Chapter 23, Problem 23.28QA

Interpretation Introduction

To find:

The name of the co-ordination compounds.

a) ${\left(N{H}_4\right)}_3\left[Co{\left(CN\right)}_6\right]$

b) $\left[Co{\left(en\right)}_{2}Cl\right]{\left(N{O}_3\right)}_2$

c) $\left[Fe{\left(H_{2}O\right)}_4{\left(OH\right)}_2\right]Cl$

Answer to Problem 23.28QA

Solution:

a) Ammonium hexacyanocobaltate(III).

b) Bis(ethylenediamine) chlorocobalt(III) nitrate

c) Tetraaquadihydroxoiron(III) chloride

Explanation of Solution

When writing the name of the coordination compound, if the counter ion of the complex ion is a cation, the cation’s name goes first, followed by the name of the anionic complex ion.

If the counter ion of the complex ion is an anion the name of the cationic complex ion goes first, followed by the name of the anion.

Taking an inventory of the ligands and counter ions in all the three compounds:

Compound		Ligand
	Counter ion	Formula	Name	Number	Prefix
${\left(N{H}_4\right)}_3\left[Co{\left(CN\right)}_6\right]$	$N{H}_4^{+}$	$C{N}^{-}$	cyano	$6$	hexa
$\left[Co{\left(en\right)}_{2}Cl\right]{\left(N{O}_3\right)}_2$	$N{O}_3^{-}$	$en$	ethylenediamine	$2$	bis
		$C{l}^{-}$	chloro	$1$	-
$\left[Fe{\left(H_{2}O\right)}_4{\left(OH\right)}_2\right]Cl$	$C{l}^{-}$	$H_{2}O$	aqua	$4$	tetra
		$O{H}^{-}$	hydroxo	$2$	di

The oxidation state of the central metal ion is calculated by setting the sum of the charges on all the ions in the compound equal to zero.

a) Oxidation state of central metal ion in ${\left(N{H}_4\right)}_3\left[Co{\left(CN\right)}_6\right]$:

$\left(3 N{H}_4^{+} ions\right)+\left(1 Co ion\right)+\left(6 C{N}^{-} ions\right)=0$

$\left(+3\right)+\left(x\right)+\left(-6\right)=0$

$x=+3$

Oxidation state of $Co$ is $+3$.

Because the counter ion, ammonium, is a cation, its name comes first. The complex ion

is an anion. To name it, we begin with the ligand cyano, to which we add the prefix

hexaand write hexacyano. This is followed by the name of the transition metal ion:

hexacyanocobalt. We add -ate to the ending of the name of the complex ion because

it is an anion: hexacyanocobaltate. We then add a Roman numeral to indicate the oxidation state of the cobalt: hexacyanocobaltate(III).

So, the full name of this co-ordination compound is ammonium hexacyanocobaltate(III).

b) Oxidation state of central metal ion in $\left[Co{\left(en\right)}_{2}Cl\right]{\left(N{O}_3\right)}_2$

$\left(1 Co ion\right)+\left(2 en\right)+\left(1 C{l}^{-} ion\right)+\left(2 N{O}_3^{-} ions\right)=0$

$\left(x\right)+0+\left(-1\right)+\left(-2\right)=0$

$x-3=0$

$x=+3$

Oxidation state of $Co$ is $+3$.

The complex ion is the cation in this compound. We begin by naming the ligands directly attached to the metal ion in alphabetical order: ethylenediamine and chloro. The number of ligands is indicated by the prefixes as bis(ethylenediamine) and chloro. The name of the metal indicating its oxidation state with a Roman numeral is written as: bis(ethylenediamine) chlorocobalt(III).  The name of the counter ion is nitrate

So, the full name of this co-ordination compound is

bis(ethylenediamine) chlorocobalt(III) nitrate.

c) Oxidation state of central metal ion in $\left[Fe{\left(H_{2}O\right)}_4{\left(OH\right)}_2\right]Cl$

$\left(1 Fe ion\right)+\left(4 H_{2}O\right)+\left(2 O{H}^{-} ions\right)+\left(1 C{l}^{-} ion\right)=0$

$\left(x\right)+0+\left(-2\right)+\left(-1\right)=0$

$x-3=0$

$x=+3$

Oxidation state of $Fe$ is $+3$

The complex ion is the cation in this compound. We begin by naming the ligands directly attached to the metal ion in alphabetical order: aqua and hydroxo. The number of ligands is indicated by the prefixes as tetraaqua and dihydroxo. The name of the metal indicating its oxidation state with a Roman numeral is written as: tetraaquadihydroxoiron(III). The name of the counter ion is chloride.

So, the full name of this co-ordination compound is tetraaquadihydroxoiron(III) chloride.

Conclusion:

The names of the given coordination compounds are written using the naming rules.