COORDINATION COMPUND CLASS XII CHEMISTRY STUDY MATERIAL

L –9

COORDINATION COMPOUNDS

BASIC CONCEPTS:

1.DEFINITION  : Coordination compounds

Coordination compounds are compounds in which a central metal atom or ion is linked to a number of ions or neutral molecules by coordinate bonds/donation of 1 or more electron pairs . Examples- K₄[Fe(CN)₆]; [ Cu(NH₃)₄]SO₄; Ni(CO)₄

2. HISTORY :   First studied by Alfred  2Werner Who proposed his “ theory of coordination compounds”.Accordingly

i) In coordination compounds metals show two types of linkages or valencies- Primary and Secondary.

ii) The primary--- ionisable and  satisfied by negative ions.

iii) The secondary valencies - non- ionisable satisfied by neutral molecules or negative ions. It is equal to the C.N and is fixed for a metal. The ions or groups bonde by secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination nos.

3.Difference between a double salt and a complex  

COMPLEXES: They do not dissociate in to simple Ions when dissolved in water Example: K₄[Fe(CN)₆] contains a central metal atom or ion surrounded by number of oppositely charged ions or neutral molecules-(Ligands).

DOUBLE SALTS: They  dissociate into simple ions completely when dissolved in water. Ex.- Karnelite KCl.MgCl₂.6H₂O, Mohr‘s salt, FeSO₄.(NH₄)₂SO₄.6H₂O, potash alum, K₂SO₄Al₂(SO4)_3.24H₂O, etc. dissociate into simple ions completely when dissolved in water.

4. IMPORTANT TERMS

(I )Coordination entity: It constitutes the central metal ion or atom bonded to a fixed number of ions or molecules represented within a square bracket. [Fe(CN)₆]^-

(ii) Central M atom/ ion: In a coordination entity, the atom/ion to which a fixed number of ions/groups are bound in a definite geometrical arrangement is called the central atom or ion.

iii) Ligands: The neutral molecules or negative ions bound to the central metal or ion in the coordination entity. These donate a pair /s of electrons to the central metal atom /ion.Ex- NH₃, CN^-

5. LIGANDS MAY BE CLASSIFIED AS-

a) Monodentate /Unidentate: Ligands bound to the central metal atom/ion through a single donor atom. Ex- Cl^- ; H₂O ; NH₃ ; NO₂^-.

b) Didentate/Bidentate: The Ligands which bind to the central metal atom/ion through two donor atoms. Ex- C₂ O₄ ^2- (ox); H₂NCH₂CH₂NH₂(en)

c) Polydentate: The Ligands which bind to the central metal atom/ion through two or more donor atoms present in a single ligand. Ex- (EDTA)4-

d) Chelating ligands: Di- or polydentate ligands that uses two or more donor atoms to bind to a single metal ion to form ring- like complexes. (Ox); (EDTA)

e) Ambidentate ligand: A ligand that can ligate through two different atoms, one at a time. Ex-CN^-,NO₂^- ; SCN^-

v) Coordination number: The no. of ligand/ donor atoms to which the metal is directly bonded through sigma bonds only.  Commonly 4 or 6.

vi) Counter ions: The ionisable groups written outside the square bracket. Ex- K+ in

K₄[Fe(CN)₆]  OR  Cl^- in [Co(NH₃)₆]Cl₃

6. COORDINATION POLYHEDRON: The spatial arrangement of the ligand atoms which are directly attached to the central metal atom/ion. They are commonly Octahedral, Square-planar or Tetrahedral

7.OXIDATION NUMBER: The charge that the central atom would carry if all the ligands are

removed along with their pairs of electrons shared with the central atom. It is represented in parenthesis.

8.  HOMOLEPTIC COMPLEXES: Complexes in which a metal is bonded to only one kind of donor groups. Ex- [Co(NH3)₆] ³⁺

9. HETEROLEPTIC COMPLEXES: Complexes in which a metal is bonded to more thanone kind of donor groups. Ex- [Co(NH₃)₄ Cl₂]⁺

10. NOMENCLATURE OF COORDINATION COMPOUNDS

Rules –

        i.            The cation is named first in both positively and negatively charged coordination entities.

      ii.            The ligands are named in an alphabetical order before the name of the central atom/ion

    iii.            The name of the anionic ligands end in –o, those of neutral and cationic ligands are

the same except aqua for H2O, ammine for NH3, carbonyl for CO and nitrosyl for NO, placed within enclosing marks .

    iv.            The prefixes mono, di, tri, etc., are used to indicate the number of the individual ligands in the coordination entity. When the names of the ligands include a numerical prefix, then the terms, bis, tris , tetrakis are used, the ligand to which they refer being placed in parenthesis.

      v.            Oxidation state of the metal in cation, anion, or neutral coordination entity is indicated by roman numeral in parenthesis.

    vi.            If the complex ion is a cation , the metal is same as the element.

  vii.            For the neutral complex molecule is named similar to that of the complex cation.

11. LIGANDS NAMES OF SOME COMMON

Anionic Ligand Name		CHARGE		NEUTRAL LIGANDS		CHARGE
CN-	Cyano		-1	NH3	Ammine		0
Cl-	Chlorido		-1	H2O	Aqua/aquo		0
Br-	Bromido		-1	NO	Nitrosyl		0
F-	Fluoride		-1	CO	Carbonyl		0
SO42-	Sulphato		-2	PH3	Phosphine		0
C2O42-	Oxalato		-4	CH2-NH2 | CH2NH2	(1,2-Ethane diamine)		0
NH2-	Amido		-1
NH2-	Imido		-2
ONO-	Nitrito		-1
NO2-	Nitro		-1
NO3-	Nitrato		-1
SCN-	Thiocyanato		-1
NCS-	Isothiocyanato		-1
EDTA			-4

12.ISOMERISM  IN COORDINATION COMPOUNDS

Two or more substances having the same molecular formula but different structural formula or spatial arrangements are called isomers and the phenomenon is called isomerism

IN Coordination compounds show -

A)    Structural Isomerism

B)      Stereoisomerism

STRUCTURAL ISOMERISM:- due to the difference in structures of coordination compounds. It is further subdivided into the following types-

        i.            Ionisation isomerism: This isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion.Example - the ionization isomers:[Co(NH₃)5SO4]Br and[Co(NH3)5Br]SO4.

      ii.            Hydrate or solvate isomerism: This isomerism is known as hydrate isomerism‘ in case where water is involved as a solvent. This is similar to ionisation isomerism. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice. An example is provided by the aqua complex [Cr(H2O)6]Cl3 (violet) and its solvate isomer [Cr(H2O)5Cl]Cl2.H2O (grey-green).

    iii.            Linkage Isomerism: Linkage isomerism arises in a coordination compound having ambidentate ligand. A simple example-Complexes containing the thiocyanate ligand SCN- which may bind through the N to give M–NCS or through S to give M–SCN.

    iv.            4). Coordination isomerism: It arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.  Example [Co(NH3)6][Cr(CN)6] &  [Cr(NH3)6][Co(CN)6]

STEREOISOMERISM: Stereo isomers have the same chemical formula and chemical bonds but they have different spatial arrangement. This isomerism is of two kinds

A. Geometrical isomerism

B. Optical isomerism

1.      GEOMETRICAL ISOMERISM- This isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. Important examples of this behaviour re found with coordination numbers 4 and 6.

Ex-1, In a square planar complex of formula [MX2L2] (X and L- unidentate), the two ligands X may be arranged adjacent to each other in a cis or opposite to each other in a trans isomer 

Ex2 Name.2,  [MABXL]-Where A,B,X,L are unidentates Two cis- and one trans isomers are possible.

 Facial (fac) and Meridional isomer Another type of geometrical isomerism occurs in octahedral coordination entities of the type [Ma3b3] like [Co(NH3)3(NO2)3]. If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face, we have the facial (fac) isomer. When the positions are around the meridian of the octahedron, we get the meridional (mer) isomer.

2.      OPTICAL ISOMERISM: Optical isomers are mirror images that cannot be superimposed on one another. These are called as enantiomers. The molecules or ions that cannot be superimposed are called chiral.

The two forms are called dextro (d) and laevo (l).Dextro rotates plane of polarised light  to the right, Laevo to the left).  Optical isomerism is common in octahedral complexes involving didentate ligands. In a coordination entity of the type [CoCl2(en)2]2+, only the cis-isomer shows optical activity.

 

Type of hybridisation	Type of hybridisation	Acquired geometry
4	sp3	Tetrahedral
4	dsp2	Square planar
5	sp3d	Trigonal bipyramidal
6	sp3d2	Octahedral
6	d2sp3	Octahedral

8.)  Valence bond Theory:- Example in which bonding explained on the basis of  VBT using the concept of hybridisation . This theory also explains low spin & high spin complexes. This theory also accounts for magnetic behaviour of complexes based on presence of unpaired  electrons ,.VBT also explains bonding in complexes using the concept of a strong & weak ligands which affects  the pairing of electron

8.CRYSTAL FIELD THEORY: CFT)

1.      The metal-ligand bond is ionic arising purely from electrostatic interactions between the metal ion and the ligand.

2.      Ligands are treated as point charges or dipoles in case of anions and neutral molecules.

3.      In an isolated gaseous metal atom or ion the five d-orbitals are degenerate.

4.      Degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal /ion.

5.      In a complex the negative field becomes asymmetrical and results in splitting of the  d-orbitals.

A)    CRYSTAL FIELD SPLLITING IN OCTAHEDRAL COORDINATION ENTITIES

1.      For d4 ions, two possible patterns of electron distribution arise:

                    i.            If Δo < P, the fourth electron enters one of the eg orbitals giving the configuration t 2g e1g . Ligands for which Δo < P are known as weak field ligands and form high spin complexes.

                  ii.            (ii) If Δo > P, it becomes energetically favourable for the fourth electron to occupy a t2g orbital with configuration t42g e0g. Ligands which produce this effect are known as strong field ligands and form low spin complexes.

B.CRYSTAL FIELD SPLLITING IN TETRAHEDRAL COORDINATION ENTITIES

1. The four surrounding ligands approach the central metal atom/ion along the planes between the axes.

2. The t2g orbitals are raised in energy (2/5) ∆t.        

3. The two eg orbitals are lowered in energy (3/5) ∆t.

4. The splitting is smaller as compared to octahedral field splitting ∆t=5/9∆_o.

5. Pairing of electrons is rare and thus complexes have generally high spin configurations.

BONDING IN METAL CARBONYLS

1.      The metal-carbon bond in metal carbonyls possess both σ and π character.

2.      The M–C σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal..

3.      The M–C π bond is formed by back the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding π* orbital of carbon monoxide.

4.      The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal .

EX-  Fe (CO)5, Ni(CO)4, Cr(CO)6 …..

IMPORTANT SUB TOPICS OF CO ORDINATION COMPOUNDS

NOMENCLATURE OF COORDINATION COMPOUNDS

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

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

LIGANDS NAMES OF SOME COMMON

NH3	Ammine	0
H2O	Aqua/aquo	0
NO	Nitrosyl	0
CO	Carbonyl	0
PH3	Phosphine	0
CH2-NH2 | CH2NH2	(1,2-Ethane diamine)	0

3.ISOMERISM IN COORDINATION COMPOUNDS

STEREOISOMERISM

1.      Geometrical isomerism

2.      Optical isomerism ( NOTE :Concept is more difficult for slow learners)

a.       Facial (fac) and Meridional isomer

4 .CRYSTAL FIELD THEORY: CFT

This concept is very high level for slow learner. He/ She gets confused in understanding splitting of  d orbitals  in tetrahedral & Octahedral  complexes

3 .SURE SHOT QUESTIONS WITH ANSWERS (5a)

Q.1. Name the following complex using IUPAC norms:

                        [Co(en)₂(ONO)Cl]Cl

Ans 1: Chlorbosis (ethylene diamine ) nitro cobalt (III) chloride

Q.2.Give names of two complexes which are used in medicines.

Ans:     (i) EDTA , used in the treatment of lead poisoning.

            (ii) cis-platin [Pt(NH₃)₂Cl₂] used in the treatment of cancer

Q.3. Write the formula of copper hexacyanoferrate  (II).

Ans: Cu₂[Fe(CN)₆]     

Q.4. Give an example of chelate complex.

Ans: [CO(en)₃]³⁺

Q.5. Name the hybridisation  and the orbitals involved in the shape of [Ni(CN)₄]^2-.

Ans:     Hybridisation =dsp2

            Orbitals involved 3d, 4s, and 4p and shape is square planar.

Q.6.     (a) Give the IUPAC name of [PtCl(NH₂CH₃)(NH₃)₂]Cl

            (b) Compare the magnetic behaviour of the complex entities [Fe(CN)₆]^4- and          [FeF₆]^3-

Ans:     (a) Diamminechloro (methyl amine) platinum (II) chloride

            (b) Oxidation state of Fe in [Fe(CN)6]4-

              O.S. of  Fe = +2

Q.7.     (a) Write the IUPAC name of the ionisation isomer of [Ni (NH₃)₅NO₃] Cl

(b) How are coordination compounds useful in biological processes ? (Give two uses).

Ans:     (a) PentaammineChloridonickel (II) nitrate

            (b) (i) Haemoglobin a complex compound of iron acts as oxygen carrier in human   body.

            (ii) Chlorophyll, a complex of Mg helps in photosynthesis.

Q.6.     Write the IUPAC name and draw the structure of coordination entities of Cl(NH₃)₅]Cl₃.

Ans: IUPAC Name Pentaamminechlorido platinum (IV) chloride

            It has octahedral structure

 

                                             NH₃                    Cl                NH_{3          3+}

Pt

 

               

                                               NH₃            NH₃                  NH₃

Q.7.     write the chemical formulae of the following complexes.

                        (i) Hexaammine platinum (IV) choride.

                        (ii) Tetraamminedichoro cobalt (III) ion.

Ans:     (i) Pt[(NH₃)₆]Cl₄                      (ii) [CoCl₂(NH₃)₄]⁺

Q.8. How would you account for the following?

            (a) [Ti(H₂O)₆]³⁺ is coloured while [SC (H₂O)₆]³⁺ is colourless.

            (b) [Fe(CN)₆]³ is weakly paramagnetic while [Fe(CN)₆]^4- is diamagnetic.

            (c) Ni(Co)₄ posses tetrahedral geometry while [Pt(NH₃)₂Cl₂] is square planer and

diamagnetic.

Ans;     (a) [Ti(H₂O)₆]³⁺ is coloured because it has unpaired electrons and it undergo d-d

transition.

             [SC (H₂O)₆]³⁺ is colourless because it has not having number of unpaired

            electrons that’s why it does not undergo d-d transition        

(b) [Fe(CN)₆]³ is weakly paramagnetic because of the presence of one unpaired

            electron.

            [Fe(CN)₆]^4- is diamagneticdue to the absence of unpaired electron. 

(c) Ni(Co)₄has sp³ hybridisation, therefore, the shape is tetrahedral.

            [Pt(NH₃)₂Cl₂] has dsp² hybridisation therefore it is square planer and due to

            absence of unpaired electron it is diamagnetic.

Q.9      What are the two types of organometallic compounds?? Give

examples.

Ans: An organometallic is a compound which contains at least one of the following bonds.

Metal  -  Carbon

Metalloid (B,  Si,  As,  Te) – Carbon.

Q.10. Write the IUPAC name of K3[Fe(C₂O₄)₃]

Ans 10 .Potassium  trioxalato ferrate III

SURE SHOT QUESTIONS (SSQ)  (5b)

Q1.. Give the electronic configuration of the,

a)      d – orbitals of Ti on [Ti(H₂O)₆]³⁺ion in the octahedral crystal field.

b)      Why is the complex coloured? Explain on the basis of distribution of electrons in the d orbitals.

c)    How does the colour change on heating [Ti(H₂O)₆]³⁺ion?

Ans: (a) In [Ti(H₂O)₆]³⁺ion

            Oxidation state is Ti=+3

            There is only one electron in the d-orbital and its configuration is t_2g¹ e_g⁰

            (b) Due to d-d transition configuration becomes t_2g⁰ e_g^1.

            (c) On heating [Ti(H₂O)₆]³⁺ion becomes colourless as there is no ligand (H2O) left in

heating. In the absence of ligand, crystal field splitting does not occure.

Q.2.     Give a suitable example for each, example the following:

                                i.            Crystal field splitting.

                              ii.            Linkage isomerism.

                            iii.            Ambidentate ligand.

Ans: (i) In most of the transition metal complexes, either six or four ligands surround the   metal/ion. It gives rise to octahedral and tetrahedral structures. In both cases, the         field produced by the ligands is not spherically symmetrical. As a result, d-orbitals   arenot equally affected by the field. It results in spiltting the d-orbital energies.

(ii) Linkage isomerism. It occurs in complexes when an ambidentateligand  occurs in         the co-ordination sphere.

e.g. [CO(NH₃)₅NO₂]²⁺            and [CO(NH₃)₅(-DNO)]²⁺

Another example is

[Cr(H₂O)₅(SCN)²⁺                   and [Cr (H₂O)₅NCS)]²⁺

(iii) Ambidentateligand : Ligands which can ligate through two different atoms

Present in it are called ambidentate ligands, e.g.

NO₂^-,  SCN^-    

Q.3      Compare the following complexes with respect to structural shapes of units, magnetic  behaviour and hybrid orbitals involved in the units.

[Co(NH₃)₆]³⁺, Cr[NH₃)₆] ³⁺, [Ni(CO)₄]

[At No: Co=27, Cr=24, Ni=28].

Complexes	Magnetic Behaviour	Hybridisation	Shape
(i)  [Co(NH3)6]3+ (ii) [Cr(NH3)6]3+ (iii) [Ni(CO)4]	Diamagnetic Paramagnetic Diamagnetic	d2sp3 d2sp3 sp3	octahedral octahedral Tetrahedral

.4         (a) Draw the structure and write the hybridisation state of Co incis – [Co(NH₃)₄Cl₂]⁺

            (b) Using the IUPAC norms name the following complex:  [Co(NH₃)₄Cl(ONO)]Cl

Ans:(a) Structure of  cis – [Co(NH₃)₄Cl₂]⁺

                       

                                                            Cl                          +

                                        NH3                             Cl

                                                              Co

                                        NH3                                     NH3

                                                           

          NH3

            (b) Cis-tetramminedichloro cobalt (III) ion.

                        We have                                 

                                                3d                       4s             4p

Fe atom (Z=26)

In the ground state

Fe²⁺ ion                                     

d²sp³ hybridised                                                       

                                                Six empty d²sp³ hybrid orbitals

××

××

××

××

××

××

Formation of

Fe(CN)6]^4- 

                                                                        six pairs of electrons from six CN^- ions

Q.5.     Square planer complexes with a co-ordination number 4 exhibits geometrical isomerism  whereas tetrahedral complexes do not, Why?

Ans: The tetrahedral complexes do not show geometrical isomerism because the relative positions of the atoms with respect to each other will be the same . The square planner  complexes on the other hand show geometrical isomerism because if same kind of  ligands occupy positions adjacent to each other it is called cis-form and if these are opposite to each other it is called trans-form.

Q.6      (a) A coordination compound has the formula CoCl₃ . 4Nh₃. It does not liberate ammonia but forms a precipitate with AgNO3. Write the structure and IUPAC name of the complex compound.

            (b) Name the ligand which is bidentate and give an example of the complex formed by this ligand.

Ans:     (a) [C0(NH3)4Cl2]Cl

            (b) Ethylene diamine (en) is bidentate ligand [Co(en)3]3+. Its IUPAC name is tris (ethylenediamine) cobalt (III) ion.

Q.7.     (a) Write the formulae of the following complexes :

            (i) Hexaammine platinum (IV) chloride

            (ii) Dichlorotetraammine cobalt (III) ion.

(b) The values of dissociation constant of [Cu(NH₃)₄]²⁺ and [Co(NH₃)₆]³⁺ are 1.0 ×10^-12 and 6.2 ×10 ^-36 respectively. Which complex would be more stable and why?

Ans:     (a) (i) [Pt(NH₃)₆]Cl⁴                (ii) [CoCl₂(NH₃)₄]⁺

            (b) [Co(NH₃)₆]³⁺is more stable because it has low value of dissociation constant, therefore ,           will have high stability constant.

Q.8.     Write the formulas for the following coordination compounds.

            (i)  potassiumtetrahydroxo zincate(II)

            (ii) potassium trioxalatoaluminate(III)

            (iii) dichloridobis cobalt III (ethane-1, 2 diamine)

Ans:     (i) K₂ [Zn(OH)₄]

            (ii) K₃[Al(C₂O₄)₃]

            (iii) [CoCl₂(en)₂]⁺

Q9.      Write the IUPAC names of the following co-ordination compounds:

            (a) K₃[Cr(C₂O₄)₃]

            (b) Hg[Co(SCN)₄]

            (c) [Co(NH₃)₅(CO)₃)]Cl

Ans:     (a) Potassium trioxalatoelectromate III

            (b) Mercuric tetrathiiocyanatocobaltate chloride.

            (c) Pentaamminecarbonatocobalt (III) chloride.

Q.10    Give an example of hexadentate ligand.

Ans: Ethylene diamminetetraacetate .

Higher Order Thinking Skills Questions (HOTS)

Q.1(a) What is a ligand? Give an example of a bidentateligand.

(b) Explain as to how the two complexes of nikel [Ni(CN)₄]^2- and [Ni(CO)₄] have different structures but do not have their different magnetic behaviour. (Ni=28)

Ans: (a) Ligands .  The ions or molecules bound to the central atom ion in the coordination

entity are called ligands. Example of bidentate ligand: H₂NCH₂CH₂NH₂ (ethylenediammine)

(b) [Ni(CN)4]2- is a square planer complex. which is diamagnetic as no unpaired electron    

is present. [Ni(CO)₄] is a tetrahedral complex which is diamagnetic due to the absence of

unpaired electron

Q.2      (a) Name two main factors that affects a metal ions forming complex.

            (b) Give an example of industrial application of forming co-ordination complex.

            (c) Write IUPAC name of [Co(en)₂ Cl(ONO)]⁺

Ans:   (a) (i) Smaller size of cation and higher charge

                (ii)Presence of vacant orbitals

            (b) Silver and gold are extracted by treating zinc with their cynide complexes.        K[Ag(CN)2] is used for electroplating of silver , K[Au(CN)2] is used for gold plating .

Q.3.     Write the name and draw the structure of each of the following complex compounds:

            (i)[Co(NH₃)₄)H₂O)₂]Cl₃          (ii)[Pt(NH₃)₄][NiCl₄]              

                                                                                               

Q4.      How  is stability of co-ordination compounds determined in aqueous solution?

Ans: The stability of a co-ordination compound is measured in terms of stability constant

Thus if we have a reaction of the type

                                    M + 4L             ML4

   [ML4]

                                                 β4 =

                                                                            [M] [L]4

Q.5      (a) Give the IUPAC name of [Crcl₂(H₂O)₄]Cl

            (b)Give the number of unpaired electrons in the following complex ions:

             [FeF₆]^4-  and  [Fe(CN)₆]^4-

                (c) Name the isomerism exhibited by the following pair of coordination compounds.

             [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br

Ans:     (a) Tetraquadichloro chromium (III) chloride

            (b) [FeF₆]^4-has four unpaired electron as F^- is a weak field ligand.

             [Fe(CN)₆]^4-has no unpaired electrons because CN- is a strong field ligand.

            (c) Ionisation isomerism. On addition of dilute HCl followed by aqueous BaCl2,

             [CO(NH3)5]SO4 will give a white precipitate while the other coordination            compound will not give any white precipitate

VALUE BASED QUESTIONS ON CORDINATION COMPOUNDS

Q1) what is the role of coordination in society that you learn from the chemistry of coordination?

     Ans 1 We learn that social progress can be concrete stable by coordination between the society members. The person who has enough wealth should donate its some part to the

Person who needs it for its existence.  : - NH3 molecules donate lone pair of electrons to the various transition metal or ion resulting in a stable & useful complex which is also helpful in analysis of chemical samples.

Q2) A person is suffering from anemia then which coordination compound will you suggest for recovery of the health?

Ans 2:- Complex compound of Iron in the form of Tablet or Liquid syrup