9 coordination compounds - shiksha houseii) potassium tetrachloridopalladate(ii) iii)...

Coordination Compounds

1) Explain the bonding in coordination compounds in terms of Werner’s postulates.

Solution

There are two types of valencies shown by central metal ion in a corrdination compound: i) Primary valency ii) Secondary valency

1. Primary valency corresponds to oxidation number and secondary valency corresponds to corrdination number.

2. In the complex [Co(NH3)]Cl3 , primary valency of Co is three and secondary valency is six.

3. Primary valency is satisfied by negative ions(anions) while secondary valency is satisfied by ligands which can be neutral and as well as negative (positive in rare case).

In [Co(III) (NH3)6 ]Cl3. six NH3 ligands satisfy secondary valencies while three Cl-ions satisfy primary valencies.

In [Co(III) (NH3)4 (H2O)2]SO4 ,four NH3 and two H2O (ligands) satisfy secondary and one SO 4

2- ions satisfy primary valencies.

4. Ligands attached by secondary valency do not ionise while ions attached by primary valency ionise.

K4[Fe(CN)6] → 4K+ + [Fe (CN)6]4-

[Pt(NH)6]Cl4 → [Pt(NH3)6]4+ + 4Cl-

5. The secondary valencies are directional and are responsible for the isomerism in complexes.

6. Primary valency in structural formula of the complex is represented by dotted lines(-----)while secondary valency is represented by thick lines (_____).

7. In all cases, metal ion should satisfy primary and secondary valencies . But in some cases negative ions may satisfy both primary as well as secondary valencies (dual nature) . In the complexes, they are represented by both dotted and thick lines.

8. To distinguish the primary valency from the secondary valency, Werner in his formula has incorporated a square bracket. The part of the complex within the square bracket is the corrdination sphere and will not ionize in water, it contains the metal ion with its secondary valencies bound by coordinate covalent bonds. The part of the complex outside the square brackets is the primary valencies and which ionise in water.

2)When FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. Explain why?

Solution

When FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the double salt FeSO4.(NH4)2SO4 . In the solution of a double salt all the ions are free. Hence, Fe2+ gives test of its presence . While CuSO4 solution when mixed with aqueous ammonia in 1:4 ratio, a complex(coordination ) compound, [Cu(NH3)4]SO4 is formed . In a solution of complex compound , complex ion [Cu(NH3)4]

2+ and SO42- ions are present. There is no

free Cu2+ ion. Hence, the test for Cu2+ is not given.

3) Explain with two examples each of the following: coordination entity, ligand , coordination number , coordination polyhedron , homoleptic and heteroleptic.

Solution

Coordination entity: A coordination entity constitutes a cetral metal atom or ion bonded to a fixed number of ions or molecules.

Ex: [Ni(CO)4] , [Co(NH3)6]3+

Ligands: The ions or molecules bound to the central metal ion in the coordination entity are called ligands.

Ex: Cl-, H2O, H2NH2C-CH2NH2 etc.

Corrdination number (CN) : The number of ligand dononr atoms to which the metal is idrec tly bound.

Ex: In[PtCl6]2- C.N. of Pt= 6

In [Ni(NH3)4]2+ C.N of Ni=4

Coordination Polyhedron: The spatial arrangement of the ligand atoms which are directly attached to the central metal ion defines a coordination polyhedron about the central atom.

Ex: [Co(NH3)6] Octahedral

[Ni(CO)4] Tetrahedral

Homoleptic complex: Complexes in which a metal is bound to only one kind of donor groups.

Ex: [Co(NH3)6]3+ , [PtCl6]

2-

Heteroleptic complex: Complexes in which a metal is bound to more than one kind of donor groups.

Ex: [Co(NH3)4Cl2]+ , [Fe(CN)5NO2]

2-

4) What is meant by unidentate. didentate and anbidentate ligands? Give two examples for each.

Solution

Unidentate: A ligand is bound to a metal ion through a single donor atom.

Ex: Cl-. NH3

Didentate: The ligand is said to be didentate when a ligand cann bind through two donor atoms.

Ex: H2N-CH2-CH2-NH2(Ethane -1,2-diamine)

C2O42- (Oxalate ion)

Ambidentate: Ligand which can ligate through two different atoms is called ambidentate ligand.

Ex: NO2-, CN-

NO2- can coordinate through nitrogen or through oxygen to a central metal ion.

CN- can coordinate either through carbon or through nitrogen atoms to the central metal ion.

5) Specify the oxidation numbers of the .0.metals in the following coordination entities:

i) [Co(H2O)(CN)(en)2]2+

ii) [CoBr2(en)2]+

Solution

i) Oxidation number of cobalt (Co) in [Co(H2O)(CN)(en)2]2+.

Let O.N of Co be x

H2O and ‘en’ are neutral ligands (charge is zero)

CN- is a negative ligand with a charge of -1.

O.N of Co, x-1 = +2

x= +3

ii) [CoBr2(en)2]+

Let O.N of Co be x.

Br- is negative ligand with a charge of -1.

en is neutral ligand and its charge is zero.

O.N of Co, x-2 = +1

x= +3

Therefore, the O.N of Co is +3.

iii) [PtCl4]2-

iv) K3[Fe(CN)6]

v) [Cr(NH3)3Cl3]

Solution

iii) [PtCl4]2-

Let O.N of Pt be x.

Cl- is negative ligand with a charge of -1.

O.N of Pt , x-4= -2

x= -2 +4

x= +2

Therefore, the O.N of Pt is +2.

iv) K3[Fe(CN)6]

Let O.N of Fe be x.

K+ is a positive ligand with charge +1, CN- is a negative ligand with charge -1, CN- is a negative ligand with charge -1.

O.N of Fe , x-6 +3=0

x= +3

Therefore, the O.N of Fe is +3.

v) [Cr(NH3)6]Cl3

Let O.N of Cr be x.

NH3 is a neutral ligand , so charge is zero.

Cl is a negative ligand with a charge of -1.

O.N of Cr , x-3=0

x=+3

Therefore, the O.N of Cr is +3.

6)Using IUPAC norms write the formulas for the following:

i) Tetrahydroxozincate(II)

ii) Potassium tetrachloridopalladate(II)

iii) Diamminedichloridoplatinum(II)

Solution

i) Tetrahydroxozincate(II)

Four negative ligands OH-.

Central metal ion is Zn2+

It is an anionic complex.

The complex is [Zn(OH)4]2-

ii) Potassium tetrachloridopalladate(II)

‘K+’ is primary valency,

Pd2+ is the central metal ion,

Cl- is a negative ligand,

It is an anionic complex.

The complex is K2[PdCl4]

iii) Diamminedichloridoplatinum(II)

Central metal ion Pt2+

2NH3 neutral ligands,

2Cl- negative ligands.

The complex is [Pt(NH3)2Cl2].

iv) Potassium tetracyanonickelate (II)

v) Pentamminenitrtrito –O-cobalt(III)

vi) Hexamminecobalt (III) sulphate.

Solution

iv) Potassium tetracyanonickelate(II)

K+ is the primary valecne,

4CN- negative ligands(secondary valency),

Ni2+ is the central metal ion.

The complex is K2[Ni(CN)4].

v) Pentamminenitrito-O-cobalt(III)


One NO2- is a negative ligand,

Co3+ is the central metal ion.

The complex is [Co(NH3)5(ONO)]2+.

vi) Hexamminecobalt (III) sulphate.

6NH3neutral ligands,

SO42- is a negative and is the primary valency , Co3+ is the central metal ion.

The complex is [Co(NH3)6]2(SO4)3.

vii) Potassium tri(oxalate)chromate (III)

viii) Hexaamineplatinum (IV)

ix) Tetrabromidocuprate (II)

x) Pentaamminenitrito –N-cobalt (III)

Solution

vii) Potassium tri(oxalate)chromate (III)

K+ is a positive ion and is the primary valecny,

3C2O42- is a negative ligand,

Cr3+ is the central metal ion.

The complex is K3[Cr(C2O4)3]

viii) Hexaamineplatinum (IV)


Pf4+ is the central metal ion.

The complex is [Pt(NH3)6]4+

ix) Tetrabromidocuprate (II)

4Br- ions . It is a negative ligand,

Cu2+ is the central metal ion

The complex is [CuBr4]2-.

x) Pentaamminenitrito –N-cobalt (III)


NO2- is a negative ligand,

Co3+ is the central metal ion.

The complex is [Co(NH3)5(NO2)]2+.

7) Using IUPAC norms write the systematic names of the following:

i) [Co(NH3)6]Cl3

ii) [Pt(NH3)2Cl(NH2CH3)]Cl

iii) [Ti(H2O)6]3+

iv) [Co(NH3)4Cl(NO2)]Cl

v) [Mn(H2O)6]2+

vi) [NiCl4]2-

vii) [Ni(NH3)6]Cl2

viii) [Co(en)3]3+

ix) [Ni(CO)4]

Solution

IUPAC names:

i) Hexamminecobalt (III) chloride

ii) Diamminechloroethylammineplatinum (II) chloride

iii) Hexaaquatitanium (III) ion

iv) Tetraamminechloronitrocobalt (III) chloride

OR

Tetraamminechloronitrito –N-cobalt (III) chloride

v)Hexaaquamanganese(II) ion

vi) Tetrachloronickelate (II)

vii) Hexaamminenickel (II) chloride

viii) Tris(ethylenediamine ) cobalt (III) ion

ix) Tetracarbonynickel (0)

8) List various types of isomerism possible for coordination compounds, giving an example of each.

Solution

i) Linkage isomerism: This isomerism arises in a coordination compound containing ambidentate ligand. These isomers differ in the ;linkage of the donar atom of the ligand to the central metal ion.

Ex: [Co(NH3)5(NO2)]Cl2 and [Co(NH3)5(ONO)]Cl2

ii) Coordination isomerism: This type of isomerism arises , from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.

Ex: [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6]

iii) Ionisation isomerism: It arises when the counter ion in a complex ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. It is due to the exchange of the ligands between the coordination sphere and ionizing sphere.

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

iv) Hydrate isomerism : It arises due to difference in the number of water molecules between the coordination sphere and ionizing sphere.

Ex: [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl] Cl2H2O

v) Geometric isomerism : It arises due to the difference in the special arrangement of the ligands around the central metal ion. The isomers are not mirror image to each other.

vi) Optical isomerism: It arises due to the difference in the special arrangement of the ligands around the central metal ion. The isomers are non-super imposable mirror images to each other.

9) How many geometrical isomers are possible in the following coordination entities?

i) [Cr(C2O4)3]3- ii) [Co(NH3)3Cl3]

Solution

i) [Cr(C2O4)3]3-

No geometrical isomers are possible for this complex as there cannot be any difference in the arrangement of the oxalate ions around the Cr3+ ions.

ii) [Co(NH3)3Cl3]

Two geometrical isomers are possible for this complex . They are fac and mer isomers.

10) Draw the structures of optical isomers of:

i) [Cr(C2O4)3]3- ii) [PtCl2(en)2]

2+ iii) [Cr(NH3)2Cl2(en)]+

Solution

i) Optical isomers of [Cr(C2O4)3]3-

ii) Optical isomers of [PtCl2(en)2]2+

iii) Optical isomers of [Cr(NH3)2Cl2(en)]+

11)Draw all the isomers (geometrical and optical ) of:

i) [CoCl2(en)2]+

Solution

Geometrical isomers:

Optical isomers:

ii) [Co(NH3)Cl(en)2]2+


Optical isomers:

iii) [Co(NH3)2Cl2(en)]+

Solution

[Co(NH3)2Cl2(en)]+


iii) [Co(NH3)2Cl2(en)]+

Optical isomers:

12) Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?

Solution

Three geometrical isomers are possible for this complex [Pt(NH3)(Br)(Cl)(py)].

Optical isomerism is not exhibited by this complex as it is a square planar complex. Square planar complex cannot exhibit optical isomerism.

13) Aqueous copper sulphate solution (blue in colour) gives:

i) a green precipitate with aqueous potassium fluoride, and

ii) a bright green solution with aqueous potassium chloride

Explain these experimental results.

Solution

Aqueous CuSO4 solution exists as [Cu(H2O)4]SO4 which has blue colour due to [Cu(H2O)4]

2+ ions.

i) When KF is added to aqueous copper sulphate solution , the H2O ligands are replaced by F- ligands forming [CuF4]

2- ions which is a green precipitate.

[Cu(H2O)4]2+ + 4F- →[CuF4]

2- + 4H2O

ii) Whe KCl is added to aqueous copper sulphate solution, Cl- ligands replace the H2O

ligands forming [CuCl4]2- ions which has bright green colour.

[Cu(H2O)4]2+ + 4Cl- - →[CuCl4]

2- + 4H2O

14) What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution?

Solution

As CN- is a strong ligand, when excess of aqueous KCN is added to an aqueous solution of copper sulphate, it forms a highly stale complex with Cu2+ ions.

[Cu(H2O)4]2+ + 4CN- →[Cu(CN)4]

2- + 4H2O

On passing H2S, free Cu2+ ions are not available to form the precipitate of CuS. Therefore, a precipitate of copper sulphide is not formed.

15) Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

i) [Fe(CN)6]4- ii) [FeF6]

3-

Solution

i) [Fe(CN)6]4-

In [Fe(CN)6]4- the oxidation state of Fe is +2 and the coordination number is 6. Fe2+ in

this complex undergoes d2sp3 hybridisation and exhibits octahedral geometry. Due to strong field strength of CN- ligand, it is an inner orbital low spin complex with diamagnetic nature.

Orbitals of Fe2+ ion:

ii) [FeF6]3-

In [FeF6]3- the oxidation state of iron is +2 and the coordination number is 6. Fe3+ in this

complex undergoes sp3d2 hybridisation and exhibits octahedral geometry. Due to weak field strength of F- ligand, it is an outer orbital high spin complex with paramagnetic nature.

iii) [Co(C2O4)3]3- iv) [CoF6]

3-

Solution

iii) [Co(C2O4)3]3-

In [Co(C2O4)3]3- the oxidation state of Co is +3 and the coordination number is 6. Co3+ in

this complex undergoes d2sp3 hybridisation and exhibits octahedral geometry. Due to strong field strength of C2O4

2- ligand, it is an inner orbital or low spin complex with diamagnetic nature.

iv) [CoF6]3-

In [CoF6]3- the oxidation state of Co is +3 and the coordination numbers is 6. Co+3 in this

complex undergoes sp3d2 hybridisation and exhibits octahedral geometry . Due to weak field strength of F- ligand, it is an outer orbital complex high spin complex with paramagnetic nature.

16) Draw figure to show the splitting of d orbitals in an octahedral crystal field.

Solution

17) What is spectrochemical series ? Explain the difference between a weak field ligand and a strong field ligand.

Solution

The arrangement of ligands in the increasing order of their strength is called spectrochemical series.

Sprectrochemical series

I-<Br-<SCN-<Cl-< S2-<F-< OH-<C2O42- <H2O <NCS-<edta4- < NH3< en < CN<CO

i) Ligands for which ∆0 or ∆t <P are known as weak field ligands and form high spin complexes.

ii) Ligands for which ∆0 or ∆t >P are known as strong field ligands and form low spin complexes.

Note: ∆0 = crystal field splitting energy for octahedral complex

∆t = crystal field splitting energy for tetrahedral complex.

P= pairing energy of the electrons.

18) What is crystal field splitting energy? How does the magnitude of ∆o decide the actual configuration of d-orbitals in a coordination entity?

Solution

Splitting of the degenerate orbitals of d-subshell due to the presence of ligands in a definite geometry is termed as crystal field splitting and the energy separation between the eg and the t2g level is called crystal field splitting energy. for octahedral complex, it is denoted by ∆0 (the subscript ‘O’ represents octahedral).

i) If ∆o < P, the fourth electron enters one of the eg orbitals giving the configuration t2g3

eg1.

ii) If ∆0 < P , it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with configuration t2g

4eg0.

∆o = crystal field splitting energy for octahedral complex.

P = pairing energy of the electrons

19) [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]

2- is diamagnetic . Explain why?

Solution

In [Cr(NH3)6]3+ complex ion, the central metal ion Cr3+ has 3 unpaired electrons in its d-

orbitals . due to the presence of 3 unpaired electrons, it is paramagnetic.

Electron distribution in d-orbitals and d2sp3 hybrid orbitals of chromium ion in [Cr(NH3)6]

3+ complex ion is

In [Ni(CN)4]2+ complex ion, the central metal ion Ni2+ has no unpaired electrons. All the

electrons are paired. Therefore , it is diamagnetic.

Electron distribution in d-orbital and dsp2 hybrid orbitals of nickel ion in [Ni(CN)4]2+

complex ion is

20) A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]

2- is colourless . Explain.

Solution

i) [Ni(H2O)6]2+(aq) is green due to the presence of unpaired electrons in d-orbital of Ni+2.

Due weak field strength of H2O ligand in [Ni(H2O)6]2+ the electrons in the d-orbital of

Ni+2 ion remain unpaired . due to d-d transition of electrons [Ni(H2O)6]2+ is green in

colour.

ii) [Ni(CN)4]2- is colourless due to non availability of unpaired electron. This is due to the

strong field strength of CN- ion in [Ni(CN)4]2- the electrons are forced to pair up resulting

in th absence of unpaired electrons.

21) [Fe(CN)6]4- and [Fe(H2O)6]

2+ are of different colours in dilute solutions. Why?

Solution

In both [Fe(CN)6]4- and [Fe(H2O)6]

2+ iron is in +2 oxidation state and has four unpaired electrons. But in [Fe(CN)6]

4- due to strong field strength of CN- ligand the electrons in 3d –orbital of Fe2+ ion are forced to pair up. Therefore, [Fe(CN)6]

4- is colourless due to the absence of unpaired electrons.

Electron distribution in d-orbital and d2sp3 hybrid orbitals of Fe2+ in [Fe(CN)6]4- complex

ion is

In [Fe(H2O)6]2+ , due to the weak field strength of H2O ligand , the electrons in 3d-

orbital of Fe2+ ion will not pair up and remain as unpaired electrons even after the formation of the complex. therefore, [Fe(H2O)6]

2+ is coloured due to the presence of unpaired electrons.

Electron distribution in d-orbital and d2sp3 hybrid orbitals of Fe2+ in [Fe(H2O)6]2+

complex ion is

22) Discuss the nature of bonding in metal carbonyls.

Solution

The M-C bond in metal carbonyls has both σ and π character. The M-C σ-bond is formed by the donation of lone pair of electrons on the carbonyl carbon into the vacant orbital of the metal. The M-C π-bond is formed by the donation of a electrons from filled d-orbital

of metal into the vacant anti-bonding π molecular orbital of carbon monoxide. This is called back bonding.

Synergic bonding in metal carbonyls

The metal to ligand bonding creates a synergic effect which strengthens the bond between carbon monoxide in the metal.

23) Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:

i) K3[Co(C2O4)3] ii) cis-[Cr(en)2Cl2]Cl

iii) (NH4)2[CoF4] iv) [Mn(H2O)6]SO4

Solution

i) K3[Co(C2O4)3]

Oxidation state of Co=+3

C.No=6

d-orbital occupation : d6(t2g6, eg

0)

ii) cis-[Cr(en)2Cl2]Cl

Oxidation state of Cr=+3

C.No=6

d-orbital occupation : d3(t2g3, eg

0)

iii) (NH4)2[CoF4]

Oxidation state of Co=+2

C.No=4

d-orbital occupation : d7(eg4 ,t2g

3)

iv) [Mn(H2O)6]SO4

Oxidation state of Mn= +2

C.No= 6

d- orbital occupation : d5(t2g3, eg

2)

24) Write down the IUPAC name for each of the following complxes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:

i) K[Cr(H2O)2(C2O4)2].3H2O ii) [Co(NH3)5Cl]Cl2

Solution

i) K[Cr(H2O)2(C2O4)2].3H2O

IUPAC name : Potassium diaquabis (oxalate) chromium (III) trihydrate

Oxidation state of Cr= +3

C.No = 6

Shape = Octahedral

Electronic configuration (d3 configuration ) : t2g3, eg

0

Electron distribution in the metal ion in the complex:

Magnetic moment (µ) = √[n(n+2)]

n= number of unpaired electrons= 3

Magnetic moment = √[3(3+2)]

= √(15)

= 3.87 BM

ii) [Co(NH3)5Cl]Cl2

IUPAC name : Pentaaminechlorocobalt (III) chloride

Oxidation state of Co= +3

C.No= 6

Shape = Octahedral

Electronic configuration (d6configuration): t2g6, eg

0


Magnetic moment (µ) = √[n(n+2)]



Magnetic moment (µ) =0

iii) CrCl3(py)3 iv) Cs[FeCl4]

Solution

iii) CrCl3(py)3

IUPAC name : Trichlorotripyridinechromium (III)

Oxidation state of Cr= +3

C.No= 6 Shape = Octahedral

Electronic configuration (d3 configuration) : t2g3, eg

0


Magnetic moment(µ) =√[n(n+2)]



= √15

= 3.87 BM

iv) Cs[FeCl4]

IUPAC name : Caesium tetrachloroferrate(III)

Oxidation state of Fe= +3

C.No= 4

Shape = Tetrahedral

Electronic configuration (d5 configuration ): eg2, t2g

3


Magnetic moment(µ) =√[n(n+2)]

n= number of unpaired electrons = 5

Magnetic moment(µ) =√[5(5+2)]

= √35

= 5.916 BM

Magnetic moment(µ) =5.916 BM

v) K4[Mn(CN)6]

Solution

K4[Mn(CN)6]

IUPAC name : Potassium hexacyanomanganate(II)

Oxidation state of Mn= +2

C.No= 6

Shape = Octahedral

Electronic configuration(d5configuration) : t2g5, eg

0


Magnetic moment(µ) = √[n(n+2)]

n = number of unpaired electrons = 1


= √3

= 1.732 BM

Magnetic moment (µ) = 1.732 BM

25) What is meant by stability of a coordination compound in solution ? State the factors which govern stability of complexes.

Solution

The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium.

The magnitude of the equilibrium constant for the association, quantitatively express the stability of the complex.

The stability of a complex is expressed in terms of its formation constant (βn).

This can be explained as follows:

[M(H2O)n] + nL ⇌ [MLn] + n H2O

The complex MLn can be explained by the following steps:

Step – I

[M(H2O)n] + L ⇌ [ML(H2O)n-1] + H2O

The formation constant for this step,

K1= [ML(H2O)n-1]/[M(H2O)n] [L]

Step – II

[ML(H2O)n-1]+ L ⇌ [ML(H2O)n-2] + H2O

The formation constant for this step ,

K2= [ML(H2O)n-2]/ ]/[M(H2O)n] [L]

Step – III

[ML2(H2O)n-2] + L ⇌ [MLn] + H2O

The formation constant for nth step,

Kn= [MLn]/ [M(H2O)n-2] [L]

βn = K1 x K2 x K3 x K4 x ........ Kn

βn = overall stability constant;

K1 , K2 , K3 , K4 ........ Kn = stepwise constants.

Factors effecting the stability of the complex

i) the nature of the central ion and

ii) the nature of ligand.

26) What is meant by the chelate effect? Give an example.

Solution

When a di-or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it is said to be a chelate ligand. Such ligands form a ring around the central metal ion. Such complexes are called chelate complexes . These complexes are more stable than similar complexes containing unidentate ligands . The stabilization of a coordination compound due to chelation is called the chelate effect.

Example:

Ni2+ (aq) + 6NH3(aq) ⇌ [Ni(NH3)6]2+ (aq) ; log β = 8.61

Ni2+ (aq) + 3en (aq) ⇌ [Ni(en)3]2+ (aq) ; log β = 18.28

From the above two examples it can be easily observed that Ni2+ forms a relatively more stable complex with bidentate ethylene diammine (en) than mondentate ammonia molecule. This is due to the formation of a chelate (ring) by ethylene diammine.

27) Discuss briefly giving an example in each case the role of coordination compounds in:

i) biological system ii) medicinal chemistry

iii) analytical chemistry iv) extraction/metallurgy of metals

Solution

i) The pigment responsible for photosynthesis , chlorophyll is a coordination compound of magnesium. Haemoglobin , the red pigment of blood which acts as oxygen carrier is a coordination compound of iron. Vitamin B12 (cyanocobalamine), is a coordination compound of cobalt.

ii) EDTA is used in the treatment of lead poisoning . Cis-platin is a platinum coordination compound effectively inhibits the growth of tumours.

iii) Hardness of water is estimated by simple titration with Na2EDTA . The Ca2+ and Mg2+ ions form stable complexes with EDTA.

iv) Silver and gold extracted from complexes.

Ex: [Au(CN)2]-

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

i) 6 ii) 4 iii)3 iv) 2

Solution

iii) 3

Co(NH3)6Cl2 forms 3 ions in its aqueous solution.

One [Co(NH3)6]2+ and two Cl- ions.

29) Amongst the following ions which one has the highest magnetic moment value?

i) [Cr(H2O)6]3+ ii) [Fe(H2O)6]

2+ iii) [Zn(H2O)6]2+

Solution

ii) [Fe(H2O)6]2+ 4 unpaired electrons are present.

Magnetic moment is related to number of unpaired electrons in the central metal ion of the complex , therefore, the complex with maximum number of unpaired electrons has highest magnetic moment. In the given complexes Cr has 3 unpaired electrons , Fe has 4 unpaired electrons and Zn has no unpaired electrons. Therefore, [Fe(H2O)6]

2+ with 4 unpaired electrons has highest magnetic moment.

30) The oxidation number of cobalt in K[Co(CO)4] is

i) +1 ii)+3 iii)-1 iv)-3

Solution

iii) -1

K[Co(CO)4]

Let the oxidation state of cobalt be x

Oxidation state of K= +1

Charge due to CO = 0 (0 as it is neutral ligand , has zero charge)

Therefore, +1 + x + 0= 0

x= -1

Oxidation state of cobalt= -1

31) Amongst the following , the most stable complex is

i) [Fe(H2O)6]3+ ii) [Fe(NH3)6]

3+ iii) [Fe(C2O4)3]3-

iv) [FeCl6]3-

Solution

iii) [Fe(C2O4)3]3-

[Fe(C2O4)3]3- is the most stable complex in the given complexes due to the presence of

chelate as oxalate (C2O42-) is a chelating ligand.

32) What will be the correct order for the wavelengths of absorption in the visible region for the following :

[Ni(NO2)6]4- , [Ni(NH3)6]

2+ , [Ni(H2O)6]2+

Solution

The order of the ligand in the spectrochemical series: H2O < NH3 < NO2-.

Hence, the wavelength of the light observed will be in the order:

[Ni(H2O)6]2+ <[Ni(NH3)6]

2+ <[Ni(NO2)6]4-

Thus, wavelengths absorbed will be in the opposite order.

[Ni(H2O)6]2+ >[Ni(NH3)6]

2+ >[Ni(NO2)6]4-

9 coordination compounds - shiksha houseii) potassium tetrachloridopalladate(ii) iii)...

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