Careers360 Logo
NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds

author
Shivani PooniaUpdated on 25 Jun 2025, 11:25 PM IST

Did you know why the colour of blood is red or how some medicines work so effectively? The answer lies in Coordination compounds.These are special compounds formed when a central metal atom combines with small atoms or ions called ligands.These compounds function like a well organised system, where each part plays a specific role to make the whole structure stable and useful. The structure and properties of these compounds that make them essential in nature and industry.

This Story also Contains

  1. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 1)
  2. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 2)
  3. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Short Answer Type
  4. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Matching Type
  5. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Assertion and Reason Type
  6. NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Long Answer Type
  7. Class 12 Chemistry NCERT Chapter 9: Higher Order Thinking Skills (HOTS) Questions
  8. Approach to Solve Questions of Chapter 9
  9. Topics Covered in NCERT Exemplar Solutions Class 12 Chemistry Chapter 9
  10. NCERT Class 12 Chemistry Exemplar Chapter 9 Important Formulas
  11. NCERT Exemplar Solutions Class 12 Chemistry Chapter-Wise
  12. NCERT Solutions for Class 12 Chemistry
  13. NCERT Solution subject-wise
  14. NCERT Exemplar Class 12 Solutions
  15. NCERT Notes subject-wise
NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 Coordination Compounds
Coordination Compounds



The important topics like crystal field theory (CFT), ligands, color and complex formation are all discussed in this chapter. The NCERT Exemplar Class 12 Chemistry Solutions chapter 9 are designed to present these concepts in the form of solved questions. These NCERT Exemplar solutions are valuable resources to enhance performance in board exams as well as in competitive exams as they provide well-explained answers to help you understand the concepts with ease. The higher-order thinking skills (HOTS) questions are also provided to develop your critical thinking. We have also added some points that will help you build a good problem-solving strategy.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 1)

The MCQ questions are covered in the NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 to enhance your knowledge. The concepts are explained in detail in class 12 chemistry chapter 9 notes available on our website.

Question 1 Which of the following complexes formed by Cu2+ ions is most stable?
(i) Cu2++4NH3[Cu(NH3)4]2+logK=11.6
(ii) Cu2++4CN[Cu(CN)4]2logK=27.3
(iii) Cu2++2en[Cu(en)2]2+logK=15.4
(iv) Cu2++4H2O[Cu(H2O)4]2+logK=8.9
Answer:

Aakash Repeater Courses

Take Aakash iACST and get instant scholarship on coaching programs.

Option (ii) is the correct answer.
Explanation:
The stability of a compound will be more when the value of log K increases.
For reaction,
Cu2++4CN[Cu(CN)4]2
K=[Cu(CN4)2][Cu2+][CN]4andlog K=27.3
Log K has the highest value for this reaction amongst all the four reactions. The value of K will also be higher so, the stability of this complex will be highest.

Question 2: The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, [Co(NH3)6]3+, [Co(CN)6]3, [Co(H2O)6]3+
(i) [Co(CN)6]3>[Co(NH3)6]3+>[Co(H2O)6]3+
(ii) [Co(NH3)6]3+>[Co(H2O)6]3+>[Co(CN)6]3
(iii) [Co(H2O)6]3+>[Co(NH3)6]3+>[Co(CN)6]3
(iv) [Co(CN)6]3>[Co(NH3)6]3+>[Co(H2O)6]3+
Answer:

Option (iii) is the correct answer.
Explanation :
Δ0 values follow the order : [Co(H2O)6]3+<[Co(NH3)6]3+<[Co(CN)6]3 and therefore,absorption wavelength follows the order:
[Co(H2O)6]3+>[Co(NH3)6]3+>[Co(CN)6]3

Question 3 When 0.1 mol CoCl3(NH3)5 is treated with an excess of AgNO3, 0.2 mol of AgCl are obtained. The conductivity of the solution will correspond to
(i) 1:3 electrolyte
(ii) 1:2 electrolyte
(iii) 1:1 electrolyte
(iv) 3:1 electrolyte
Answer:

Option (ii) is the correct answer.
Explanation:
One mole of chloride ion gets precipitated by one mole of AgNO3. AgCl, is obtained when 0.1 mole of CoCl3(NH3)5 is treated with AgNO3 in excess. So this leaves two free chloride ions in the solution of electrolyte for every 1 mole reaction.
Hence the molecular formula should be [Co(NH3)5Cl]Cl2 and the solution of electrolyte should contain [Co(NH3)5Cl]2+ and two Cl as their constituent ions. Therefore, it is an 1:2 electrolyte.
[Co(NH3)5Cl]Cl2[Co(NH3)5Cl]2+(aq)+2Cl(aq)

Question 4 When 1 mol CrCl3.6H2O is treated with an excess of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is:
(i) [CrCl3(H2O)3].3H2O
(ii) [CrCl2(H2O)4]Cl.2H2O
(iii) [CrCl(H2O)5]Cl2.H2O
(iv) [Cr(H2O)6]Cl3
Answer:

Option (iv) is the correct answer.
Explanation: 3 mol of AgCl indicates that 3Cl ions are given in the solution. Therefore, the formula should be [Cr(H2O)6]Cl3.

Question 5: The correct IUPAC name of [Pt(NH3)2Cl2] is
(i) Diamminedichloridoplatinum (II)
(ii) Diamminedichloridoplatinum (IV)
(iii) Diamminedichloridoplatinum (0)
(iv) Dichloridodiammineplatinum (IV)
Answer:

Option (i) is the correct answer.
Explanation: diamminedichloridoplatinum (II) is [Pt(NH3)2Cl2].

Question 6 The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species?
(i) [Fe(CO)5]
(ii) [Fe(CN)6]3
(iii) [Fe(C2O4)3]3
(iv) [Fe(H2O)6]3+
Answer:

Option (iii) is the correct answer.
Explanation: Chelation is the formation of cycle of linkages between the metal ion and ligands is responsible for stabilizing the coordination compounds. A ligand which chelates any metal ion is known as chelating ligand.
Here, oxalate ion is the chelating ligand and [Fe(C2O4)3]3 is the coordination compound. Hence, the oxalate ions stabilize the coordination compound by chelating Fe3+ ions.

Question 7 Indicate the complex ion which shows geometrical isomerism.
(i) [Cr(H2O)4Cl2]+
(ii) [Pt(NH3)3Cl]
(iii) [Co(NH3)6]3+
(iv) [Co(CN)5(NC)]3
Answer:

Option (i) is the correct answer.
Explanation :
(a) [Cr(H2O)4Cl2]+
isomers

Question 8 The CFSE for octahedral [CoCl6]4 is 18,000 cm–1. The CFSE for tetrahedral [CoCl4]2 will be
(i) 18,000 cm1
(ii) 16,000 cm1
(iii) 8,000 cm1
(iv) 20,000 cm1
Answer:

Option (iii) is the correct answer.
Explanation :
(c) CFSE for tetrahedral complex is Δt=(49)Δ0
Δt=49×18000=8,000cm1

Question 9 Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium complexes of the type [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are
(i) linkage isomers
(ii) coordination isomers
(iii) ionisation isomers
(iv) geometrical isomers
Answer:

Option (i) is the correct answer.
Explanation: Ambident ligands are those ligands which have two different bonding sites. For example: NCS,NO2 etc.
In this case, NCS contains two binding sites at N and S. Therefore, NCS (thiocyanate) will be bonding to the metal in these two ways; MNCS or MSNC
Hence, the coordination compounds containing NCS as ligands can show linkages isomerism; [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are linkage isomers.

Question 10 The compounds [Co(SO4)(NH3)5]Br and [Co(SO4)(NH3)5]Cl represent
(i) linkage isomerism
(ii) ionisation isomerism
(iii) coordination isomerism
(iv) no isomerism
Answer:

Option (iv) is the correct answer.
Explanation:[Co(SO4)(NH3)5]Br and [Co(SO4)(NH3)5]Cl are different compounds and show no isomerism.

Question 11 A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following is not a chelating agent?
(i) thiosulphate
(ii) oxalate
(iii) glycinato
(iv) ethane-1,2-diamine
Answer:

Option (i) is the correct answer.
Explanation: Thiosulphate or S2O32 is a monodentate ligand and not a chelating agent.

Question 12 Which of the following species is not expected to be a ligand?
(i) NO
(ii) NH4+
(iii) NH2CH2CH2NH2
(iv) CO
Answer:

Option (ii) is the correct answer.
Explanation: To form M-L bonds, the ligand must donate a pair of electrons or any loosely held electron pair to the metal.
eg:


Among all NH4+ does not have any pair of electrons.
Hence NH4+ is not a ligand.

Question 13 What kind of isomerism exists between [Cr(H2O)6]Cl3(violet) and [Cr(H2O)5Cl]Cl2.H2O(greyish-green) ?
(i) linkage isomerism
(ii) solvate isomerism
(iii) ionisation isomerism
(iv) coordination isomerism
Answer:

Option (ii) is the correct answer.
Explanation: The compound has different water molecules in number inside and outside of the coordination sphere.

Question 14 IUPAC name of [Pt(NH3)2Cl(NO2)] is :
(i) Platinum diaminechloronitrite
(ii) Chloronitrito-N-ammineplatinum (II)
(iii) Diamminechloridonitrito-N-platinum (II)
(iv) Diamminechloronitrito-N-platinate (II)
Answer:

Option (iii) is the correct answer.
Explanation: diamminechloridonitrito-N-platinum (II) is [Pt(NH3)2Cl(NO2)]

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: MCQ (Type 2)

The class 12 chemistry chapter 9 questions are provided here with simple explanations. Learn more through these advanced MCQs

Question 15 The atomic number of Mn, Fe and Co are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complexions are diamagnetic?
(i) [Co(NH3)6]3+
(ii) [Mn(CN)6]3
(iii) [Fe(CN)6]4
(iv) [Fe(CN)6]3
Answer:

Option (i) and (iii) are the correct answers.
Explanation :
(a,c)
(a) Molecular orbital electronic configuration of Co3+ in [Co(NH3)6]3+is
Electronic configuration
Number of unpaired electron = 0
Magnetic property = Diamagnetic
(b) Molecular orbital electronic configuration of Mn3+ in [Mn(CN)6]3
Electronic configuration
Number of unpaired electron = 2
Magnetic property = Paramagnetic
(c) Molecular orbital electronic configuration of Fe2+ in [Fe(CN)6]4
Electronic configuration
Number of unpaired electron = 0
Magnetic property = Diamagnetic
(d) Molecular orbital electronic configuration of Fe3+ in [Fe(CN)6]3
Electronic configuration
Number of unpaired electron = 1
Magnetic property = Paramagnetic

Question 16 The atomic number of Mn, Fe, Co and Ni are 25, 26 27 and 28 respectively. Which of the following outer orbital octahedral complexes have the same number of unpaired electrons?
(i) [MnCl6]3
(ii) [FeF6]3
(iii) [CoF6]3
(iv) [Ni(NH3)6]2+
Answer:

Option (i) and (iii) are the correct answer.
Explanation :
(a,c)
[MnCl6]3:Mn3+(3d4)orbital
[CoF6]3:Co3+(3d6)orbital

Question 17 Which of the following options are correct for [Fe(CN)6]3 complex?
(i) d2sp3hybridisation
(ii) sp3d2hybridisation
(iii) paramagnetic
(iv) diamagnetic
Answer:

Option (i) and (iii) are the correct answers.
Explanation :
(a,c) [Fe(CN)6]3
electronic configuration
Magnetic nature -paramagnetic.

Question 18 An aqueous pink solution of cobalt(II) chloride changes to deep blue on the addition of an excess of HCl. This is because____________.
(i) [Co(H2O)6]2+ is transformed into [CoCl6]4
(ii) [Co(H2O)6]2+ is transformed into [CoCl4]2
(iii) tetrahedral complexes have smaller crystal field splitting than octahedral complexes.
(iv) tetrahedral complexes have larger crystal field splitting than octahedral complex.
Answer:

Option (ii) and (iii) are the correct answers.
Explanation: The transition of electrons from t2g to eg energy level gives an aqueous pink solution of cobalt (II) chloride in [Co(H2O)6]2+ complex and when excess HCl is added to the solution:
(i) [Co(H2O)6]2+ is transformed into [CoCl4]2
(ii) Tetrahedral complexes have smaller crystal field spliting than octahedral complexes because Δr=49Δ0
Hence, options (b) and (c) are correct.

Question 19 Which of the following complexes is homoleptic?
(i) [Co(NH3)6]3+
(ii) [Co(NH3)4Cl2]+
(iii) [Ni(CN)4]2
(iv) [Ni(NH3)4Cl2]
Answer:

Option (i) and (iii) are the correct answers.
Explanation: Both Co and Ni are attached the same kind of ligands in the complexes [Co(NH3)6]3+ and [Ni(CN)4]2. Hence they are homoleptic.

Question 20 Which of the following complexes are heteroleptic?
(i) [Cr(NH3)6]3+
(ii) [Fe(NH3)4Cl2]+
(iii) [Mn(CN)6]4
(iv) [Co(NH3)4Cl2]
Answer:

Option (ii) and (iv) are the correct answers.
Explanation: Metal is bonded to more than one kind of ligands in complexes [Fe(NH3)4Cl2]+and [Co(NH3)4Cl2] therefore they are heteroleptic.

Question 21 Identify the optically active compounds from the following: -
(i) [Co(en)3]3+
(ii) trans[Co(en)2Cl2]+
(iii) cis[Co(en)2Cl2]+
(iv) [Cr(NH3)5Cl]

Answer:

Option (i) and (iii) are the correct answers.
[Co(en)3]3+ and cis[Co(en)2Cl2]+ are optically active compounds because their mirror images are non superimposable isomer.
isomers
Non-superimposable isomers of [Co(en)3]3+
isomers
Non-superimosable isomers of [Co(en)2Cl2]+

Question 22 Identify the correct statements for the behaviour of ethane-1, 2-diamine as a ligand.
(i) It is a neutral ligand.
(ii) It is a didentate ligand.
(iii) It is a chelating ligand.
(iv) It is a unidentate ligand.
Answer:

Option (i), (ii) and (iii) are the correct answers.
Explanation :
Ethane 1, 2-diamine is a neutral and didentate ligand. It is a chelating agent.
didentate ligand

Question 23 Which of the following complexes show linkage isomerism?
(i) [Co(NH3)5(NO2)]2+
(ii) [Co(H2O)5CO]3+
(iii) [Cr(NH3)5SCN]2+
(iv) [Fe(en)2Cl2]+

Answer:

Option (i) and (iii) are the correct answers.
Explanation:NO2 and SCN show linkage isomerism because they are ambidentate ligands.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Short Answer Type

The short-answer type questions are also given here in the NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 coordination compounds for practice. This section contains chemitsry chapter 9 important questions that are asked in the exams. Practice short answer types from the questions below

Question 24 Arrange the following complexes in the increasing order of conductivity of their solution: [Co(NH3)3Cl3],[Co(NH3)4Cl2]Cl,[Co(NH3)6]Cl3,[Cr(NH3)5Cl]Cl2
Answer:

The correct order of increase in conductivity is given below:
[Co(NH3)3Cl3]<[Co(NH3)4Cl2]Cl<[Cr(NH3)5Cl]Cl2<[Co(NH3)6]Cl3

Question 25 A coordination compound CrCl3.4H2O precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write the structural formula of the compound and name it.
Answer:

There should be one free chlorine atom left outside the coordination sphere if silver chloride is formed in the reaction. [Cr(H2O)4Cl2]Cl has to be the formula and it is called tetraaquadichlorido chromium(III) chloride.

Question 26 A complex of the type [M(AA)2X2]n+is known to be optically active. What does this indicate about the structure of the complex? Give one example of such complex.
Answer:

If the complex is optically active then its structure has to be cis-octahedral. [Co(en)2Cl2]+ is one of the prominent examples of an optically stable compound.

Question 27 The magnetic moment of [MnCl4]2 is 5.92BM. Explain giving reason.
Answer:

A magnetic moment of 5.92 BM indicates that there are at least 5 unpaired electrons. The formula for Magnetic Moment =n(n+2)
As four ligands are attached to Mn2+, therefore the geometry tetrahedral with 5 unpaired electrons gives a magnetic moment of 5.92 BM.

Question 28 Based on crystal field theory explain why Co(III) forms a paramagnetic octahedral complex with weak field ligands whereas it forms a diamagnetic octahedral complex with strong field ligands.
Answer:

t2g4eg2 will be the electronic configuration where it indicates that it has 4 unpaired electrons and is paramagnetic, with a weak ligand Δ0<p. The configuration will be t2g6eg0 with strong field ligand and there will not be any unpaired electrons. Hence it is diamagnetic.

Question 29 Why are low spin tetrahedral complexes not formed?
Answer:

The crystal field splitting energy is very low for tetrahedral complexes. The crystal field splitting energy is lower than pairing energy, therefore pairing of electrons is not favorable. Therefore, the complexes may not form low spin complexes

Question 31 Explain why [Fe(H2O)6]3+ has a magnetic moment value of 5.92 BM whereas [Fe(CN)6]3 has a value of only 1.74 BM.
Answer:

[Fe(CN)6]3- involves d2sp3 hybridisation with one unpaired electron and [Fe(H2O)6]3+ involves sp3d2 hybridisation with five unpaired electrons. This difference is due to the presence of strong ligand CN and weak ligand H2O in these complexes

Question 32 Arrange following complex ions in increasing order of crystal field splitting energy (Δo) :
[Cr(Cl)6]3,[Cr(CN)6]3,[Cr(NH3)6]3+.

Answer:

The correct increasing order of crystal field is:
[Cr(Cl)6]3<[Cr(NH3)6]3+<[Cr(CN)6]3
According to the spectrochemical series, this is also the correct increasing of order of the field strength.

Question 33 Why do compounds having similar geometry have a different magnetic moment?
Answer:

The difference lies in the number of paired and unpaired electrons. Strong field ligand can easily cause the pairing of electrons while the weak field ligands are not able to form pairs. The magnetic moment of a compound depends on the number of unpaired or paired electrons. Hence it is different for compounds having similar geometry.

Question 34 CuSO4.5H2O is blue while CuSO4 is colourless. Why?
Answer:

In CuSO4.5H2O, the extra 5 water act as ligands. They apparently excite the electrons to the higher d orbital which shows the blue color. In CuSO4, there are no water molecule that can act as ligands, so there is no crystal field slitting happening, hence there is no color.

Question 35 Name the type of isomerism when ambidentate ligands are attached to a central metal ion. Give two examples of ambidentate ligands.
Answer:

The ligands which have two binding sites are known as Ambidendate ligands. Some of the examples are: Nitrite-N, Nitrito-O and Isothiocyanato, Thiocyanato.
Linkage Isomerism is a type of isomerism when ambidentate ligands get attached to central metal ions. This happens because it only differs in the atom that they are linked the central metal ion.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Matching Type

The chemistry chapter 9 important questions are discussed below. These are generally asked in exams to test your knowledge. The NCERT exemplar Class 12 chemistry solutions chapter 9 is quite helpful for competitive exams.

Question 36 In the following question match the items given in Columns I and II.
Match the complex ions given in Column I with the colours given in Column II and assign the correct code :

Column I (Complex ion)

Column II (Colour)

(A) [Co(NH3)6]3+

1. Violet

(B) [Ti(H2O)6]3+

2. Green

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

3. Pale blue

(D) [Ni(H2O)4(en)]2+(aq)

4.Yellowish Orange

5.Blue

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

(i) A (1) B (2) C (4) D (5)
(ii) A (4) B (3) C (2) D (1)
(iii) A (3) B (2) C (4) D (1)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (ii) is the correct answer.

(a) [Co(NH3)6]3+ is Yellowish Orange because the cobalt(III) ammine complex typically appears yellowish orange due to ligand field transitions.

(b) [Ti(H2O)6]3+ is Pale Blue.

(c) [Ni(H2O)6]2+ is Green, the nickel(II) aqua complex is characteristically green in solution.

(d) [Ni(H2O)4(en)]2+ is Violet because mixed-ligand nickel (II) complex can show a violet colour due to stronger ligand field splitting by ethylenediamine.

Question 37 In the following question, match the items given in Columns I and II.
Match the coordination compounds given in Column I with the central metal atoms given in Column II and assign the correct code :

Column I (Coordination Compound)

Column II ( Central metal atom)

A. Chlorophyll

1. rhodium

B. Blood pigment

2. Cobalt

C.Wilkinson catalyst

3.Calcium

D Vitamin B12

4.Iron

5. magnesium

(i) A (5) B (4) C (1) D (2)
(ii) A (3) B (4) C (5) D (1)
(iii) A (4) B (3) C (2) D (1)
(iv) A (3) B (4) C (1) D (2)

Answer:

Option (i) is the correct answer.

(a) Chlorophyll contains a central magnesium (Mg2+) ion.

(b) The blood pigment haemoglobin contains a central iron (Fe2+/Fe3+) ion.

(c) Wilkinson's catalyst [RhCl(PPh3)3] contains rhodium as the central atom.

(d) Vitamin B12 contains a central cobalt (Co) atom.

Question 38 In the following question match the items given in Columns I and II.
Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code

Column I (Complex ion)

Column II (Hybridisation, number of unpaired electrons)

A. [Cr(H2O)6]3+

1. dsp2,1

B. [Co(CN)4]2

2.sp3d2,5

C. [Ni(NH3)6]2+

3. d2sp3,3

D. [MnF6]4

4. sp3,4

5.sp3d2,2

(i) A (3) B (1) C (5) D (2)
(ii) A (4) B (3) C (2) D (1)
(iii) A (3) B (2) C (4) D (1)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (i) is the correct answer.
Explanation : (i) Strong field ligand forms an inner orbital complex with hybridization d2sp3
(ii) Weak field ligand forms an outer orbital complex with hybridization d2sp3.
According to VBT, hybridization and the number of unpaired electrons of coordination compounds can be calculated as
(a) [Cr(H2O)6]3+
MOEC (Molecular orbital electronic configuration) of Cr3+ in [Cr(H2O)6]3+ is
electronic configuration
Hybridisation = d2sp3
n (number of unpaired electrons) = 3
(b) [Co(CN)4]2 is
MOEC of Co2+ in [Co(CN)4]2 is
electronic configuration
Hybridisation = dsp2
n (number of unpaired electrons) = 1
(c) [Ni(NH3)6]2+
MOEC of Ni2+ in [Ni(NH3)6]2+ is
electronic configuration
Hybridisation = sp3d2
n (number of unpaired electrons) = 2
(d) [MnF6]4
MOEC of Mn2+ in [MnF6]4 is
electronic configuration
Hybridisation = sp3d2
n (number of unpaired electrons) = 5
Hence , correct choice can be represented by (a).

Question 39 In the following question match the items given in Columns I and II.
Match the complex species given in Column I with the possible isomerism given in Column II and assign the correct code :

Column I (Complex species)

Column II (Isomerism)

A. [Co(NH3)4Cl2]+

1. Optical

B. cis[Co(en)2Cl2]+

2. Ionisation

C. [Co(NH3)5(NO2)]Cl2

3. Coordination

D. [Co(NH3)6][Cr(CN)6]

4. Geometrical

5. Linkage

(i) A (1) B (2) C (4) D (5)
(ii) A (4) B (3) C (2) D (1)
(iii) A (4) B (1) C (5) D (3)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (iv) is the correct answer.
Explanation : The types of ligands, arrangement of ligands and the geometry of coordination decide the isomerism in coordination compounds.
(a) [Co(NH3)4Cl2]+ shows geometrical isomerism due to presence of two types of ligand whoes [Co(NH3)4Cl2]+ arrangement around central metal ion
isomers
(b) cis[Co(en)2Cl2]+ shows an optical isomer due to its non - superimposable mirror image relationship.
isomers
(c) [Co(NH3)5(NO2)]Cl2 shows an ionization isomer due to its interchanging ligand from outside the ionization sphere.
(d) [Co(NH3)6][Cr(CN)6] shows coordination isomer due to interchanging of ligand in between two metal ions from one coordination sphere to another coordination sphere.
Hence, correct choice is (iv)

Question 40 In the following question match the items given in Columns I and II.
Match the compounds given in Column I with the oxidation state of cobalt present in it (given in Column II) and assign the correct code.

Column I (Compound)

Column II (Oxidation state of Co)

A. [Co(NCS)(NH3)5](SO3)

1. +4

B. [Co(NH3)4Cl2]SO4

2. 0

C. Na4[Co(S2O3)3]

3. +1

D. [Co2(CO)8]

4. +2

5. +3

(i) A (1) B (2) C (4) D (5)
(ii) A (4) B (3) C (2) D (1)
(iii) A (5) B (1) C (4) D (2)
(iv) A (4) B (1) C (2) D (3)

Answer:

Option (iii) is the correct answer.
Explanation: The oxidation state of the whole molecule is considered to calculate the oxidation state of CMI (central metal ion).
(a) [Co(NCS)(NH3)5](SO3)
Let oxidation state of Co be x.
x1+5×0=+2
x=+2+1=+3
(b) [Co(NH3)4Cl2]SO4
Let oxidation state of Co = x.
x+4×0+2×(1)=+2
x2=+2
x=4
(c) Na4[Co(S2O3)3]
Let oxidation state of Co = x.
x+3×(2)=4
x6=4
x=4+6=+2
(d) [Co2(CO)8]
Let oxidation state of Co = x.
x8×0=0
x=0

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Assertion and Reason Type

This is one of the most important sections covered in the NCERT exemplar Class 12 chemistry solutions chapter 9. These questions will improve your critical thinking.

Question 41 In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Toxic metal ions are removed by the chelating ligands.
Reason: Chelate complexes tend to be more stable.

(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (i) is correct answer.
Explanation: Ligands chelate the metal ions by forming stable complex when solution of chelating ligands is added to a solution containing toxic metals.

Question 42 In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion:[Cr(H2O)6]Cl2 and [Fe(H2O)6]Cl2 are reducing in nature.
Reason: Unpaired electrons are present in their d-orbitals.

(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (ii) is correct answer.
Explanation: In both the complexes, Cr and Fe Coexist as Cr2+ and Fe2+ respectively, these both complexes become stable when oxidation of metal ion happens to Cr3+ and Fe3+.

Question 43 In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Linkage isomerism arises in coordination compounds containing ambidentate ligand.
Reason: The Ambidentate ligand has two different donor atoms.

(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, and the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (i) is correct answer.
Explanation: Two different donor atoms in ambidentate ligand gives rise to linkage isomerism.

Question 44 In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Complexes of MX6 and MX5L type (X and L are unidentate) do not show geometrical isomerism.
Reason: Geometrical isomerism is not shown by complexes of coordination number 6.

(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (iii) is correct answer.
Explanation: A different arrangement of ligands is not possible for complexes of MX6 and MX5L type. The complexes MA4B2, M(AA)2B2 and MA3B3 with coordination number 6 show geometrical isomerism.

Question 45 In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion:[Fe(CN)6]3 ion shows magnetic moment corresponding to two unpaired electrons.
Reason: Because it has d2sp3 type hybridisation.

(i) Assertion and reason both are true, the reason is the correct explanation of assertion.
(ii) Assertion and reason both are true, but the reason is not the correct explanation of assertion.
(iii) An assertion is true, the reason is false.
(iv) The assertion is false, the reason is true.
Answer:

Option (iv) is correct answer.
Explanation: Corresponding to one unpaired ion, [Fe(CN)6]3 shows magnetic moment.

NCERT Exemplar Class 12 Chemistry Solutions Chapter 9: Long Answer Type

The following are the long-answer type questions that needs more practice and learning. These are the chemistry chapter 9 important questions that are asked in the exams.

Question 46 Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following :(i) [CoF6]3,[Co(H2O)6]2+,[Co(CN)6]3
(ii) [FeF6]3,[Fe(H2O)6]2+,[Fe(CN)6]4

Answer:

(i)
[CoF6]3 :

electronic configuration
Co3+=3d6
Number of unpaired electrons =4
Magnetic moment
=n(n+2)=4(4+2)=4.9B.M
[Co(H2O)6]2+
electronic configuration
Co2+=3d7
Number of unpaired electrons =3
Magnetic moment =3(3+2)=3.87B.M
[Co(CN)6]3
electronic configuration
Co3+=3d6
Number of unpaired electrons =0
Diamagnetic .
(ii)
[FeF6]3
electronic configuration
Fe3+=3d5
Number of unpaired electrons =5
Magnetic moment = 5(5+2)=5.92B.M
[Fe(H2O)6]2+
electronic configuration
Fe2+=3d6
Number of unpaired electrons =4
Magnetic moment = 4(4+2)=4.9B.M
[Fe(CN)6]4:
electronic configuration
Fe2+=3d6
Diamagnetic.

Question 47 Using valence bond theory, explain the following in relation to the complexes gives below:
[Mn(CN)6]3,[Co(NH3)6]3+,[Cr(H2O)6]3+,[FeCl6]4
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.

Answer:

[Mn(CN)6]3:
Mn3+=3d4electronic configuration
(i) Hybridisation - d2sp3
(ii) Inner orbital complex
(iii) Paramagnetic
(iv) Magnetic moment 2(2+2)=2.87B.M
[Co(NH3)6]3+:
Co3+=3d6
electronic configuration
(i) Hybridisation - d2sp3
(ii) Inner orbital complex
(iii) Diamagnetic
(iv) Magnetic moment=0
[Cr(H2O)6]3+
Cr3+=3d3
electronic configuration
(i) Hybridisation - d2sp3
(ii) Inner orbital complex
(iii) Paramagnetic
(iv) Magnetic moment=3(3+2)=3.87B.M
[FeCl6]4
Fe2+=3d6
electronic configuration
(i) Hybridisation - sp3d2
(ii) Outer orbital complex
(iii) Paramagnetic
(iv) Magnetic moment=4(4+2)=4.9B.M

Question 48 CoSO4Cl.5NH3 exists in two isomeric forms ‘A’ and ‘B’. Isomer ‘A’ reacts with AgNO3 to give white precipitate, but does not react with BaCl2 . Isomer ‘B’ gives white precipitate with BaCl2 but does not react with AgNO3. Answer the following questions.

(i) Identify ‘A’ and ‘B’ and write their structural formulas.
(ii) Name the type of isomerism involved.
(iii) Give the IUPAC name of ‘A’ and ‘B’.

Answer:

CoSO4Cl.5NH3 :
(i). The first isomer A reacts with AgNO3 and not with BaCl2, that means it has a Cl ion outside the coordination sphere. Therefore, A is [Co(NH3)5SO4]Cl
B reacts with BaCl2 and does not react with AgNO3, that means it has SO4 outside the coordination sphere. Therefore B is [Co(NH3)5Cl]SO4
(ii). Ionisation Isomerism
(iii). A is Pentaamminesulphatocobalt (III) chloride while B is Pentaamminechlorocobalt (III) sulphate.

Question 49 What is the relationship between the observed color of the complex and the wavelength of light absorbed by the complex?
Answer:

Some part of the white light is absorbed when it falls on a complex compound. The crystal field splitting is inversely proportional to the wavelength absorbed by the complex i.e. higher the crystal field splitting, lower is the wavelength absorbed by the complex. The observed color by the complex is the color generated from the wavelength left over.

Question 50 Why are different colors observed in octahedral and tetrahedral complexes for the same metal and same ligands?
Answer:

In an octahedral complex, the lower wavelength of light is absorbed more than tetrahedral complex for the same metal and ligand. The formation of the d-orbital splitting is inverted in the tetrahedral coordination entity and it is smaller than the octahedral field splitting. It can be shown below for the same metal ligand.
Δt=(49)Δo.

Class 12 Chemistry NCERT Chapter 9: Higher Order Thinking Skills (HOTS) Questions

HOTS-type questions are covered to improve your problem-solving ability and conceptual thinking. The class 12 chemistry chapter 9 questions and answers are given below that will help you tackle complex problems. Students can follow coordination compounds class 12 notes to learn the concepts in detail. The questions below will help you evaluate your understanding of the concepts.

Question: Given below are two statements :
Statement (I) : In octahedral complexes, when Δo<P high spin complexes are formed. When Δo>P low spin complexes are formed.
Statement (II) : In tetrahedral complexes because of Δt<P, low spin complexes are rarely formed.
In the light of the above statements, choose the most appropriate answer from the options given below :

(1) Statement I is correct but Statement II is incorrect.

(2) Both Statement I and Statement II are incorrect

(3) Statement I is incorrect but Statement II is correct

(4) Both Statement I and Statement II are correct

Answer:

In octahedral complex (CN=6)
If Δ0< P.E. , then high-spin complexes are formed
If Δ0> P.E. , then low spin complexes are formed
But in tetrahedral complex (CN=4)
Δt< P.E. , then mainly high spin complexes are formed and rarely low spin complexes are formed.

Hence, the correct answer is the option (4).

Question: Identify the homoleptic complexes with odd number of d electrons in the central metal.
(A) [FeO4]2
(B) [Fe(CN)6]3
(C) [Fe(CN)5NO]2
(D) [CoCl4]2
(E) [Co(H2O)3 F3]

Choose the correct answer from the options given below :

(1) (B) and (D) only

(2) (C) and (E) only

(3) (A), (B) and (D) only

(4) (A), (C) and (E) only

Answer:

To determine the complexes that are both homoleptic and have an odd number of d-electrons, we analyze each option based on the oxidation state and electron count of the central metal.

1. [FeO4]2 : Fe is in +6 oxidation state (since Fe=[Ar]3 d6, and losing six electrons leaves d2 ). This is not an odd number

2. [Fe(CN)6]3: Iron in +3(Fe3+,d5). Since all ligands are CN, the complex is homoleptic and has an odd number (5) of d-electrons.

3. [Fe(CN)5NO]2: The NO ligand introduces π-back bonding and redox ambiguity, making it a nonhomoleptic complex. Hence, it is not considered.

4. [CoCl4]2: Cobalt is in +2(Co2+,d7). Since 7 is odd and all ligands are Cl(making it homoleptic), this satisfies both conditions.

5. [Co(H2O)3 F3] : This complex contains two different ligands (H2O and F), so it is not homoleptic and does not qualify.

Hence, the correct answer is option (1).

Question: The number of paramagnetic metal complex species among [Co(NH3)6]3+,[Co(C2O4)3]3, [MnCl6]3,[Mn(CN)6]3,[CoF6]3,[Fe(CN)6]3 and [FeF6]3 with same number of unpaired electrons is___________.

Answer:

Number of paramagnetic complexes with same number of unpaired electrons = 2 pairs

2 complexes have 0 unpaired electrons; they are

[Co(NH3)6]3+,[Co(C2O4)3]3

Now,

[MnCl6]3
Cl:6×(1)=6Mn=+3
Mn atomic number =25Mn3+=3 d4
Clis a weak-field ligand high-spin
3d4 high-spin 4 unpaired electrons
Paramagnetic, 4 unpaired

[Mn(CN)6]3
CN:6×(1)=6Mn=+3
Mn3+=3 d4
CN=strong field low-spin
3 d4 low-spin 2 unpaired electrons
Paramagnetic, 2 unpaired

[CoF6]3
- F:6×(1)=6Co=+3
Co3+=3 d6
F=weak field high-spin
3d6 high-spin 4 unpaired electrons
Paramagnetic, 4 unpaired

[Fe(CN)6]3
CN:6×(1)=6Fe=+3
Fe atomic number =26Fe3+=3 d5
CN=strong field low-spin
- 3d5 low-spin 1 unpaired electron
Paramagnetic, 1 unpaired

[FeF6]3
F:6×(1)=6Fe=+3
-Fe3+=3 d5
F=weak field high-spin
3 d5 high-spin 5 unpaired electrons Paramagnetic, 5 unpaired

2 paramagnetic species have the same number of unpaired electrons. [Mn(Cl)6]3,[CoF6]3

No other pairs have the same number.

Hence, the answer is 2.

Approach to Solve Questions of Chapter 9

To effectively solve questions from Chapter 9, students can follow the approaches given below:

1. Understand the basic terminology

This is the basic yet important step. Some of the important terminologies are given below:

  • Types of ligands
  • Coordination number
  • Coordination complex/entity
  • Central metal ion
  • Counter ion

2. Remember IUPAC nomenclature rules for coordinate complexes.

While solving questions related to nomenclature, identify the ligands, metal, and their oxidation states, and follow IUPAC rules like:

  • Firstly, name the ligand
  • Name the ligands in alphabetical order, and if the complex is an anion, then end the naming with 'ate'.
  • Oxidation states of metals are written in Roman numerals.

Learn more from coordination compounds class 12 notes.

3. Theories of coordination compounds

Theories like Werner's theory, valence bond theory, the crystal field theory, forms the basis of this chapter and are frequently asked in exams.

4. Memorise some basic geometries

It's important to memorise some of the general and basic geometries like Linear, Trigonal planar, Tetrahedral, Square planar, Square pyramidal, Octahedral. he NCERT Exemplar Class 12 Chemistry Solutions Chapter 9 will help you understand these concepts better.

5. Practice questions

A Proper understanding of basic concepts and practice helps students clear their doubts and solve questions effectively. You can get these type of questions in the NCERT Class 12 solutions available on our website. Students can refer NCERT Exemplar textbooks for practice

Topics Covered in NCERT Exemplar Solutions Class 12 Chemistry Chapter 9

The following are the important topics in the NCERT chemistry chapter 9. Practice more at NCERT Class 12 chemistry chapter 9 exemplar solutions.

  • Coordination Compounds
  • Werner's Theory of Coordination Compounds
  • Definitions of Some Important Terms Pertaining to Coordination Compounds
  • Nomenclature of Coordination Compounds
  • Isomerism In Coordination Complexes
  • Bonding in Coordination Compounds
  • Bonding in Metal Carbonyls
  • Stability of Coordination Compounds
  • Importance and Applications of Coordination Compounds

NCERT Class 12 Chemistry Exemplar Chapter 9 Important Formulas

The following are the important trends covered in the Class 12 NCERT Chemistry chapter 9. Learn more from class 12 chemistry chapter 9 notes

1. Effective Atomic Number Formula:

EAN=Zx+2n

2. Oxidation Number of Central Atom:

Oxidation number = Total charge charges of ligands

3. Crystal field Stabilisation Energy

For octahedral complexes

CFSE=(0.4×nt2g+0.6×neg)Δ0

Tetrahedral complexes

CFSE=(0.6×nt2+0.4×ne)Δt

5. Some basic geometries according to Valence Bond Theory

Coordination NumberType of HybridisationShape
4sp3Tetrahedral
4dsp2Square Planar
5sp3dTrigonal Bipyramidal
6sp3d2Octahedral
6d2sp3Octahedral

7. Magnetic moment

μ=n(n+2)

NCERT Solution subject-wise

The NCERT subject-wise solutions will help you broaden your concepts and will also help in revision. Learn more from Class 12 NCERT notes.

NCERT Exemplar Class 12 Solutions

Students can refer to the links given below for the NCERT Exemplar subject-wise solutions for Class 12.

NCERT Notes subject-wise

You can follow the links given in the table below to get access to the Class 12 NCERT notes.

NCERT Books and NCERT Syllabus

You can find links to the Class 12 NCERT chemistry book and syllabus for the respective subjects.

Frequently Asked Questions (FAQs)

Q: What are coordination compounds, and how are they formed?
A:

Coordination compounds are complexes consisting of a central metal atom or ion bonded to surrounding molecules or ions called ligands. They are formed through coordinate covalent bonds, where ligands donate a pair of electrons to the metal to form these bonds, resulting in a stable structure.

Q: What is the role of CFT in coordination compounds ?
A:

Crystal field theory explains the electronic structure of transition metal complexes by considering the interactions between the central metal ion and surrounding ligands. The presence of ligands alters the energy levels of the d-orbitals, leading to the splitting of these orbitals based on the geometry of the complex. Learn morefrom our class 12 NCERT notes.

Q: Why do ligands play a significant role in Coordination compounds as discussed in NCERT class 12 solutions?
A:

Properties of coordination compounds are determined with the help of ligands. Ligands affect properties like stability, colour of compounds and their reactivity.

Q: How are coordination number of metals determined in coordination compounds?
A:

 The coordination number refers to the number of ligand atoms or ions that are directly bonded to the central metal atom in a coordination compound. It is usually determined by the spatial arrangement of ligands around the metal atom and can vary based on the size and charge of the metal ion as well as the type of ligands involved.

Q: What is the difference between strong field and weak field ligands?
A:

Strong field ligands create a larger splitting of d-orbitals while weak field ligands creates small splitting.Strong field ligands like CN⁻ and CO, tend to lead to low-spin complexes, while weak field ligands, such as I⁻ and Br⁻, lead to high-spin complexes.

Articles
|
Next
Upcoming School Exams
Ongoing Dates
Maharashtra SSC Board Application Date

1 Aug'25 - 31 Aug'25 (Online)

Ongoing Dates
Maharashtra HSC Board Application Date

1 Aug'25 - 31 Aug'25 (Online)

Certifications By Top Providers
Explore Top Universities Across Globe

Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?

Hello

Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.

Hello Aspirant,

Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.

Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.

From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .

Hello Prisha

As you have compartment in Mathematics and you wish to change your stream to Humanities you have 2 options from here on:

1. You could try to study and clear your compartment of Mathematics.

2. You can change your stream to Humanities but you will need to repeat from Class 11th so you will need 2 years from now to clear Class 12th.

My personal suggestion will be to choose Option 1. Trust me I have dealt with this feeling myself and wanted to choose Option 2 but I worked hard and cleared my exams. I was scared not only in Mathematics but Physics, Chemistry and even Computer Science.
At last, it's on you what you want to choose.

Thank You!

HELLO,

The GUJCET merit list is calculated by considering 60% of the marks obtained in Class 12 and 40% of the GUJCET score. CBSE students are also eligible to compete with GSEB students, as the merit list is prepared by combining the Class 12 scores with the GUJCET performance.

Merit Calculation:
  • The merit list is calculated by combining the scores from both class and the GUJCET exam
  • This means that both your performance in the class 12 board exams and the GUJCET exam are considered when determining your rank

CBSE Students and GUJCET:

  • CBSE students can appear for the GUJCET exam, even though they are not from the Gujrat Board
  • For CBSE students, the 60% weightage for class 12 marks will be based on their performance in Physics, Chemistry, and Mathematics in their class 12 board exams
  • The remaining 40% weightage will be given to their GUJCET score.

Hope this Helps!