NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds presented here help students learn about “coordination compounds” and understand the concepts related to it. However, this chapter which is included in Class 12 chemistry syllabus is an important concept as we talk about atoms and chemical reactions. As such, coordination compounds include detailed solutions for all the questions of the NCERT textbooks. Students need to understand the topics clearly in order to score good marks and to avoid any difficulty in the future.

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The chemistry of coordination compounds is an important and challenging area of modern inorganic chemistry. The solutions are prepared by our subject experts with the objective of helping students in preparing well for their exams. Apart from the structured questions, this solution has detailed explanations to important questions from previous year question papers, exemplary questions, MCQs and HOTS (Higher Order Thinking Skills) are also provided to help students take coordination chemistry notes and to understand coordination compounds concepts seamlessly.

Class 12 Chemistry NCERT Solutions for Coordination Compounds

This NCERT solution comes in a PDF for easy access and free download to use it for their preparations. All in all, the chemistry solutions for Class 12 has been designed to help students to have a quick revision of a complete chapter along with the important questions.

Subtopics of Chemistry Chapter 9 – Coordination Compounds

  1. Werner’s Theory of Coordination Compounds
  2. Definitions of Some Important Terms Pertaining to Coordination Compounds
  3. Nomenclature of Coordination Compounds
  4. Isomerism in Coordination Compounds
  5. Bonding in Coordination Compounds
  6. Bonding in Metal Carbonyls
  7. Stability of Coordination Compounds
  8. Importance and Applications of Coordination Compounds.

Class 12 Chemistry NCERT Solutions (Coordination Compounds) – Important Questions

Q1. Explain the bonding in coordination compounds in terms of Werner’s postulates.

Ans :

( a ) A metal shows two kinds of valencies viz primary valency and secondary valency. Negative ions satisfy primary valencies and secondary valencies are filled by both neutral ions and negative ions.
( b ) A metal ion has a fixed amount of secondary valencies about the central atom. These valencies also 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 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?

Ans :

FeSO4 solution when 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 with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.

Ans :

( a ) Ligands – they are neutral molecules or negative ions bound to a metal atom in the coordination entity. Example- ClOH
( b ) Coordination entity –  they are electrically charged radicals or species. They constitute 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 kind of 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. What is meant by unidentate, bidentate and ambidentate ligands? Give two examples for each.

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 ) Bidentate – these are ligands with two donor sites.
Example – Ethane-1,2-diamine , Oxalate ion ( C2O42- )

Q5. Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H2O)(CN)(en)2]2+

(ii) [CoBr2(en)2]+

(iii) [PtCl4]2–

(iv) K3[Fe(CN)6]

(v) [Cr(NH3)3Cl3]  

Ans :

( i ) [Co (H2O) (CN) (en)2 ] 2+
=>  x + 0  + ( -1)  + 2 (0) = +2
x -1 = +2
x = +3

( ii )  [ Co Br2 (en)2 ] 2+
=> x + 2 ( -1 ) + 2 ( 0 ) = +1
x  – 2 = +1
x = +3

( iii) [ PtCl4 ] 2-
=> x + 4 ( -1 ) = -2x
x = +2

( iv ) K3 [Fe (CN)6 ]
=>  [ Fe ( CN )6]3-
=> x + 6 ( -1 ) = -3
x = +3

( v ) [ Cr(NH3)3Cl3 ]
=> x + 3( 0 ) + 3 ( -1 ) = 0
x – 3 = 0
x = 3

Q6. Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxidozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanidonickelate(II)
(v) Pentaamminenitrito-O-cobalt(III)

(vi) Hexaamminecobalt(III) sulphate

(vii) Potassium tri(oxalato)chromate(III)

(viii) Hexaammineplatinum(IV)

(ix) Tetrabromidocuprate(II)

(x) Pentaamminenitrito-N-cobalt(III)

Ans :

( i ) [Zn(OH)4]2 –

( ii ) K2[ Pd Cl4]

( iii ) [ Pt ( NH3)2Cl2]

( iv ) K2[ Ni(CN )4]

( v ) [ Co (NO2) ( NH3)5] 2+

( vi) [ Co( NH3)6]2 (SO4)3

( vii ) K3 [ Cr ( C2O4)3 ]

( viii ) [ Pt (NH3)6] 4+

( ix ) [ Cu (Br)4] 2−

( x ) [Co ( ONO )( NH3)5] 2+

Q7. 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]

Ans :

( i ) Hexaamminecobalt(III) chloride

( ii ) Diamminechlorido(methylamine) platinum(II) chloride

( iii ) Hexaquatitanium(III) ion

( iv ) Tetraamminichloridonitrito-N-Cobalt(III) chloride

( v ) Hexaquamanganese(II) ion

( vi ) Tetrachloridonickelate(II) ion

( vii ) Hexaamminenickel(II) chloride

( viii ) Tris(ethane-1, 2-diammine) cobalt(III) ion

( ix ) Tetracarbonylnickel(0)

 Q8.  List various types of isomerism possible for coordination compounds, giving an example of each.
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. How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C2O4)3]3–

(ii) [Co(NH3)3Cl3]

Ans :

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


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



Q10. Draw the structures of optical isomers of:
(i) [Cr(C2O4)3]3–

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

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

Ans :

( i ) [ Cr( C2O4 )3 ] 3−


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


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


Q11. Draw all the isomers (geometrical and optical) of:
(i) [CoCl2(en)2]+

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

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


( i ) [ CoCl2 (en)2 ] +


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


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


Q12. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?


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


None of the above isomers will exhibit optical isomerism.

Q13. 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


The blue colour 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 coloured [  CuF4 ] 2+ ions.
[Cu(H2O)4]2++4F[CuF4]2+4H2O[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 coloured [ CuCl4 ] 2+ ions.
[Cu(H2O)4]2++4Cl[CuCl4]2+4H2O[Cu(H_{2}O)_{4}]^{2+} + 4Cl^{-}\rightarrow [CuCl_{4}]^{2-} + 4H_{2}O

Q14. What is the coordination entity formed when an 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?


2[CuSO4](aq)+10KCN(aq)2K2[Cu(CN)4](aq)+2K2SO4(aq)+(CN)22[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. 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–

(iii) [Co(C2O4)3]3–

(iv) [CoF6]3


( i ) [ 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 :


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


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


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



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

( ii ) [FeF6] 3−
In this coordinate entity, the oxidation state of iron is  +3.
Orbitals of Fe +3 ion:


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 :


Thus, the geometry of this coordinate entity is octahedral.

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


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+ :


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


Thus, the complex shows octahedral geometry.

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


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 :


Thus, the complex has a geometric configuration of an octahedral and it is paramagnetic.

Q16. Draw figure to show the splitting of d orbitals in an octahedral crystal field.


Ans: 16.1

Q17. What is the spectrochemical series? Explain the difference between a weak
field ligand and a strong field ligand.


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

Q18. What is crystal field splitting energy? How does the magnitude of ∆o decide the actual configuration of d orbitals in a coordination entity?


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 symbolized 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 an ∆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. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic. Explain why?


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 diamagnetic in 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.

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

Q20. A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2– is colourless. Explain


[ 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 colour 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 colour.

Q21. [Fe(CN)6]4– and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?


[ 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 colours in a solution.

Q22.  Discuss the nature of bonding in metal carbonyls.

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 are donated to the empty orbital of the metal by the carbonyl carbon. A π bond forms when a pair of electrons are 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 are represented as :


Q23. Give the oxidation state, d orbital occupation and coordination number of
the central metal ion in the following complexes:
(i) K3[Co(C2O4)3] (iii) (NH4)2[CoF4]
(ii) cis-[CrCl2(en)2]Cl (iv) [Mn(H2O)6]SO4


( i )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.
( ii ) 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.
( iii ) ( 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.
( iv ) [ 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. Write down the IUPAC name for each of the following complexes 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

(iii) [CrCl3(py)3]

(iv) Cs[FeCl4]

(v) K4[Mn(CN)6]


(i) 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 :


Magnetic moment, µ = n(n+2)\sqrt{n(n +2)}              [ n = unpaired electrons ]
= 3(3+2)\sqrt{3(3 +2)}
= 15\sqrt{15}≈ 4BM
( ii ) 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 :


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


n = 3
Magnetic moment, µ = n(n+2)\sqrt{n(n +2)}
= 3(3+2)\sqrt{3(3 +2)}
= 15\sqrt{15}≈ 4BM
( iv )      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 = n(n+2)\sqrt{n(n +2)}
= 5(5+2)\sqrt{5(5 +2)}
= 35\sqrt{35}≈ 6 BM
( v )      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 = n(n+2)\sqrt{n(n +2)}
= 1(1+2)\sqrt{1(1 +2)}
= 3\sqrt{3}= 1.732  BM

Q25.  Explain the violet colour of the complex [Ti(H2O)6]3+ on the basis of crystal field theory

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 = [ML3][M][L]3\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 meant by the chelate effect? Give an example.


When a polydentate or a bidentate ligand 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. Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems (iii) analytical chemistry
(ii) medicinal chemistry and (iv) extraction/metallurgy of metals.


( i ) 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, chlorophyll pigment is a coordination compound of magnesium.
( ii ) 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.
( iii ) Role in analytical chemistry:
Determination of hardness of the water.
( iv) 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. How many ions are produced from the complex Co(NH3)6Cl2 in solution?
(i) 6 (ii) 4 (iii) 3 (iv) 2


(iii) 3

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

Q29. 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+


( i ) [ Cr( H2O)6 ] 3+
number of unpaired electrons, n  = 3
Magnetic moment, µ = 3(3+2)\sqrt{3(3 +2)}                                                                                                                                                  = 3(3+2)\sqrt{3(3 +2)}
= 15\sqrt{15}≈ 4BM
( ii ) [ Fe ( H2O)6] 2+
number of unpaired electrons, n  = 4
Magnetic moment, µ = 4(4+2)\sqrt{4(4 +2)}                                                                                                                                                  = 4(4+2)\sqrt{4(4 +2)}
= 24\sqrt{24}≈ 5 BM
( iii ) [ 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]?


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

Q31. 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–

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

Q32. 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+?


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

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

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Frequently Asked Questions on NCERT Solutions for Class 12 Chemistry Chapter 9

What can I learn from the Chapter 9 of NCERT Solutions for Class 12 Chemistry?

From the Chapter 9 of NCERT Solutions for Class 12 Chemistry, you can learn about topics/subtopics like –
1. Werner’s Theory of Coordination Compounds
2. Definitions of Some Important Terms Pertaining to Coordination Compounds
3. Nomenclature of Coordination Compounds
4. Isomerism in Coordination Compounds
5. Bonding in Coordination Compounds
6. Bonding in Metal Carbonyls
7. Stability of Coordination Compounds
8. Importance and Applications of Coordination Compounds.

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Students should make use of a proper study material while preparing the Chapter 9 of Class 12 Chemistry. The chapter should be revised more than three times to get a clear view about the concepts and the method of solving numericals. By referring to the solutions from BYJU’S, students will be able to get their doubts clarified instantly. It will also improve problem solving and logical thinking skills among students which will help them to perform well in the board exams. The solutions are designed by the experts having conceptual knowledge about the fundamental topics to help students score well in the Class 12 exam.

How to answer the questions from the Chapter 9 of NCERT Solutions for Class 12 Chemistry?

The Chapter 9 of NCERT Solutions for Class 12 Chemistry basically has questions related to naming the chemical formula of compounds. The other important concept is drawing the orbital for the various compounds. Several reasoning questions like why certain compounds have different colors are explained in brief under this chapter. If you are worried about the perfect study material to score more marks in the Class 12 exam, the NCERT Solutions at BYJU’S is the best option. The solutions are created in an interactive manner to make learning fun for the students.

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