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Coordination Compounds - Ligands and coordination number

Grade 12ICSEChemistry

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

🔑Concepts

Ligands are neutral molecules, anions, or cations which can donate a pair of electrons to the central metal atom or ion to form a coordinate covalent bond. Common examples include H2OH_2O (aqua), NH3NH_3 (ammine), and ClCl^- (chlorido).

Ligands are classified by denticity: Unidentate ligands (one donor atom, e.g., FF^-), Didentate/Bidentate ligands (two donor atoms, e.g., C2O42C_2O_4^{2-} or ethane-1,2-diamine represented as enen), and Polydentate ligands (multiple donor atoms, e.g., EDTA4EDTA^{4-} which is hexadentate).

Ambidentate ligands are unidentate ligands that contain more than one coordination site but bind through only one atom at a time. Examples include NO2NO_2^- (can bind via NN or OO) and SCNSCN^- (can bind via SS or NN).

Chelating Ligands are multidentate ligands that bind to a single metal ion through two or more donor atoms, forming a ring structure. This 'Chelate Effect' significantly increases the stability of the complex.

The Coordination Number (C.N.C.N.) of a metal ion in a complex is defined as the total number of ligand donor atoms to which the metal is directly bonded. It is not necessarily equal to the number of ligands if polydentate ligands are present.

The Oxidation State of the central metal is the charge it would carry if all the ligands were removed along with the electron pairs that are shared with the central atom.

📐Formulae

C.N.=(Number of ligands×Denticity of each ligand)C.N. = \sum (\text{Number of ligands} \times \text{Denticity of each ligand})

Net Charge of Complex=Oxidation State of Metal+(Charges of Ligands)\text{Net Charge of Complex} = \text{Oxidation State of Metal} + \sum (\text{Charges of Ligands})

EAN=Z(Oxidation State)+2×(C.N.)EAN = Z - (\text{Oxidation State}) + 2 \times (C.N.)

💡Examples

Problem 1:

Calculate the coordination number and oxidation state of the central metal in [Co(en)2Cl2]+[Co(en)_2Cl_2]^+.

Solution:

Coordination Number = 66; Oxidation State of Co=+3Co = +3.

Explanation:

Ethylenediamine (enen) is a bidentate ligand and ClCl^- is a unidentate ligand. Total C.N.=(2×2)+(2×1)=6C.N. = (2 \times 2) + (2 \times 1) = 6. For oxidation state (xx): x+2(0)+2(1)=+1x2=1x=+3x + 2(0) + 2(-1) = +1 \Rightarrow x - 2 = 1 \Rightarrow x = +3.

Problem 2:

Identify the coordination number of FeFe in [Fe(C2O4)3]3[Fe(C_2O_4)_3]^{3-}.

Solution:

C.N.=6C.N. = 6

Explanation:

The oxalate ion (C2O42C_2O_4^{2-}) is a didentate (bidentate) ligand. Since there are three such ligands, the total number of coordinate bonds is 3×2=63 \times 2 = 6.

Problem 3:

What is the coordination number of Ca2+Ca^{2+} in [Ca(EDTA)]2[Ca(EDTA)]^{2-}?

Solution:

C.N.=6C.N. = 6

Explanation:

EDTA4EDTA^{4-} (ethylenediaminetetraacetate) is a hexadentate ligand, meaning it occupies 6 coordination sites on a single metal ion. Therefore, C.N.=1×6=6C.N. = 1 \times 6 = 6.

Ligands and coordination number - Revision Notes & Key Formulas | ICSE Class 12 Chemistry