Coordination Compounds

According to Werner’s postulates, a metal in a coordination compound shows two kinds of valence:

1. Primary valence = oxidation state of the metal. It is satisfied by anions and is ionisable.
2. Secondary valence = coordination number. It is satisfied by anions or neutral molecules directly attached to the metal and is non-ionisable.

The groups satisfying secondary valence remain within the coordination sphere (inside square brackets) and are arranged in a definite spatial geometry such as octahedral, tetrahedral or square planar.

So, bonding in coordination compounds is explained as:
- primary valence bonds with counter ions outside the bracket,
- secondary valence bonds with ligands inside the bracket,
- and the secondary valences give a fixed geometry to the complex.

$\mathrm{FeSO_4}$ mixed with $(\mathrm{NH_4})_2\mathrm{SO_4}$ gives the test of $\mathrm{Fe^{2+}}$ because this is a double salt. Double salts dissociate completely in water into simple ions, so $\mathrm{Fe^{2+}}$ is free in solution and gives its normal test.

But $\mathrm{CuSO_4}$ mixed with aqueous ammonia forms a complex. In a complex, the metal ion is not present as free $\mathrm{Cu^{2+}}$; instead it remains inside the coordination entity, so it does not give the ordinary test for $\mathrm{Cu^{2+}}$.

Thus:
- Double salt → dissociates completely into ions in water.
- Complex → does not dissociate into the central metal ion and ligands completely.

- Coordination entity: A central metal atom/ion bonded to a fixed number of ions or molecules.
- Example 1: $[\mathrm{CoCl_3(NH_3)_3}]$
- Example 2: $[\mathrm{Fe(CN)_6}]^{4-}$

- Ligand: Ions or molecules bound to the central atom/ion.
- Example 1: $\mathrm{NH_3}$
- Example 2: $\mathrm{Cl^-}$

- Coordination number: Number of ligand donor atoms directly bonded to the metal.
- Example 1: In $[\mathrm{PtCl_6}]^{2-}$, CN = 6
- Example 2: In $[\mathrm{Ni(NH_3)_4}]^{2+}$, CN = 4

- Coordination polyhedron: The spatial arrangement of ligand atoms around the metal.
- Example 1: Octahedral in $[\mathrm{Co(NH_3)_6}]^{3+}$
- Example 2: Square planar in $[\mathrm{PtCl_4}]^{2-}$

- Homoleptic complex: Metal bound to only one kind of donor group.
- Example 1: $[\mathrm{Co(NH_3)_6}]^{3+}$
- Example 2: $[\mathrm{Ni(CO)_4}]$

- Heteroleptic complex: Metal bound to more than one kind of donor group.
- Example 1: $[\mathrm{Co(NH_3)_4Cl_2}]^+$
- Example 2: $[\mathrm{PtCl_2(NH_3)_2}]$

- Unidentate ligand: A ligand that binds through one donor atom.
- Examples: $\mathrm{Cl^-}$, $\mathrm{NH_3}$

- Didentate ligand: A ligand that binds through two donor atoms.
- Examples: ethane-1,2-diamine (en), $\mathrm{C_2O_4^{2-}}$ (oxalate)

- Ambidentate ligand: A ligand having two different donor atoms, but binding through only one at a time.
- Examples: $\mathrm{NO_2^-}$, $\mathrm{SCN^-}$

Find oxidation state by using the total charge of the complex and ligand charges.

(i) $[\mathrm{Co(H_2O)(CN)(en)_2}]^{2+}$
- $\mathrm{H_2O}$ neutral, $\mathrm{CN^-}=-1$, en neutral
- Let Co = $x$
- $x-1=+2\Rightarrow x=+3$

(ii) $[\mathrm{CoBr_2(en)_2}]^+$
- $\mathrm{Br^-}\times2=-2$, en neutral
- $x-2=+1\Rightarrow x=+3$

(iii) $[\mathrm{PtCl_3}]^{2-}$
- $\mathrm{Cl^-}\times3=-3$
- $x-3=-2\Rightarrow x=+1$

(iv) $\mathrm{K_3[Fe(CN)_6]}$
- Complex ion charge = $-3$
- $\mathrm{CN^-}\times6=-6$
- $x-6=-3\Rightarrow x=+3$

(v) $[\mathrm{Cr(NH_3)_3Cl_3}]$
- Neutral complex
- $\mathrm{NH_3}$ neutral, $\mathrm{Cl^-}\times3=-3$
- $x-3=0\Rightarrow x=+3$

Write each formula using the rules given in the chapter.

(i) Tetrahydroxidozincate(II)
- Zn is +2 and four OH⁻ ligands give charge −4
- Formula: **$[\mathrm{Zn(OH)_4}]^{2-}$

(ii) Potassium tetrachloridopalladate(II)**
- Anionic complex $[\mathrm{PdCl_4}]^{2-}$
- Formula: **$\mathrm{K_2[PdCl_4]}$

(iii) Diamminedichloridoplatinum(II)
- Neutral complex with 2 NH₃ and 2 Cl⁻ around Pt(II)
- Formula: $[\mathrm{Pt(NH_3)_2Cl_2}]$

(iv) Potassium tetracyanidonickelate(II)**
- Ni(II) with four CN⁻ gives $[\mathrm{Ni(CN)_4}]^{2-}$
- Formula: **$\mathrm{K_2[Ni(CN)_4]}$

(v) Pentaamminenitrito-O-cobalt(III)
- One nitrito-O ligand and five ammines around Co(III)
- Formula: $[\mathrm{Co(ONO)(NH_3)_5}]$

(vi) Hexaamminecobalt(III) sulphate**
- Cation is $[\mathrm{Co(NH_3)_6}]^{3+}$
- Sulphate is $\mathrm{SO_4^{2-}}$
- Balance charges: 2 cations and 3 anions
- Formula: **$[\mathrm{Co(NH_3)_6}]_2(\mathrm{SO_4})_3$

(vii) Potassium tri(oxalato)chromate(III)**
- $[\mathrm{Cr(C_2O_4)_3}]^{3-}$
- Formula: **$\mathrm{K_3[Cr(C_2O_4)_3]}$

(viii) Hexaammineplatinum(IV)
- Neutral complex with six NH₃
- Formula: $[\mathrm{Pt(NH_3)_6}]^{4+}$

(ix) Tetrabromidocuprate(II)**
- Cu(II) with four Br⁻ gives $[\mathrm{CuBr_4}]^{2-}$
- Formula: **$[\mathrm{CuBr_4}]^{2-}$

(x) Pentaamminenitrito-N-cobalt(III)**
- Nitrito-N means bound through N, written as $\mathrm{NO_2}$
- Formula: **$[\mathrm{Co(NH_3)_5(NO_2)}]$**

Name each compound by IUPAC rules: cation first, ligands in alphabetical order, then metal with oxidation state in Roman numerals.

(i) $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$ → hexaamminecobalt(III) chloride

(ii) $[\mathrm{Co(NH_3)_5Cl}]\mathrm{Cl_2}$ → pentaamminechloridocobalt(III) chloride

(iii) $\mathrm{K_3[Fe(CN)_6]}$ → potassium hexacyanidoferrate(III)

(iv) $\mathrm{K_3[Fe(C_2O_4)_3]}$ → potassium trioxalatoferrate(III)

(v) $\mathrm{K_2[PdCl_4]}$ → potassium tetrachloridopalladate(II)

(vi) $[\mathrm{Pt(NH_3)_2Cl(NH_2CH_3)}]\mathrm{Cl}$ → diamminechlorido(methanamine)platinum(II) chloride

The main types of isomerism in coordination compounds are:

### 1. Stereoisomerism
These have the same formula and bonding, but different spatial arrangement.
- Geometrical isomerism: Example, cis- and trans-$[\mathrm{Pt(NH_3)_2Cl_2}]$
- Optical isomerism: Example, $[\mathrm{Co(en)_3}]^{3+}$

### 2. Structural isomerism
These have different bonds.
- Linkage isomerism: Example, $[\mathrm{Co(NH_3)_5(NO_2)}]\mathrm{Cl_2}$ showing nitro and nitrito forms
- Coordination isomerism: Example, $[\mathrm{Co(NH_3)_6}][\mathrm{Cr(CN)_6}]$ and $[\mathrm{Cr(NH_3)_6}][\mathrm{Co(CN)_6}]$
- Ionisation isomerism: Example, $[\mathrm{Co(NH_3)_5SO_4}]\mathrm{Br}$ and $[\mathrm{Co(NH_3)_5Br}]\mathrm{SO_4}$
- Solvate isomerism: Example, $[\mathrm{Cr(H_2O)_6}]\mathrm{Cl_3}$ and $[\mathrm{Cr(H_2O)_5Cl}]\mathrm{Cl_2}\cdot \mathrm{H_2O}$

Thus, coordination compounds show geometrical, optical, linkage, coordination, ionisation, and solvate isomerism.

(i) $[\mathrm{Cr(C_2O_4)_3}]^{3-}$
- Three identical didentate ligands arranged in octahedral geometry do not give geometrical isomers.
- So number of geometrical isomers = 1.

(ii) $[\mathrm{Co(NH_3)_3Cl_3}]$
- This is an octahedral complex of the type $\mathrm{[Ma_3b_3]}$.
- It shows fac and mer forms.
- So number of geometrical isomers = 2.

For $[\mathrm{Pt(NH_3)(Br)(Cl)(py)}]$, the metal is square planar and all four ligands are different. Therefore, the possible geometrical arrangements are the three distinct square planar isomers obtained by choosing which ligand is trans to which.

So the geometrical isomers are the three arrangements of the four different ligands in a square plane.

For optical isomerism, square planar complexes generally do not show optical isomerism unless the arrangement is chiral. In this case, none of the geometrical isomers is optically active.

So:
- Geometrical isomers = 3
- Optical isomers exhibited = 0

Aqueous $\mathrm{CuSO_4}$ gives different colours because of ligand substitution and crystal field effects.

1. With aqueous $\mathrm{KF}$, fluoride ions replace water around $\mathrm{Cu^{2+}}$ to form a different coordination entity, giving a green precipitate.
2. With aqueous $\mathrm{KCl}$, chloride ions similarly form a complex with copper(II), producing a bright green solution.

Thus, the different colours arise due to formation of different coordination compounds of copper(II) with different ligands.

When excess aqueous $\mathrm{KCN}$ is added to aqueous copper sulphate, copper(II) forms a stable cyanide complex rather than free $\mathrm{Cu^{2+}}$ ions. Hence the coordination entity formed is the copper cyanide complex.

Because copper is tied up in this complex, passing $\mathrm{H_2S}$ does not produce a precipitate of copper sulphide. The free $\mathrm{Cu^{2+}}$ concentration is too low for $\mathrm{CuS}$ to precipitate.

So, complex formation suppresses the ordinary sulphide precipitation.

Using valence bond theory:

1. **$[\mathrm{Fe(CN)_6}]^{4-}$**
- $\mathrm{CN^-}$ is a strong field ligand.
- It causes pairing of electrons in $3d$ orbitals.
- The complex uses **$d^2sp^3$ hybridisation.
- It is an inner orbital complex.

2. $[\mathrm{FeF_6}]^{3-}$**
- $\mathrm{F^-}$ is a weak field ligand.
- It does not force pairing of $3d$ electrons.
- The complex uses **$sp^3d^2$ hybridisation.
- It is an outer orbital complex and is paramagnetic.

3. $[\mathrm{Co(C_2O_4)_3}]^{3-}$**
- Oxalate is a ligand that produces pairing in $\mathrm{Co^{3+}}$.
- The complex uses **$d^2sp^3$ hybridisation.
- It is an inner orbital complex.

4. $[\mathrm{CoF_6}]^{3-}$**
- $\mathrm{F^-}$ is weak field.
- No pairing of $3d$ electrons occurs.
- The complex uses **$sp^3d^2$ hybridisation.
- It is an outer orbital** complex and paramagnetic.

Thus, strong ligands give inner orbital complexes, while weak ligands give outer orbital complexes.