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Coordination Compounds Complete NCERT Resource Guide
Welcome back to examspark.in! Aaj hum CBSE Class 12 Chemistry Chapter 9, Coordination Compounds cover karenge. Yeh chapter aapke 2026-27 board exams aur competitive exams (JEE/NEET) dono ke liye super important hai. Agar concepts clear hain, toh isme marks score karna bahut easy ho jata hai. We have compiled the updated NCERT solutions and important questions to make your preparation smooth. Chaliye shuru karte hain!
Chapter NameCoordination Compounds
SubjectChemistry
ClassClass 12
BoardCBSE
Important TopicsWerner's Theory, VBT, CFT, Isomerism, IUPAC Nomenclature
Difficulty LevelModerate to High
Exam WeightageAround 7 Marks
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Learning Objectives
After completing this chapter, students will be able to:
- Understand the postulates of Werner's theory of coordination compounds.
- Write the formulas and IUPAC names of mononuclear coordination entities.
- Explain the different types of isomerism in coordination compounds.
- Understand the bonding in coordination compounds using Valence Bond Theory (VBT) and Crystal Field Theory (CFT).
- Appreciate the importance and applications of coordination compounds in everyday life.
Key Concepts, Definitions & Formulas
- Coordination Entity: Constitutes a central metal atom or ion bonded to a fixed number of ions or molecules (ligands). Example: $[\text{Co}(\text{NH}_3)_6]^{3+}$.
- Central Atom/Ion: The atom/ion to which a fixed number of ligands are bound in a definite geometrical arrangement. It acts as a Lewis acid.
- Ligands: The ions or molecules bound to the central atom. They can be unidentate (e.g., $\text{Cl}^-$, $\text{H}_2\text{O}$), didentate (e.g., ethane-1,2-diamine), or polydentate (e.g., EDTA).
- Coordination Number (CN): The number of ligand donor atoms to which the metal is directly bonded.
- Werner's Theory: Metals exhibit two types of valencies: primary (ionizable, corresponds to oxidation state) and secondary (non-ionizable, corresponds to coordination number).
- Crystal Field Splitting ($\Delta_o$): The splitting of degenerate d-orbitals due to the presence of ligands in a definite geometry.
Full NCERT Solutions (Step-by-Step)
(Note for ExamSpark readers: We have detailed the most critical exercise questions step-by-step for board exams)
Question 9.1: Explain the bonding in coordination compounds in terms of Werner's postulates.
Step 1: Differentiate Valencies. According to Werner's theory, metals show two types of linkages (valencies): Primary valency and Secondary valency.
Step 2: Define Primary Valency. Primary valencies are normally ionizable and are satisfied by negative ions. They represent the oxidation state of the central metal atom.
Step 3: Define Secondary Valency. Secondary valencies are non-ionizable. These are satisfied by neutral molecules or negative ions (ligands). The secondary valency is equal to the coordination number and is fixed for a metal.
Step 4: Spatial Geometry. The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers, defining the geometry of the complex.
Step 2: Define Primary Valency. Primary valencies are normally ionizable and are satisfied by negative ions. They represent the oxidation state of the central metal atom.
Step 3: Define Secondary Valency. Secondary valencies are non-ionizable. These are satisfied by neutral molecules or negative ions (ligands). The secondary valency is equal to the coordination number and is fixed for a metal.
Step 4: Spatial Geometry. The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers, defining the geometry of the complex.
Question 9.2: $\text{FeSO}_4$ solution mixed with $(\text{NH}_4)_2\text{SO}_4$ solution in 1:1 molar ratio gives the test of $\text{Fe}^{2+}$ ion but $\text{CuSO}_4$ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of $\text{Cu}^{2+}$ ion. Explain why.
Step 1: Analyze the Iron Mixture (Double Salt). When $\text{FeSO}_4$ and $(\text{NH}_4)_2\text{SO}_4$ are mixed, they form a double salt called Mohr's salt, $\text{FeSO}_4 \cdot (\text{NH}_4)_2\text{SO}_4 \cdot 6\text{H}_2\text{O}$. In an aqueous solution, double salts dissociate completely into their constituent simple ions ($\text{Fe}^{2+}$, $\text{NH}_4^+$, $\text{SO}_4^{2-}$). Hence, it gives the test for $\text{Fe}^{2+}$ ions.
Step 2: Analyze the Copper Mixture (Coordination Complex). When $\text{CuSO}_4$ is mixed with ammonia, it forms a coordination complex, $[\text{Cu}(\text{NH}_3)_4]\text{SO}_4$. Coordination complexes do not dissociate into simple constituent ions in solution. The $\text{Cu}^{2+}$ is tightly bound inside the coordination sphere as $[\text{Cu}(\text{NH}_3)_4]^{2+}$. Therefore, it does not give the standard test for free $\text{Cu}^{2+}$ ions.
Step 2: Analyze the Copper Mixture (Coordination Complex). When $\text{CuSO}_4$ is mixed with ammonia, it forms a coordination complex, $[\text{Cu}(\text{NH}_3)_4]\text{SO}_4$. Coordination complexes do not dissociate into simple constituent ions in solution. The $\text{Cu}^{2+}$ is tightly bound inside the coordination sphere as $[\text{Cu}(\text{NH}_3)_4]^{2+}$. Therefore, it does not give the standard test for free $\text{Cu}^{2+}$ ions.
Question 9.3: Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Step 1: Coordination Entity. A central metal atom/ion bonded to ligands. Examples: $[\text{Ni}(\text{CO})_4]$, $[\text{PtCl}_2(\text{NH}_3)_2]$.
Step 2: Ligand. An ion/molecule bonded to the central atom. Examples: $\text{Cl}^-$ (chloride), $\text{H}_2\text{O}$ (aqua).
Step 3: Coordination Number. The total number of coordinate bonds formed by ligands with the central metal. Examples: In $[\text{Co}(\text{NH}_3)_6]^{3+}$, CN is 6. In $[\text{Ni}(\text{CO})_4]$, CN is 4.
Step 4: Coordination Polyhedron. The spatial arrangement of the ligands defining the shape. Examples: Octahedral $[\text{Co}(\text{NH}_3)_6]^{3+}$, Tetrahedral $[\text{Ni}(\text{CO})_4]$.
Step 5: Homoleptic vs Heteroleptic. A Homoleptic Complex is one in which a metal is bound to only one kind of donor group (Examples: $[\text{Co}(\text{NH}_3)_6]^{3+}$, $[\text{Fe}(\text{CN})_6]^{4-}$). A Heteroleptic Complex is bound to more than one kind of donor group (Examples: $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+$, $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$).
Step 2: Ligand. An ion/molecule bonded to the central atom. Examples: $\text{Cl}^-$ (chloride), $\text{H}_2\text{O}$ (aqua).
Step 3: Coordination Number. The total number of coordinate bonds formed by ligands with the central metal. Examples: In $[\text{Co}(\text{NH}_3)_6]^{3+}$, CN is 6. In $[\text{Ni}(\text{CO})_4]$, CN is 4.
Step 4: Coordination Polyhedron. The spatial arrangement of the ligands defining the shape. Examples: Octahedral $[\text{Co}(\text{NH}_3)_6]^{3+}$, Tetrahedral $[\text{Ni}(\text{CO})_4]$.
Step 5: Homoleptic vs Heteroleptic. A Homoleptic Complex is one in which a metal is bound to only one kind of donor group (Examples: $[\text{Co}(\text{NH}_3)_6]^{3+}$, $[\text{Fe}(\text{CN})_6]^{4-}$). A Heteroleptic Complex is bound to more than one kind of donor group (Examples: $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+$, $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$).
Question 9.4: What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Step 1: Unidentate Ligands. Ligands bounding to the metal through a single donor atom. Examples: $\text{H}_2\text{O}$, $\text{NH}_3$.
Step 2: Didentate Ligands. Ligands binding through two donor atoms. Examples: ethane-1,2-diamine ($\text{en}$), oxalate ion ($\text{C}_2\text{O}_4^{2-}$).
Step 3: Ambidentate Ligands. Ligands with two different donor atoms, either of which can coordinate to the metal, but only one at a time. Examples: $\text{NO}_2^-$ (can bind through N or O), $\text{SCN}^-$ (can bind through S or N).
Step 2: Didentate Ligands. Ligands binding through two donor atoms. Examples: ethane-1,2-diamine ($\text{en}$), oxalate ion ($\text{C}_2\text{O}_4^{2-}$).
Step 3: Ambidentate Ligands. Ligands with two different donor atoms, either of which can coordinate to the metal, but only one at a time. Examples: $\text{NO}_2^-$ (can bind through N or O), $\text{SCN}^-$ (can bind through S or N).
Question 9.5: Specify the oxidation numbers of the metals in the following coordination entities: (i) $[\text{Co}(\text{H}_2\text{O})(\text{CN})(\text{en})_2]^{2+}$ (ii) $[\text{CoBr}_2(\text{en})_2]^+$
Step 1: Analyze Complex (i). For $[\text{Co}(\text{H}_2\text{O})(\text{CN})(\text{en})_2]^{2+}$:
Let oxidation number of Co be $x$.
$x + 0 \text{ (for } \text{H}_2\text{O}) + (-1) \text{ (for CN)} + 2(0) \text{ (for en)} = +2$
$x - 1 = +2 \implies x = +3$. The oxidation number is +3.
Step 2: Analyze Complex (ii). For $[\text{CoBr}_2(\text{en})_2]^+$:
Let oxidation number of Co be $x$.
$x + 2(-1) \text{ (for Br)} + 2(0) \text{ (for en)} = +1$
$x - 2 = +1 \implies x = +3$. The oxidation number is +3.
Let oxidation number of Co be $x$.
$x + 0 \text{ (for } \text{H}_2\text{O}) + (-1) \text{ (for CN)} + 2(0) \text{ (for en)} = +2$
$x - 1 = +2 \implies x = +3$. The oxidation number is +3.
Step 2: Analyze Complex (ii). For $[\text{CoBr}_2(\text{en})_2]^+$:
Let oxidation number of Co be $x$.
$x + 2(-1) \text{ (for Br)} + 2(0) \text{ (for en)} = +1$
$x - 2 = +1 \implies x = +3$. The oxidation number is +3.
Extra Important Questions (Board Style)
Yahan kuch extra practice questions hain jo CBSE Class 12 Chemistry board exam 2026 mein aa sakte hain.
Section A: Multiple Choice Questions (MCQs)
Q1. Which of the following is an ambidentate ligand?
a) $\text{en}$
b) EDTA
c) $\text{SCN}^-$
d) $\text{C}_2\text{O}_4^{2-}$
Correct Answer Choice: c
Step 1: Explanation. $\text{SCN}^-$ is ambidentate because it can coordinate through either Sulfur or Nitrogen. (Difficulty: Easy)
Step 1: Explanation. $\text{SCN}^-$ is ambidentate because it can coordinate through either Sulfur or Nitrogen. (Difficulty: Easy)
Q2. The geometry of $[\text{Ni}(\text{CO})_4]$ is:
a) Square planar
b) Tetrahedral
c) Octahedral
d) Linear
Correct Answer Choice: b
Step 1: Explanation. $\text{Ni}$ is in $sp^3$ hybridization state due to the strong field CO ligand pairing up the 4s electrons into 3d. This results in a Tetrahedral geometry. (Difficulty: Medium)
Step 1: Explanation. $\text{Ni}$ is in $sp^3$ hybridization state due to the strong field CO ligand pairing up the 4s electrons into 3d. This results in a Tetrahedral geometry. (Difficulty: Medium)
Q3. The crystal field stabilization energy (CFSE) for a high spin $d^4$ octahedral complex is:
a) $-0.6 \Delta_o$
b) $-1.8 \Delta_o$
c) $-1.6 \Delta_o + P$
d) $-1.2 \Delta_o$
Correct Answer Choice: a
Step 1: Explanation. High spin $d^4$ means configuration is $t_{2g}^3 e_g^1$. $\text{CFSE} = (3 \times -0.4) + (1 \times 0.6) = -1.2 + 0.6 = -0.6 \Delta_o$. (Difficulty: Hard)
Step 1: Explanation. High spin $d^4$ means configuration is $t_{2g}^3 e_g^1$. $\text{CFSE} = (3 \times -0.4) + (1 \times 0.6) = -1.2 + 0.6 = -0.6 \Delta_o$. (Difficulty: Hard)
Section B: Short Answer Questions
Q4. Write the IUPAC name of $\text{K}_3[\text{Fe}(\text{CN})_6]$.
Step 1: Identify components. Potassium is the counter ion. Hexa (six) cyano ($\text{CN}$) ferrate (metal in anionic complex).
Step 2: Determine Oxidation State. $3(+1) + x + 6(-1) = 0 \implies x = +3$.
Step 3: Combine. Potassium hexacyanoferrate(III).
Step 2: Determine Oxidation State. $3(+1) + x + 6(-1) = 0 \implies x = +3$.
Step 3: Combine. Potassium hexacyanoferrate(III).
Q5. Why are low spin tetrahedral complexes rarely observed?
Step 1: Analyze Splitting Energy. The orbital splitting energy for tetrahedral complexes ($\Delta_t$) is much smaller than for octahedral complexes ($\Delta_t = \frac{4}{9}\Delta_o$).
Step 2: Compare to Pairing Energy. Because $\Delta_t$ is usually less than the pairing energy ($P$), the electrons prefer to occupy higher energy orbitals rather than pair up, resulting in high spin complexes.
Step 2: Compare to Pairing Energy. Because $\Delta_t$ is usually less than the pairing energy ($P$), the electrons prefer to occupy higher energy orbitals rather than pair up, resulting in high spin complexes.
Q6. Draw the geometrical isomers of $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$.
Step 1: Identify the isomer types. It shows cis and trans isomerism.
Step 2: Describe Cis-isomer. In the cis-isomer, the two chloride ligands are adjacent to each other.
Step 3: Describe Trans-isomer. In the trans-isomer, they are opposite each other.
Step 2: Describe Cis-isomer. In the cis-isomer, the two chloride ligands are adjacent to each other.
Step 3: Describe Trans-isomer. In the trans-isomer, they are opposite each other.
Section C: Long Answer & Case-Based Questions
Q7. Discuss the nature of bonding in $[\text{CoF}_6]^{3-}$ on the basis of Valence Bond Theory.
Step 1: Determine Oxidation State and Configuration. Oxidation state of Co is +3. Configuration of $\text{Co}^{3+}$ is $[\text{Ar}] 3d^6$.
Step 2: Analyze Ligand Strength. Fluoride ($\text{F}^-$) is a weak field ligand, so it cannot force the pairing of 3d electrons.
Step 3: Determine Hybridization. Therefore, it uses outer 4d orbitals for hybridization, forming $sp^3d^2$ hybridization.
Step 4: Conclude Geometry and Magnetism. The geometry is octahedral, and since it has 4 unpaired electrons, it is highly paramagnetic and is called an outer orbital complex or high spin complex.
Step 2: Analyze Ligand Strength. Fluoride ($\text{F}^-$) is a weak field ligand, so it cannot force the pairing of 3d electrons.
Step 3: Determine Hybridization. Therefore, it uses outer 4d orbitals for hybridization, forming $sp^3d^2$ hybridization.
Step 4: Conclude Geometry and Magnetism. The geometry is octahedral, and since it has 4 unpaired electrons, it is highly paramagnetic and is called an outer orbital complex or high spin complex.
Q8. What is the spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Step 1: Define the series. A spectrochemical series is the arrangement of ligands in increasing order of their crystal field splitting energy ($\Delta_o$).
Step 2: Explain Strong Field Ligands. (e.g., $\text{CO}$, $\text{CN}^-$) Cause a large splitting of d-orbitals. Here $\Delta_o > P$ (pairing energy), leading to electron pairing and forming low-spin complexes.
Step 3: Explain Weak Field Ligands. (e.g., $\text{I}^-$, $\text{Cl}^-$) Cause small splitting. Here $\Delta_o < P$, so electrons do not pair up easily, forming high-spin complexes.
Step 2: Explain Strong Field Ligands. (e.g., $\text{CO}$, $\text{CN}^-$) Cause a large splitting of d-orbitals. Here $\Delta_o > P$ (pairing energy), leading to electron pairing and forming low-spin complexes.
Step 3: Explain Weak Field Ligands. (e.g., $\text{I}^-$, $\text{Cl}^-$) Cause small splitting. Here $\Delta_o < P$, so electrons do not pair up easily, forming high-spin complexes.
Q9. Context: Crystal Field Theory (CFT) considers the metal-ligand bond as purely ionic arising from electrostatic interactions.
(a) What happens to the degeneracy of d-orbitals in an octahedral crystal field?
(b) Calculate the CFSE for a $d^5$ low spin complex.
Step 1: Answer Part (a). The five degenerate d-orbitals split into two sets: a lower energy $t_{2g}$ set ($d_{xy}$, $d_{yz}$, $d_{xz}$) and a higher energy $e_g$ set ($d_{x^2-y^2}$, $d_{z^2}$).
Step 2: Answer Part (b). For a $d^5$ low spin complex, configuration is $t_{2g}^5 e_g^0$.
Step 3: Calculate CFSE. $\text{CFSE} = [5 \times (-0.4)] \Delta_o + 2P = -2.0 \Delta_o + 2P$.
Step 2: Answer Part (b). For a $d^5$ low spin complex, configuration is $t_{2g}^5 e_g^0$.
Step 3: Calculate CFSE. $\text{CFSE} = [5 \times (-0.4)] \Delta_o + 2P = -2.0 \Delta_o + 2P$.
Section D: Assertion-Reason Questions
Directions:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Q10. Assertion (A): $[\text{Ni}(\text{CN})_4]^{2-}$ is diamagnetic.
Reason (R): It involves $dsp^2$ hybridization and has no unpaired electrons.
Correct Answer Choice: A
Step 1: Explanation. Both statements are true and the reason correctly explains why the complex exhibits diamagnetism.
Step 1: Explanation. Both statements are true and the reason correctly explains why the complex exhibits diamagnetism.
Q11. Assertion (A): Linkage isomerism arises in coordination compounds containing ambidentate ligands.
Reason (R): Ambidentate ligands have two different donor atoms.
Correct Answer Choice: A
Step 1: Explanation. Linkage isomers form precisely because the ambidentate ligand can attach via two different donor atoms.
Step 1: Explanation. Linkage isomers form precisely because the ambidentate ligand can attach via two different donor atoms.
Q12. Assertion (A): Toxic metal ions are removed by chelating ligands.
Reason (R): Chelate complexes tend to be more stable.
Correct Answer Choice: B
Step 1: Explanation. Both are true, but stability is a general property; removal is specifically due to the formation of soluble, excretable chelates.
Step 1: Explanation. Both are true, but stability is a general property; removal is specifically due to the formation of soluble, excretable chelates.
Q13. Assertion (A): According to crystal field theory, $d_{z^2}$ and $d_{x^2-y^2}$ orbitals have higher energy in an octahedral field.
Reason (R): These orbitals are oriented directly towards the point charges (ligands).
Correct Answer Choice: A
Step 1: Explanation. The direct head-on repulsion between the ligand electron clouds and these axial orbitals raises their energy level.
Step 1: Explanation. The direct head-on repulsion between the ligand electron clouds and these axial orbitals raises their energy level.
Q14. Assertion (A): Optical isomerism is common in square planar complexes.
Reason (R): Square planar complexes possess a plane of symmetry.
Correct Answer Choice: D
Step 1: Explanation. Assertion is false because square planar complexes rarely show optical isomerism precisely because they possess a plane of symmetry (which the reason correctly states).
Step 1: Explanation. Assertion is false because square planar complexes rarely show optical isomerism precisely because they possess a plane of symmetry (which the reason correctly states).
Q15. Assertion (A): Oxidation state of Iron in $[\text{Fe}(\text{H}_2\text{O})_5\text{NO}]\text{SO}_4$ (brown ring complex) is +1.
Reason (R): NO acts as a neutral ligand in this complex.
Correct Answer Choice: C
Step 1: Explanation. The assertion is true, but the reason is false because $\text{NO}$ acts as an $\text{NO}^+$ cation in this specific complex, not as a neutral ligand.
Step 1: Explanation. The assertion is true, but the reason is false because $\text{NO}$ acts as an $\text{NO}^+$ cation in this specific complex, not as a neutral ligand.
Common Mistakes Students Make
Exam prep mein silly mistakes se bachna zaroori hai. Yahan kuch common points hain jahan maximum students galti karte hain:
- IUPAC Naming: Forgetting to write the oxidation state in Roman numerals inside parentheses.
- Anionic vs. Cationic Complexes: Not adding the suffix "-ate" to the metal when the complex part is an anion (e.g., writing Iron instead of Ferrate).
- VBT vs CFT: Mixing up the rules. Remember, VBT talks about hybridization ($sp^3$, $dsp^2$), whereas CFT talks about orbital splitting ($t_{2g}$, $e_g$).
- Strong vs Weak Field Ligands: Ratta maarne ki jagah spectrochemical series ka logic samjho. $\text{CO}$, $\text{CN}^-$ are strong; halogens are weak.
Exam Preparation Tips
- Understand IUPAC Rules: Practice writing names and drawing structures backwards (Name -> Structure).
- Focus on Isomerism: Practice drawing Geometrical (cis/trans) and Optical (d/l) isomers.
- VBT & CFT Practice: Always write the electronic configuration of the metal ion first, then apply ligand field strength.
- PYQs are Gold: CBSE loves repeating VBT reasoning questions. Practice Previous Year Questions!
Frequently Asked Questions (FAQs)
1. Is Chapter 9 Coordination Compounds important for Class 12 CBSE boards?
Yes, it is highly important and carries a weightage of around 7 marks. It's also a high-scoring chapter if your concepts are clear.
2. What is the difference between double salt and a coordination complex?
A double salt completely dissociates into simple ions in water (e.g., Mohr's salt), while a coordination complex retains its identity in solution and does not dissociate completely.
3. Which topics are most frequently asked in Board Exam Questions 2026?
IUPAC nomenclature, finding hybridization and magnetic behavior using VBT, and Crystal Field Theory (CFT) splitting diagrams are guaranteed questions.
4. Where can I download NCERT PDF for Class 12 Chemistry?
You can download the official PDFs from the NCERT website or get simplified notes right here on examspark.in.
5. How do I memorize the spectrochemical series?
Use mnemonics! Start from Halides (weakest) -> Oxygen donors -> Nitrogen donors -> Carbon donors (strongest).
Conclusion: Coordination Compounds might seem tricky at first, but with clear visualizations of VBT and CFT, you can easily secure full marks. Lucky, managing these complex VBT and CFT visualizations is key as you prepare for your upcoming board and entrance exams at 19. Make sure to revise your notes regularly, practice IUPAC naming daily, and solve Previous Year Questions (PYQs). Good luck for your boards—prepare confidently and let examspark.in be your study partner!
More Class 12 Chemistry Chapters
Ch 1: The Solid State
Ch 2: Solutions
Ch 3: Electrochemistry
Ch 4: Chemical Kinetics
Ch 5: Surface Chemistry
Ch 6: Isolation of Elements
Ch 7: The p-Block Elements
Ch 8: The d- & f-Block Elements
Ch 9: Coordination Compounds
Ch 10: Haloalkanes
Ch 11: Alcohols & Phenols
Ch 12: Aldehydes & Ketones
Ch 13: Amines
Ch 14: Biomolecules
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