NCERT Solutions For Class 12 Chemistry Chapter 9

NCERT Solutions Class 12 Chemistry Coordination Compounds

NCERT solutions for class 12 chemistry chapter 9 coordination compounds include detailed solutions for all the questions of the NCERT textbooks. These solutions are prepared by our subject experts with an objective of helping students in their exam preparations.

NCERT solutions for class 12 chemistry chapter 9 provides complete knowledge about the topic and all the solutions are framed in an easily understandable language for the students. Students can easily download these pdf files and use it for their preparations. NCERT solutions for class 12 chemistry chapter 9 pdf help students to have a quick revision of a complete chapter along with its important questions.

Q1. Using Werner’s postulates describe the bonding present in coordination compounds.

Ans :

( a ) A metal shows two kinds of valencies viz primary valency and secondary valency.Negative ionssatisfy primary valenciesand secondaryvalencies are filled by bothneutral ions and negative ions.
( b ) A metal ion has a fixed amount of secondary valencies about the central atom. These valenciesalso orient themselves in a particular direction in the space provided to the definite geometry of the coordination compound.
( c ) Secondary valencies cannot be ionized , while primary valencies can usually be ionized.

 

Q2. FeSO4 solution is mixed with (NH4)2SO4solution in the molar ratio of 1:1 molar. It gives a positive test for Fe2+ ion. However, when CuSO4 solution is mixed with aqueous ammonia in the molar ratio of 1:4 , it does not test positive for Cu2+ ion. Why?

Ans :

FeSO4 solutionwhen mixed with (NH4)2SO4in 1 : 1 molar ratio produces a double salt FeSO4(NH4)2SO4-6H2O. This salt is responsible for giving the Fe2+.
CuSO4 mixed with aqueous ammonia in the ratio of 1:4 gives a complex salt. The complex salt does not ionize to give Cu2+, hence failing the test.


Q3. Explain the following giving two examples for each of them :ligand, coordination entity, coordination polyhedron , coordination number,heteroleptic and homoleptic.

Ans :

( a ) Ligands – they are neutral molecules or negative ions bound to a metal atom in coordination entity. Example- Cl, OH
( b ) Coordination entity – they are electrically charged radicals or species. They constitute of a central ion or atom surrounded by neutral molecules or ions. Example – [ Ni(CO)4] , [COCL3(NH3)3]
( c ) Coordination number– it is the number of bonds formed between ligands and central atom/ion.
Example : ( i ) In K2[PtCl6], 6 chloride ions are attached to Pt in the coordinate sphere. Thus, 6 is the coordination number of Pt.
( ii ) In [Ni(NH3)4]Cl2, the coordination number of the central metal ion (Ni) is 4.
( d ) Coordination polyhedron – it is the spatial positioning of ligands that are directly connected to the central atom in the coordination sphere. Example –
Square Planner

( v )Heteroleptic : they are complexes with their metal ion being bounded to more than one kindof donor group. Example – [ Co(NH3)4Cl2]+ , [ Ni(CO)4 ]
( vi ) Homoleptic : they are complexes with their metal ion being bounded to only one type of donor. Example – [ PtCl4]2- , [ Co(NH3)6]3+

 

Q4.Providing two examples in each case,explain the followingterms :unidentate, ambidentate and didentate ligands?

Ans :

( i ) Unidentate ligands : these are ligands with one donor site. Example Cl , NH3
( ii ) Ambidentate ligands : these are ligands that fasten themselves to the central metal ion/ atom via two different atoms.
Example NO2or ONO , CN or NC
( iii ) Didentate – these are ligands with two donor sites.
Example – Ethane-1,2-diamine , Oxalate ion ( C2O42- )

Q5. Present below are coordination entities, state the oxidation number of their metals :
( i ) [Cr(NH3)3Cl3]
( ii ) [PtCl4]2-
( iii ) K3 [Fe(CN)6]
( iv ) [ Co Br2 (en)2 ] +
( v ) [Co (H2O) (CN) (en)2 ] 2+

 

Ans :

( i ) [ Cr(NH3)3Cl3 ]
=> x + 3( 0 ) + 3 ( -1 ) = 0
x – 3 = 0
x = 3
( ii ) [ PtCl4 ] 2-
=> x + 4 ( -1 ) = -2x
x = +2
( iii ) K3 [Fe (CN)6 ]
=> [ Fe ( CN )6]3-
=> x + 6 ( -1 ) = -3
x = +3
( iv ) [ Co Br2 (en)2 ] 2+
=> x + 2 ( -1 ) + 2 ( 0 ) = +1
x – 2 = +1
x = +3
( v ) [Co (H2O) (CN) (en)2 ] 2+
=> x + 0 + ( -1) + 2 (0) = +2
x -1 = +2
x = +3

Q6. Making use of standard IUPAC norms give the formulae of the following compounds :
( a ) Potassiumtetrachloridopalladate(II)
( b ) Tetrahydroxozincate(II)
( c )Hexaammineplatinum (IV)
( d ) Potassium tetracyanonickelate(II)
( e ) Tetrabromidocuprate(II)
( f ) Hexaamminecobalt(III) sulphate
( g ) Pentaamminenitrito-N-cobalt(III)
( h ) Diamminedichloridoplatinum(II)
( i ) Pentaamminenitrito-O-cobalt(III)
( j ) Potassium tri(oxalato)chromate(III)

Ans :

( a ) K2[ Pd Cl4]
( b ) [Zn (OH) ]2-
( c ) [ Pt (NH3)6] 4+
( d ) K2[ Ni(CN )4]
( e ) [ Cu (Br)4] 2−
( f ) [ Co( NH3)6]2 (SO4)3
( g ) [ Co (NO2) ( NH3)5] 2+
( h ) [ Pt ( NH3)2Cl2]
( i ) [Co ( ONO )( NH3)5] 2+
( j ) K3 [ Cr ( C2O4)3 ]

Q7.Making use of standard IUPAC norms give the systematic names of the compounds given below :
( i ) [ Pt( NH3 )2Cl( NH2CH3) ]Cl
( ii ) [ Co ( NH3 )6 ]Cl3
( iii ) [ Ti (H2O)6 ]3+
( iv ) [ NiCl4 ] 2−
( v ) [ Ni( NH3 )6 ]Cl2
(vi) [ Co( NH3 )4Cl( NO2 ) ]Cl
(vii) [ Mn(H2O )6 ]2+
(viii) [ Ni( CO)4 ]
(ix) [ Co( en)3 ] 3+

Ans :
( i ) Diamminechlorido(methylamine) platinum(II) chloride
( ii ) Hexaamminecobalt(III) chloride
( iii ) Hexaquatitanium(III) ion
( iv ) Tetrachloridonickelate(II) ion
( v ) Hexaamminenickel(II) chloride
( vi ) Tetraamminichloridonitrito-N-Cobalt(III) chloride
( vii ) Hexaquamanganese(II) ion
( viii ) Tetracarbonylnickel(0)
( ix ) Tris(ethane-1, 2-diammine) cobalt(III) ion


Q8. What are the various kinds of isomerism present in coordination compounds ? ( Provide an example for each case )

Ans :

The various types of isomerism present in coordination compounds are :
( i ) Geometrical isomerism : 8.1

( ii ) Optical isomerism : Optical Isomerism

(iii) Linkage isomerism : This is found in complexes that have ambidentate ligands. For e.g. :

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

( iv ) Coordination isomerism :
This kind of isomerism comes up when ligands are interchanged between anionic and cationic entities of different metal ions present in the complex.
Example – [ Cr( NH3)6] [ Co( CN )6]
( v ) Ionisation isomerism :
This is the kind of isomerism where a counter ion takes the place of a ligand inside the coordination sphere.. For e.g., [ Co( NH3)5Br ]SO4and [ Co( NH3)5SO4 ]Br
( vi ) Solvate isomerism :
[ Cr( H2O)5 Cl ]Cl.H2O

Q9. Find the number of geometrical isomers possible in the coordination entities given below :
( i ) [ Co( NH3)3 Cl3 ]
( ii ) [ Cr(C2O4)3] 3−

Ans :

( i ) In [ Co( NH3)3 Cl3 ]two isomers are possible.

Isomers

( ii ) In [ Cr(C2O4)3] 3− no geometric isomers are present because it is a bidentate ligand.

9.2

 

 

Q10. Draw the structures of the optical isomers given below :
( i ) [ Cr( NH3)2 Cl2( en ) ]+
( ii ) [ Cr( C2O4 )3 ] 3−
( iii ) [ PtCl2( en )2 ] 2+

Ans :

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

10.1

( ii ) [ Cr( C2O4 )3 ] 3−

10.2

( iii ) [ PtCl2( en )2 ] 2+

10.3

 

Q11. Sketch all the isomers ( optical and geometrical) of:
( i ) [ Co( NH3 )2 Cl2( en) ]+
( ii ) [ Co(NH3) Cl( en)2 ] 2+
( iii ) [ CoCl2 (en)2 ] +

Ans:

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

11.1

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

11.2

( iii ) [ CoCl2 (en)2 ] +

11.3

 

Q12. What are the geometrical isomers of [ Pt( NH3)( Br)(Cl)(py ) ] ? Also , what number of these isomers will show optical isomerism ?

Ans:

[ Pt( NH3)( Br)(Cl)(py ) ]

12.1

None of the above isomers will exhibit optical isomerism.

 


Q13. Explain why :
Aqueous CuSO4solution,which is blue in color, yields :
( a ) a green precipitate if aqueous KF is added to it , and
( b ) a bright green solution if aqueous KCL is added to it .

Ans:

The blue color of aqueous CuSO4 solution is because of the presence of [ Cu( H2O)4 ] 2+ ions.
( i ) So when KF is added, H2O ligands are replaced by F ligands which yield green colored [ CuF4 ] 2+ ions.
\([Cu(H_{2}O)_{4}]^{2+} + 4F^{-}\rightarrow [CuF_{4}]^{2-} + 4H_{2}O\)

( ii ) So when KCL is added, H2O ligands are replaced by Cl ligands which yield bright green colored [ CuCl4 ] 2+ ions.
\([Cu(H_{2}O)_{4}]^{2+} + 4Cl^{-}\rightarrow [CuCl_{4}]^{2-} + 4H_{2}O\)

 

 

Q14. Give the coordination entity produced when excess of aqueous KCN is put into an aqueous solution ofCuSO4 ? When hydrogen sulphidegas isbubbled through this solution no precipitate of copper sulphide is formed. Why ?

Ans:

\(2[CuSO_{4}](aq)+ 10KCN(aq)\rightarrow 2K_{2}[Cu(CN)_{4}](aq) + 2K_{2}SO_{4}(aq) + (CN)_{2}\)
Therefore, the coordination entity obtained in the above process is K2[ Cu(CN)4 ] .
As the above coordination entity is highly stable it does not ionize to yield Cu2+ ions. Thus, no precipitate is obtained when hydrogen Sulphide gas is bubbled through it.

 

 

Q15. On the grounds of Valence bond theory, write about the nature of bonding in the given coordination entities :
( i ) [ Co(C2O4)3 ]3-
( ii ) [ CoF6 ] 3−
( iii ) [ Fe( CN)6 ] 4−
( iv ) [ FeF6] 3−

Ans:

( i ) [ Co( C2O4)3] 3−
In this complex, the oxidation state of cobalt is +3.
Orbitals of Co 3+ ion :

15.1

Oxalate is a weak field ligand. Thus, it will not cause the pairing of the 3d orbital electrons.
As there are 6 ligands, hybridization has to be either sp3d 2 or d 2sp3 hybridization.
sp3d2 hybridization of Co 3+ :

15.2

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp3d 2 orbitals :

15.3

Thus, the complex shows octahedral geometry .

( ii )) [CoF6] 3−
In this complex Cobalt has an oxidation state of +3.
Orbitals of Co3+ ion:

15.4

As, fluoride ion is a weak field ligand it will not cause the 3d electrons to pair. Hence, the Co3+ ion will go through sp3d2 hybridization.
sp3d2 hybridized orbitals of Co3+ ion are :

15.5

Thus, the complex has a geometric configuration of an octahedral and it is paramagnetic.
( iii ) [ Fe(CN)6] 4−
In this coordination complex, the oxidation state of Fe is +3.
Fe 2+ : Electronic configuration is 3d6
Orbitals of Fe2+ ion :

15.6

Since CN is a strong field ligand, it causes the unpaired 3d electrons to pair up:

15.7

As there are six ligands around the central metal ion, the most practical hybridization is d2sp3 , d2sp3 hybridized orbitals of Fe2+ are:

15.8

6 electron pairs from CN ions take the place of the six hybrid d2sp3 orbitals.

Then,

15.9

 

 

 

Thus, the geometry of the complex is octahedral and it is a diamagnetic complex ( since all the electrons are paired ).

( iv ) [FeF5] 3−
In this coordinate entity the oxidation state of iron is +3.
Orbitals of Fe +3 ion:

15.10

There are 6 F ions. Hence, it will go through d2sp3 or sp3d2 hybridization.
Since F is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Thus, the most practical hybridization is sp3d2 . sp3d 2 hybridized orbitals of Fe are :

15.11

Thus, the geometry of this coordinate entity is octahedral.

 

Q16. Sketch a figure depicting the split of d orbitals within an octahedral crystal field.

 

Ans: 16.1


 

Q17.What do you understand by the term “spectrochemical series” ? Differentiate between a strong field ligand and a weak field ligand.

Ans:

Series of common ligands in an ascending order of their crystal-field splitting energy (CFSE) is termed as Spectrochemical series.
Strong field ligands have larger values of CFSE. Whereas, weak field ligands have smaller values of CFSE.

Q18. Define crystal field splitting energy. In what way is the d-orbital configuration depending upon the magnitude of ∆o ?

Ans:

Crystal-field splitting energy is the difference in the energy between the two levels ( i.e., t2g and eg ) that have split from a degenerated d orbital because of the presence of a ligand. It is symbolised as ∆o.
Once the orbitals split up, electrons start filling the vacant spaces. An electron each goes into the three t2g orbitals, the fourth electron, however, can enter either of the two orbitals:

( 1 ) It can go to the eg orbital ( producing t2g3eg 1 like electronic configuration), or
( 2 ) it can go to the t2g orbitals ( producing t2g4eg 0 like electronic configuration).

This filling of the fourth electron is based on the energy level of ∆o. If a ligand has a ∆o value smaller than the pairing energy, then the fourth electron enters the eg orbital. However, if the value of ∆o is greater than the value of pairing energy, the electron enters t2gorbital.

Q19. Explain why [ Ni ( CN)4 ]2− is diamagnetic while [ Cr( NH3 )6 ] 3+ is paramagnetic.

Ans:

In [ Ni ( CN)4 ] 2−, Ni has an oxidation state of +2. Thus, it has d 8configuration.
Ni 2+ : 19.1

CN being a strong field ligand causes the electrons in 3d orbitals to pair. This causes, Ni 2+ to undergo dsp2 hybridization.19.2

Since all the electrons are paired, it is diamagneticin nature.

Cr has an oxidation state of +3. Thus, it has a d 3 configuration. As, NH3 is not a strong field ligand it does not cause the electrons in the 3d orbital to pair.
Cr3+ : 19.3

It undergoes d2sp3 hybridization and the 3d orbital electrons remain unpaired. Thus, [ Ni ( CN)4 ] 2− is paramagnetic in nature.

Q20. Why is a [ Ni ( H2O)6 ] 2+ solution green when a [ Ni ( CN)4 ] 2− solution is colorless?

Ans:

[ Ni ( H2O)6 ] 2+ consists of Ni+2 ion with 3d8 electronic configuration. In this configuration, there are two unpaired electrons which cannot pair up because H2O is a weak ligand. Thus, the d – d transition absorbs the incoming light and it emits a green light. Thereby, giving a green color to the solution.
[ Ni ( CN)4] 2- consists of Ni+2 ion with 3d8 electronic configuration. But, CN is present here, which is a strong ligand and in its presence, the unpaired electrons pair up. Thus, there is no d –d transition so no color.

 

 

Q21. Why do [Fe(H2O)6] 2+ and [Fe(CN)6] 4− have different colors in dilute solutions?

Ans:

[ Fe ( H2O)6 ] 2+ and [ Fe( CN)6 ] 4−have two different ligands H2O and CN­­ –. CN­ – being a strong field ligand has a higher value of CFSE ( crystal field splitting energy ) than water. As a result, the d-d transitions absorb and give back different wavelengths of light. Thus, they have different colors in a solution.

 

 

Q22. Write about the bonding present in metal carbonyls.
Ans:

In metal carbonyls, the metal-carbon bond contains both the σ and π bond characters.A σ bond forms when a lone pair of electrons is donated to the empty orbital of the metal by the carbonyl carbon. A π bond forms when a pair of electrons is donated to the empty antibonding π* orbital by the filled d orbital of the metal. This in entirety stabilizes and strengthens the metal-ligand bonding.
The above two types of bonding is represented as :

22.1

 

 

Q23. Given below are complex entities, provide the coordination number, oxidation state and d-orbital occupation of their central metal ion:
( a ) K3[ Co ( C2O4)3 ]
( b ) cis – [ Cr( en)2 Cl2 ]Cl
( c ) ( NH4)2[ CoF4 ]
( d ) [ Mn (H2O)6 ]SO4

Ans:

( a )K3[ Co ( C2O4)3 ]
Central metal ion: Co.
Coordination number = 6.
We know,
Oxidation state is :
x − 6 = −3
x = + 3
The d orbital occupation : t2g6eg 0.
( b ) cis – [ Cr( en)2 Cl2 ]Cl
Central metal ion : Cr.
Coordination number = 6.
We know,
Oxidation state is :
x +2(0) + 2 ( -1) = +1
x -2 = -1
x = + 3
The d orbital occupation : t2g3.
( c ) ( NH4)2[ CoF4 ]
Central metal ion : Co.
Coordination number = 4.
We know,
Oxidation state is :
x – 4 = -2
x = + 2
The d orbital occupation : eg4t2g3.
( d ) [ Mn (H2O)6 ]SO4
Central metal ion : Mn.
Coordination number = 6.
We know,
Oxidation state is :
x + 0 = 2
x = + 2
The d orbital occupation : t2g3 eg4.

 

 

Q24. State the IUPAC name for each of the complexes given below and provide their coordination number, oxidation state, and electronic configuration.
Also, indicate the stereochemistry and magnetic moment of these complexes :
( a ) K[ Cr( H2O)2( C2O4)2 ].3H2O
( b ) [ Co( NH3)5Cl ]Cl2
( c ) CrCl3( py)3
( d ) Cs[ FeCl4 ]
( e ) K4[ Mn( CN)6 ]

Ans:

( a ) IUPAC name = Potassium diaquadioxalatochromate (III) trihydrate.
Coordination number = 6
Oxidation state of chromium :
x + 0 +2( -2) = -1
x = 3
Electronic configuration: 3d 3 = t2g3
Shape : Octahedral
Stereochemistry :

24.1

Magnetic moment, µ = \(\sqrt{n(n +2)}\) [ n = unpaired electrons ]
= \(\sqrt{3(3 +2)}\)
= \(\sqrt{15}\)≈ 4BM
( b ) IUPAC name = Pentaamminechloridocobalt(III) chloride
Coordination number = 6
Oxidation state of Co :
x + 0 – 1 = + 2
x = 3
Electronic configuration: 3d 6 = t2g6
Shape : Octahedral
Stereochemistry :

24.2

n = 0.
Thus, Magnetic moment = 0
( c ) IUPAC name = Trichloridotripyridinechromium (III)
Coordination number = 6
Oxidation state of Cr :
x – 3 + 0 = 0
x = 3
Electronic configuration: 3d3 = t2g3
Shape : Octahedral
Stereochemistry :

24.3

n = 3
Magnetic moment, µ = \(\sqrt{n(n +2)}\)
= \(\sqrt{3(3 +2)}\)
= \(\sqrt{15}\)≈ 4BM
( d ) IUPAC name = Caesiumtetrachloroferrate (III)
Coordination number = 4
Oxidation state of Fe :
x – 4 = -1
x = 3
Electronic configuration: d6 = eg2 t2g3
Shape : Tetrahedral
Stereochemistry :- optically inactive
n = 5
Magnetic moment, µ = \(\sqrt{n(n +2)}\)
= \(\sqrt{5(5 +2)}\)
= \(\sqrt{35}\)≈ 6 BM
( e ) IUPAC name = Potassium hexacyanomanganate(II)
Coordination number = 6
Oxidation state of Mn :
x – 6 = -4
x = + 2
Electronic configuration: 3d5 = t2g5
Shape : Octahedral
Stereochemistry :- optically inactive
n = 1
Magnetic moment, µ = \(\sqrt{n(n +2)}\)
= \(\sqrt{1(1 +2)}\)
= \(\sqrt{3}\)= 1.732 BM

 


Q25. “Stability of a coordination compound in solution”, what do you understand by it? What are the factors responsible for the stability of a complex?

Ans:

Stability of a coordination compound in a solution is the degree/level of association among the species involved in a state of equilibrium.
Stability can also be written quantitatively in terms of formation constant or stability constant.
M + 3L ML3
Stability constant , β = \(\frac{[ML_{3}]}{[M][L]^{3}}\)
Greater the value of β , stronger is the metal –ligand bond.
Factors responsible for the stability of a complex:
( 1 ) Charge on the central metal ion – bigger the charge, more stable is the complex.
( 2 ) Nature of ligand – chelating ligand produces a more stable complex.
( 3 ) The basic strength of ligand- more basic a ligand, more stable its complex.

 

 

Q26. What is the chelate effect? Provide an example

Ans:

When a polydentate or a didentateligand fastens itself toa metal ion in such a way that it assumes the shape of a ring, the metal-ligand bond becomes more stable. These rings are called chelate rings.
From here we can infer that complexes with chelate rings are more stable than complexes without the rings. This phenomenon is termed the chelate effect.
Ni2+ (aq) + 6NH3 [ Ni( NH3)6 ]2+ (aq)
logβ = 7.99

Ni2+ (aq) + 3en (aq) [ Ni( en)3 ]2+ (aq)
logβ = 18.1 ( more stable )

Chelate Effect

 

Q27. Write short notes on the role of coordination compounds in the following fields :
( Also cite example in each case )
( a ) biological system
( b ) medicinal chemistry
( c ) analytical chemistry
( d ) metallurgy/extraction of metals

Ans:

( a ) Role in biological systems:
In the body of animals, there are several very important coordination compounds e.g. hemoglobin is a coordination compound of iron.
In plants, the chlorophyll pigment is a coordination compound of magnesium.
( b ) Role in medicinal chemistry:
So many coordinate compounds are used for curing purposes. For e.g., a coordination compound of platinum, cis-platin is used for checking the growth of tumors.
( c ) Role in analytical chemistry:
Determination of hardness of water.
( d ) Role in metallurgy or extraction:
During metal extraction from ores, complexes are formed. For e.g. gold combines with cyanide ions in an aqueous solution. Gold is then extracted from this complex using zinc.

 

 

Q28. When you put the complex Co(NH3)6Cl2 in a solution, how many ions do you get?

Ans:

The given complex [ Co( NH3)6 ]Cl2 ionizes to give three ions ,viz one [ Co( NH3)6] + and two Cl ions.

 


Q29. From the ions given below which one has the largest value of magnetic moment ?
( a ) [ Cr( H2O)6 ] 3+
( b ) [ Fe( H2O)6 ]
( c ) [Zn ( H2O)6 ] 2+

Ans:

( a ) [ Cr( H2O)6 ] 3+
number of unpaired electrons, n = 3
Magnetic moment, µ = \(\sqrt{3(3 +2)}\) = \(\sqrt{3(3 +2)}\)
= \(\sqrt{15}\)≈ 4BM
( b ) [ Fe ( H2O)6] 2+
number of unpaired electrons, n = 4
Magnetic moment, µ = \(\sqrt{4(4 +2)}\) = \(\sqrt{4(4 +2)}\)
= \(\sqrt{24}\)≈ 5 BM
( c ) [ Zn ( H2O)6] 2
n = 0
Thus, [ Zn ( H2O)6] 2has the highest magnetic moment value.

Q30. What is the oxidation number of cobalt in K[Co( CO )4] ?

Ans:

K[Co( CO )4] = K+[Co( CO )4]
We know,
=> x + 0 = – 1 [ Where x is the oxidation number.]
x = -1

Q31. Identify the most stable complex from among the following :
( a ) [ Fe ( H2O)6 ] 3+
( b ) [ Fe( NH3)6 ] 3+
( c ) [ Fe( C2O4)3 ] 3−
( d ) [ FeCl6 ] 3−

Ans:

In all the cases Fe has an oxidation state of +3. ( C2O4)3is a didentate chelating ligand and it forms chelating rings. Thus, ( c ) is the most stable complex.

 

 

Q32. Re arrange the following complexes in an ascending order of wavelengths they absorb from the visible spectrum :[ Ni ( NO2)6 ] 4−, [ Ni( H2O )6 ] 2+ , [ Ni ( NH3)6 ] 2+

Ans:

All of the complexes have the same metal ion, so the energy absorption depends upon the CFSE values of the ligands. According to the spectro-chemical series, the CFSE values of the ligands are in the order of : H2O < NH3< NO2

As,
E =hc / λ
=> E ∝ 1/ λ
Therefore, the values of the absorbed wavelength in an ascending order would be :
[ Ni ( NO2)6 ] 4−<[ Ni ( NH3)6 ] 2+ < [ Ni ( NO2)6 ] 4−