NCERT Solutions | Class 12 Chemistry Chapter 9 | Coordination Compounds

CBSE Solutions | Chemistry Class 12
Check the below NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds Pdf free download. NCERT Solutions Class 12 Chemistry were prepared based on the latest exam pattern. We have Provided Coordination Compounds Class 12 Chemistry NCERT Solutions to help students understand the concept very well.
NCERT | Class 12 Chemistry
Book: | National Council of Educational Research and Training (NCERT) |
---|---|
Board: | Central Board of Secondary Education (CBSE) |
Class: | 12th |
Subject: | Chemistry |
Chapter: | 9 |
Chapters Name: | Coordination Compounds |
Medium: | English |
Coordination Compounds | Class 12 Chemistry | NCERT Books Solutions
NCERT Exercises
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 1.
- Tetraamminediaquacobalt (lll) chloride
- Potassium tetracyanonickelate (ll)
- Tris(ethane-1,2-diamine) chromium (lll) chloride
- Amminebromidochloridonitrito-N- platinate (ll)
- Dichloridobis(ethane-1,2-diamine) platinum (IV) nitrate
- Iron (lll) hexacyanoferrate (ll)
Solution.

NCERT Solutions for Class 12 Chemistry Chapter 9, Question 2.

Solution.
- Hexaamminecobalt (III) chloride
- Pentaamminechloridocobalt (III) chloride (HO3)2
- Potassium hexacyanoferrate (III)
- Potassium trioxalatoferrate (III)
- Potassium tetrachloridopalladate (II)
- Diamminechlorido (methylamine) platinum (II) chloride
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 3.


Solution.


NCERT Solutions for Class 12 Chemistry Chapter 9, Question 4.
Solution.
The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents :
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 5.
Solution.

NCERT Solutions for Class 12 Chemistry Chapter 9, Question 6.
Solution.
In Ni(CO)4, Ni is in zero oxidation state whereas in NiCl42-, it is in +2 oxidation state. In the presence of CO ligand, the unpaired d-electrons of Ni pair up but Cl– being a weak ligand is unable to pair up the unpaired electrons.NCERT Solutions for Class 12 Chemistry Chapter 9, Question 7.
Solution.
In presence of CN–, (a strong ligand), the 3d electrons pair up leaving only one unpaired electron. The hybridisation is d2 sp3 forming inner orbital complex. In the presence of H2O, (a weak ligand), 3d electrons do not pair up. The hybridisation is sp3 d2 forming an outer orbital complex containing five unpaired electrons hence, it is strongly paramagnetic.NCERT Solutions for Class 12 Chemistry Chapter 9, Question 8.
Solution.
in the presence of NH3, the 3d electrons pair up leaving two d orbitals empty to be involved in d2 sp3 hybridisation forming inner orbital complex in case of [CO(NH3)6]3+.In Ni(NH3)6]2+, Ni is in +2 oxidation state and has d3 configuration, the hybridisation involved is sp3 d2 forming outer orbital complex.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 9.
Solution.
For square planer shape, the hybridisation is dsp2 . Hence, the unpaired electrons in 5d orbital pair up to make one f-orbital empty for dsp2 hybridisation. Thus there is no unpaired electron.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 10.
Solution.
Mn(II) has 3d5 electronic configuration. Water is a weak field ligand and therefore ∆0 is small. Therefore, the hexaaqua complex will be high spin complex containing 5 unpaired electrons. On the other hand, CN– is a strong field ligand and therefore, ∆0 is large. Therefore, in its cyano complex, the electrons pair up and have only one unpaired electron.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 11.
Solution.


NCERT Exercises
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 1.
Solution.
The main postulates are :- In coordination compounds metals show two types of linkages (valencies)-primary and secondary.
- The primary valencies are normally ionisable and are satisfied by negative ions.
- The secondary valencies are non ionisable. These are satisfied by neutral molecules or negative ions. The secondary valency is equal to the coordination number and is fixed for a metal.
- The ions/groups bound by the secondary linkages to the metal have characteristic spatial .arrangements corresponding to different coordination numbers.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 2.
Solution.
FeSO4 solution when mixed with (NH4)2SO4 solution in 1:1 molar ratio forms a double salt FeSO4(NH4)2SO4-6H2O which when dissolved in water dissociates into simple ions to give tests for its constituent ions. When CuSO4 is mixed with aqueous ammonia a complex ion [Cu(NH3)4]2+ is formed which does not give Cu2+ in the solution.NCERT Solutions for Class 12 Chemistry Chapter 9, Question 3.
Solution.
(i) Coordination entity : A coordination entity constitutes a central metal atom or ion bonded loa fixed number of ions or molecules. For example, [COCl3(NH3)3] is a coordination entity in which the cobalt ion is surrounded by three ammonia molecules and three chloride ions. Other examples are [Ni(CO)4], [PtCl2(NH3)2], [Fe(CN)6]4-, [CO(NH3)6]3+.(ii) Ligand : The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be simple ions such as Cl–, small molecules such as H2O or NH3, larger molecules such as H2NCH2CH2NH2 or N(CH2CH2NH2)3 or even macromolecules, such as proteins.
(iii) Coordination number : The coordination number [C.N.] of a metal ion in a complex can be defined as the number of ligand or donor atoms to which the metal is directly bonded. For example, in the complex ions, [PtCl6]2- and [Ni(NH3)4]2+, the coordination number of Pt and Ni are 6 and 4 respectively. Similarly, in the complex ions, [Fe(C2O4)3]3- and [CO(en)3]3+,the coordination number of both, Fe and Co, is 6 because C2O42-and en (ethane-1,2-diamine) are bidentate ligands.
(iv) Coordination polyhedron : The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. The most common coordination polyhedra are octahedral, square planar and tetrahedral. For example, [CO(NH3)6]3+ is octahedral, [Ni(CO)4] is tetrahedral and [PtCl4]2- is square planar.
(v) Homoleptic and heteroleptic complexe : Complexes in which a metal is bound to only one kind of donor groups, e.g., [CO(NH3)6]3+, are known as homoleptic. Complexes in which a metal is bound to more than one kind of donor groups, e.g., [CO(NH3)4Cl2]+, are known as heteroleptic.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 4.
Solution.
Unidentate ligands are those which bind to the metal ion through a single donor atom, e.g., Cl– , H2O.Bidentate ligands are those which bind to the metal ion through two donor atoms. e.g., ethane-1,2-diamine (H2NCH2CH2NH2), oxalate (C2O42-) ion.
Ambidentate ligands are those which can bind to metal ion through two different donor atoms, e.g., NO2– and SCN– ion.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 5.

Solution.

NCERT Solutions for Class 12 Chemistry Chapter 9, Question 6.
- Tetrahydroxozincate (ll)
- Potassium tetrachloridopalladate (ll)
- Diamminedichloridoplatinum (ll)
- Potassium tetracyanonickelate (ll)
- Pentaamminenitritio-O-cobalt (lll)
- Hexaamminecobalt (lll) sulphate
- Potassium tri(oxalato)chromate (lll)
- Hexaammineplatinum (IV)
- Tetrabromidocuprate (ll)
- Pentaamminenitrito-N-cobalt (lll)
Solution.


NCERT Solutions for Class 12 Chemistry Chapter 9, Question 7.

Solution.
- Hexaamminecobalt (III) chloride
- Diamminechloridomethylamine platinum (II) chloride
- Hexaaquatitanium (III) ion
- Tetraamminechloridonitrito-N-cobalt (III) chloride
- Hexaaquamanganese (II) ion
- Tetrachloridonickelate (II) ion
- Hexaammine nickel (II) chloride
- Tris (ethane-1,2-diamine) cobalt(III) ion
- Tetracarbonylnickel (0)
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 8.
Solution.
(i) Geometrical isomerism : This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. Important examples of this behaviour are found with coordination numbers 4 and 6. In a square planar complex of formula [MX2L2] (X and L are unidentate), the two identical ligands may be arranged adjacent to each other in a cis- isomer, or opposite to each other in a trans 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.
(ii) Optical isomerism : Optical isomerism is common in octahedral complexes involving bidentate ligands. In a coordination entity of the [PtCl2(en)2]2+, only the ds-isomer shows optical activity.
(iii) Linkage isomerism : Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand, NCS–, which may bind through the nitrogen to give M-N CS or through sulphur to give M-SCN. This behaviour was seen in the complex [CO(NH3)5(NO2)]Cl2, which is obtained as the red form, in which the nitrite ligand is bound through oxygen ( ONO), and as the yellow form, in which the nitrite ligand is bound through nitrogen (-NO2).
(iv) 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. An example is provided by [Co(NH3)6] [Cr(CN)6], in which the NH3 ligands are bound to CO3+ and the CN– ligands to Cr3+. In its coordination isomer [Cr(NH3)6][CO(CN)6], the NH3 ligands are bound to Cr3+ and the CN– ligands to CO3+.
(v) Ionisation isomerism : This form of 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. An example is provided by the ionisation isomers [CO(NH3)5SO4]Br and [CO(NH3)5Br]SO4.
(vi) Solvate isomerism : This form of 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).
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 9.
- [Cr(C2O4)3]3-
- [CO(NH3)3CI3]
Solution.
- Zero
- Two – Facial and meridional.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 10.
Solution.


NCERT Solutions for Class 12 Chemistry Chapter 9, Question 11.

Solution.


NCERT Solutions for Class 12 Chemistry Chapter 9, Question 12.
Solution.

NCERT Solutions for Class 12 Chemistry Chapter 9, Question 13.
- a green precipitate with aqueous potassium fluoride and
- a bright green solution with aqueous potassium chloride. Explain these experimental results.
Solution.

NCERT Solutions for Class 12 Chemistry Chapter 9, Question 14.
Solution.

CN– is a strong field ligand, therefore, [Cu(CN)4]2- is highly stable and has large value of stability constant. On passing H2S, CuS is not formed because this coordination entity does not give Cu2+ ion.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 15.

Solution.


NCERT Solutions for Class 12 Chemistry Chapter 9, Question 16.
Solution.
Let us assume that the six ligands are positioned symmetrically along the Cartesian axes, with metal atom at the origin.
As the ligands approach, first there is an increase in energy of d-orbitals relative to that of the free ion just as would be the case in a spherical field.
The orbitals lying along the axes (dz2 and dx2 – y2) get repelled more strongly than dxy, dyz and dzx orbitals which have lobes directed between the axes.
The dz2 and dx2-y2 orbitals get raised in energy and dxy dyz, dxz orbitals are lowered in energy relative to the average energy in the spherical crystal field. Thus, the degenerate set of d-orbitals get split into two sets : the lower energy orbitals set, t2g and the higher energy orbitals set, eg. The energy is separated by ∆0
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 17.
Solution.
The crystal field splitting, ∆0, depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong fields in which, the splitting will be large whereas others produce weak fields and consequently result in small splitting of d-orbitals. In general, ligands can be arranged in a series in the order of increasing field strength as given below :
Such a series is termed as spectrochemical series.
If ∆0 < P, the fourth electron enters one of the eg orbitals giving the configuration t23g g e1g. Ligands for which ∆0 < P are known as weak field ligands and form high spin complexes. If ∆0 > P, it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with configuration t2g 4e0g. Ligands which produce this effect are known as strong field ligands and form low spin complexes.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 18.
Solution.

NCERT Solutions for Class 12 Chemistry Chapter 9, Question 19.
Solution.

NCERT Solutions for Class 12 Chemistry Chapter 9, Question 20.
Solution.
[Ni(H2O)6]2+has unpaired electrons due to weak H2O ligands which absorb light from visible region and radiate complementary colour i.e., green whereas [Ni(CN)4]2- does not have any unpaired electron due to strong CN– ligand, therefore, does not absorb light from visible region hence, it is colourless.NCERT Solutions for Class 12 Chemistry Chapter 9, Question 21.
Solution.


Both ligands show different magnitude of crystal field splitting energy due to different nature hence, absorb different wavelengths ans show different colours.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 22.
Solution.

The metal-carbon bond in metal carbonyls possess both s and p character. 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. The M-C π bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant antibonding π* orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 23.

Solution.

NCERT Solutions for Class 12 Chemistry Chapter 9, Question 24.

Solution.



NCERT Solutions for Class 12 Chemistry Chapter 9, Question 25.
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 (stability or formation) equilibrium constant for the association, quantitatively expresses the stability. Thus, if we have a reaction of the type :

Factors affecting stability of complexes :
- Smaller the size of cation, greater will be the stability of complex e.g., Fe3+ forms more stable complex than Fe2+.
- Greater the charge on central metal ion, more stable will be the complex e.g., Pt4+ forms more stable complex than Pt2+.
- Stronger the ligand, more stable will be the complex formed e.g., CN forms more stable complex then NH3.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 26.
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. The number of such ligating groups is called the denticity of the ligand. Such complexes, called chelate complexes tend to be more stable than similar complexes containing unidentate ligands. Example : EDTA, DMG, etc.NCERT Solutions for Class 12 Chemistry Chapter 9, Question 27.
- biological systems
- medicinal chemistry
- analytical chemistry
- extraction/metallurgy of metals.
Solution.
(i) Coordination compounds are of great importance in biological systems. The pigment responsible for photosynthesis, chlorophyll, is a coordination compound of magnesium. Haemoglobin, the red pigment 1 of blood which acts as oxygen carrier is a coordination compound of iron. Vitamin B12, cyanocobalamine, the anti- pernicious anaemia factor, is a coordination compound of cobalt. Among the other compounds of biological importance with coordinated metal ions are the enzymes like, carboxypeptidase A and carbonic anhydrase (catalysts of biological systems).(ii) There is growing Interest in the use of chelate therapy in medicinal chemistry. An example is the treatment of problems caused by the presence of metals in toxic proportions in plant/animal systems. Thus, excess of copper and iron are removed by the chelating ligands D-penicillamine and desferrioxime B via the formation of coordination compounds.
EDTA is used in the treatment of lead poisoning. Some coordination compounds of platinum effectively inhibit the growth of tumours. Examples are: ds-platin and related compounds.
(iii) Coordination compounds find use in many qualitative and quantitative chemical analysis. The familiar colour reactions given by metal ions with a number of ligands (especially chelating ligands), as a result of formation of coordination entities, form the basis for their detection and estimation by classical and instrumental methods of analysis. Examples of such reagents include EDTA, DMG (dimethylglyoxime), a-nitroso- β-naphthol, cupron, etc.
(iv) Some important extraction processes of metals, like those of silver and gold, make use of complex formation. Gold, for example, combines with cyanide in the presence of oxygen and water to form the coordination entity [Au(CN)2]- in aqueous solution. Gold can be separated in metallic form from this solution by the addition of zinc.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 28.
(a) 6
(b) 4
(c) 3
(d) 2
Solution.

NCERT Solutions for Class 12 Chemistry Chapter 9, Question 29.
(a) [Cr(H2O)6]3+
(b) [Fe(H2O)6]2+
(c) [Zn(H2O)6]2+
Solution.

NCERT Solutions for Class 12 Chemistry Chapter 9, Question 30.
(a) +1
(b) +3
(c) -1
(d) -3
Solution.
(c) : + 1 + x + 4(0) = 0∴ x = – 1.
NCERT Solutions for Class 12 Chemistry Chapter 9, Question 31.
(a) [Fe(H2O)6]3+
(b) [Fe(NH3)6]3+
(c) [Fe(C2O4)3]3-
(d) [FeCI6]3-
Solution.
(c) : Since C2O42- is a bidentate ligand, it forms the most stable complex.NCERT Solutions for Class 12 Chemistry Chapter 9, Question 32.

Solution.
In spectrochemical series the order of the given ligands is H2O < NH3 < NO2– . Hence, the wave length of light will be absorbed in the opposite order since E = \(\frac { hc }{ \lambda } \)
NCERT Class 12 Chemistry
Class 12 Chemistry Chapters | Chemistry Class 12 Chapter 9
Chapterwise NCERT Solutions for Class 12 Chemistry
-
NCERT Solutions For Class 12 Chemistry Chapter 1 The Solid State
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions
NCERT Solutions For Class 12 Chemistry Chapter 3 Electro chemistry
NCERT Solutions For Class 12 Chemistry Chapter 4 Chemical Kinetics
NCERT Solutions For Class 12 Chemistry Chapter 5 Surface Chemistry
NCERT Solutions For Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements
NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements
NCERT Solutions For Class 12 Chemistry Chapter 8 The d and f Block Elements
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds
NCERT Solutions For Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes
NCERT Solutions For Class 12 Chemistry Chapter 11 Alcohols Phenols and Ethers
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life
NCERT Solutions for Class 12 All Subjects | NCERT Solutions for Class 10 All Subjects |
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