh. 22 - Prob. 22.54QPCh. 22 - Prob. 22.55QPCh. 22 - Prob. 22.56QPCh. 22 - Prob. 22.57QPCh. 22 - Prob. 22.58QPCh. 22 - Prob. 22.59QPCh. 22 - Prob. 22.60QPCh. 22 - Prob. 22.61QPCh. 22 - Prob. 22.62QPCh. 22 - Prob. 22.63QPCh. 22 - Prob. 22.64QPCh. 22 - Prob. 22.65QPCh. 22 - Prob. 22.66QPCh. 22 - Prob. 22.67QPCh. 22 - Prob. 22.68QPCh. 22 - There are only two geometric isomers of the...Ch. 22 - Prob. 22.70QPCh. 22 - Prob. 22.71QPCh. 22 - Prob. 22.72QPCh. 22 - Prob. 22.73QPCh. 22 - Prob. 22.74QPCh. 22 - Prob. 22.75QPCh. 22 - Prob. 22.76QPCh. 22 - Consider the complex ion [CoCO3(NH3)4], where the...Ch. 22 - Prob. 22.78QPCh. 22 - Prob. 22.79QPCh. 22 - Prob. 22.80QPCh. 22 - Prob. 22.81QPCh. 22 - Prob. 22.82QPCh. 22 - What is the name of K2[MoOCl4]?Ch. 22 - Write the formula and draw the structure of...Ch. 22 - Prob. 22.85QPCh. 22 - Prob. 22.86QPCh. 22 - Is it possible to have a paramagnetic...Ch. 22 - Prob. 22.88QPCh. 22 - Prob. 22.89QPCh. 22 - Prob. 22.90QPCh. 22 - Prob. 22.91QPCh. 22 - Prob. 22.92QP
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