Chapter 23, Problem 23.22QA

Interpretation Introduction

To write:

The names of coordination compounds of chromium(III) given in the problem 23.20.

Answer to Problem 23.22QA

Solution:

a) Hexaaquachromium(III) chloride

b) Pentaaquachlorochromium(III) chloride

c) Tetraaquadichlorochromium(III) chloride

Explanation of Solution

Based on the information given in Problem 23.20, the formulas of the chromium(III) coordination compounds are:

a) $[Cr{\left(H_{2 }O\right)}_6{]Cl}_3$

b) $[Cr{\left(H_{2 }O\right)}_5{Cl]Cl}_2$

c) $\left[Cr{\left(H_{2 }O\right)}_{4}C{l}_2\right]{Cl}_{ }$

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
$[Cr{\left(H_{2 }O\right)}_6{]Cl}_3$	$C{l}^{-}$	$H_{2}O$	aqua	$6$	Hexa-
$[Cr{\left(H_{2 }O\right)}_5{Cl]Cl}_2$	$C{l}^{-}$	$H_{2}O$	aqua	$5$	Penta-
		$C{l}^{-}$	chloro	1	-
$\left[Cr{\left(H_{2 }O\right)}_{4}C{l}_2\right]{Cl}_{ }$	$C{l}^{-}$	$H_{2}O$	aqua	$4$	Tetra-
		$C{l}^{-}$	chloro	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: $\left[Cr{\left(H_{2 }O\right)}_6\right]{Cl}_3$

$\left(1 Cr ion\right)+\left(6 H_{2}O\right)+\left(3 C{l}^{-} ions\right)=0$

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

$x=+3$

Oxidation state of Cr is $+3$.

The complex ion is the cation in this compound. We begin by naming the ligand directly attached to the metal ion: aqua. The number of ligands is indicated by the prefix as hexaaqua. The name of the metal indicating its oxidation state with a Roman numeral is written as: hexaaquachromium(III). The name of the counter ion is chloride.

So, the full name of compound is hexaaquachromium(III) chloride.

b) Oxidation state of central metal ion:$\left[Cr{\left(H_{2 }O\right)}_{5}Cl\right]{Cl}_2$

$\left(1 Cr ion\right)+\left(5 H_{2}O\right)+\left(3 C{l}^{-} ions\right)=0$

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

$x=+3$

Oxidation state of Cr is $+3$.

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

So, the full name of compound is pentaaquachlorochromium(III) chloride.

c) Oxidation state of central metal ion:$\left[Cr{\left(H_{2 }O\right)}_{4}C{l}_2\right]Cl$

$\left(1 Cr ion\right)+\left(4 H_{2}O\right)+\left(3 C{l}^{-} ions\right)=0$

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

$x=+3$

Oxidation state of Cr is $+3$.

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

So, the full name of compound is tetraaquadichlorochromium(III) chloride.

Conclusion:

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