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| Chapter | 9. Coordination Compounds |
| Subject | Chemistry |
| Textbook | Chemistry I |
| Class | Twelve |
| Category | NCERT Solutions for Class 12 |
NCERT Solutions for Class 12 Chemistry is an excellent tool to revise your syllabus quickly. These solutions are made by expert faculties and teachers of Chemistry. It provides you with complete information, making it easier for you to solve the problems. The NCERT textbook includes questions based on concepts and theories. They are important for testing students’ understanding of the topics. On this page, we have provided you with the NCERT Solutions for Class 12 Chemistry Chapter 9: Coordination Compounds.
NCERT Solutions for Class 12 Chemistry Chapter 9
Coordination Compounds Solutions
Q1) Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer) Werner’s postulates explain the bonding in coordination compounds as follows:
- 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.
- 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.
- 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?
Answer) 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.
Answer)
(a) Ligands: they are neutral molecules or negative ions bound to a metal atom in the coordination entity. Example- Cl–, –OH
(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.
ex: (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 –
(i) Square Planar
(ii) Tetrahedral
(e) 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, didentate and ambidentate ligands? Give two examples for each.
Answer) A ligand may contain one or more unshared pairs of electrons which are called the donor sites of ligands. Now, depending on the number of these donor sites, ligands can be classified as follows:
(a) Unidentate ligands: Ligands with only one donor sites are called unidentate ligands. For e.g., 
, Cl− etc.
(b) Didentate ligands: Ligands that have two donor sites are called didentate ligands. Ex- Ethane-1,2-diamine, Oxalate ion
(c) Ambidentate ligands: Ligands that can attach themselves to the central metal atom through two different atoms are called ambidentate ligands. For example: NO–2or ONO–, CN– or NC–
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]
Answer)
( i ) [Co (H2O) (CN) (en)2 ] 2+
=> x + 0 + ( -1) + 2 (0) = +2
x -1 = +2
x = +3
( ii ) [ Co Br2 (en)2 ] 1+
=> x + 2 ( -1 ) + 2 ( 0 ) = +1
x – 2 = +1
x = +3
( iii) [ PtCl4 ] 2-
=> x + 4 ( -1 ) = -2
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) Tetrahydroxozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanonickelate(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)
Answer)
(i) [Zn(OH)4]2−
(ii) K2[PdCl4]
(iii) [Pt(NH3)2Cl2]
(iv) K2[Ni(CN)4]
(v) [Co(ONO) (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[NO2)(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]
Answer)
(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.
Answer)
(a) Geometric isomerism: This type of isomerism is common in heteroleptic complexes. It arises due to the different possible geometric arrangements of the ligands. For example:
(b) Optical isomerism: This type of isomerism arises in chiral molecules. Isomers are mirror images of each other and are non-superimposable.
(d) Coordination isomerism: This type of isomerism arises when the ligands are interchanged between cationic and anionic entities of differnet metal ions present in the complex. example- [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6]
(e) Ionization isomerism: This type of isomerism arises when a counter ion replaces a ligand within the coordination sphere. Thus, complexes that have the same composition, but furnish different ions when dissolved in water are called ionization isomers. For e.g., Co(NH3)5SO4)Br and Co(NH3)5Br]SO4.
(f) Solvate isomerism: Solvate isomers differ by whether or not the solvent molecule is directly bonded to the metal ion or merely present as a free solvent molecule in the crystal lattice. example- [Cr[H2O)6]Cl3 [Cr(H2O)5Cl]Cl2⋅H2O [Cr(H2O)5Cl2]Cl⋅2H2O
Q9) How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C2O4)3]3− (ii) [Co(NH3)3Cl3]
Answer i) In [ Cr(C2O4)3] 3− no geometric isomers are present because it is a bidentate ligand.
Answer 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)]+
Answer i) [Cr(C2O4)3]3−
Answer ii) [PtCl2(en)2]2+
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)]+
Answer i) [ CoCl2 (en)2 ] +
Answer ii) [ Co(NH3) Cl( en)2 ] 2+
Answer 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?
Answer) [Pt(NH3)(Br)(Cl)(py)
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.
Answer) Aqueous CuSO4 exists as [Cu(H2O)4]SO4. It is blue in colour due to the presence of [Cu[H2O)4]2+ ions.
Q14) 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?
Answer)
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−
Answer i) [Fe(CN)6]4−
In the above coordination complex, iron exists in the +II oxidation state.
Fe 2+ : Electronic configuration is 3d6
Orbitals of Fe2+ ion :
Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).
Answer ii) [FeF6]3−
In this complex, the oxidation state of Fe is +3.
Orbitals of Fe+3 ion:
Hence, the geometry of the complex is found to be octahedral.
Answer iii) [Co(C2O4)3]3−
Cobalt exists in the +3 oxidation state in the given complex.
Orbitals of Co3+ ion:
Hence, the geometry of the complex is found to be octahedral.
Answer iv) [CoF6]3−
Cobalt exists in the +3 oxidation state.
Orbitals of Co3+ ion:
Hence, the geometry of the complex is octahedral and paramagnetic.
Q16) Draw figure to show the splitting of d orbitals in an octahedral crystal field.
Answer)
Q17) What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer) A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy (CFSE) values. The ligands present on the R.H.S of the series are strong field ligands while that on the L.H.S are weak field ligands. Also, strong field ligands cause higher splitting in the d orbitals than weak field ligands.
I− < Br− < S2− < SCN− < Cl−< N3 < F− < OH− < C2O42− ∼ H2O < NCS− ∼ H− < CN− < NH3 < en ∼ SO32− < NO2− < phen < CO
Q18) What is crystal field splitting energy? How does the magnitude of Δo decide the actual configuration of d-orbitals in a coordination entity?
Answer) The degenerate d-orbitals (in a spherical field environment) split into two levels i.e., eg and t2g in the presence of ligands. The splitting of the degenerate levels due to the presence of ligands is called the crystal-field splitting while the energy difference between the two levels (eg and t2g) is called the crystal-field splitting energy. It is denoted by Δo.
After the orbitals have split, the filling of the electrons takes place. After 1 electron (each) has been filled in the three t2g orbitals, the filling of the fourth electron takes place in two ways. It can enter the eg orbital (giving rise to t2g3 eg1 like electronic configuration) or the pairing of the electrons can take place in the t2g orbitals (giving rise to t2g4 eg0 like electronic configuration). If the Δo value of a ligand is less than the pairing energy (P), then the electrons enter the eg orbital. On the other hand, if the Δo value of a ligand is more than the pairing energy (P), then the electrons enter the t2g orbital.
Q19) [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2− is diamagnetic. Explain why?
Answer) Cr is in the +3 oxidation state i.e., d3 configuration. Also, NH3 is a weak field ligand that does not cause the pairing of the electrons in the 3d orbital.
As there are no unpaired electrons, it is diamagnetic.
Q20) A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2− is colourless. Explain.
In [Ni(H2O)6]2+, 
is a weak field ligand. Therefore, there are unpaired electrons in Ni2+. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i.e., the possibility of d−d transition is present. Hence, Ni(H2O)6]2+ is coloured.
In [Ni(CN)4]2−, the electrons are all paired as CN– is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)4]2−. Hence, it is colourless.
Q21) [Fe(CN)6]4− and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
Answer) The colour of a particular coordination compound depends on the magnitude of the crystal-field splitting energy, Δ. This CFSE in turn depends on the nature of the ligand. In case of [Fe(CN)6]4− and [Fe(H2O)6]2+, the colour differs because there is a difference in the CFSE. Now, CN− is a strong field ligand having a higher CFSE value as compared to the CFSE value of water. This means that the absorption of energy for the intra d-d transition also differs. Hence, the transmitted colour also differs.
Q22) Discuss the nature of bonding in metal carbonyls.
Answer) 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]
(ii) cis-[Cr(en)2Cl2]Cl
(iii) (NH4)2[CoF4]
(iv) [Mn(H2O)6]SO4
Answer i) K3[Co(C2O4)3]
The central metal ion is Co.
Its coordination number is 6.
The oxidation state can be given as:
x − 6 = −3
x = + 3
The d orbital occupation for Co3+ is t2g6eg0.
Answer ii) cis-[Cr(en)2Cl2]Cl
The central metal ion is Cr.
The coordination number is 6.
The oxidation state can be given as:
x + 2(0) + 2(−1) = +1
x − 2 = +1
x = +3
The d orbital occupation for Cr3+ is t2g3.
Answer iii) (NH4)2[CoF4]
The central metal ion is Co.
The coordination number is 4.
The oxidation state can be given as:
x − 4 = −2
x = + 2
The d orbital occupation for Co2+ is eg4 t2g3.
Answer iv) [Mn(H2O)6]SO4
The central metal ion is Mn.
The coordination number is 6.
The oxidation state can be given as:
x + 0 = +2
x = +2
The d orbital occupation for Mn is t2g3 eg2.
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]
Answer i) Potassium diaquadioxalatochromate (III) trihydrate.
Oxidation state of chromium = 3
Electronic configuration: 3d3: t2g3
Coordination number = 6
Shape: octahedral
Stereochemistry:
Answer ii) [Co(NH3)5Cl]Cl2
IUPAC name: Pentaamminechloridocobalt(III) chloride
Oxidation state of Co = +3
Coordination number = 6
Shape: octahedral.
Electronic configuration: d6: t2g6.
Stereochemistry:
Magnetic Moment = 0
iii) CrCl3(py)3
IUPAC name: Trichloridotripyridinechromium (III)
Oxidation state of chromium = +3
Electronic configuration for d3 = t2g3
Coordination number = 6
Shape: octahedral.
Stereochemistry:
(iv) Cs[FeCl4]
IUPAC name: Caesium tetrachloroferrate (III)
Oxidation state of Fe = +3
Electronic configuration of d6 = eg2t2g3
Coordination number = 4
Shape: tetrahedral
Stereochemistry: optically inactive
Magnetic moment:
Answer v) K4[Mn(CN)6]
Potassium hexacyanomanganate(II)
Oxidation state of manganese = +2
Electronic configuration: d5+: t2g5
Coordination number = 6
Shape: octahedral.
Streochemistry: optically inactive
Magnetic moment, μ = √n(n + 2)
Q25) Explain the violet colour of the complex [Ti(H2O)6]3+ on the basis of crystal field theory
Answer) 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
Factors responsible for the stability of a complex:
- Charge on the central metal ion – bigger the charge, more stable is the complex.
- Nature of ligand – chelating ligand produces a more stable complex.
- The basic strength of ligand- more basic a ligand, more stable its complex.
Q26) What is meant by the chelate effect? Give an example.
Answer) When a ligand attaches to the metal ion in a manner that forms a ring, then the metal- ligand association is found to be more stable. In other words, we can say that complexes containing chelate rings are more stable than complexes without rings. This is known as 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 )
Q27) 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
Answer i) Role of coordination compounds in biological systems: We know that photosynthesis is made possible by the presence of the chlorophyll pigment. This pigment is a coordination compound of magnesium. In the human biological system, several coordination compounds play important roles. For example, the oxygen-carrier of blood, i.e., haemoglobin, is a coordination compound of iron.
Answer ii) Role of coordination compounds in medicinal chemistry: Certain coordination compounds of platinum (for example, cis-platin) are used for inhibiting the growth of tumours.
Answer iii) Role of coordination compounds in analytical chemistry: During salt analysis, a number of basic radicals are detected with the help of the colour changes they exhibit with different reagents. These colour changes are a result of the coordination compounds or complexes that the basic radicals form with different ligands.
Answer iv) Role of coordination compounds in extraction or metallurgy of metals: The process of extraction of some of the metals from their ores involves the formation of complexes. For example, in aqueous solution, gold combines with cyanide ions to form [Au(CN)2]. From this solution, gold is later extracted by the addition of zinc metal.
Q28) How many ions are produced from the complex Co(NH3)6Cl2 in solution?
(i) 6 (ii) 4 (iii) 3 (iv) 2
Answer) (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+
Answer)
Q30) The oxidation number of cobalt in K[Co(CO)4] is
Answer) We know that CO is a neutral ligand and K carries a charge of +1.
Therefore, the complex can be written as K+[Co(CO)4]−. Therefore, the oxidation number of Co in the given complex is −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−
Answer) We know that the stability of a complex increases by chelation. Therefore, the most stable complex is [Fe(C2O4)3]3−.
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+
Answer) 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 ascending order would be :
[Ni(H2O)6]2+ <[ Ni(NH3)6] 2+ < [ Ni(NO2)6] 4−
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