Friday, 15 November 2013

PICNIC and ENGLISH EXAM

Dear Friends,
                   The exam which was to be conducted on Monday has been postponed to Tuesday and the Physics Exam Remain on the same day.
Also the last date for submission of monet for PICNIC is on Tuesday. If you are interested, please bring it on that day.

PICNIC
   Malampuzha, Fantasy Park, Snake Garden
FEES
    Rs.500

Revised Time table
    Date          XII
19-11-13      English
20-11-13        Phy

Friday, 27 September 2013

Co ordination Compounds


 

Coordination Compounds

Introduction:

We have already seen the ability of transition elements to form coordination compounds. This property is however not restricted to transition elements alone. It is also exhibited by certain other metals although to a lesser extent. In this unit we will study the different aspects of coordination compounds including their application. We will also study about organometallic compounds

Double Salt and a Complex Ion:

Double salt: Ionizes completely in solution.    Example:       Mohr’s salt   Description: double salt ionizes completely

The complex ion: The complex ion does not dissociate.   Example:                  Description: complex ion does not dissociate

Co-ordination compounds:

Co-ordination compounds are compounds in which the central metal atom is linked to a number of ions or neutral molecules by co-ordinate bonds i.e., donation of electron pairs. They retain their identities even when dissolved in water

Example: Ni(CO)4,  [Co(NH3)6]Cl3,   K[Ag (CN)2]

Types of Complexes

1. Cationic complexes- Those in which the complex ion has a positive charge.  Example: [Co(NH3)6]3+

 

2. Anionic complexes - Those in which the complex carries a negative charge. Example: [Ag (CN)2]-

 

3. Neutral complex- The complex does not carry a charge.                                  Example: [Ni(CO)4]

 

Ligands

Ligand (ligare is Latin, to bind): A ligand is a molecule or ion that is directly bonded to a metal ion in a coordination complex. Molecule or ion having a lone electron pair that can be used to form a bond to a metal ion. A ligand uses a lone pair of electrons (Lewis base) to bond to the metal ion (Lewis acid)

The donor atoms, molecules or anions which donate an electron pair to the metal atom or the atoms or group of atoms that surround the central metal atom.  Example: NH3, H2O and CN-

 

 

 

 

 

 

 

 

Classification of Ligands:

 

1.    Based on denticity:

Unidentate/ monodentate - If one donor atom is present. Forms one bond to metal ion

 Example: NH3, H2O.

Bidentate - If two donor atoms are present on the ligand. Forms two bond to metal ion

            Example: NH2 - CH2 - CH2 - NH2     ethylene diamine.

 

Multidentate/ polydentate –If more than two atoms are present on the lignd. more than two bonds to a metal ion possible

 Example: Ethylene diamine tetra acetic acid (EDTA) is hexadentate

Some multidentate ligands

Description: http://images.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img16.gif

Description: http://images.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img17.gif

Description: http://images.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img18.gif

Description: http://images.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img19.gif

Description: http://images.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img20.gif

2.    Based on charge

Name
Formula
Charge
Name of Ligand
Ammonia
NH3
Zero
ammine
Water
H2O
Zero
aqua
Phosphine
PH3
Zero
Phosphine
Nitrogen oxide
NO
Zero
Nitrosyl
Carbon monoxide
CO
Zero
Carbonyl
Ethylene diammine
NH2-CH2-CH2-NH2 or (en)
Zero
Ethane-1,2-diamine
Pyridine
C5H5N
Zero
Pyridine(Py)
Triphenyl phosphine
(C6H5)3P or P(PH3)3
Zero
Triphenyl phosphine
Thiocarbonyl
CS
Zero
Thiocarbonyl
Thio urea
H2NCSNH2
Zero
Thiourea(tu)
Halide ion
X-(X=Cl,Br,I)
-1
halido
Hydroxide ion
:OH -
-1
hydroxo
Cyanide ion
:CN -
-1
Cyanato
Nitrite ion
NO2-
-1
Nitrito-N-
Nitrito ion
ONO-
-1
Nitrito-O-
Thiocyanate ion
SCN-
- 1
Thiocyanato-S-
Isothiocyanato
NCS-
-1
Thiocyanato-N-
Acetate ion
CH3COO-
- 1
Acetato
Nitrate ion
NO3-
-1
Nitrato
Amide ion
NH2-
-1
Amido
Oxide ion
O2 -
- 2
Oxo
Peroxide ion
O22 -
- 2
Peroxo
Carbonate ion
CO32 -
- 2
Carbonato
Sulphate ion
SO42 -
-2
Sulphato
Sulphide ion
S2 -
- 2
Sulphido
Thiosulphate
S2O32 -
- 2
Thiosulphato
Sulphite ion
SO32-
-2
Sulphito
Imide ion
NH2 -
- 2
Imido
Nitrosonium
NO+
+1
Nitrosonium
Nitronium
NO2+
+1
Nitronium

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3.    Chelate ligands: The ligands which when attach to central metal atom form a cyclic/ ring structures are called chelate ligands. It bind through more than one donor atom per ligand.

Example: en, ox, gly  etc

 

          The octahedral [Co(en)3]3+ is a typical en complex.

          Chelate effect: More stable complexes are formed with chelating agents than the equivalent number of monodentate ligands.


 

Co-ordination Number (CN):

The number of co-ordinate bonds formed with the central metal ion by the ligand or the no of ligands     surrounding the central metal atom.

Examples:

[Ag(CN)2]-                   [Cu(NH3)4]2+                           [Cr(H2O)6]3+   

C.N = 2                         C.N = 4                       C.N = 6

Co-ordination Sphere

The central atom and the ligands which are directly attached are enclosed in square brackets called the coordination sphere. The ionisable groups are written outside the brackets.

Example:               Description: examples for co ordination sphere

FG22_10-02UN 

Common Structures


Octahedral                                                                         Tetrahedral                              Square planar

 

The formation of a coordinate complex is a Lewis acid-base reaction

img120

 

 

Rules for Writing Chemical Formula of a complex:

Rules for formula writing:

1) Formula of the cation must be written first followed by anion.

2) The Co-ordination sphere is written in square brackets.

3) Within the co-ordination sphere the sequence is (Metal atom, anionic, neutral, cationic, ligand).

4) Polyatomic ligands are enclosed in parentheses.

Rules for Naming Co-ordination Compounds (IUPAC Nomenclature)

(1)  The cation is named first (whether simple or complex) then the anion.

                        [Co(NH3)6]Cl3                         hexaammine cobalt (III) chloride.

K2[PtCl6]                      potassium hexachloroplatinate (IV).

(2)  Name is started with a small letter and the complex part is written as one word.

 

(3)  For non-ionic or molecular complexes one word name is given

 

[Co(NO2)3 (NH3)3]      triamminetrinitrocobalt (III)

 


 name of the central metal atom along with its oxidation state in Roman numerals within a bracket.

(5)  Naming of ligands:

 Negative ligands end in - O.                          Example: CN- (cyano),  Cl- (chlorido)

Neutral ligands have no special ending: NH3 (ammine), H2O (aqua), CO (Carbonyl) ,NO (Nitrosyl).

Positive ligands end in-ium.    Example: NO2+ (nitronium)    NO+ (nitrosonium)

(6)   In unidentate ligands with more than one co-ordinating atoms (ambidentate ligands) the point of attachment is designated. 

Example: - SCN thiocyanato-S;         - NCS isothiocyanato-N

    - NO2 nitrito-N                     - ONO nitrito-O

     (7) If several ligands of the same type are present the prefix di, tri, tetra, penta and hexa are used.

          For complex ligands like ethane-1,2-diamine the prefix bis, tris and tetrakis are used.

Preference order

All ligands are named in alphabetical order. The prefix di, tri are not considered.

Example: [PtCl(NO2)(NH3)4] SO4

Tetra ammine chloronitro platinum (IV) sulphate.

 

 

 

 

Naming the complex

Ligands are named first followed by the metal atom. The ending of the name of the metal depends on the nature of the complex ion.

a) If the complex ion is a cation or non-ionic, the name of the central metal ion is written as such followed by its oxidation state, indicated by roman numerals (II, III, IV).

Example:

i) [Cu(NH3)4]SO4 Tetramminecopper (II) sulphate

ii) [Pt Cl4 (NH3)2] Diammine tetrachloroplatinum (IV).

b) If the complex ion is an anion the name of the central metal atom is made to end in -ate.

Example: Na3 [AlF6] sodium hexafluoro aluminate (III)

K2[Co(NH3)2Cl4]-              potassium diamminetetrachlorocobaltate(II)

[Co(NH3)4Cl2]Cl -              tetraamminedichlorocobalt(III) chloride

[Cr(NH3)6]3+ -                   hexamine Chromium (III) ion

[Ru(NH3)4(HSO3)2]-         bis(bisulfite)tetraamineruthenium(II)

K3[Fe(CN)6]-                       potassium hexacyanoferrate(III)

[Cu(NH3)4]SO4 -                                tetraaminecopper(II) sulfate

[Ti(H2O)6][CoCl6]-            hexaaquatitanium(III)hexachlorocobaltate(III)

 

Isomerism in Coordination Compounds

Two or more substances having the same molecular formula but different structural or spatial arrangement are called isomers.

These are of two types:          a) Structural isomerism          b) Stereo isomerism

Structural Isomerism

Two or more substances having the same molecular formula but different structural arrangement are called isomers.

(a)  Ionisation isomerism: Compounds, which give different ions in solution although they have same composition are called ionization isomers. There is an exchange in groups between the coordination sphere of the metal ion and ions outside the sphere.

Example:                                 Description: http://image.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img26.gif     Description: http://image.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img27.gif

Description: http://image.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img28.gif

               Description: example for ionization isomerism

The first compound gives a white precipitate with Barium chloride. The second compound gives a precipitate with AgNO3 and not with BaCl2.

(b)  Hydrate Isomerism:Compounds, which have the same composition but differ in the number of water molecules present as ligands (in the co-ordination sphere) and as molecules of hydration (present outside the coordination sphere). Example:

                                    Description: example for hydrate isomerism

 

 

 

 

 

 

 

(c)  Linkage Isomerism: It occurs when more than one atom in a mono-dentate ligand may function as a donor. Example:

Description: http://image.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img31.gif

Description: example for linkage isomerismDescription: http://image.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img32.gif

(d)  Coordination Isomerism: This type of isomerism is possible when both positive and negative ions of a salt are complex ions and the two isomers differ in the distribution of ligands in the cation and anion. Example:

Description: coordination isomerism in hexam min ecobalt III

Description: coordination isomerism in hexacyanocobaltate III

Stereo Isomerism or Space Isomerism

It arises because of the different positions and arrangements of ligands in space around the metal ion.

(a)  Geometrical Isomerism or Cis-trans Isomerism: When two identical groups (ligands) occupy adjacent positions, the isomer is called cis and when arranged opposite to one another the isomer   is called trans.

It is not possible for co-ordination number of 2 and 3 and for tetrahedral complexes of co-ordination number 4.

It is common in square planar and octahedral complexes.

Square planar complexes

Example: [Pt Cl2(NH3)2] diammine dichloro platinum (II).

Description: cis-trans isomerism in diammine dichloro platinum II

 

Octahedral complexes

Example: [CoCl2(NH3)4]+-      tetraammine dichloro cobalt (III) ion.

Description: cis-trans isomerism in tetraammine dichloro cobalt III ion

[CoCl2 (en)2]+

Description: cis-trans isomerism in CoCl2 en2

            [Co(NH3)4Cl2]+


 

 

 

 

 

0751

          In case of complexes of the type Ma3b3

          fac (facial) => three identical ligands occupying the corners of a common triangular

          mer (meridional) => three identical ligands occupying three consecutive corners of a square plane


(b)  Optical Isomerism: Optical isomers rotate the plane of polarised light in opposite directions.        The two isomers are mirror images of each other. The two mirror images are non superimposable, do not possess a plane of symmetry. They are called dextro and laevo      (d and l) rotatory depending upon the direction in which plane polarised light is rotated.

It occurs in only octahedral complexes with coordination number 6 with 2 or 3 bidentated ligands.

Examples:[CoCl2(en)2 ]+, [Co(en)3]3+ and [Cr(C2O4)3 ]3-

[CoCl2(en)2]+

It can exist in cis and trans forms. The trans form is symmetrical and optically inactive. Cis form exists in d and e forms.

Description: Non superimposable mirror images

Description: Non superimposable mirror images

Non superimposable mirror images

Bonding in Co-ordination Compounds

The first theory was called the Werner's theory of co-ordination compounds.

1) Metals possess two types of valencies:

a) Primary valency or ionizable valency. It is also referred to oxidation state.

b) Secondary valency which a metal atom or cation exercises towards neutral molecules or negative        groups (ligands) in the formation of complex ions.The secondary valency is also called the coordination number.

Example: In [Pt(NH3)6]Cl4 secondary valency of Pt is 6

2) Primary valencies are satisfied by negative ions, secondary valencies may be satisfied by negative ions or neutral molecules.

3) Ligands satisfying secondary valencies are directed towards fixed positions in space giving a definite geometry to the complex, but the primary valencies are non-directional.

Six valencies are directed towards a regular octahedron while four are directed towards either a tetrahedral manner or square planar.

Complex Structure using Werner's Theory

Example: In the complex between CoCl3 and NH3 the ionizable chloride ions are found by precipitation with AgNO3. The remaining Cl and NH3 are present around the central Co in such directions so as to minimize repulsion and are linked by secondary valencies.

a) Primary valencies - represented by dotted lines.

b) Secondary valencies - by solid lines.

c) Groups satisfying both primary and secondary valency - by solid and dotted lines.

            Description: structure of Co NH3 6 Cl3                        Description: structure of CoCl NH3 5 Cl2

 

 

               Description: structure of CoCl2 NH3 Cl                                        Description: structure of CoCl3 NH3 3



Werner was given the Nobel Prize in 1913 for his work in complexes.

Valence Bond (VB) Theory

I) The metal ligand bond arises by donation of pair of electrons by ligands to the central metal atom.

II) To accommodate these electrons the metal ion must possess requisite number of vacant orbitals of equal energy. These orbitals of the metal atom undergo hybridisation to give hybrid orbitals.

III) Sometimes the unpaired (n-1)d orbitals pair up before bond formation making (n-1)d orbitals vacant. The central metal atom makes available number of d-orbitals equal to its co-ordination number.

iv) The metal ligand bonds are thus formed by donation of electron pairs by the ligands to the empty hybridized orbitals. These bonds are equal in strength and directional in nature.

v) Octahadral, square planar and tetrahedral complexes are formed as a result of d2sp3 (or sp3 d2), dsp2 and sp3 hybridization respectively.

Square Planar Complexes

Description: http://image.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img53.gif

Description: square planar complexes

dsp2 hybridization - square planar

Diamagnetic - No unpaired electrons.

 

 

Description: dsp2 hybridization of diamagnetic

Inner and Outer Orbital Complexes

In the octahedral structure the central metal atom uses inner (n -1) d - orbitals or outer (n)d-orbitals for hybridization. This results in

Inner orbital complex:  involving (n-1)d orbitals for d2sp3 hybridization.

The electrons in the metal pair, so the complex is either diamagnetic or will have lesser number of unpaired electrons. Example: [V(H2O)6]3+, [Co(NH3)6]3+

Outer orbital complex: When the complex formed involves use of outer nd - orbitals sp3d2. It is called outer orbital complex. It has a large number of unpaired electrons.

Example: [CoF6]3-, [MnF6]3-, [Fe F6]3- because F- is a weak ligand, it cannot cause force pairing.

 

 

Drawbacks of Valence Bond Theory

1) It is a qualitative approach describing bonding in coordination compounds.

2) The theory fails to explain the optical absorption spectra and magnetic properties of coordination compounds.

3) The theory does not provide an answer to the origin of characteristic colors of complex ions.

4) It does not explain why [Co(NH3)6]3+ is an inner orbital complex and [CoF6]3- is an outer orbital complex

 

 

 

 

 

 

 

 

Crystal Field Theory:

Bonding in a complex ion is due to electrostatic interactions between the positively charged nucleus of the central metal ion and electrons in the ligands i.e., attractive as well as repulsive interactions.

MAIN POSTULATESOFCRYSTALSPLITTING FIELD THEORY:

a.            It considers the bond to be ionic arising purely from electrostatic interactions between the metal ion and the                 ligands.

 b.           It treats each ligand as a point of negative charge.

c.             Ligands arranges around metal ion in such a way so as to minimise repulsions

d.            In a free transition metal ion all the five d orbitals have equal energies DEGENERATE

e.            The attractive forces will arrive due to the positive metal ion and the negatively charged ligand.

f.          The repulsive forces arise between the lone pairs on the ligand and the electrons in the d-orbital of the metal.

d-orbitals and Crystal Field Splitting

The five d-orbitals can be divided into two groups depending upon the nature of their orientation in space.

(i) The three d-orbitals (dxy, dyz, dzx) which orient in the regions in between the coordinate axes are designated as t2g orbitals.

(ii) The two other orbitals (dx2 - y2 and dz2) which orient along the axes are labeled as e.g., orbitals.

Description: Orientation of five d orbitals in octahedral environment

Description: Orientation of five d orbitals in octahedral environment

Orientation of five d orbitals in octahedral environment

In the case of a free ion, all the five d-orbitals have the same energy, i.e., they are degenerate (meaning energetically alike).

When the ligand approaches the central metal ion, the electrons in d-orbital of the central metal ion will be repelled by the lone pairs of the ligands. As a result of these interactions the degeneracy of d-orbitals of metal ions is lost and these split into two set of orbitals having different energies. This is known as Crystal Field Splitting (D) and it forms the basis of crystal field theory.

The crystal field splitting depends upon the number of ligands approaching the central metal ions. That is, crystal field splitting will be different in different structures with different coordination numbers.

Crystal Field Splitting in Tetrahedral Complexes

The tetrahedral arrangement of four ligands surrounding the metal ions is as shown in the figure.

Description: Tetrahedral arrangement of four ligands

Tetrahedral arrangement of four ligands surrounding the metal ions

It is clear from the figure that none of the d-orbitals point exactly towards the ligands. The three d-orbitals dxy, dyz and dzx are pointing close to the direction in which ligands are approaching.

As a result of this, the energy of these three orbitals increases much more than the other two d-orbitals    (dx2 - y2 and dz2).

The d-orbitals will thus split as shown below:

Description: stages of d orbitals

Octahedral Complex Crystal Field Splitting

The octahedral arrangement of six ligands surrounding the central metal ion is as shown in the figure.

Description: Octahedral arrangement of six ligands surrounding the central metal ion

Octahedral arrangement of six ligands surrounding the central metal ion

 

 

Looking at the figure, we observe that as the ligands approach the x,y and z axis, the two d-orbitals lying along the axis namely dx2 - y2 and dz2 will suffer more electrostatic repulsions and hence their energy will be greater than the other three d-orbitals dxy, dyz and dzx,

Accordingly, the d-orbitals will thus split as follows.

Description: stages of splitting in d orbitals

fig 10.5

Magnitude of Crystal Field Splitting

Different ligands differ in their ability to produce a splitting of the d-orbitals.

For e.g., strong lewis bases, such as CN- and NH3, produce a strong field. They repel d electrons most strongly and cause a greater separation of the two groups (t2g and eg) of d-orbitals than do weak bases like H2O and F-.

The crystal field splitting is measured in terms of energy difference between t2g and eg orbitals and is denoted by symbol D. It is generally measured in terms of a parameter, Dq.

When a d-electron enters or resides in one of the t2g orbital, the energy of that orbital becomes 4Dq or 0.4D less or more than that of hypothetical degenerate orbitals.

Similarly, for each electron entering e.g., orbital, the energy of that orbital becomes 6 Dq or 0.6 D less or more, than that of degenerate orbitals.


 


         Spectrochemical Series:

         Ligands can be arranged into a spectrochemical series according to the magnitude of splitting of the d-orbitals

         Large splitting is associated with strong field ligands

         Small splitting is associated with weak field ligands

CN- > en > NH3 > H2O > F- >SCN- > Cl- > Br- > I-

 

Co-ordination Compounds Stability

Although a complex ion does not dissociate generally, it may sometimes dissociate to a small extent.

Example:

Description: dissociation of complex ion

The complex is in equilibrium with its ions in solution.

The instability constant

Description: instability constant of complex ion

The stability constant

Description: stability constant of complex ion

For a general expression

Description: formation of complex ion

a+, x- and b+ are the charge on the metal, the ligand and the complex ion respectively.

(a+) + n (x-) = (b+)

Description: stability of complex ion

Greater the value of K more is the stability.

Factors Affecting Stability of a Complex Ion

1) Charge on the central metal ion: Greater the charge, more the stability.

2) Basic nature of ligand: More the basic nature, more is the stability.

3) Presence of chelate rings: Formation of chelate rings increases the stability.

Example:

Description: http://image.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img67.gif

Description: http://image.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img68.gif

 

 

 

 

 

 

 

Preparation of Co-ordination Compounds

In this method a stronger ligand replaces a weaker ligand.

Description: preparation of co ordination compounds by substitution method

Direct combination

 

                                    Description: preparation of co ordination compounds direct combination

 

 

Redox method: In this method, a complex with the metal atom in a higher oxidation state than the starting material is obtained.

Example:

Description: preparation of co ordination compounds by redox method

Pentaamminenitro cobalt (III) nitrate is prepared by adding NH3 and NH4NO3 to cobalt (II) nitrate solution in the presence of H2O2.

 

Applications of Co-ordination Compounds

1) Estimation of hardness in water:    Ca2+ and Mg2+ ions water can be estimated using EDTA

2) Animal and plant world:      Chlorophyll - Contains Mg complex

Haemoglobin - Contains Fe complex

Vitamin B12 - Contains Co complex

3) Electroplating of metal: Co-ordination compounds of gold and silver are used in electroplating both for controlled deposition of metal ions.

Example: K[Ag(CN)2]

4)Extraction of metals

Au and Ag are extracted from their ores through formation of cyanide complexes, [Ag(CN)2]- and       [Au(CN)4]-

5) Estimation of Ni(II) as red glyoxime using dimethyl glyoxime.

 

 

 

 

 

 

 

 

Synthesis of Organometallic Compounds

i) Organometallic compounds and Grignard reagent.

Description: preparation of s - Bonded Complexes

Description: http://images.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img78.gif

ii) Tetra ethyl lead and tetra alkyl lead obtained by reaction of metal halide with organometallic compound.

Description: preparation of s - Bonded Complexes

Description: http://images.tutorvista.com/contentimages/chemistry_12/content/us/class12chemistry/chapter05/images/img80.gif

(Bu = C4Hg)

(Et = C2H5)

TEL (Tetra Ethyl Lead) used as an antiknock agent is prepared as follows:

Description: preparation of tetra ethyl lead

Preparation of p-Complexes

 

1) Zeise's salt

Description: preparation of zeise s salt

2) Ferrocene

Description: preparation of ferrocene

3) Dibenzene chromium

Description: preparation of dibenzene chromium


Direct combination

Description: preparation of nickel tetracarbonyl

Description: preparation of ferric pentacarbonyl

Reaction between Ni and CO is used in the Mond's process for purification of Ni.

 

 

Applications of Organometallic Compounds

Many reactions in solution are catalyzed by organometallic compounds or transition metal complexes.

Example: Wilkinson's catalyst

(Ph3P)3RhCl - Catalyst in hydrogenation of alkenes.

Heterogenous Catalysts

Ziegler - Natta, example trialkyl aluminium + TiCl4 used for low temperature polymerization of alkenes.

Ethyl mercury chloride - To prevent infection in small plants.

 

Coordination Compounds Conclusion

Transition elements are known for their ability to form many complex compounds. The complex compounds in which the metallic ion is surrounded by two or more ions or molecules are called coordination compounds, also known as complex compounds. The chapter has covered the nomenclature, isomerism, bonding, applications and stability of coordination compounds and organometallic compounds.

 

VERY SHORT ANSWER TYPE QUESTIONS:   ( 1  Mark each)

 

01)  What is the coordination number of central metal ion in  [Fe(C2O4)3]3-  ?

       Ans =  6

 

02)  Write IUPAC name of [ Pt Cl2 (NH3)4 ] 2+.

       Ans =  dichloride  tetraannine  platinum (IV) ion.

 

03) Name the type of isomerism that occurs in complex in which both cation and anion are complex ions.

      Ans = Co-Ordination isomerism.

 

04)  Write IUPAC name for ionization isomer [Co (NH3)5 SO4]Br.

      Ans =  The ionization isomer of the given complex compound is     [Co (NH3)5 Br] SO4.

 Its IUPAC name is pentaamminebromidocobalt.(III) sulphate

 

05)  How many moles of AgCl will be precipitated when an excess of AgNO3 is  added to a molar solution of [Cr(H2O)5Cl]Cl2.

       Ans =  2 moles of AgCl will be precipitated .

 

06)   If the geometry of  (PtCl4)2- is square planar, which orbitals of Pt are involved in the bonding ?

        Ans =   dsp2 hybridization is taking place involving 5d, 6s and 6p.

 

07)   Which of the two, K4 [Fe(CN)6] or K3 [Fe(CN)6] is more stable ?

        Ans =  K3[Fe(CN)6] is more stable.

 

08)   What is the coordination number of a central metal ion in an octahedral complex ?

         Ans =    6

 

09)   How many ions are produced from the complex [Co(NH3)6]Cl2 in solution ?

        Ans = 03 ions are produced.

 

10)   Amongst the following ions which one has the highest magnetic moment values ?

 

        a)      [Cr(H2O)6]3+             (b) [Fe(H2O)6]2+             (c) [Zn(H2O)6]2+.

 

        Ans =      a)   Cr3+     :   [Ar]3d3        µ  =   √3 (3+2)   =    √15  = 

 

                       b)    Fe 2+    :   [Ar]3d6        µ   =  √4 (4+2)   =    √24  = 

     

                       c)    Zn2+   :   [Ar] 3d10      µ   =   0

 

 

SHORT ANSWER TYPE QUESTIONS ( 02 Marks Each).

 

01)       A coordination  compound has the formula CoCl3. 4NH3. It does not librated NH3 but precipitates chloride ion as AgCl. Write the IUPAC name of the complex and write its formula.

 

Ans =  [CO(NH3)4Cl2]Cl

            IUPAC name :             Tetraamminedichloridocobalt(III) chloride.

 

 

02)       Using valence bond theory of complexes, explain the geometry and magnetic nature of [Cr(H2O)6]3+.

Ans =  Cr  has oxidation state of +3

            Its configuration is [Ar]3d3.

 

                         3d                              4s                     4p                

1
1
1
 
 

 

 

 

 
 
 

  Cr3+

 

As a weak ligand, H2O approaches towards it, the pairing against Hund’s rule does not take place and thus 3d orbitals participate in hybridization.

 

                                    3d                        4s                     4p                

1
1
1
xx
xx

 

xx

 

xx
xx
xx

[Cr(H2O)6]3+

 

                                                            d2sp3 hybridization.

 

As the hybridization is  d2sp3, the shape of the ion is octahedral since, it contains 3 unpaired e-s is paramagnetic in nature.

 

03)       Explain the following :

a)      [Co(NH3)6]3+ is diamagnetic, while [CoF6]3- is paramagnetic.

b)      [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.

 

     Ans =  a)  [ Co (NH3)6]3+, co exists as Co3+.

               

                 Co3+  -  [ Ar] 3d6.

 

                                                   3d                                           4s                   4p


 

 

 

 
 
 

               Co3+    

 

 

               As strong ligand NH3 approaches, the  e-s tend to pair up against the

               Hunds rule, thus.

                                                              3d                       4s               4p                             4d

 
 

 

 
 
 

 

 

 

 
 
 
 
 

           [Co(NH3)6]3+

           

                                      

                                                                           d2sp3

             But in [CoF6]3-, a weak ligand  F- approaches, thus pairing against Hund,s

             rule does not, take place.                                         

          

                                                        3d                               4s               4p                      4d


 

 
 
 

 

 

 

 
 
 
 
 

           [CoF6]3-

           

                                      

                                                                          sp3d2.

 

            Thus due to absence of unpaired e-s in [Co(NH3)6]3+, it is diamagnetic but

            dues to presence of unpaired e-s in [CoF6]3-, it is paramagnetic.

 

        Ans =   b)   [Ti (H2O)6]3+, Ti  exists as Ti3+

                           Ti 3+    [Ar]3d1

 

                           While Sc exists as Sc3+ in [Sc(H2O)6]3+ with no unpaired e-.

Due to presence of un paired e- in [Ti (H2O)6]3+, it undergoes d-d transition & hence appears coloured.

 

  04)  Mention application of coordination compound in

              a)  metallurgical operations.

              b)  medicine.

    Ans =  a)  metals like Ag & Au are extracted by the formation of their cyanide complexes.

 

                     Ag2S + NaCN                                  Na  [ Ag (CN)2]  + Na2S

                     2Na [ Ag (CN)2] + Zn                     Na2 [Zn (CN)4     + 2 Ag.

 

      Ans =  b)   cis – platin  i.e, cis [ PtCl2 (NH3)2] is used in cancer chemotherapy.

 

    05)   Name the isomerism associated with these coordination compounds.

            [Cr(CH3)5 Cl]SO4  and [Cr(NH3)5  SO4]Cl. State a chemical test to distinguish between these two

isomers.  

 

    Ans =  They are ionisation isomers. They can be differentiated by adding BaCl2  solution in their

aqueous solutions.

  

                        [Cr(NH3)5Cl]SO4     aq    [ Cr(NH3)5Cl]2+ +   SO42-    

         

                                                                                         BaCl2                                                                                                                                                                                                                                                                                 

 

                                                                                              BaSO4

                                                                                                                                        (White  ppt)

 

             While [Cr(NH3)5 SO4 ] Cl does not form white ppt when ag  BaCl2 is added to its aq. solution.

 

  06)     Draw fac and mer isomers of [Co(NH3)3Cl3].

           

                                              NH3                                                                             NH3

   Ans                 Cl                                    NH3                                    Cl                                    NH3

 

                                              CO                                                                             CO

 

                          Cl                                   NH3                                       Cl                                    Cl

                                      

                                               Cl                                                                               NH3

 

           Fac – isomer                                                           Mer – isomer

 

 

 

 

 

SHORT ANSWER TYPE QUESTIONS  ( 03 Marks )

 

01)    Draw a figure to show splitting of degenerate d orbitals in an octahedral crystal field. How does the  magnitude of  0  decide the  configuration of d orbitals in a complex entity ?

      

Ans : Refer note                                                                                                                                                   

                          

      02)   The value  of  dissociation  constant  of  [Cu(NH3)4]2+  and [Co(NH3)6]3+  are 1 x 10-12,  and                   

             6.2 x 10-36 respectively, which complex would be more stable and why ?

 

Ans =  [Cu(NH3)6]3+ would be more stable as its dissociation constant is less and  dissociation constant 

            is  reciprocal  of  stability  constant.  Thus  ß 4 (stability constant )  =  1/Kd   (dissociation const)

 

 

HIGHER ORDER THINKING QUESTIONS:

 

01)   A  cationic complex  has 2  isomers  (A) and  (B) Each  one  has  one Co3+, five NH3,   One Br- and

one SO42-.  (A) gives a white ppt with BaCl2 solution while  (B) gives a yellow ppt with AgNO3 solution.

         a) What are the possible structures of (A) and (B) ?

         b) Will the two complexes have same colour ?

 

Ans   a)   (A)  gives a white ppt with BaCl2, SO42- ion is present in its ionization

                sphere   and thus  its structure is [ Co(NH3)5 Br]SO4.

 

                (B)gives a  yellow ppt with AgNO3 solution,  … Br- ion must be

                 present in its ionization sphere…. Its structure is [Co(NH3)5 SO4]Br.

 

         b)    No, the two complexes will not have the same colour.

 

02)    A student prepared following coordination complexes containing pt  having the following

characteristics.

 

          Formula                                              Cations present in solution per   formula   unit.

         (A)   Pt Cl4   . 6 NH3                                                                  4

         (B)   Pt  Cl4  . 5 NH3                                                                  3

         (C)    Pt  Cl4  . 4 NH3                                                                 2

 

  a)  Write the structures of these comnplexes.

  b)  Write IUPAC name of the  complex (B)  & (C).

 

Ans  =   a)   A –  [ Pt  (NH3)6] Cl4.

                     B –  [ Pt  (NH3)5 Cl ]  Cl3.

                     C –  [ Pt  (NH3)4 Cl2 ] Cl2.

 

            b)   IUPAC name of :

                   B -   pentaamminechloridoplatinum    (IV) chloride

                   C –  tetraammine dichloridoplatinum  (IV) chloride.

 

03)   A metal ion Mn+ having d5 valence electronic configuration combines with 3 didebtateligands to form a complex compound. Assuming      0 >  P,

  

i)        draw the diagram assuming d – orbital splitting during the complex formation.

         ii)    Write the electronic configuration of valence electrons of the metals Mn+  ion in  terms of                       

                    t2g & eg.

iii)                What type of hybridization will Mn+  ion have  ?

Ans = (i) figure                                                                                                                                                                                                                                                            


 


ii)   t2g5eg0    :.       0   >  P,  as   P ( Pairing energy ) is less than ∆0   pairing   of e- takes place in lower 

      energy t2g level.

           

iii)   Mn+  ion  will have d2sp3  hybridisation as its coordination no. is 6 due to its tendency to   

     combine with 3 bidentate ligands and also due to presence of  vacant orbitals.

 

04)   Arrange the following in increasing order of electrical conductivity.

         [Co(NH3)3 Cl3],  [Co(NH3)5 Cl]Cl2,   [Co(NH3)6]Cl3,    [Co(NH3)5Cl]Cl.

 

Ans =   [Co(NH3)3Cl3] <   (Co(NH3)5 Cl ]Cl <   [Co(NH3)5Cl]Cl2 <   [Co(NH3)6]Cl3.

 

 

 

 

*************