Chapter 22, Problem 22.86QP

(a)

Interpretation Introduction

Interpretation:

Name of the low spin complex ions ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$ and ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$ has  to be determined.

Concept Introduction:

Complex compounds are primarily formed by transition elements which are d-block elements.  A co-ordination compound is known as a complex compound is made of metal atom/ion and ligand(s).  Ligands are considered as Lewis bases and the central metal atom is Lewis acid. Ligands donate a pair of electron to metal ion and establishes bonding with metal ion which is known as co-ordinate bond and hence these compounds are named as co-ordination compounds.  The ligands represented inside the square brackets adjacent to the central metal ion, which is termed as co-ordination sphere.

IUPAC has devised rules to name a co-ordination compound in chemical terminology.  The rules are stated below -

• Name of the cation must be written before the name of the anion.
• Metal ions are named first followed by ligand names.
• If the ligand is negatively charged, the name of the ligand carries suffix 'o'.  example - $CN$ is a negative ligand and it must be named as 'cyano' not cyanide.
• Number of simple ligands present in a co-ordination compound is denoted by prefixes - mono-, di-, tetra-, etc., so on.
• The prefixes bis-, tris-, tetrakis,- etc are used to denote the number of complicated ligands.
• Oxidation state of the central metal ion must be mentioned using a Roman numeral in parentheses.
• The ligands are named in alphabetical order if a complex contains more than one ligand.
• The suffix 'ate' is added to the name of the central metal atom in case of negatively charged complex ion.  While naming some of the metals, their Latin name is preferred.

Answer to Problem 22.86QP

Name of the complex ion ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$ is hexaaquachromium(III)ion.

Name of the complex ion ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$ is hexacyanomanganate(II)

Explanation of Solution

Oxidation state of Chromium in the complex ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$,

$\begin{array}{l}oxidation\text{ state of Cr}\text{= charge on complex - charge of ligands}\\ \text{= +3-6}\left(0\right)\text{ = +3}\end{array}$

In the complex ion ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$ six aqua ligands are co-ordinated to $C{r}^{3+}$ ion that the term ''hexa'' has to be used to name the ligand.  Chromium is in $+3$ oxidation state and the complex is ionic and so the name of the complex according to IUPAC conventions is hexaaquachromium(III)ion.

Oxidation state of Manganese in the complex ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$,

$\begin{array}{l}oxidation\text{ state of Mn}\text{= charge on complex - charge of ligands}\\ \text{= -4-6}\left(-1\right)\text{ = +2}\end{array}$

In the complex ion ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$ six cyano ligands are co-ordinated to $M{n}^{2+}$ ion that the term ''hexa'' has to be used to name the ligand.  Manganese is in $+2$ oxidation state and the complex is negatively charged so that the name of the metal should end with ''ate''.  So the name of the complex according to IUPAC conventions is hexacyanomanganate(II).

(b)

Interpretation Introduction

Interpretation:

Number of unpaired electrons in the low spin complex ions ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$ and ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$ has  to be determined.

Concept Introduction:

Complex compounds are primarily formed by transition elements which are d-block elements.  A co-ordination compound is known as a complex compound is made of metal atom/ion and ligand(s).  Ligands are considered as Lewis bases and the central metal atom is Lewis acid. Ligands donate a pair of electron to metal ion and establishes bonding with metal ion which is known as co-ordinate bond and hence these compounds are named as co-ordination compounds.  The ligands represented inside the square brackets adjacent to the central metal ion, which is termed as co-ordination sphere.

Answer to Problem 22.86QP

Number of unpaired electrons in ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$  is three.

Number of unpaired electrons in ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$  is one.

Explanation of Solution

As per spectrochemical series aqua ligand is a weak ligand and so the complex ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$ would be high-spin complex.  Chromium is in $+3$ oxidation state.  Electronic configuration of Chromium is $\left[Ar\right]3d^{5}4s^1$.  So electronic configuration of $C{r}^{3+}$ would be $\left[Ar\right]3d^{3}4s^0$.  Thus there three d-electrons in the complex and these are unpaired electrons.  Cyano ligand is a strong ligand  that ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$ would be low-spin complex.  Manganese is in  $+2$ oxidation state.  Electronic configuration of Manganese is $\left[Ar\right]3d^{5}4s^2$.  So electronic configuration of $M{n}^{2+}$ would be $\left[Ar\right]3d^5$.  Thus there five d-electrons in the complex and these are paired up as cyano ligand is stronger ligand.  Thus there would be one unpaired electron in the complex.

(c)

Interpretation Introduction

Interpretation:

Among the two complex ions ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{3+}$ and ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$ the complex ion that absorb the highest frequency light has to be identified.

Concept Introduction:

• Spectrochemical series refers to the arrangement of ligands with respect to crystal field splitting they cause in the d-orbitals of a metal ion. 
• The ligands are arranged in increasing order of crystal field splitting they produce.

  $\begin{array}{l}{I}^{-}<B{r}^{-}<S^{2-}<SC{N}^{-}<C{l}^{-}<N{O}_3{}^{-}<N_3{}^{-}<F^{-}<O{H}^{-}<C_2{O}_4{}^{2-}<H_{2}O<NC{S}^{-}\\ <C{H}_{3}CN<py<N{H}_3<en<bipy<phen<N{O}_2{}^{-}<PP{h}_3<C{N}^{-}<CO\end{array}$

• The ligand that causes larger crystal field splitting is strong ligand and has higher crystal field splitting energy.  The ligand that causes smaller splitting is termed as weak ligand and has lower crystal field splitting energy.  If the complex has lower  crystal field splitting energy it would absorb in lower frequency region as frequency and energy are directly proportional to each other.  If the complex has higher crystal field splitting energy it would absorb in higher frequency region.

Answer to Problem 22.86QP

The complex ion ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$ would absorb in highest frequency light.

Explanation of Solution

According to spectrochemical series, cyano ligands are stronger than aqua ligands.  Thus cyano ligands have higher crystal field splitting energy that the complex ${\left[Mn{\left(CN\right)}_6\right]}^{4-}$ absorbs in higher frequency light.