Chapter 26, Problem 26.50P

(a)

Interpretation Introduction

Interpretation:

Complex ion ${[{\text{Mn(NH}}_{\text{3}}{\text{)}}_6]}^{3+}$ having two unpaired electrons has to be classified as high-spin or low-spin complex.

Explanation of Solution

Given complex ion is ${[{\text{Mn(NH}}_{\text{3}}{\text{)}}_6]}^{3+}$.  Ammonia molecules do not possess any charge.  The oxidation state of manganese can be calculated as shown below.  Let the charge of manganese atom be $x$.

    $\begin{array}{c}x+6(0)=+3\\ x=+3\end{array}$

Given complex contains $\text{Mn(III)}$ 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, $\text{Mn(III)}$ will have $22$ electrons.

In the octahedral complex, the $d$ orbitals splits into two energy levels and they are $t_{2g}$ and $e_g$.  In this $t_{2g}$ is the lower energy degenerate orbital and $e_g$ is the higher energy degenerate orbital.

Electronic configuration of cobalt is given as shown below;

  $\text{Mn}:[\text{Ar}]4s^2 3d^5$

Electronic configuration of $\text{Mn(III)}$ is given as shown below;

  $\text{Mn}:[\text{Ar}]4s^2 3d^5\to \text{Mn(III)}:[\text{Ar}]4s^0 3d^4+3e^{-}$

Therefore, there are four electrons in $d$-orbital.  If the ${\Delta }_0$ is greater than the electron pairing energy, then the electrons will enter $t_{2g}$ orbital and pairing starts.  If the ${\Delta }_0$ is lesser than the electron pairing energy, then the electrons will enter $t_{2g}$ orbital and $e_g$ orbital until all are singly occupied and then the pairing starts.

Given complex is said to contain two unpaired electrons.  This means that the pairing of electrons takes place in $t_{2g}$ orbital before filling of $e_g$ orbital.  Therefore, ${[{\text{Mn(NH}}_{\text{3}}{\text{)}}_6]}^{3+}$ is a low-spin complex.

(b)

Interpretation Introduction

Interpretation:

Complex ion ${[{\text{Rh(CN)}}_6]}^{3-}$ having no unpaired electrons has to be classified as high-spin or low-spin complex.

Explanation of Solution

Given complex ion is ${[{\text{Rh(CN)}}_6]}^{3-}$.  Cyanide ligands possess a negative charge.  The oxidation state of iron can be calculated as shown below.  Let the charge of rhodium atom be $x$.

    $\begin{array}{c}x+6(-1)=-3\\ x-6=-3\\ x=+3\end{array}$

Given complex contains $\text{Rh(III)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of rhodium is $45$.  Therefore, $\text{Rh(III)}$ will have $42$ electrons.

In the octahedral complex, the $d$ orbitals splits into two energy levels and they are $t_{2g}$ and $e_g$.  In this $t_{2g}$ is the lower energy degenerate orbital and $e_g$ is the higher energy degenerate orbital.

Electronic configuration of rhodium is given as shown below;

  $\text{Rh}:[\text{Kr}]5s^1 4d^8$

Electronic configuration of $\text{Rh(III)}$ is given as shown below;

  $\text{Rh}:[\text{Kr}]5s^1 4d^8\to \text{Rh(III)}:[\text{Kr}]5s^0 4d^6+3e^{-}$

Therefore, there are six electrons in $d$-orbital.  If the ${\Delta }_0$ is greater than the electron pairing energy, then the electrons will enter $t_{2g}$ orbital and pairing starts.  If the ${\Delta }_0$ is lesser than the electron pairing energy, then the electrons will enter $t_{2g}$ orbital and $e_g$ orbital until all are singly occupied and then the pairing starts.

Given complex is said to contain no unpaired electrons.  This means all the six $d$ electrons are present in $t_{2g}$ orbital.  Therefore, ${[{\text{Rh(CN)}}_{\text{6}}]}^{3-}$ is a low-spin complex.

(c)

Interpretation Introduction

Interpretation:

Complex ion ${[{\text{Co(C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{)}}_3]}^{4-}$ having three unpaired electrons has to be classified as high-spin or low-spin complex.

Explanation of Solution

Given complex ion is ${[{\text{Co(C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{)}}_3]}^{4-}$.  Oxalate ion possess two negative charges.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt atom be $x$.

    $\begin{array}{c}x+3(-2)=-4\\ x-6=-4\\ x=+2\end{array}$

Given complex contains $\text{Co(II)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of cobalt is $27$.  Therefore, $\text{Co(II)}$ will have $25$ electrons.

In the octahedral complex, the $d$ orbitals splits into two energy levels and they are $t_{2g}$ and $e_g$.  In this $t_{2g}$ is the lower energy degenerate orbital and $e_g$ is the higher energy degenerate orbital.

Electronic configuration of iron is given as shown below;

  $\text{Co}:[\text{Ar}]4s^2 3d^7$

Electronic configuration of $\text{Fe(III)}$ is given as shown below;

  $\text{Co}:[\text{Ar}]4s^2 3d^7\to \text{Co(II)}:[\text{Ar}]4s^0 3d^7+2e^{-}$

Therefore, there are seven electrons in $d$-orbital.  If the ${\Delta }_0$ is greater than the electron pairing energy, then the electrons will enter $t_{2g}$ orbital and pairing starts.  If the ${\Delta }_0$ is lesser than the electron pairing energy, then the electrons will enter $t_{2g}$ orbital and $e_g$ orbital until all are singly occupied and then the pairing starts.

Given complex is said to contain three unpaired electrons.  This means all the orbitals are singly filled before pairing takes place.  Therefore, ${[{\text{Co(C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{)}}_3]}^{4-}$ is a high-spin complex.

(d)

Interpretation Introduction

Interpretation:

Complex ion ${[{\text{IrBr}}_6]}^{4-}$ having three unpaired electrons has to be classified as high-spin or low-spin complex.

Explanation of Solution

Given complex ion is ${[{\text{IrBr}}_6]}^{4-}$.  Bromide ion possess a negative charge.  The oxidation state of iridium can be calculated as shown below.  Let the charge of iridum ion be $x$.

    $\begin{array}{c}x+6(-1)=-4\\ x=+2\end{array}$

Given complex contains $\text{Ir(II)}$ 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, $\text{Ir(II)}$ will have $75$ electrons.

In the octahedral complex, the $d$ orbitals splits into two energy levels and they are $t_{2g}$ and $e_g$.  In this $t_{2g}$ is the lower energy degenerate orbital and $e_g$ is the higher energy degenerate orbital.

Electronic configuration of cobalt is given as shown below;

  $\text{Ir}:[\text{Xe}]6s^2 4f^{14} 5d^7$

Electronic configuration of $\text{Ir(II)}$ is given as shown below;

  $\text{Ir}:[\text{Xe}]6s^2 4f^{14} 5d^7\to \text{Ir(II)}:[\text{Xe}]6s^0 4f^{14} 5d^7+2e^{-}$

Therefore, there are seven electrons in $d$-orbital.  If the ${\Delta }_0$ is greater than the electron pairing energy, then the electrons will enter $t_{2g}$ orbital and pairing starts.  If the ${\Delta }_0$ is lesser than the electron pairing energy, then the electrons will enter $t_{2g}$ orbital and $e_g$ orbital until all are singly occupied and then the pairing starts.

Given complex is said to contain three unpaired electrons.  This means the five $d$ electrons are present in $t_{2g}$ orbital and two $d$ electrons are present in $e_g$ orbital.  Therefore, ${[{\text{IrBr}}_6]}^{4-}$ is a high-spin complex.

(e)

Interpretation Introduction

Interpretation:

Complex ion ${[{\text{Ru(NH}}_{\text{3}}{\text{)}}_6]}^{3+}$ having one unpaired electron has to be classified as high-spin or low-spin complex.

Explanation of Solution

Given complex ion is ${[{\text{Ru(NH}}_{\text{3}}{\text{)}}_6]}^{3+}$.  Ammonia molecules do not possess any charge.  The oxidation state of ruthenium can be calculated as shown below.  Let the charge of ruthenium ion be $x$.

    $\begin{array}{c}x+6(0)=+3\\ x=+3\end{array}$

Given complex contains $\text{Ru(III)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of ruthenium is $44$.  Therefore, $\text{Ru(III)}$ will have $41$ electrons.

In the octahedral complex, the $d$ orbitals splits into two energy levels and they are $t_{2g}$ and $e_g$.  In this $t_{2g}$ is the lower energy degenerate orbital and $e_g$ is the higher energy degenerate orbital.

Electronic configuration of cobalt is given as shown below;

  $\text{Ru}:[\text{Kr}]5s^1 4d^7$

Electronic configuration of $\text{Ru(III)}$ is given as shown below;

  $\text{Ru}:[\text{Kr}]5s^1 4d^7\to \text{Ru(III)}:[\text{Kr}]5s^0 4d^5+3e^{-}$

Therefore, there are five electrons in $d$-orbital.  If the ${\Delta }_0$ is greater than the electron pairing energy, then the electrons will enter $t_{2g}$ orbital and pairing starts.  If the ${\Delta }_0$ is lesser than the electron pairing energy, then the electrons will enter $t_{2g}$ orbital and $e_g$ orbital until all are singly occupied and then the pairing starts.

Given complex is said to contain one unpaired electron.  This means the five $d$ electrons are present in $t_{2g}$ orbital.  Therefore, ${[{\text{Ru(NH}}_{\text{3}}{\text{)}}_6]}^{3+}$ is a low-spin complex.