krit.club logo

Coordination Compounds - Introduction and terminology

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Coordination compounds are those in which a central metal atom or ion is bonded to a fixed number of ions or molecules (ligands) via coordinate covalent bonds. Example: [Fe(CN)6]4[Fe(CN)_6]^{4-}.

Werner's Theory: It postualtes two types of valencies. Primary Valency is ionizable, corresponds to the oxidation state, and is satisfied by negative ions. Secondary Valency is non-ionizable, corresponds to the coordination number, and is satisfied by neutral molecules or negative ions.

Ligands: Ions or molecules capable of donating an electron pair to the central metal atom. They are classified as unidentate (ClCl^-, H2OH_2O), didentate (enen, C2O42C_2O_4^{2-}), or polydentate (EDTA4EDTA^{4-}).

Ambidentate Ligands: Ligands which can ligate through two different atoms but use only one at a time, such as NO2NO_2^- (can bond via NN or OO) and SCNSCN^- (can bond via SS or NN).

Chelate Effect: When a di- or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it forms a ring-like structure called a chelate. Chelated complexes are more stable than similar complexes with unidentate ligands.

Coordination Number (C.N.): The total number of coordinate bonds formed between the central metal ion and the ligands. For example, in [PtCl6]2[PtCl_6]^{2-}, the C.N. of PtPt is 66.

Homoleptic and Heteroleptic Complexes: Complexes with only one type of ligand are homoleptic (e.g., [Co(NH3)6]3+[Co(NH_3)_6]^{3+}), while those with more than one type are heteroleptic (e.g., [Co(NH3)4Cl2]+[Co(NH_3)_4Cl_2]^+).

📐Formulae

Oxidation Number Calculation: Total Charge=(x)+(Charge of Ligands)\text{Oxidation Number Calculation: } \text{Total Charge} = (x) + \sum \text{(Charge of Ligands)}

Effective Atomic Number (EAN)=Z(O.S.)+2(C.N.)\text{Effective Atomic Number (EAN)} = Z - (O.S.) + 2(C.N.)

Coordination Number (C.N.)=(Number of ligands×denticity)\text{Coordination Number (C.N.)} = \sum (\text{Number of ligands} \times \text{denticity})

💡Examples

Problem 1:

Calculate the oxidation state of the central metal in K3[Fe(C2O4)3]K_3[Fe(C_2O_4)_3].

Solution:

+3+3

Explanation:

In K3[Fe(C2O4)3]K_3[Fe(C_2O_4)_3], the complex ion is [Fe(C2O4)3]3[Fe(C_2O_4)_3]^{3-}. Let the oxidation state of FeFe be xx. Each oxalate ion (C2O42)(C_2O_4^{2-}) has a charge of 2-2. Therefore, x+3(2)=3x6=3x=+3x + 3(-2) = -3 \Rightarrow x - 6 = -3 \Rightarrow x = +3.

Problem 2:

Identify the coordination number of PtPt in [Pt(en)2Cl2][Pt(en)_2Cl_2].

Solution:

66

Explanation:

In [Pt(en)2Cl2][Pt(en)_2Cl_2], enen (ethylenediamine) is a didentate ligand and ClCl^- is a unidentate ligand. C.N.=(2×2)+(2×1)=4+2=6C.N. = (2 \times 2) + (2 \times 1) = 4 + 2 = 6.

Problem 3:

Calculate the EAN of FeFe in [Fe(CN)6]4[Fe(CN)_6]^{4-}, given ZZ for FeFe is 2626.

Solution:

3636

Explanation:

First, find the oxidation state: x+6(1)=4x=+2x + 6(-1) = -4 \Rightarrow x = +2. Using the formula EAN=Z(O.S.)+2(C.N.)EAN = Z - (O.S.) + 2(C.N.), we get EAN=262+2(6)=24+12=36EAN = 26 - 2 + 2(6) = 24 + 12 = 36.

Introduction and terminology - Revision Notes & Key Formulas | ICSE Class 12 Chemistry