Chapter - 5 Coordination Compounds

CHAPTER - 5 COORDINATION COMPOUNDSCHAPTER - 5COORDINATION COMPOUNDSMEMORY CARD DEFINITIONS:

o COMPLEX COMPOUNDS: The compounds in which ions or molecules (called ligands) are coordinated to the central metal atom or ion are called complex compounds.

o COMPLEX ION: An electrically charged species formed by the union of a central metal ion with two or more ligands is called a complex ion. It may have a positive charge or negative charge.

o CATIONIC COMPLEX: A complex which forms positively charged complex ion in solution is called cationic complex.

Eg: [Cu(NH3)4]SO4 [Cu(NH3)4]+2 + SO4–2

o ANIONIC COMPLEX: A complex which forms negatively charged complex ion in solution is called anionic complex.

Eg: K4[Fe(CN)6] 4K+ + [Fe(CN)6]–4 o LIGANDS: The ions or molecules which can donate one or more pairs of electrons to a central

metal atom or ion in complex are called ligands.o TYPES OF LIGANDS:MONODENTATE LIGAND: The ligand which is capable of donating only one pair of electrons to the

central metal atom. These may be neutral molecules or negatively charged or positively charged.NEUTRAL LIGANDS:

FORMULA :NH3 : CO :NO H2O: C5H5N: NAME ammine carbonyl nitrosyl aqua or aquo pyridylNEGATIVELY CHARGED LIGANDS:

FORMULA X – : CN – :SCN – SCN: :OH –

NAME halo cyano thiocyanato isothiocyanato hydroxo :O2 –2 :O –2 :NO2

– : ONO – :NO3 – :NH2–

peroxo oxo nitro nitrito nitrato amido :NH –2 :S –2 CO3 –2 : SO4 –2 : SO3 –2 : CH3COO– :

imido sulphido carbonato sulphato sulphito acetatoPOSITIVELY CHARGED LIGNADS:FORMULA NO2 + NO + NH2-NH3

+ H3O +

NAME nitronium nitrosonium hydrazinium hydronium

BIDENTATE LIGAND: The ligand which is capable of donating two pairs of electrons to the central metal atom.

a. ethylenediammine (en) b. oxalate (ox) c. glycinato (gly) d. dimethyl glyoxime (dmg)

TRIDENTATE LIGAND: The ligand which is capable of donating three pairs of electrons to the central metal atom.

a. diethylene triammine(dien) b. 2,2’,2”-terpyridine (terpy) c. iminodiacetato ion

POLYDENTATE LIGANDS: The ligand which is capable of donating many pairs of electrons to the central metal atom.

a. triethylenediammine (trien) b. ehylenediamminetriacetato ion c. EDTA (ethylenediammine tetraacetate)

1

H2N NH2

H2C CH2

| |H2C CH2

HN NH H2C ––– CH2

H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


CH2COO:–

CH2-N | CH2 CH2COO:–

|HN –– CH2COO:–

CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



CH2COO:–



H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

O O O || || C – CH2

C ––––CH2 O N

| CH2

| CH2

O N C –––– CH2 |

|| C– CH2 O O || O

N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N N

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

O = C – O–: |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H2C –– NH2 |

O = C – O–:

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

O = C – O:– –:O – C = O | | H2C CH2 N H

CHAPTER - 5 COORDINATION COMPOUNDS

o COORDINATION NUMBER (C.N): The total number of coordination bonds formed by monodentate ligands with a central metal atom or ion in a complex is called the coordination number of central metal.

WERNER’S THEORY OF COORDINATION COMPOUNDS: Alfred Werner proposed the theoryPostulates of the theory are;

1. A metal possesses two types of valencies: viz, primary valency and secondary valency2. The primary valency of a metal is variable & corresponds to its oxidation number (charge on

metal ion).The secondary valency of a metal is fixed & corresponds to its coordination number.

3. The primary valency of a metal is satisfied only by negative ions. The secondary valency of a metal is satisfied either by negative ions and/or by neutral molecules which act as ligands.

4. Secondary valencies are directional and these give a definite geometrical shape to the complex. SIDGWICK’S ELECTRONIC THEORY: According to Sidgwick, “The ligands donate electron pair (or

pairs) to the central metal atom or ion and form a number of coordinate bonds with the central metal atom or ion.

Eg, In the formation of cuprammonium ion, each of the four NH3 ligands donate an electron pair to the Cu+2 ion forming four coordinate bonds.

Cu+2 + 4 :NH3

EFFECTIVE ATOMIC NUMBER (EAN): The total number of electrons around the central metal ion

including those gained by coordination of the ligands is called the effective atomic number of the central metal atom in the complex. It is calculated as;i.e, EAN = Z – X + Y, where Z = Atomic No. of the metal

X = No. of e– lost in the formation of ionY = No. of e– donated by ligands.

EAN Rule: A metal atom or ion continues to accept electron pairs from the ligands until the total number of electrons around the metal atom or ion in the complex becomes equal to the atomic number of the nearest noble gas.

NOMENCLATURE OF COORDINATION COMPOUNDS: According to IUPAC, name of complex should give information about;

a. Number of ligands: The prefixes like mono, di, tri, tetra, penta and hexa are used to indicate the number of ligands 1, 2, 3, 4, 5 and 6 resply. The prefixes like bis, tris, tetrakis, pentakis and hexakis are used to indicate the number of complex ligands.

b. Name of the ligand: Names of negative ligands end in “o”. Names of positive ligands end in “ium”. Neutral ligands have either special names or named as it is.

c. Name of central metal atom/ion: In neutral and cationic complexes the name of central metal atom or ion remains the same. However, in anionic complexes, the name of the central metal atom ends in “ate”.

d. Oxidation state of the central metal atom/ion: Oxidation state of central metal atom/ion is indicated by Roman numbers 0, I, II, III, IV, V and IV etc in parenthesis.

NAMING OF IONIC COMPLEXES: If the complex is ionic, the cation is named first & then the anion. NAMING OF NON-IONIC COMPLEXES: Non-ionic complex is named as one word. ISOMERISM IN COMPLEXES:

o IONISATION ISOMERISM: Complexes having same molecular formula but furnish different ions in solution are called ionisation isomers & the phenomenon is ionisation isomerism.

Eg; [CoBr(NH3)5]SO4 and [CoSO4(NH3)5]Br are ionisation isomerso LINKAGE ISOMERISM: Complexes having same molecular formula but differ in the mode of

attachment of the ligands to the central metal ion are called linkage isomers & the phenomenon is linkage isomerism.

Eg;[CoNO2(NH3)5]Cl2 and [CoONO(NH3)5]Cl2 are linakge isomers

2

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

COMPLEX CENTRAL METAL ATOM/ION C.NK4[Fe(CN)6] Fe+2 6K3[Fe(CN)6] Fe+3 6

[Cu(NH3)4]SO4 Cu+2 4[Co(en)3]Cl3 Co+3 6

[PtCl4]–2 Pt+2 4[PtCl6]–2 Pt+4 6

CHAPTER - 5 COORDINATION COMPOUNDSo HYDRATE ISOMERISM: Complexes having same molecular formula but differ in the number of

water molecules present as ligands or as water of hydration are called hydrate isomers & the phenomenon is hydrate isomerism.

Eg; [Co(H2O)6]Cl3, [CoCl(H2O)5]Cl2.H2O, [CoCl2(H2O)4]Cl.2H2O and [CoCl3(H2O)3].3H2O are hydrate isomers.

VELANCE BOND THEORY (VBT): Proposed by Linus Pauling to explain bonding in coordination compounds. The main features of VBT are;

1. A central metal atom or ion in a complex provides a definite number of empty orbitals for the formation of coordinate bonds.

2. Under the influence of strong ligands like NH3, CN–, CO etc unpaired electrons present in the metal atom or ion are forced to pair up (spin pairing).

3. These empty orbitals hybridise & form the same number of identical hybrid orbitals.4. Each ligand has at least one orbital containing a lone pair of electrons.5. The hybrid orbitals of the metal atom or ion overlap with the filled orbitals of the ligands to form

coordinate bonds between metal and ligand6. The geometrical shape of the complex depends on the type of hybridisation.

HYBRIDISATION STRUCTURE HYBRIDISATION STRUCTUREsp Linear dsp2 Square planarsp2 Triagonal dsp3 Triagonal bipyramidalsp3 Tetrahedral d2sp3 Octahedral

d3sp3 Pentagonal bipyramidal7. When a complex compound contains one or more unpaired electrons, it exhibits paramagnetic

property. If no unpaired electrons are present, the complex is diamagnetic.STRUCTURE OF COMPLEX COMPOUNDS BASED ON VBT: Ground state electronic configuration of central metal atom. Electronic configuration of central metal ion. Rearrangement of electrons only in the presence of strong ligands like NH3, CN–, CO. Hybridisation of required number of vacant orbitals of central metal ion. Overlapping of hybrid orbitals of central metal ion with filled orbitals of ligands to form

coordinate bonds. Based on hybridisation predict the shape of the complex. Based on presence or absence of unpaired electrons predict the magnetic property of the

complex.

ONE MARK EACH1. What is a coordination compound? OR

What are complex compounds?A. The compounds in which ions or molecules (called ligands) are coordinated to the central metal

atom or ion are called complex compounds.2. Give the names of two complexes found in biological systems.A. Haemoglobin and Chlorophyll.3. Write the molecular formula of (a) cuprammonium sulphate (b) potassium ferrocyanide

(c) potassium ferricyanide.A. (a) [Cu(NH3)4]SO4 (b) K4[Fe(CN)6] (c) K3[Fe(CN)6]4. How many ions are formed per mole of potassium ferrocyanide when dissolved in water?A. 5 ions. (i.e, four K+ ions and one [Fe(CN)6]– 4 complex ion) 5. Define a complex ion.A. An electrically charged species formed by the union of a central metal ion with two or more

ligands is called a complex ion. It may have a positive charge or negative charge.6. What are ligands?A. The ions or molecules which can donate one or more pairs of electrons to a central metal atom or

ion in a complex are called ligands.7. Give an example of (a) neutral ligand (b) negative ligand (c) positive ligand (d) bidentate ligand

(e) tridentate ligand (f) polydentate ligand.A. (a) neutral ligand – : NH3 (b) negative ligand – : CN – (c) positive ligand – NO2

+

(d) bidentate ligand – (f) polydentate ligand

3

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |

H2C –– NH –– CH2

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O

EDTA

O O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––– CH2 | || C– CH2

O O || O


(e) tridentate ligand –

8. Why are ligands called Lewis bases?A. According to Lewis, a substance which donates a pair of electron to the other substance for the

formation of a coordinate bond. Ligands also form coordinate bond with central metal atom or ion in a complex by donating a lone pair of electron. Hence, ligands are called Lewis bases.

9. Write the structures of the following ligands: (i) en (ii) dien (iii) trien (iv) ox (v) gly (vi) dmg (vii) terpy (viii) EDTA (ix) ethylenediaminetriacetate ion (x) iminodiacetate ion

A. i. ii. iii. iv.

v. vi. vii.

viii. ix.

x.

10. How many coordination sites present in ethylenediamine?A. Two11. Define coordination number.A. The total number of coordination bonds formed by monodentate ligands with a central metal

atom or ion in a complex is called the coordination number of central metal.12. What is the coordination number of iron in (a) potassium ferrocyanide (b) potassium ferricyanide?A. Coordination number of iron in potassium ferrocyanide and potassium ferricyanide is 6.13. What is a cationic complex? Give an example.A. A complex which forms positively charged complex ion in solution is called cationic complex.

Eg: [Cu(NH3)4]SO4 [Cu(NH3)4]+2 + SO4–2

14. What is an anionic complex? Give an example.A. A complex which forms negatively charged complex ion in solution is called anionic complex.

Eg: K4[Fe(CN)6] 4K+ + [Fe(CN)6]–4

15. What type of ligands satisfies the primary valencies of a metal ion?A. Only negatively charged ligands can satisfy the primary valencies of a metal ion.16. What is primary valency of a metal ion?A. Primary valency of a metal ion is its oxidation state.17. What is secondary valency of a metal ion?A. Secondary valency of a metal ion is its coordination number.18. Give the primary & secondary valencies of (a) copper in cuprammonium sulphate (b) iron in both

potassium ferrocyanide & potassium ferricyanide.A. Primary & secondary valencies of copper in cuprammonium sulphate is 2 & 4 resply.

Primary & secondary valencies of iron in potassium ferrocyanide is 2 & 6 resply.Primary & secondary valencies of iron in potassium ferricyanide is 3 & 6 resply.

19. What is the coordination number of the metal ion if the complex ion has(i) square planar structure (ii) octahedral structure.

4

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

terpy N N N

ethylenediaminetriacetate ion CH2COO:–


| HN –– CH2COO:–









































































(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

(en)

H2C –– CH2 | |

H2N: :NH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

dien

H2N NH2 | | H2C CH2

| |H2C –– NH –– CH2

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

EDTA O

O O || || C – CH2

C –––– CH2 O N

| CH2

| CH2

O N C –––––CH2 | || C– CH2

O O ||

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

ox

O = C – O–: |O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

gly

H2C –– NH2

|O = C – O–:

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

dmg

H3C – C ––– C – CH3 || ||

HO – N: :N – OH

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

iminodiacetate ion

O = C – O:– –:O – C = O | | H2C CH2 N H

trien H2N NH2

H2C CH2

| | H2C CH2

HN NH H2C –– CH2

trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


trien H2N NH2

H2C CH2

| | H2C CH2


CHAPTER - 5 COORDINATION COMPOUNDSA. (i) 4 (ii) 620. What is the oxidation state of nickel in nickel carbonyl?A. zero21. What type of ligand satisfies (a) primary valency of a metal ion (b) secondary valency of a metal

ion?A. (a) Only negatively charged ligands can satisfy the primary valencies of a metal ion.

(b) secondary valency of a metal ion is satisfied either by negative ions or neutral molecules.22. What is EAN?A. The total number of electrons around the central metal ion including those gained by

coordination of the ligands is called the effective atomic number of the central metal atom in the complex. OR

EAN = Z – X + Y, where Z = Atomic No. of the metalX = No. of e– lost in the formation of ionY = No. of e– donated by ligands.

23. What is EAN rule?A. A metal atom or ion continues to accept electron pairs from the ligands until the total number of

electrons around the metal atom or ion in the complex becomes equal to the atomic number of the nearest noble gas.

24. Give an example of a complex compound which (a) obeys EAN rule (b) does not obey EAN rule.A. K4[Fe(CN)6] obeys EAN rule and K3[Fe(CN)6] does not obey EAN rule.25. What is (a) ionisation isomerism (b) linkage isomerism (c) hydrate isomerism?A. (a) IONISATION ISOMERISM: Complexes having same molecular formula but furnish different ions

in solution are called ionisation isomers & the phenomenon is ionisation isomerism. (b) LINKAGE ISOMERISM: Complexes having same molecular formula but differ in the mode of attachment of the ligands to the central metal ion are called linkage isomers & the phenomenon is linkage isomerism.(c) HYDRATE ISOMERISM: Complexes having same molecular formula but differ in the number of water molecules present as ligands or as water of hydration are called hydrate isomers & the phenomenon is hydrate isomerism.

26. What is an ambidentate ligand?A. When more than one atom in a monodentate ligand functions as a donor atom, such ligands are

called ambidentate ligands.27. Give an example for (a) ionisation isomerism (b) linkage isomerism (c) hydrate isomerism (d)

ambidentate ligand.A. IONISATION ISOMERISM:

Eg; [CoBr(NH3)5]SO4 and [CoSO4(NH3)5]Br are ionisation isomersLINKAGE ISOMERISM:

Eg; [CoNO2(NH3)5]Cl2 and [CoONO(NH3)5]Cl2 are linakge isomersHYDRATE ISOMERISM:

Eg; [Co(H2O)6]Cl3, [CoCl(H2O)5]Cl2.H2O, [CoCl2(H2O)4]Cl.2H2O and [CoCl3(H2O)3].3H2O are hydrate isomers.

28. Name the following complexes according to IUPAC:a. [Cu(NH3)2(H2O)2Cl2] b. [Fe(H2O)6]Cl3 c. [Co(NH3)6][Co(NO2)6] d. (NH4)3[Co(C2O4)3]

e. K3[Fe(CN)5NO] f. Na2[Ni(EDTA)] g. [CoBr(ONO)(en)2]+ h. [PtNO2(NH3)5]+3

i. [CoCl2(NH3)4]Cl j. [PtCl(NH3)5]Cl3 k. [Cr(NO2)2(NH3)3H2O]NO2 l. [Co(en)3]+3

m. [Cr(NH3)3(H2O)3]Cl3 n. Fe4[Fe(CN)6]3 o. Na3[Fe(ox)3] p. K2[Pt F6]

A. a. [Cu(NH3)2(H2O)2Cl2] diamminediaaquadichlorocopper(II)

b. [Fe(H2O)6]Cl3 hexaaquairon(III) chloride

c. [Co(NH3)6][Co(NO2)6] hexaamminecobalt(III) hexanitrocobalt(III)

d. (NH4)3[Co(C2O4)3] ammonium trioxalatocobaltate(III)

e. K3[Fe(CN)5NO] potassium pentacyanonitrosylferrate(II)

f. Na2[Ni(EDTA)] sodium ethylenediamminetetraacetatonickelate(IV)

g. [CoBr(ONO)(en)2]+ monobromobis(ethylenediammine)nitritocobalt(III) ion

h. [PtNO2(NH3)5]+3 pentaamminenitroplatinum(IV) ion

i. [CoCl2(NH3)4]Cl tetraamminedichlorocobalt(III) chloride

j. [PtCl(NH3)5]Cl3 pentaamminechloroplatinum(IV) chloride

k. [Cr(NO2)2(NH3)3H2O]NO2 triammineaquadinitrochromium(III) nitrate

5

CHAPTER - 5 COORDINATION COMPOUNDSl. [Co(en)3]+3 tris(ethylenediammine)cobalt(III) ion

m .[Cr(NH3)3(H2O)3]Cl3 triamminetriaquachromium(III) chloride

n. Fe4[Fe(CN)6]3 ferric hexacyanoferrate(II)

o. Na3[Fe(ox)3] sodium trioxalatoferrate(III)

p. K2[Pt F6] potassium hexafluoroplatinate(IV)

6


29. Write the formula of the following compounds:a. Hexaamminecobalt(III) bromide b. tetraamminecopper(II) ion

c. diamminetriaquahydroxochromium(III) nitrate d. potassium hexacyanoferrate(II)

e. bis(en)chloronitrocobalt(III) ion f. ammoniumtrioxalatocobaltate(III)

g. tetraamminechloronitritochromium(III) chloride h. hexafluoroferrate(II) ion

i. tetrakis(pyridine)platinum(II) tetrachloroplatinate (II)

A. a. hexaamminecobalt(III) bromide [Co(NH3)6]Br3

b. tetraamminecopper(II) ion [Cu(NH3)4]+2

c. diamminetriaquahydroxochromium(III) nitrate [Cr(OH)(H2O)3(NH3)2](NO2)2

d. potassium hexacyanoferrate(II) K4[Fe(CN)6]

e. bis(en)chloronitrocobalt(III) ion [Co(Cl)(NO2)(en)2]+1

f. ammoniumtrioxalatocobaltate(III) (NH4)3[Co(ox)3]

g. tetraamminechloronitritochromium(III) chloride [Cr(Cl)(ONO)(NH3)4]Cl

h. hexafluoroferrate(II) ion [FeF6] –4

i. tetrakis(pyridine)platinum(II) tetrachloroplatinate (II) [Pt(C5H5N)4][PtCl4]30. What is (a) an inner orbital complex (b) an outer orbital complex.A. (a) If inner d-orbitals i.e, (n – 1)d orbitals are used for hybridisation along with outer ns- and np-

orbitals, then complex formed is called inner orbital complex.(b) If outer d-orbitals i.e, nd-orbitals are used for hybridisation along with outer ns- and np- orbitals, then complex formed is called outer orbital complex.

TWO MARKS EACH1. The constituents lose their identity in coordination compounds. Explain with an example.A. Coordination compounds are quite stable & retain their properties in the solid state & in their

solution. In these compounds, the constituents lose their identity & entirely new ions are formed in the solution.Eg. [Cu(NH3)4]SO4, cuprammonium sulphate complex does not form Cu+2 and NH3 in solution but forms cuprammonium complex ion [Cu(NH3)4]+2. The properties of [Cu(NH3)4]+2 are entirely different from those of Cu+2 and NH3

[Cu(NH3)4]SO4 [Cu(NH3)4]+2 + SO4–2

2. What is a cationic complex? Give an example.A. A complex which forms positively charged complex ion in solution is called cationic complex.

Eg: [Cu(NH3)4]SO4 [Cu(NH3)4]+2 + SO4–2

3. What is an anionic complex? Give an example.A. A complex which forms negatively charged complex ion in solution is called anionic complex.

Eg: K4[Fe(CN)6] 4K+ + [Fe(CN)6]–4

4. What is a ligand? Give an example of polydentate ligand?A. The ions or molecules which can donate one or more pairs of electrons to a central metal atom or

ion in complex are called ligands.Example of polydentate ligand.

5. Explain with an example:a. monodentate ligand b. bidentate ligand c. tridentate ligand d. polydentate ligand.

A. a. monodentate ligand: The ligand which is capable of donating only one pair of electrons to the central metal atom. These may be neutral molecules or negatively charged or positively charged.

Eg. :NH3 ammine b. bidentate ligand: The ligand which is capable of donating two pairs of electrons to the central metal atom.

Eg. ethylenediammine (en)

7












































































H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

H2C –– CH2 | |

H2N: :NH2

CHAPTER - 5 COORDINATION COMPOUNDS c. tridentate ligand: The ligand which is capable of donating three pairs of electrons to the central metal atom.

Eg. diethylene triammine(dien)

d. polydentate ligand: The ligand which is capable of donating many pairs of electrons to the central metal atom.

Eg. triethylenediammine (trien)

6. Calculate the EAN of the central metal in the following:a. [Ni(CO)4] b. [Cu(NH3)4]SO4 c. K4[Fe(CN)6] d. K3[Fe(CN)6]

A. a. [Ni(CO)4] EAN = Z – X + Y Where, Z = 28, X = 0 and Y = 8EAN = 28 – 0 + 8 = 36

b. [Cu(NH3)4]SO4 EAN = Z – X + Y Where, Z = 29, X = 2 and Y = 8EAN = 29 – 2 + 8 = 35

c. K4[Fe(CN)6] EAN = Z – X + Y Where, Z = 26, X = 2 and Y = 12EAN = 26 – 2 + 12 = 36

d. K3[Fe(CN)6] EAN = Z – X + Y Where, Z = 26, X = 3 and Y = 12EAN = 26 – 3 + 12 = 35

7. Explain with an example for (a) ionisation isomerism (b) linkage isomerism (c) hydrate isomerism.A. IONISATION ISOMERISM: Complexes having same molecular formula but furnish different ions in

solution are called ionisation isomers & the phenomenon is ionisation isomerism.Eg; [CoBr(NH3)5]SO4 and [CoSO4(NH3)5]Br are ionisation isomersLINKAGE ISOMERISM: Complexes having same molecular formula but differ in the mode of attachment of the ligands to the central metal ion are called linkage isomers & the phenomenon is linkage isomerism.Eg; [CoNO2(NH3)5]Cl2 and [CoONO(NH3)5]Cl2 are linakge isomersHYDRATE ISOMERISM: Complexes having same molecular formula but differ in the number of water molecules present as ligands or as water of hydration are called hydrate isomers & the phenomenon is hydrate isomerism.Eg; [Co(H2O)6]Cl3, [CoCl(H2O)5]Cl2.H2O, [CoCl2(H2O)4]Cl.2H2O and [CoCl3(H2O)3].3H2O are hydrate isomers.

8. How many Cl – ions can be ppted by adding AgNO3 solution to CoCl3.4NH3? Give reason.A. One Cl – is ppted by adding AgNO3 solution to CoCl3.4NH3 which is having the formula

[CoCl2(NH3)4]Cl and in which out of three Cl – ions, two are satisfying both primary valency & secondary valency of the central metal ion and other one is outside the coordination sphere to fulfil the remaining primary valency of the central metal ion.

9. Give a test to distinguish between [CoBr(NH3)5]SO4 and [CoSO4(NH3)5]Br. What kind of isomerism do they exhibit?

A. The aq. soln. of the complex [CoBr(NH3)5]SO4 when treated with a solution of BaCl2 gives a white ppt of BaSO4 indicating that SO4

–2 ion is in ionic sphere. But will not give ppt with AgNO3 solution as Br – is not in ionic sphere.The aq. soln. of the complex [CoSO4(NH3)5]Br when treated with a solution of AgNo3 gives a pale yellow ppt of AgBr indicating that Br – ion is in ionic sphere. But will not give ppt with BaCl2

solution as SO4–2 is not in ionic sphere. These are ionisation isomers.

10. Why is potassium ferrocyanide complex diamagnetic?A. In potassium ferrocyanide complex, K4[Fe(CN)6] the unpaired electrons of central metal ion Fe+2

are forced to pair in the presence of strong CN – ligands. Hence, there are no unpaired electrons in the complex & it is diamagnetic according to VBT.

11. Why is cuprammonium sulphate complex paramagnetic?

8

H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2

H2C CH2

| |H2C CH2


H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

H2N NH2| |

H2C CH2 | |H2C –– NH –– CH2

CHAPTER - 5 COORDINATION COMPOUNDSA. In cuprammonium sulphate complex, [Cu(NH3)4]SO4 the one unpaired electron in 3d-orbital of

central metal ion Cu+2 is excited to one of the 4p-orbitals in the presence of strong NH3 ligands. Hence, there is one unpaired electron in the complex & it is paramagnetic according to VBT.

12. Give any two differences between inner orbital and outer orbital complexes.A.

INNER ORBITAL COMPLEX OUTER ORBITAL COMPLEX1. These are formed by hybridisation of inner

(n – 1)d-orbital.1. These are formed by hybridisation of

outer nd-orbital.2. These are formed in the presence strong

ligands.2. These are formed in the presence weak

ligands.3. These are spin paired complexes. 3. These are spin free complexes.

13. Mention two limitations of VBT.A. Limitations of VBT:

1. It fails to distinguish between strong and weak ligands.2. It cannot explain the detailed magnetic properties of complexes.

14. Explain with example: (a) inner orbital complexes are called spin paired complexes.(b) outer orbital complexes are called spin free complexes.

A. (a) In the presence of strong ligands, unpaired electrons in inner (n – 1)d - orbitals are forced to pair up and contain less number of unpaired electrons. Hence, inner orbital complexes are called spin paired complexes.(b) In the presence of weak ligands, unpaired electrons in inner (n – 1)d - orbitals are not forced to pair up and contain more number of unpaired electrons. Hence, outer orbital complexes are called spin free complexes.

15. Explain Sidgwick’s Theory with an example.A. Sidgwick extended the Electronic theory of Lewis for the formation of coordination compounds.

According to Sidgwick,“ The ligands donate electron pair(s) to the central metal atom or ion and form a number

coordinate bonds with the central metal atom or ion”.For eg, In the formation of cuprammonium ion, each of the four NH3 ligands donate an electron pair to the Cu+2 ion forming four coordinate bonds.

Cu+2 + 4 :NH3

THREE MARKS EACH1. Write the postulates of Werner’s Theory. OR

Explain the structure of the following complexes on the basis of Werner’s Theory: (a) cuprammonium sulphate (b) potassium ferrocyanide (c) potassium ferricyanide.

A. WERNER’S THEORY OF COORDINATION COMPOUNDS: Alfred Werner proposed the theoryPostulates of the theory are;

1. A metal possesses two types of valencies: viz, primary valency and secondary valency.Primary valency is ionisable and secondary valency is non-ionisable.

2. The primary valency of a metal is variable & corresponds to its oxidation number (charge on metal ion).The secondary valency of a metal is fixed & corresponds to its coordination number.

Eg: In cuprammonium sulphate the central metal copper possesses +2 as primary valency and 4 as secondary valency.

In potassium ferrocyanide the central metal iron possesses +2 as primary valency and 6 as secondary valency.

In potassium ferricyanide the central metal iron possesses +3 as primary valency and 6 as secondary valency.

3. The primary valency of a metal is satisfied only by negative ions. The secondary valency of a metal is satisfied either by negative ions and/or by neutral molecules which act as ligands.

Eg: The primary valency of copper in cuprammonium sulphate is satisfied by two negative charges of sulphate ion and the secondary valency is satisfied by four NH3 molecules as ligands.

The primary valency & secondary valency of iron in potassium ferrocyanide is satisfied by cyanide ions. Out of six cyanide ions, two cyanide ions satisfy both primary & secondary valencies and other four cyanide ions satisfy the remaining secondary valencies.

The primary valency & secondary valency of iron in potassium ferricyanide is satisfied by cyanide ions. Out of six cyanide ions, three cyanide ions satisfy both primary & secondary valencies and other three cyanide ions satisfy the remaining secondary valencies.

4. Secondary valencies are directional and these give a definite geometrical shape to the complex.

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H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

H3N NH3 +2

Cu+2

H3N NH3

CHAPTER - 5 COORDINATION COMPOUNDSEg: The shape of cuprammonium sulphate is square planar and shapes of both

potassium ferrocyanide and potassium ferricyanide are octahedral.

10

CHAPTER - 5 COORDINATION COMPOUNDS2. Explain the rules of IUPAC nomenclature of coordination compounds.A. NOMENCLATURE OF COORDINATION COMPOUNDS: According to IUPAC, name of a complex

compound should give information about;i. Number of ligands: The prefixes like mono, di, tri, tetra, penta and hexa are used to indicate

the number of ligands 1, 2, 3, 4, 5 and 6 resply. The prefixes like bis, tris, tetrakis, pentakis and hexakis are used to indicate the number of complex ligands.

ii. Name of the ligand: Names of negative ligands end in “o”. Names of positive ligands end in “ium”. Neutral ligands have either special names or named as it is. If more than one kind of ligands are present, arrange names of ligands in the alphabetical order of their names. However, while writing the formula of complex compound, ligands are arranged in the order of negative ligand, neutral ligand and then positive ligand.

iii. Name of central metal atom/ion: In neutral and cationic complexes the name of central metal atom or ion remains the same. However, in anionic complexes, the name of the central metal atom ends in “ate”.

iv. Oxidation state of the central metal atom/ion: Oxidation state of central metal atom/ion is indicated by Roman numbers 0, I, II, III, IV, V and IV etc in parenthesis.

v. Naming of Ionic complexes: If the complex is ionic, the name of the cation is named first and then the anion.Eg. K4[Fe(CN)6] potassium hexacyanoferrate(II)

[Cu(NH3)4]SO4 tetraamminecopper(II) sulphatevi. Naming of Non-ionic complexes: Non-ionic complex is named as one word.

Eg. [Ni(CO)4] tetracarbonylnickel(O)3. Explain the main features of VBT.A. VELANCE BOND THEORY (VBT): Proposed by Linus Pauling to explain bonding in coordination

compounds. The main features of VBT are;1. A central metal atom or ion in a complex provides a definite number of empty orbitals for the

formation of coordinate bonds.2. Under the influence of strong ligands like NH3, CN–, CO etc unpaired electrons present in the metal

atom or ion are forced to pair up (spin pairing).3. These empty orbitals hybridise & form the same number of identical hybrid orbitals.4. Each ligand has at least one orbital containing a lone pair of electrons.5. The hybrid orbitals of the metal atom or ion overlap with the filled orbitals of the ligands to form

coordinate bonds between metal and ligand6. The geometrical shape of the complex depends on the type of hybridisation.

HYBRIDISATION STRUCTURE HYBRIDISATION STRUCTUREsp Linear dsp2 Square planarsp2 Triagonal dsp3 Triagonal bipyramidalsp3 Tetrahedral d2sp3 Octahedral

d3sp3 Pentagonal bipyramidal7. When a complex compound contains one or more unpaired electrons, it exhibits paramagnetic

property. If no unpaired electrons are present, the complex is diamagnetic.4. On the basis of VBT, explain the hybridisation, geometrical shape and magnetic properties of the

following: (a) [Ni(CO)4] (b)[Cu(NH3)4]SO4 (c) K3[Fe(CN)6]A. (a) sp3 HYBRIDISATION: Structure of [Ni(CO)4]

At. No. of Ni is 28, its ground state electronic configuration is [Ar] 3d8 4s2

3d 4s 4pNi [Ar]When strong CO ligands approach the nickel atom, 4s electrons are forced to pair up with 3d electrons. The resulting vacant 4s orbital and three 4p orbitals hybridise to give four equivalent sp3 hybrid orbitals.

3d 4s 4pNi [Ar]

sp3 hybridisationThese four sp3 hybrid orbitals of Ni overlap with filled orbitals of carbonyl ligands & form four coordinate bonds.

3d 4s 4p[Ni(CO)4] [Ar]

sp3 hybrid orbitalsThus, 1. [Ni(CO)4] molecule with sp3 hybridisation has tetrahedral structure.

2. The complex has no unpaired electrons, hence it is diamagnetic.

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CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO

CO

Ni CO CO CO


(a) dsp2 HYBRIDISATION: Structure of [Cu(NH3)4]SO4 At. No. of Cu is 29, its ground state electronic configuration is [Ar] 3d 10 4s1. The configuration of Cu+2 is [Ar] 3d9 4s0.

3d 4s 4pCu+2 [Ar]When strong NH3 ligands approach the Cu+2 ion, the unpaired 3d electron gets promoted to the 4Pz orbital. Thus,

3d 4s 4pCu+2 [Ar]

dsp2 hybridisationThe resulting vacant one 3d, one 4s and two 4p orbitals hybridise to give four equivalent dsp 2

hybrid orbitals. These four dsp2 hybrid orbitals of Cu+2 ion overlap with filled orbitals of NH3

ligands & form four coordinate bonds. 3d 4s 4p

[Cu(NH3)4]SO4 [Ar] dsp2 hybrid orbitals

Thus, 1. [Cu(NH3)4]SO4 molecule with dsp2 hybridisation

has square planar structure. 2. The complex has one unpaired electron, hence it is paramagnetic.

(b) d2sp3 HYBRIDISATION: Structure of K4[Fe(CN)6] At. No. of Fe is 26, its ground state electronic configuration is [Ar] 3d6 4s2. The configuration of Fe+2 is [Ar] 3d6 4s0.

3d 4s 4pFe+2 [Ar]When strong CN– ligands approach the Fe+2 ion, the unpaired 3d electrons are forced to pair up. Thus,

3d 4s 4p Fe+2 [Ar]

d2sp3 hybridisationThe resulting vacant two 3d, one 4s and three 4p orbitals of Fe+2 ion hybridise to give six equivalent d2sp3 hybrid orbitals. These six d2sp3 hybrid orbitals of Fe+2 ion overlap with filled orbitals of CN– ligands & form six coordinate bonds.

3d 4s 4pK4[Fe(CN)6] [Ar]

d2sp3 hybrid orbitals

Thus,

1. K4[Fe(CN)6] molecule with d2sp3 hybridisation has octahedral structure.

2. The complex has no unpaired electron, hence it is diamagnetic.

5. Write a note on inner orbital complex and outer orbital complex.A. INNER ORBITAL COMPLEX: During the complex formation, if the d-orbital involved in hybridisation

belongs to the inner (n – 1)d-sub shell, then it is called inner orbital complex.Eg: hexacyanoferrate ion [Fe(CN)6] –4

At. No. of Fe is 26, its ground state electronic configuration is [Ar] 3d6 4s2.The configuration of Fe+2

is [Ar] 3d6 4s0. 3d 4s 4p

Fe+2 [Ar]

CN – being strong ligands, force the unpaired 3d-electrons to pair up. The resulting vacant two 3d, one 4s and three 4p orbitals of Fe+2 ion hybridise to give six equivalent d2sp3 hybrid orbitals. These six d2sp3 hybrid orbitals of Fe+2 ion overlap with filled orbitals of CN– ligands & form six coordinate bonds.

3d 4s 4p

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H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

H3N NH3 +2

Cu+2 SO4–2

H3N NH3

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CN –

NC – CN – –4

4K+ Fe+2

NC – CN –

CN –

CHAPTER - 5 COORDINATION COMPOUNDS Fe+2 [Ar]

d2sp3 hybridisation

3d 4s 4p[Fe(CN)6] –4 [Ar]

d2sp3 hybrid orbitals

The complex has no unpaired electron, hence, it is diamagnetic.

OUTER ORBITAL COMPLEX: During the complex formation, if the d-orbital involved in hybridisation belongs to the outer nd-sub shell, then it is called outer orbital complex.

Eg: hexafluoroferrate ion [FeF6] –4 At. No. of Fe is 26, its ground state electronic configuration is [Ar] 3d6 4s2.The configuration of Fe+2

is [Ar] 3d6 4s0. 3d 4s 4p

Fe+2 [Ar]

F – being weak ligands, are unable to pair up the unpaired 3d-electrons. Hence, one 4s and three 4p and two 4d-orbitals of Fe+2 ion hybridise to give six equivalent sp3d2 hybrid orbitals. These six sp3d2 hybrid orbitals of Fe+2 ion overlap with filled orbitals of F – ligands & form six coordinate bonds.

3d 4s 4p 4d Fe+2 [Ar]

sp3d2 hybridisation

3d 4s 4p 4d[FeF6] –4 [Ar]

sp3d2 hybrid orbitals

The complex has four unpaired electron, hence, it is paramagnetic.

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Chapter - 5 Coordination Compounds