Chapter 26, Problem 26.61P

(a)

Interpretation Introduction

Interpretation:

The complex ion ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$ has more number of $d$ electrons than ${[{\text{CoCl}}_{\text{6}}]}^{4-}$ has to be indicated as true or false.

Explanation of Solution

Complex ion ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$:

Given complex ion is ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$.  Cyanide ion possess a negative charge.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt ion 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.

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.

Complex ion ${[{\text{CoCl}}_{\text{6}}]}^{4-}$:

Given complex ion is ${[{\text{CoCl}}_{\text{6}}]}^{4-}$.  Chloride ion possess a negative charge.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt ion be $x$.

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

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.

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;

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

Therefore, there are seven electrons in $d$-orbital.

Comparing the $d$ electrons in ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$ and ${[{\text{CoCl}}_{\text{6}}]}^{4-}$, it is found that ${[{\text{CoCl}}_{\text{6}}]}^{4-}$ contains more $d$ electrons than ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$.

Therefore, the given statement is false.

(b)

Interpretation Introduction

Interpretation:

The complex ion ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$ has same number of $d$ electrons than ${[{\text{CoCl}}_{\text{6}}]}^{4-}$ has to be indicated as true or false.

Explanation of Solution

Complex ion ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$:

Given complex ion is ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$.  Cyanide ion possess a negative charge.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt ion 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.

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.

Complex ion ${[{\text{CoCl}}_{\text{6}}]}^{4-}$:

Given complex ion is ${[{\text{CoCl}}_{\text{6}}]}^{4-}$.  Chloride ion possess a negative charge.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt ion be $x$.

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

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.

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;

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

Therefore, there are seven electrons in $d$-orbital.

Comparing the $d$ electrons in ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$ and ${[{\text{CoCl}}_{\text{6}}]}^{4-}$, it is found that they both do not contain same number of $d$ electrons.

Therefore, the given statement is false.

(c)

Interpretation Introduction

Interpretation:

The complex ion ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$ is paramagnetic while ${[{\text{CoCl}}_{\text{6}}]}^{4-}$ is diamagnetic has to be indicated as true or false.

Explanation of Solution

Complex ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$:

Given complex ion is ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$.  Cyanide ion possess a negative charge.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt ion 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.  Cyanide ligand is a strong field ligand and hence it results in formation of low-spin complex.  In low spin complex, the electron pairing takes place in $t_{2g}$ orbital before $e_g$ orbital is filled singly.  Therefore, the $d$-orbital diagram of $\text{Co(III)}$ can be drawn as shown below.

Therefore, there are no unpaired electrons in ${{\text{[Co(CN)}}_6]}^{3-}$ complex ion.  Hence, it is diamagnetic.

Complex ${[{\text{CoCl}}_{\text{6}}]}^{4-}$:

Given complex ion is ${[{\text{CoCl}}_{\text{6}}]}^{4-}$.  Chloride ion possess a negative charge.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt ion be $x$.

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

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 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.  Chloride ligand is a weak field ligand and hence it results in formation of high-spin complex.  In high spin complex, the electron pairing takes place in $t_{2g}$ orbital after $e_g$ orbital is filled singly.  Therefore, the $d$-orbital diagram of $\text{Co(II)}$ can be drawn as shown below.

Therefore, there are three unpaired electrons in ${{\text{[CoCl}}_6]}^{4-}$ complex ion.  Hence, it is paramagnetic.

Thus the given statement is false.

(d)

Interpretation Introduction

Interpretation:

The complex ion ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$ is diamagnetic while ${[{\text{CoCl}}_{\text{6}}]}^{4-}$ is paramagnetic has to be indicated as true or false.

Explanation of Solution

Complex ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$:

Given complex ion is ${[{\text{Co(CN)}}_{\text{6}}]}^{3-}$.  Cyanide ion possess a negative charge.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt ion 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.  Cyanide ligand is a strong field ligand and hence it results in formation of low-spin complex.  In low spin complex, the electron pairing takes place in $t_{2g}$ orbital before $e_g$ orbital is filled singly.  Therefore, the $d$-orbital diagram of $\text{Co(III)}$ can be drawn as shown below.

Therefore, there are no unpaired electrons in ${{\text{[Co(CN)}}_6]}^{3-}$ complex ion.  Hence, it is diamagnetic.

Complex ${[{\text{CoCl}}_{\text{6}}]}^{4-}$:

Given complex ion is ${[{\text{CoCl}}_{\text{6}}]}^{4-}$.  Chloride ion possess a negative charge.  The oxidation state of cobalt can be calculated as shown below.  Let the charge of cobalt ion be $x$.

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

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 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.  Chloride ligand is a weak field ligand and hence it results in formation of high-spin complex.  In high spin complex, the electron pairing takes place in $t_{2g}$ orbital after $e_g$ orbital is filled singly.  Therefore, the $d$-orbital diagram of $\text{Co(II)}$ can be drawn as shown below.

Therefore, there are three unpaired electrons in ${{\text{[CoCl}}_6]}^{4-}$ complex ion.  Hence, it is paramagnetic.

Thus the given statement is true.