Chapter 22, Problem 45PS

(a)

Interpretation Introduction

Interpretation: The electronic configuration, magnetic character and the number of unpaired electrons for the given set of high spin tetrahedral complexes has to be determined.

Concept introduction: The transition metal atoms have tendency to form complex compounds that are linked to the certain neutral or ionic species which leads to the formation of coordination compounds. There exists a large number of coordination compounds that have a large number of applications in the chemical industry as well as in daily life.

The properties of the coordination compounds depend upon the primary and secondary valency of the metal ion in the coordination sphere. The electrical conductivity depends upon the number of ions that are produced by complex.

The strong-field ligands results in pairing of electrons present in the complex and leads to diamagnetic species , while the low-field ligand do not have tendency to pair up the electrons therefore forms paramagnetic species.

The strong field ligands lead to splitting to a higher extent than the weak field ligands and the wavelength of light absorbed depends on the energy gap that is produced by a particular ligand.

The five d orbitals get divided into two sets that is ${\text{d}}_{\text{xy}}{\text{, d}}_{\text{yz}}{\text{ and d}}_{\text{xz}}$orbitals forms one set and ${\text{d}}_{{\text{x}}^{\text{2}}{\text{-y}}^{\text{2}}}{\text{and d}}_{{\text{z}}^{\text{2}}}$ forms another set. The first set are oriented between the x, y and z axes whereas the second set gets oriented along the axis.

Electronic configuration: It is defined as the distribution of electrons present in the atom over orbitals following certain rules like electrons starts filling the lower energy orbital to higher energy, pairing of electrons does not occur until all the orbitals are singly filled and finally no electrons present in orbital can have same set of quantum numbers.

Answer to Problem 45PS

The given complex contains 4 unpaired electrons since the given complex is high spin complex where electrons present in higher energy d orbitals. Therefore, the given complex is paramagnetic.

Explanation of Solution

Examining the given complex shows that $Fe$ serves as the central metal atom since it is surrounded by 4 ligands with it. The periodic table clearly shows that the outer electronic configuration of $Fe$ is$4s^{2}3d^6$.

The oxidation state for $Fe$in given complex is $+2$ since given complex has $4Cl$ and the charge of $Cl$ is $-1$ which is showed as follows,

$\begin{array}{l}{\left[{\text{FeCl}}_{\text{4}}\right]}^{2-}\\ \text{Let x be charge of Fe}\\ \text{x+}\left(\text{-1}\times \text{4}\right)\text{= -2}\\ x-4=-2\\ \text{x = 2}\\ \text{Therefore,}\text{oxidation state of Fe is +2}\text{.}\end{array}$

Now considering the charge of metal atom the outer electrons present in the given complex is six distributed over d orbitals as follows,

$\begin{array}{l}Given\Rightarrow Highspincomplex\\ \underset{¯}{\uparrow }\underset{¯}{\uparrow }\underset{¯}{\uparrow }\\ d_{xy}{d}_{yz}{d}_{xz}\\ \underset{¯}{\uparrow \downarrow }\underset{¯}{\uparrow }\\ d_{x^2-y^2}{d}_{z^2}\end{array}$

The above configuration clearly shows that there exist 4 unpaired electrons since the given condition says that it is high spin complex where pairing of electrons does not occur and the electrons tends to present at higher level of d orbitals hence the given complex is paramagnetic.

(b)

Interpretation Introduction

Interpretation: The electronic configuration, magnetic character and the number of unpaired electrons for the given set of high spin tetrahedral complexes has to be determined.

Concept introduction: The transition metal atoms have tendency to form complex compounds that are linked to the certain neutral or ionic species which leads to the formation of coordination compounds. There exists a large number of coordination compounds that have a large number of applications in the chemical industry as well as in daily life.

The properties of the coordination compounds depend upon the primary and secondary valency of the metal ion in the coordination sphere. The electrical conductivity depends upon the number of ions that are produced by complex.

The strong-field ligands results in pairing of electrons present in the complex and leads to diamagnetic species , while the low-field ligand do not have tendency to pair up the electrons therefore forms paramagnetic species.

The strong field ligands lead to splitting to a higher extent than the weak field ligands and the wavelength of light absorbed depends on the energy gap that is produced by a particular ligand.

The five d orbitals get divided into two sets that is ${\text{d}}_{\text{xy}}{\text{, d}}_{\text{yz}}{\text{ and d}}_{\text{xz}}$orbitals forms one set and ${\text{d}}_{{\text{x}}^{\text{2}}{\text{-y}}^{\text{2}}}{\text{and d}}_{{\text{z}}^{\text{2}}}$ forms another set. The first set are oriented between the x, y and z axes whereas the second set gets oriented along the axis.

Electronic configuration: It is defined as the distribution of electrons present in the atom over orbitals following certain rules like electrons starts filling the lower energy orbital to higher energy, pairing of electrons does not occur until all the orbitals are singly filled and finally no electrons present in orbital can have same set of quantum numbers.

Answer to Problem 45PS

The given complex contains 3 unpaired electrons and it is paramagnetic.

Explanation of Solution

Examining the given complex shows that $Co$ serves as the central metal atom since it is surrounded by 4 ligands with it. The periodic table clearly shows that the outer electronic configuration of $Co$ is$4s^{2}3d^7$.

The oxidation state for $Co$in given complex is $+2$ since given complex has $4Cland2Na$ where the charge of $Cl$ is $-1$ and of $Na=+1$ which is showed as follows,

$\begin{array}{l}N{a}_2\left[\text{Co}C{l}_4\right]\\ \text{Let x be charge of Co}\\ \left(2\times 1\right)\text{+x+}\left(-1\times 4\right)\text{= 0}\\ \text{2+x- 4 = 0}\\ \text{x-2=0}\\ \text{x=2}\\ \text{Therefore,}\text{oxidation state of Co is +2}\text{.}\end{array}$

Now considering the charge of metal atom the outer electrons present in the given complex is 7 distributed over d orbitals as follows,

$\begin{array}{l}Given\Rightarrow Highspincomplex\\ \underset{¯}{\uparrow }\underset{¯}{\uparrow }\underset{¯}{\uparrow }\\ d_{xy}{d}_{yz}{d}_{xz}\\ \underset{¯}{\uparrow \downarrow }\underset{¯}{\uparrow \downarrow }\\ d_{x^2-y^2}{d}_{z^2}\end{array}$

The above configuration clearly shows that there exist 3 unpaired electrons since the given condition says that it is high spin complex where pairing of electrons does not occur and the electrons tends to present at higher level of d orbitals hence the given complex is paramagnetic.

(c)

Interpretation Introduction

Interpretation: The electronic configuration, magnetic character and the number of unpaired electrons for the given set of high spin tetrahedral complexes has to be determined.

Concept introduction: The transition metal atoms have tendency to form complex compounds that are linked to the certain neutral or ionic species which leads to the formation of coordination compounds. There exists a large number of coordination compounds that have a large number of applications in the chemical industry as well as in daily life.

The properties of the coordination compounds depend upon the primary and secondary valency of the metal ion in the coordination sphere. The electrical conductivity depends upon the number of ions that are produced by complex.

The strong-field ligands results in pairing of electrons present in the complex and leads to diamagnetic species, while the low-field ligand do not have tendency to pair up the electrons therefore forms paramagnetic species.

The strong field ligands lead to splitting to a higher extent than the weak field ligands and the wavelength of light absorbed depends on the energy gap that is produced by a particular ligand.

The five d orbitals get divided into two sets that is ${\text{d}}_{\text{xy}}{\text{, d}}_{\text{yz}}{\text{ and d}}_{\text{xz}}$orbitals forms one set and ${\text{d}}_{{\text{x}}^{\text{2}}{\text{-y}}^{\text{2}}}{\text{and d}}_{{\text{z}}^{\text{2}}}$ forms another set. The first set are oriented between the x, y and z axes whereas the second set gets oriented along the axis.

Electronic configuration: It is defined as the distribution of electrons present in the atom over orbitals following certain rules like electrons starts filling the lower energy orbital to higher energy, pairing of electrons does not occur until all the orbitals are singly filled and finally no electrons present in orbital can have same set of quantum numbers.

Answer to Problem 45PS

The given complex contains 5 unpaired electrons and it is paramagnetic.

Explanation of Solution

Examining the given complex shows that $Mn$ serves as the central metal atom since it is surrounded by 4 ligands with it. The periodic table clearly shows that the outer electronic configuration of $Mn$ is$4s^{2}3d^5$.

The oxidation state for $Mn$in given complex is $+2$ since given complex has $4Cl$ and the charge of $Cl$ is $-1$ which is showed as follows,

$\begin{array}{l}{\left[{\text{MnCl}}_4\right]}^{2-}\\ \text{Let x be charge of Mn}\\ \text{x+}\left(\text{-1}\times 4\right)\text{= -2}\\ x-4=-2\\ \text{x = 2}\\ \text{Therefore,}\text{oxidation state of Mn is +2}\text{.}\end{array}$

Now considering the charge of metal atom the outer electrons present in the given complex is five distributed over d orbitals as follows,

$\begin{array}{l}Given\Rightarrow Highspincomplex\\ \underset{¯}{\uparrow }\underset{¯}{\uparrow }\underset{¯}{\uparrow }\\ d_{xy}{d}_{yz}{d}_{xz}\\ \underset{¯}{\uparrow }\underset{¯}{\uparrow }\\ d_{x^2-y^2}{d}_{z^2}\end{array}$

The above configuration clearly shows that there exist 5 unpaired electrons since the given condition says that it is high spin complex where pairing of electrons does not occur and the electrons tends to present at higher level of d orbitals hence the given complex is paramagnetic.

(d)

Interpretation Introduction

Interpretation: The electronic configuration, magnetic character and the number of unpaired electrons for the given set of high spin tetrahedral complexes has to be determined.

Concept introduction: The transition metal atoms have tendency to form complex compounds that are linked to the certain neutral or ionic species which leads to the formation of coordination compounds. There exists a large number of coordination compounds that have a large number of applications in the chemical industry as well as in daily life.

The properties of the coordination compounds depend upon the primary and secondary valancy of the metal ion in the coordination sphere. The electrical conductivity depends upon the number of ions that are produced by complex.

The strong-field ligands results in pairing of electrons present in the complex and leads to diamagnetic species , while the low-field ligand do not have tendency to pair up the electrons therefore forms paramagnetic species.

The strong field ligands lead to splitting to a higher extent than the weak field ligands and the wavelength of light absorbed depends on the energy gap that is produced by a particular ligand.

The five d orbitals get divided into two sets that is ${\text{d}}_{\text{xy}}{\text{, d}}_{\text{yz}}{\text{ and d}}_{\text{xz}}$orbitals forms one set and ${\text{d}}_{{\text{x}}^{\text{2}}{\text{-y}}^{\text{2}}}{\text{and d}}_{{\text{z}}^{\text{2}}}$ forms another set. The first set are oriented between the x, y and z axes whereas the second set gets oriented along the axis.

Electronic configuration: It is defined as the distribution of electrons present in the atom over orbitals following certain rules like electrons starts filling the lower energy orbital to higher energy, pairing of electrons does not occur until all the orbitals are singly filled and finally no electrons present in orbital can have same set of quantum numbers.

Answer to Problem 45PS

The given complex contains 0 unpaired electron hence it is diamagnetic.

Explanation of Solution

Examining the given complex shows that $Zn$ serves as the central metal atom since it is surrounded by 4 ligands with it. The periodic table clearly shows that the outer electronic configuration of $Zn$ is$4s^{2}3d^{10}$.

The oxidation state for $Zn$in given complex is $+3$ since given complex has $4Cl$ and $2N{H}_4$ where the charge of $Cl$ is $-1$ and charge of $N{H}_4\text{}\text{}\text{}\text{}\text{}$is $+1$ which is showed as follows,

$\begin{array}{l}{\left(N{H}_4\right)}_2\left[ZnC{l}_4\right]\\ \text{Let x be charge of Zn}\\ \left(2\times -1\right)\text{+x+}\left(\text{-1}\times \text{4}\right)\text{= 0}\\ \text{x-4 =}\text{-2}\\ \text{x=4-2}\\ \text{x=2}\\ \text{Therefore,}\text{oxidation state of Zn is +2}\text{.}\end{array}$

Now considering the charge of metal atom the outer electrons present in the given complex is four distributed over d orbitals as follows,

$\begin{array}{l}Given\Rightarrow Highspincomplex\\ \underset{¯}{\uparrow \downarrow }\underset{¯}{\uparrow \downarrow }\underset{¯}{\uparrow \downarrow }\\ d_{xy}{d}_{yz}{d}_{xz}\\ \underset{¯}{\uparrow \downarrow }\underset{¯}{\uparrow \downarrow }\\ d_{x^2-y^2}{d}_{z^2}\end{array}$

The above configuration clearly shows that there are 0 unpaired electron since the given condition says that it is high spin complex where pairing of electrons does not occur and the electrons tends to present at higher level of d orbitals but the given metal has more electrons which has to be paired in order to accommodate them into the orbitals hence the given complex is diamagnetic.