Class 12 Chemistry Chapter 9 Coordination Compounds

Chapter 9 of Class 12 Chemistry, titled “Coordination Compounds”, dives into one of the most important and conceptually rich areas of inorganic chemistry. These compounds form the basis of advanced topics in chemistry and are commonly encountered in both theoretical and practical applications. Understanding coordination compounds is not only crucial for scoring well in CBSE Class 12 board exams, but also essential for excelling in competitive exams like JEE Main, NEET, and other entrance tests.

To support your preparation, we provide comprehensive, accurate, and step-by-step NCERT Solutions for Class 12 Chemistry Chapter 9 – Coordination Compounds, completely based on the latest CBSE syllabus. These solutions simplify complex concepts, decode tricky questions, and offer clear reasoning for better retention.

What You Will Learn in Chapter 9 – Coordination Compounds

This chapter focuses on the unique and fascinating world of coordination compounds — a class of compounds in which central metal atoms or ions are bonded to surrounding ligands. You’ll explore how these compounds form, their structure, bonding, nomenclature, and properties.

Key Topics Covered:

Basic concepts: Ligands, coordination number, coordination sphere
Nomenclature of coordination compounds as per IUPAC rules
Bonding in coordination compounds: Valence Bond Theory (VBT), Crystal Field Theory (CFT)
Isomerism in coordination compounds: Structural and stereoisomerism
Stability of complexes and factors affecting it
Importance and applications of coordination compounds in biological systems, medicines, and industries
Werner’s theory of coordination compounds

Why Use Our NCERT Solutions for Chapter 9?

Coordination compounds often involve a mix of chemical theories, spatial understanding, and rigorous nomenclature rules. Our NCERT solutions are crafted to make these topics approachable and exam-ready, offering clarity and confidence to every student.

Key Features:

Step-by-step explanations with clear logic
Proper use of chemical equations, structures, and diagrams
Answers written in simple, lucid language
Fully solved in-text and end-of-chapter exercises
Useful for quick revisions and in-depth studies
Aligned with the latest CBSE Class 12 Chemistry curriculum

Download NCERT Solutions for Class 12 Chemistry Chapter 9 PDF

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What to Know Before Studying Chapter 9:

Before starting this chapter, it’s helpful to review:

Lewis acid-base theory and basic bonding concepts
Hybridization and VSEPR theory
Oxidation numbers and types of chemical compounds
Basics of naming inorganic compounds

Start Mastering Coordination Compounds Today!

Scroll down to access the complete NCERT Solutions for Class 12 Chemistry Chapter 9 – Coordination Compounds. Whether you’re preparing for board exams or entrance tests, these solutions are designed to strengthen your understanding, eliminate confusion, and help you succeed.

Let’s explore the intriguing world of ligands, metal ions, and complex ions — where coordination chemistry opens doors to real-world applications and advanced scientific understanding.

Section Name	Topic Name
9	Coordination Compounds
9.1	Werner’s Theory of Coordination Compounds
9.2	Definitions of Some Important Terms Pertaining to Coordination Compounds
9.3	Nomenclature of Coordination Compounds
9.4	Isomerism in Coordination Compounds
9.5	Bonding in Coordination Compounds
9.6	Bonding in Metal Carbonyls
9.7	Stability of Coordination Compound
9.8	Importance and Applications of Coordination Compounds

NCERT Solutions Class 12 Chemistry Chapter 9 – Coordination Compounds

Intext Question 9.1

Write the formulas for the following coordination compounds:
(i) Tetraamminediaquacobalt(III) chloride
(ii) Potassium tetracyanidonickelate(II)
(iii) Tris(ethane-1,2-diamine) chromium(III) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
(v) Dichloridobis(ethane-1,2-diamine) platinum(IV) nitrate
(vi) Iron(III) hexacyanidoferrate(II)

Answer:
(i) $Co(NH_3)_4(H_2O)_2]Cl_3$
(ii) $K_2[Ni(CN)_4]$
(iii) $Cr(en)_3]Cl_3$
(iv) $Pt(NH_3)(Br)(Cl)(NO_2)]^-$
(v) $PtCl_2(en)_2](NO_3)_2$
(vi) $Fe_4[Fe(CN)_6]_3$

Intext Question 9.2

Write IUPAC names of the following coordination compounds:
(a) $Co(NH_3)_6]Cl_3$
(b) $Co(NH_3)_5Cl]Cl_2$
(c) $K_3[Fe(CN)_6]$
(d) $K_3[Fe(C_2O_4)_3]$
(e) $K_2[PdCl_4]$
(f) $Pt(NH_3)_2Cl(NH_2CH_3)]Cl$

Answer:
(a) Hexaamminecobalt(III) chloride
(b) Pentaamminechloridocobalt(III) chloride
(c) Potassium hexacyanidoferrate(III)
(d) Potassium trioxalato ferrate(III)
(e) Potassium tetrachloridopalladate(II)
(f) Diamminechlorido(methylamine)platinum(II) chloride

Intext Question 9.3

Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:
(i) $K[Cr(H_2O)_2(C_2O_4)_2]$
(ii) $Co(en)_3]Cl_3$
(iii) $Co(NH_3)_5(NO_2)](NO_3)_2$
(iv) $Pt(NH_3)(H_2O)Cl_2]$

Answer:

(i) Exhibits geometrical isomerism (cis and trans forms due to arrangement of bidentate oxalato and monodentate water ligands).
(ii) Shows optical isomerism due to the presence of three chiral bidentate ligands (en).
(iii) Shows linkage isomerism as NO₂ can bind via nitrogen (nitro) or oxygen (nitrito).
(iv) Exhibits geometrical isomerism (cis and trans forms with different positions of ligands).

Intext Question 9.4

Give evidence that $Co(NH_3)_5Cl]SO_4$ and $Co(NH_3)_5SO_4]Cl$ are ionisation isomers.

Answer:
Ionisation isomers differ in the counter ion outside the coordination sphere.

In $Co(NH_3)_5Cl]SO_4$ , the chloride is bonded to cobalt and sulfate is outside.
In $Co(NH_3)_5SO_4]Cl$ , sulfate is coordinated while chloride is the counter ion.
On adding BaCl₂ to an aqueous solution of the first complex, white precipitate of BaSO₄ forms, proving SO₄²⁻ is free.
In the second, AgNO₃ gives a white precipitate of AgCl, indicating Cl⁻ is free.
This confirms they are ionisation isomers.

Intext Question 9.5

Explain using valence bond theory why $Ni(CN)_4]^{2-}$ is diamagnetic and square planar, whereas $NiCl_4]^{2-}$ is paramagnetic and tetrahedral.

Answer:
Nickel has the configuration: $3d^8 4s^2$ .
In Ni²⁺: $3d^8$

For $Ni(CN)_4]^{2-}$ :
CN⁻ is a strong field ligand. It causes pairing of 3d electrons, leaving no unpaired electrons. Hybridisation: $dsp^2$ → square planar. Result: Diamagnetic.
For $NiCl_4]^{2-}$ :
Cl⁻ is a weak field ligand. No pairing occurs, so two unpaired electrons remain. Hybridisation: $sp^3$ → tetrahedral. Result: Paramagnetic.

Intext Question 9.6

$NiCl_4]^{2-}$ is paramagnetic, while $Ni(CO)_4]$ is diamagnetic, although both are tetrahedral. Why?

Answer:
In $NiCl_4]^{2-}$ , Cl⁻ is a weak field ligand. It does not cause electron pairing, so Ni²⁺ remains with two unpaired electrons – paramagnetic.

In $Ni(CO)_4]$ , CO is a strong field ligand and causes pairing of electrons in 3d orbitals. All electrons are paired – diamagnetic.

Intext Question 9.7

$Fe(H_2O)_6]^{3+}$ is strongly paramagnetic, but $Fe(CN)_6]^{3-}$ is weakly paramagnetic. Explain.

Answer:
Fe³⁺ = $3d^5$ configuration.

In $Fe(H_2O)_6]^{3+}$ , H₂O is a weak ligand. No pairing occurs → 5 unpaired electrons → strongly paramagnetic.
In $Fe(CN)_6]^{3-}$ , CN⁻ is a strong ligand. It causes pairing, leaving only one unpaired electron → weakly paramagnetic.

Intext Question 9.8

Explain why $Co(NH_3)_6]^{2+}$ is an inner orbital complex, while $Ni(NH_3)_6]^{2+}$ is an outer orbital complex.

Answer:

In $Co(NH_3)_6]^{2+}$ , Co²⁺ (3d⁷) with strong field NH₃ causes pairing, allowing the use of inner $d^2sp^3$ orbitals → inner orbital (low spin) complex.
In $Ni(NH_3)_6]^{2+}$ , Ni²⁺ (3d⁸) does not undergo pairing due to weaker ligand effect. Uses $sp^3d^2$ orbitals from outer shell → outer orbital (high spin) complex.

Intext Question 9.9

Predict the number of unpaired electrons in the square planar $Pt(CN)_4]^{2-}$ ion.

Answer:
Pt²⁺ has configuration: $5d^8$ .
CN⁻ is a strong ligand, causing electron pairing in $d$ -orbitals. All electrons pair up in $dsp^2$ hybridisation.
So, no unpaired electrons → 0 unpaired electrons.

Intext Question 9.10

Explain why hexaaquamanganese(II) has five unpaired electrons, but the hexacyanomanganate(II) has only one.

Answer:
Mn²⁺ has configuration: $3d^5$

In $Mn(H_2O)_6]^{2+}$ , H₂O is a weak field ligand → high spin complex → no pairing → 5 unpaired electrons
In $Mn(CN)_6]^{4-}$ , CN⁻ is strong field → causes pairing → low spin complex → 1 unpaired electron

Intext Question 9.11

Calculate the overall dissociation equilibrium constant for $Cu(NH_3)_4]^{2+}$ , given β₄ = $2.1 \times 10^{13}$ .

Answer:
Overall stability constant (β₄) is the reciprocal of the dissociation constant (K_d):

$Kd=1β4=12.1×1013≈4.76×10−14K_d = \frac{1}{β₄} = \frac{1}{2.1 \times 10^{13}} ≈ 4.76 \times 10^{-14}$

So, dissociation constant = $4.76 \times 10^{-14}$

NCERT Exercise Solutions for Class 12 Chemistry Chapter 9 – Coordination Compounds

Q9.1 Explain the bonding in coordination compounds in terms of Werner’s postulates.

Answer: Werner proposed a theory to explain the structure and bonding of coordination compounds. His key postulates are:

Two types of valencies: Metals show:
- Primary valency (ionisable): Corresponds to the oxidation state. These are satisfied by negatively charged ions.
- Secondary valency (non-ionisable): Corresponds to the coordination number and is satisfied by ligands (neutral or anionic).
Primary valencies are satisfied by ions and are shown outside the coordination sphere, while secondary valencies are directional and lie within the coordination sphere.
Every metal has a fixed number of secondary valencies, which determine the shape of the complex (e.g., 6 for octahedral geometry).

Example: In $CoCl_3·6NH_3]$ , three Cl⁻ ions satisfy the primary valency and six NH₃ molecules satisfy the secondary valency. The complex ion is $Co(NH_3)_6]^{3+}$ , and the three Cl⁻ ions lie outside the square brackets.

Q9.2 FeSO₄ solution mixed with $(NH4)2SO4(NH₄)_2SO₄$ in a 1:1 ratio gives the test for Fe²⁺ ion, but CuSO₄ with aqueous ammonia (1:4) does not give the test for Cu²⁺. Why?

Answer:

Mixing FeSO₄ and $(NH4)2SO4(NH₄)_2SO₄$ forms a double salt, Mohr’s salt $FeSO_4·(NH_4)_2SO_4·6H_2O]$ . It dissociates completely in water into Fe²⁺, $NH_4^+$ , and $SO_4^{2-}$ ions. Hence, Fe²⁺ is detectable.
CuSO₄ with 4 equivalents of NH₃ forms a complex $Cu(NH_3)_4]SO_4$ , where Cu²⁺ becomes part of a stable coordination entity and does not dissociate freely. Therefore, no test for Cu²⁺ is observed.

Q9.3 Explain with two examples each:

(i) Coordination entity
(ii) Ligand
(iii) Coordination number
(iv) Coordination polyhedron
(v) Homoleptic
(vi) Heteroleptic

Answer:

Coordination entity: A central atom/ion bonded to a group of ligands.
Examples: $Co(NH_3)_6]^{3+}$ , $Fe(CN)_6]^{3-}$
Ligand: Ions or molecules donating a lone pair to the metal.
Examples: NH₃, Cl⁻, CN⁻
Coordination number: Number of ligand donor atoms bonded to the metal.
Examples: 6 in $Co(NH_3)_6]^{3+}$ , 4 in $Ni(CO)_4]$
Coordination polyhedron: The 3D arrangement of ligands.
Examples: Octahedral in $Co(NH_3)_6]^{3+}$ , Tetrahedral in $Ni(CO)_4]$
Homoleptic complex: Only one kind of ligand present.
Example: $Ni(CO)_4]$
Heteroleptic complex: More than one type of ligand.
Example: $CoCl_2(en)_2]^+$

Q9.4 What are unidentate, didentate, and ambidentate ligands? Give two examples of each.

Answer:

Unidentate ligand: Donates through a single donor atom.
Examples: Cl⁻, NH₃
Didentate ligand: Has two donor atoms that can simultaneously bind to the metal.
Examples: Ethylenediamine (en), C₂O₄²⁻ (oxalate)
Ambidentate ligand: Has two donor atoms, but only one bonds at a time.
Examples: NO₂⁻ (via N or O), SCN⁻ (via S or N)

Q9.5 Determine the oxidation number of the metal in:

(i) $Co(H_2O)(CN)(en)_2]^{2+}$
(ii) $CoBr_2(en)_2]^+$
(iii) $PtCl_4]^{2-}$
(iv) $K_3[Fe(CN)_6]$
(v) $Cr(NH_3)_3Cl_3]$

Answer: (i) Co = +3
(ii) Co = +3
(iii) Pt = +2
(iv) Fe = +3
(v) Cr = +3

Q9.6 Write formulae for:

(a) Tetrahydroxozincate(II) → $Zn(OH)_4]^{2-}$
(b) Hexaammineplatinum(IV) → $Pt(NH_3)_6]^{4+}$
(c) Potassium tetrachloridopalladate(II) → $K_2[PdCl_4]$
(d) Tetrabromidocuprate(II) → $CuBr_4]^{2-}$
(e) Hexaamminecobalt(III) sulfate → $Co(NH_3)_6]_2(SO_4)_3$
(f) Potassium tetracyanonickelate(II) → $K_2[Ni(CN)_4]$
(g) Potassium trioxalatochromate(III) → $K_3[Cr(C_2O_4)_3]$
(h) Pentaamminenitrito-O-cobalt(III) → $Co(NH_3)_5(ONO)]^{2+}$
(i) Diamminedichloridoplatinum(II) → $Pt(NH_3)_2Cl_2]$
(j) Pentaamminenitrito-N-cobalt(III) → $Co(NH_3)_5(NO_2)]^{2+}$

Q9.7 Write IUPAC names of the following:

(i) $Co(NH_3)_6]Cl_3$ → Hexaamminecobalt(III) chloride
(ii) $Pt(NH_3)_2Cl(NH_2CH_3)]Cl$ → Diamminechlorido(methylamine)platinum(II) chloride
(iii) $Ti(H_2O)_6]^{3+}$ → Hexaaquatitanium(III) ion
(iv) $Co(NH_3)_4Cl(NO_2)]Cl$ → Tetraamminechloridonitrito-N-cobalt(III) chloride
(v) $Mn(H_2O)_6]^{2+}$ → Hexaaquamanganese(II) ion
(vi) $NiCl_4]^{2-}$ → Tetrachloridonickelate(II) ion
(vii) $Ni(NH_3)_6]Cl_2$ → Hexaamminenickel(II) chloride
(viii) $Co(en)_3]^{3+}$ → Tris(ethane-1,2-diamine)cobalt(III) ion
(ix) $Ni(CO)_4]$ → Tetracarbonylnickel(0)

Q9.8 What types of isomerism are shown by coordination compounds? Give one example for each.

Answer: 1. Structural Isomerism:

Ionisation Isomerism: $Co(NH_3)_5SO_4]Br$ and $Co(NH_3)_5Br]SO_4$
Linkage Isomerism: $Co(NH_3)_5NO_2]^{2+}$ and $Co(NH_3)_5ONO]^{2+}$

2. Stereoisomerism:

Geometrical Isomerism: $Pt(NH_3)_2Cl_2]$ → cis and trans
Optical Isomerism: $Co(en)_3]^{3+}$ → d- and l-forms

Q9.9 How many geometrical isomers are possible for:

(i) $Cr(C_2O_4)_3]^{3-}$
(ii) $CoCl_3(NH_3)_3]$

Answer: (i) No geometrical isomerism (due to symmetrical bidentate ligands).
(ii) Two isomers: facial (fac) and meridional (mer).

Q9.10 Draw optical isomers of:

(i) $Cr(C_2O_4)_3]^{3-}$
(ii) $PtCl_2(en)_2]^{2+}$
(iii) $Cr(NH_3)_2Cl_2(en)]^+$

Answer:
These three complexes can exhibit optical isomerism due to their chiral arrangements. (To be illustrated with diagrams)

Q9.11 Draw all the isomers (geometrical and optical) of:

(i) $CoCl_2(en)_2]^+$
(ii) $Co(NH_3)Cl(en)_2]^{2+}$
(iii) $Co(NH_3)_2Cl_2(en)]^+$

Answer:
Each of these exhibits both geometrical (cis/trans) and optical isomerism. Exact diagrams show different arrangements of ligands and chiral centers.

Q9.12 Draw all geometrical isomers of $Pt(NH_3)(Br)(Cl)(Py)]$ . How many are optically active?

Answer:

This square planar complex has 3 geometrical isomers.
Due to symmetry, none of these are optically active. Optical isomerism is rare in square planar complexes unless special conditions are met.

Q9.13 Aqueous CuSO₄ gives:

(i) Green precipitate with KF
(ii) Bright green solution with KCl. Explain.

Answer:

In water, CuSO₄ forms $Cu(H_2O)_4]SO_4$ (blue solution).
(i) KF displaces H₂O ligands forming $CuF_4]^{2-}$ , a green precipitate.
(ii) KCl replaces H₂O ligands with Cl⁻ forming $CuCl_4]^{2-}$ , giving a bright green solution.

Question 9.14

What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is no precipitate of copper sulphide formed when H₂S(g) is passed through this solution?

Answer:

When excess potassium cyanide (KCN) is added to an aqueous solution of copper sulphate (CuSO₄), an initial reaction forms unstable cupric cyanide (Cu(CN)₂). This compound quickly decomposes into cuprous cyanide (CuCN) and cyanogen gas (CN)₂:

CuSO₄ + 2KCN → Cu(CN)₂ + K₂SO₄
Cu(CN)₂ → CuCN + (CN)₂↑

Now, the resulting CuCN reacts with excess KCN to form a stable coordination complex:

CuCN + 3KCN → K₄[Cu(CN)₄]

Thus, the coordination entity formed is [Cu(CN)₄]³⁻.

Why no precipitate with H₂S?
In this complex, copper is present in the form of a stable complex ion and not as free Cu²⁺ or Cu⁺ ions. Since CN⁻ is a strong field ligand, it forms a very stable complex with copper. Therefore, no free copper ions are available to react with H₂S, and thus, no black precipitate of CuS is formed.

Question 9.15

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⁻

Fe is in +2 oxidation state → [Ar]3d⁶
CN⁻ is a strong field ligand: causes pairing of electrons in lower energy orbitals
Hybridization: d²sp³ (inner orbital complex)
Magnetic behaviour: Diamagnetic (all electrons paired)
Geometry: Octahedral

(ii) [FeF₆]³⁻

Fe is in +3 oxidation state → [Ar]3d⁵
F⁻ is a weak field ligand, no pairing
Hybridization: sp³d² (outer orbital complex)
Magnetic behaviour: Paramagnetic (5 unpaired electrons)
Geometry: Octahedral

(iii) [Co(C₂O₄)₃]³⁻

Co is in +3 oxidation state → [Ar]3d⁶
C₂O₄²⁻ is a bidentate ligand (chelating), moderate field strength
Electrons pair up due to chelation effect
Hybridization: d²sp³
Magnetic behaviour: Diamagnetic
Geometry: Octahedral

(iv) [CoF₆]³⁻

Co is in +3 oxidation state → [Ar]3d⁶
F⁻ is a weak field ligand, no pairing
Hybridization: sp³d²
Magnetic behaviour: Paramagnetic (4 unpaired electrons)
Geometry: Octahedral

Question 9.16

Draw a diagram to show the splitting of d-orbitals in an octahedral crystal field.

Answer:

In an octahedral field, the five degenerate d-orbitals split into two energy levels:

t₂g set (lower energy): dxy, dyz, dzx
eg set (higher energy): dx²–y², dz²

The difference between these energy levels is called the crystal field splitting energy (Δ₀).

Question 9.17

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

Answer:

The spectrochemical series is an arrangement of ligands in order of their ability to split the d-orbitals of the central metal atom in a coordination complex. It is based on the crystal field splitting energy (Δ₀).

Spectrochemical series (partial):
I⁻ < Br⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO

Weak field ligands (e.g., Cl⁻, H₂O): cause small Δ₀; do not pair electrons → high spin complexes
Strong field ligands (e.g., CN⁻, CO): cause large Δ₀; pair electrons in lower orbitals → low spin complexes

Question 9.18

What is crystal field splitting energy? How does the magnitude of Δ₀ decide the actual configuration of d-orbitals in a coordination entity?

Answer:

Crystal field splitting energy (Δ₀) is the energy difference between the two sets of d-orbitals (t₂g and eg) in an octahedral crystal field due to interaction with ligands.

If Δ₀ < P (pairing energy): Electron goes to higher eg orbital → high spin configuration
If Δ₀ > P: Electron pairs in t₂g orbital → low spin configuration

Thus, the value of Δ₀ and the nature of the ligand determine whether the complex is high or low spin.

Question 9.19

[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]²⁻ is diamagnetic. Explain why.

Answer:

In [Cr(NH₃)₆]³⁺, Cr³⁺ has configuration 3d³. NH₃ is a neutral, weak-to-moderate field ligand. So, electrons remain unpaired → paramagnetic.
In [Ni(CN)₄]²⁻, Ni²⁺ is 3d⁸. CN⁻ is a strong field ligand, so it causes pairing of electrons. All electrons pair → diamagnetic.

Question 9.20

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

Answer:

In [Ni(H₂O)₆]²⁺, Ni²⁺ has two unpaired electrons → d-d transitions → absorbs red light → appears green.
In [Ni(CN)₄]²⁻, CN⁻ causes pairing of electrons → no unpaired electrons → no d-d transition → colourless.

Question 9.21

[Fe(CN)₆]⁴⁻ and [Fe(H₂O)₆]²⁺ have different colours in solution. Why?

Answer:

Although both have Fe²⁺ (3d⁶), the ligands are different:

CN⁻ is a strong field ligand → large Δ₀ → absorbs different wavelength.
H₂O is weaker → smaller Δ₀.

Different Δ₀ leads to absorption of different light → different colours.

Question 9.22

Discuss the nature of bonding in metal carbonyls.

Answer:

In metal carbonyls:

σ-bond: Formed by donation of lone pair from CO to metal.
π-back bonding: Metal donates electrons from filled d-orbitals into empty π* anti-bonding orbital of CO.

This synergic bonding strengthens the M–CO bond and stabilizes the complex.

Question 9.23

Give oxidation state, d-orbital occupation, and coordination number of the central metal ion in:

(i) K₃[Co(C₂O₄)₃]

Oxidation state of Co = +3
d⁶ configuration
Coordination number = 6 (each oxalate is bidentate)

(ii) cis-[Cr(en)₂Cl₂]Cl

Oxidation state of Cr = +3
d³ configuration
Coordination number = 6 (2 en + 2 Cl⁻ ligands)

(iii) (NH₄)₂[CoF₄]

Oxidation state of Co = +2
d⁷ configuration
Coordination number = 4

(iv) [Mn(H₂O)₆]SO₄

Oxidation state of Mn = +2
d⁵ configuration
Coordination number = 6

Question 9.24

Write IUPAC name, oxidation state, configuration, coordination number, stereochemistry, and magnetic moment:

(i) K[Cr(H₂O)₂(C₂O₄)₂]·3H₂O

Name: Potassium diaquadioxalatochromate(III) trihydrate
Oxidation state: +3
Coordination number: 6
Configuration: d³
Geometry: Octahedral
Magnetic moment: 3.87 BM

(ii) [Co(NH₃)₅Cl]Cl₂

Name: Pentaamminechloridocobalt(III) chloride
Oxidation state: +3
Configuration: d⁶ (low spin)
Coordination number: 6
Geometry: Octahedral
Magnetic moment: 0 (diamagnetic)

(iii) CrCl₃(py)₃

Name: Trichloridotripyridinechromium(III)
Oxidation state: +3
Configuration: d³
Coordination number: 6
Geometry: Octahedral
Magnetic moment: ~3.87 BM

Question 9.25

What is meant by stability of a coordination compound? Factors affecting it?

Answer:

Stability refers to the extent to which a complex resists dissociation into its components. It is quantified using the stability constant (β).

Factors influencing stability:

Nature of central metal ion (charge/size)
Oxidation state of metal
Nature of ligands (chelating ligands are more stable)
Basicity of ligands (more basic → stronger bond)

Question 9.26

What is chelate effect? Give an example.

Answer:

The chelate effect is the increased stability of complexes with polydentate ligands forming ring structures.

Example: [Cu(en)₂]²⁺ is more stable than [Cu(NH₃)₄]²⁺ due to two five-membered rings formed by ethylenediamine (en).

Question 9.27

Roles of coordination compounds in:

(a) Biological systems:

Chlorophyll (Mg²⁺ complex), hemoglobin (Fe²⁺), vitamin B12 (Co³⁺)

(b) Analytical chemistry:

Colorimetric detection via ligands like EDTA, DMG

EDTA for lead poisoning, cisplatin for cancer treatment

(d) Metallurgy:

Gold and silver extraction via cyanide complexes

Question 9.28

How many ions are produced from Co(NH₃)₆Cl₂ in solution?

Answer:
[Co(NH₃)₆]Cl₂ dissociates into:

Correct option: (iii) 3

Question 9.29

Which ion has highest magnetic moment:

(i) [Cr(H₂O)₆]³⁺ → d³ → 3 unpaired
(ii) [Fe(H₂O)₆]²⁺ → d⁶ → 4 unpaired
(iii) [Zn(H₂O)₆]²⁺ → d¹⁰ → 0 unpaired

Answer:
(ii) [Fe(H₂O)₆]²⁺ has 4 unpaired electrons → highest μ

Question 9.30

Oxidation number of Co in K[Co(CO)₄]?

Let x = oxidation state of Co
CO is neutral ligand
K⁺ gives +1

Answer: (iii) –1

Question 9.31

Most stable complex?

(i) [Fe(H₂O)₆]³⁺
(ii) [Fe(NH₃)₆]³⁺
(iii) [Fe(C₂O₄)₃]³⁻
(iv) [FeCl₆]³⁻

Answer:
(iii) is most stable due to chelation effect of oxalate → forms stable rings.

Question 9.32

Correct order of absorption wavelengths for:

[Ni(NO₂)₆]⁴⁻
[Ni(NH₃)₆]²⁺
[Ni(H₂O)₆]²⁺

Stronger field → larger Δ → absorbs shorter λ

Order of ligand field strength:
H₂O < NH₃ < NO₂⁻

So, absorption wavelength:
[Ni(H₂O)₆]²⁺ > [Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]⁴⁻