Chapter 26, Problem 26.49P

(a)

Interpretation Introduction

Interpretation:

Complex ion ${[{\text{Fe(CN)}}_6]}^{4-}$ having no unpaired electrons has to be classified as high-spin or low-spin complex.

Explanation of Solution

Given complex ion is ${[{\text{Fe(CN)}}_6]}^{4-}$.  Cyanide ligands possess a negative charge.  The oxidation state of iron can be calculated as shown below.  Let the charge of iron atom be $x$.

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

Given complex contains $\text{Fe(II)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of iron is $26$.  Therefore, $\text{Fe(II)}$ will have $24$ 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;

  $\text{Fe}:[\text{Ar}]4s^2 3d^6$

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

  $\text{Fe}:[\text{Ar}]4s^2 3d^6\to \text{Fe(II)}:[\text{Ar}]4s^0 3d^6+2e^{-}$

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, ${[{\text{Fe(CN)}}_{\text{6}}]}^{4-}$ is a low-spin complex.

(b)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

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

Given complex contains $\text{Fe(III)}$ as metal ion.  The Roman numeral that is present in the parenthesis indicates the oxidation state of the metal ion.  Atomic number of iron is $26$.  Therefore, $\text{Fe(III)}$ will have $23$ 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;

  $\text{Fe}:[\text{Ar}]4s^2 3d^6$

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

  $\text{Fe}:[\text{Ar}]4s^2 3d^6\to \text{Fe(III)}:[\text{Ar}]4s^0 3d^5+3e^{-}$

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 all the five $d$ electrons are present in $t_{2g}$ orbital.  Therefore, ${[{\text{Fe(CN)}}_{\text{6}}]}^{2-}$ is a low-spin complex.

(c)

Interpretation Introduction

Interpretation:

Complex ion ${[{\text{Co(NH}}_{\text{3}}{\text{)}}_6]}^{2+}$ having three unpaired electrons has to be classified as high-spin or low-spin complex.

Explanation of Solution

Given complex ion is ${[{\text{Co(NH}}_{\text{3}}{\text{)}}_6]}^{2+}$.  Ammonia molecules do not possess any charge.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt atom be $x$.

    $\begin{array}{c}x+6(0)=+2\\ x=+2\end{array}$

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 cobalt is given as shown below;

  $\text{Co}:[\text{Ar}]4s^2 3d^7$

Electronic configuration of $\text{Co(II)}$ 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 the five $d$ electrons are present in $t_{2g}$ orbital and two $d$ electrons are present in $e_g$ orbital.  Therefore, ${[{\text{Co(NH}}_{\text{3}}{\text{)}}_6]}^{2+}$ is a high-spin complex.

(d)

Interpretation Introduction

Interpretation:

Complex ion ${[{\text{CoF}}_6]}^{3-}$ having four unpaired electrons has to be classified as high-spin or low-spin complex.

Explanation of Solution

Given complex ion is ${[{\text{CoF}}_6]}^{3-}$.  Fluoride ion possesses a negative charge.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt atom be $x$.

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

Given complex contains $\text{Co(III)}$ 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, $\text{Co(III)}$ will have $24$ 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;

  $\text{Co}:[\text{Ar}]4s^2 3d^7$

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

  $\text{Co}:[\text{Ar}]4s^2 3d^7\to \text{Co(III)}:[\text{Ar}]4s^0 3d^6+3e^{-}$

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 four unpaired electrons.  This means the four $d$ electrons are present in $t_{2g}$ orbital and two $d$ electrons are present in $e_g$ orbital.  Therefore, ${[{\text{CoF}}_6]}^{3-}$ is a high-spin complex.

(e)

Interpretation Introduction

Interpretation:

Complex ion ${[{\text{Mn(H}}_{\text{2}}{\text{O)}}_6]}^{2+}$ having five unpaired electrons has to be classified as high-spin or low-spin complex.

Explanation of Solution

Given complex ion is ${[{\text{Mn(H}}_{\text{2}}{\text{O)}}_6]}^{2+}$.  Water molecules do not possess any charge.  The oxidation state of manganese can be calculated as shown below.  Let the charge of manganese atom be $x$.

    $\begin{array}{c}x+6(0)=+2\\ x=+2\end{array}$

Given complex contains $\text{Mn(II)}$ 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, $\text{Mn(II)}$ will have $23$ 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 $\text{Mn(II)}$ is given as shown below;

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

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 five unpaired electrons.  This means all orbitals are single filled before pairing takes place and this happens in case of high-spin complex only.  Therefore, ${[{\text{Mn(H}}_{\text{2}}{\text{O)}}_6]}^{2+}$ is a high-spin complex.