6.3 Formation of Complex Compounds

Introduction to Complex Compounds

A complex compound (or coordination compound) is formed between a transition metal ion and one or more ligands.

• Ligands: These are molecules or ions that donate a lone pair of electrons to the transition metal. They act as Lewis bases (electron pair donors).
• Transition Metals: These metals accept electron pairs in their vacant orbitals, forming coordinate covalent bonds. They act as Lewis acids (electron pair acceptors).

Transition metal complexes can be positively charged (cationic), negatively charged (anionic), or neutral.

Reasons for Transition Metal Complex Formation

Transition metals readily form complex ions due to two primary reasons:

1. Small Size and High Charge Density: Transition metal ions have a relatively small size and high positive charge, allowing them to strongly attract electron-donating ligands.

   • Do You Know? Transition metal ions have high charge density, making them highly polarizing compared to s-block metals, which contributes to their higher tendency to form complexes.

2. Accessible Vacant d-orbitals: They possess low-energy 3d sub-shells with vacant d orbitals that can easily accept lone pairs of electrons from ligands and can bond to the p-orbitals of ligands.

The formation process typically involves the transition metal losing electrons to become a cation. This cation then undergoes hybridization in its vacant d orbitals to produce new orbitals of equal energy. Ligands then donate electron pairs to these hybridized vacant orbitals, forming dative (coordinate covalent) bonds and resulting in a complex ion.

Complexes of Copper(II) with Water and Chloride Ligands

The copper(II) ion ($\mathrm{C}{\mathrm{u}}_{(\mathrm{aq})}^{2+}$) has an electronic configuration of $[\text{Ar}]3{\text{d}}^9$.

i. With Water Ligands

When six water ligands approach copper(II) ions, they form six coordinate covalent bonds, typically resulting in an octahedral complex. Before bonding, the vacant orbitals of copper(II) ions undergo $\mathrm{s}{\mathrm{p}}^3{\mathrm{d}}^2$ hybridization, making six hybrid orbitals available for the six water molecules.

• Example: Hexaaquacopper(II) ion, $[\mathrm{Cu}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$.

ii. With Chloride Ligands

With chloride ligands, copper(II) ions typically form a tetrahedral complex, tetrachlorocuprate(II) ion, $[{\mathrm{CuCl}}_4{]}^{2-}$. In this case, four vacant orbitals (e.g., 4s and 4p) of copper(II) ions intermix to give four $\mathrm{s}{\mathrm{p}}^3$ hybrid orbitals. Four chloride ligands then coordinatively bond to these hybrid orbitals.

The tetrahedral geometry arises because chloride is a relatively large ligand, making it sterically unfavorable for six chloride ligands to surround the small copper(II) ion.

Complexes of Cobalt(II) ion with Water and Chloride Ligands

The condensed electronic configuration of cobalt(II) ion ($\mathrm{C}{\mathrm{o}}^{2+}$) is $[\text{Ar}]3{\text{d}}^7$.

i. With Water Ligands

Cobalt(II) forms an octahedral complex by forming dative bonds with six water ligands. The six empty orbitals available with cobalt(II) ion undergo $\mathrm{s}{\mathrm{p}}^3{\mathrm{d}}^2$ hybridization to form six coordinate bonds.

• Example: Hexaaquacobalt(II) ion, $[\mathrm{Co}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$.

ii. With Chloride Ligands

Similarly, cobalt(II) ion reacts with four chloride ligands to form a tetrahedral complex, tetrachlorocobaltate(II) ion, $[{\mathrm{CoCl}}_4{]}^{2-}$. Before bonding, four vacant orbitals of cobalt(II) ions undergo hybridization (implied $\mathrm{s}{\mathrm{p}}^3$) to accommodate the four dative bonds.

Terminology of Complex Compounds

1. Coordination Number The coordination number is the total number of coordinate covalent bonds formed by the metal ion with the ligands. The most common coordination numbers are 4 and 6, though 2 is also possible.

   • Examples:
     • $[\mathrm{Cu}({\mathrm{NH}}_3{)}_4{]}^{2+}$: Coordination number = 4
     • $[\mathrm{Ni}(\mathrm{CO}{)}_4]$: Coordination number = 4
     • ${\mathrm{Na}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6]$: Coordination number = 6
     • $[\mathrm{Co}({\mathrm{NO}}_2{)}_3({\mathrm{NH}}_3{)}_3]$: Coordination number = 6

2. Coordination Sphere The coordination sphere consists of the central metal atom or ion and its attached ligands, enclosed within square brackets. The coordination sphere can be neutral, anionic, or cationic.

   • Examples:
     • ${\mathrm{K}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6]$: $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$ is an anionic coordination sphere.
     • $[\mathrm{Cu}({\mathrm{NH}}_3{)}_4]{\mathrm{SO}}_4$: $[\mathrm{Cu}({\mathrm{NH}}_3{)}_4{]}^{2+}$ is a cationic coordination sphere.
     • $[\mathrm{Ni}(\mathrm{CO}{)}_4]$: $[\mathrm{Ni}(\mathrm{CO}{)}_4]$ is a neutral coordination sphere.

3. Charge on Coordination Sphere The charge on the coordination sphere is the algebraic sum of the oxidation state of the central transition metal ion and the total charges of all the ligands.

   • Example i: ${\mathrm{Na}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6]$
     • Oxidation state of iron $=+2$
     • Total charge on six cyanide ligands $(6\times -1)=-6$
     • Charge on coordination sphere $=(+2)+(-6)=-4$
   • Example ii: $[\mathrm{Cu}({\mathrm{NH}}_3{)}_4]{\mathrm{SO}}_4$
     • Oxidation state of copper $=+2$
     • Total charge on four ammonia ligands $(4\times 0)=0$
     • Charge on coordination sphere $=(+2)+(0)=+2$

   The charge on the coordination sphere is equal in magnitude but opposite in sign to the charge of the ions outside the coordination sphere (counter ions).

Naming Complex Compounds (IUPAC Nomenclature)

The IUPAC rules for naming complex compounds are as follows:

1. Cation before Anion: Name the cation first, followed by the anion, regardless of whether they are simple ions or complex ions.
2. Naming Coordination Sphere: Within the coordination sphere, ligands are named first in alphabetical order, followed by the transition metal ion.
3. Suffix for Anionic Complexes: If the coordination sphere is anionic, the suffix "-ate" is added to the name of the central metal atom (e.g., ferrate for iron, cuprate for copper, argentate for silver). Otherwise, the metal's name remains unchanged.
4. Prefixes for Ligand Numbers:
   • For simple ligands, use prefixes: di- (2), tri- (3), tetra- (4), penta- (5), hexa- (6).
   • For polydentate ligands or ligands with prefixes already in their name (e.g., ethylenediamine), use prefixes: bis- (2), tris- (3), tetrakis- (4), pentakis- (5).
5. Naming Ligands:
   • Anionic ligands: Suffix "-o" (e.g., ${\mathrm{Cl}}^{-}$ as chloro, ${\mathrm{OH}}^{-}$ as hydroxo, ${\mathrm{CO}}_3^{2-}$ as carbonato).
   • Neutral ligands: Usually retain their common names, but some have special names (e.g., ${\mathrm{H}}_2\mathrm{O}$ as aqua or aquo, ${\mathrm{NH}}_3$ as ammine, $\mathrm{CO}$ as carbonyl).
6. Oxidation State: The oxidation state of the central metal ion is represented by Roman numerals in parentheses immediately after the name of the metal atom.

Examples of Complex Compounds

Working Out the Charge on Complexes and Their Formulae

The total charge on a complex ion depends on the oxidation state of the metal ion and the number and nature of ligands. The oxidation state of the metal ion can be calculated from the charge on the complex ion using the following relationship: $\text{Oxidation State of Metal}=\text{Charge on Complex Ion}-\sum (\text{Charges on Ligands})$ Ions outside the coordination sphere that neutralize the complex ion are called counter ions.

Worked Examples

1. Deduce the charge on the tetraamminedichlorocobalt(III) complex ion. What is its formula?

   1. Given values:
      • Ligands: two chloride ions (${\mathrm{Cl}}^{-}$, charge $-1$ each) and four ammonia molecules (${\mathrm{NH}}_3$, charge $0$ each).
      • Central metal: Cobalt with an oxidation state of $+3$.
   2. Calculate total ligand charge: $\text{Total ligand charge}=2\times (-1)+4\times (0)=-2+0=-2$
   3. Calculate the net charge on the complex ion: $\text{Net charge}=(+3)+(-2)=+1$
   4. Determine the formula: The formula is therefore $[{\mathrm{CoCl}}_2({\mathrm{NH}}_3{)}_4{]}^{+}$.

2. Deduce the oxidation state of copper in the $[{\mathrm{CuCl}}_4{]}^{2-}$ complex ion. What is its name?

   1. Given values:
      • Complex ion charge: $-2$.
      • Ligands: four chloride ions (${\mathrm{Cl}}^{-}$, charge $-1$ each).
   2. Calculate total ligand charge: $\text{Total ligand charge}=4\times (-1)=-4$
   3. Calculate the oxidation state of copper: Let the oxidation state of copper be $x$. $x+\text{Total ligand charge}=\text{Complex ion charge}$ $x+(-4)=-2$ $x=-2+4=+2$ The oxidation state of copper is $+2$.
   4. Determine the name: The complex ion is an anion with copper in the $+2$ oxidation state and four chloro ligands, so its name is the tetrachlorocuprate(II) ion.

Possible Questions/Answers

Concept-Assessment Exercise 6.1

1. Determine the oxidation states of transition metals in the following complexes. i. $[\mathrm{Co}({\mathrm{NH}}_3{)}_6]{\mathrm{Cl}}_3$

   • A: $\mathrm{Cl}$ is $-1$, so $3\mathrm{Cl}$ is $-3$. The complex ion $[\mathrm{Co}({\mathrm{NH}}_3{)}_6{]}^{3+}$ has a $+3$ charge. ${\mathrm{NH}}_3$ is neutral (0 charge). Thus, $\text{Co}+6(0)=+3\Rightarrow \text{Co}=+3$. ii. ${\mathrm{Na}}_2[{\mathrm{MnCl}}_5]$
   • A: $\mathrm{Na}$ is $+1$, so $2\mathrm{Na}$ is $+2$. The complex ion $[{\mathrm{MnCl}}_5{]}^{2-}$ has a $-2$ charge. $\mathrm{Cl}$ is $-1$, so $5\mathrm{Cl}$ is $-5$. Thus, $\text{Mn}+5(-1)=-2\Rightarrow \text{Mn}=+3$. iii. $[\mathrm{Co}(\text{en}{)}_2({\mathrm{H}}_2\mathrm{O}{)}_2{]}^{3+}$
   • A: $\text{en}$ is neutral (0 charge). ${\mathrm{H}}_2\mathrm{O}$ is neutral (0 charge). The overall charge is $+3$. Thus, $\text{Co}+2(0)+2(0)=+3\Rightarrow \text{Co}=+3$. iv. $[{\mathrm{CuCl}}_4{]}^{2-}$
   • A: $\mathrm{Cl}$ is $-1$, so $4\mathrm{Cl}$ is $-4$. The complex ion has a $-2$ charge. Thus, $\text{Cu}+4(-1)=-2\Rightarrow \text{Cu}=+2$. v. ${\mathrm{Na}}_4[\mathrm{Fe}(\mathrm{CN}{)}_2(\mathrm{OH}{)}_4]$
   • A: $\mathrm{Na}$ is $+1$, so $4\mathrm{Na}$ is $+4$. The complex ion $[\mathrm{Fe}(\mathrm{CN}{)}_2(\mathrm{OH}{)}_4{]}^{4-}$ has a $-4$ charge. $\mathrm{CN}$ is $-1$, so $2\mathrm{CN}$ is $-2$. $\mathrm{OH}$ is $-1$, so $4\mathrm{OH}$ is $-4$. Thus, $\text{Fe}+2(-1)+4(-1)=-4\Rightarrow \text{Fe}-6=-4\Rightarrow \text{Fe}=+2$.

2. Name the following complex ions. i. $[\mathrm{Co}({\mathrm{NH}}_3{)}_4{\mathrm{Cl}}_2]\mathrm{Cl}$

   • A: Tetraamminedichlorocobalt(III) chloride. ii. $[\mathrm{Zn}(\mathrm{OH}{)}_4{]}^{2-}$
   • A: Tetrahydroxozincate(II) ion. iii. $[{\mathrm{CrCl}}_2({\mathrm{NH}}_3{)}_4{]}^{+}$
   • A: Tetraamminedichlorochromium(III) ion.