Coordination Compounds — Class 12 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Coordination Compounds" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Coordination Compounds" — 6 important questions with detailed answers for CBSE board…
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NCERT Solutions for Class 12 Chemistry — Coordination Compounds. Step-by-step answers to all 6 textbook questions from this chapter, written for CBSE board preparation and free to use.
Coordination Compounds — All 6 Questions Solved
Q1. Define coordination compounds. What are ligands and coordination number?
Coordination Compound:
A compound formed by combination of a central metal atom/ion with a certain number of anions or neutral molecules (called ligands) surrounding it.
General formula: [M(L)_n]^(m+/-)
where M = central metal, L = ligand, n = coordination number, m = charge
Example: [Cu(NH₃)₄]²⁺, [Fe(CN)₆]⁴⁻, K₃[Fe(CN)₆]
Ligand:
A molecule or anion that donates a pair of electrons to the central metal ion.
Characteristics:
1. Contains at least one lone pair of electrons
2. Acts as Lewis base (electron donor)
3. Coordinates to central metal atom (acts as Lewis acid)
Types of Ligands:
1. Monodentate ligands: Donate one electron pair
Examples: NH₃, H₂O, Cl⁻, CN⁻, CO
2. Bidentate ligands: Donate two electron pairs (chelating)
Examples: Ethylenediamine (en) H₂N-CH₂-CH₂-NH₂, Oxalate (C₂O₄²⁻), Dimethylglyoxime (DMG)
3. Polydentate ligands: Donate more than two electron pairs
Example: EDTA (ethylenediaminetetraacetate) - hexadentate
4. Bridging ligands: Connect two metal centers
Examples: OH⁻, NH₃ (in polynuclear complexes)
Coordination Number (CN):
Total number of bonds formed between the central metal ion and the ligands.
Or: Number of donor atoms attached to central metal.
Common coordination numbers:
- CN = 2: Linear geometry
Examples: [Ag(NH₃)₂]⁺, [Cu(CN)₂]⁻
- CN = 4: Tetrahedral or square planar
Tetrahedral: [Zn(NH₃)₄]²⁺, [CoCl₄]²⁻
Square planar: [Ni(CN)₄]²⁻, [Pt(NH₃)₄]²⁺
- CN = 6: Octahedral (most common)
Examples: [Fe(CN)₆]⁴⁻, [Cr(H₂O)₆]³⁺, [Co(NH₃)₆]³⁺
- CN = 8: Cubic or dodecahedral
Examples: [Mo(CN)₈]⁴⁻
Factors affecting coordination number:
1. Size of central metal ion (larger ion → higher CN)
2. Charge on metal ion (higher charge → higher CN)
3. Nature and size of ligands (small ligands → higher CN)
4. Electronic configuration of metal
Example analysis:
[Cu(NH₃)₄]²⁺
Central metal: Cu²⁺
Ligand: NH₃ (monodentate, neutral)
Coordination number: 4
Geometry: Tetrahedral or square planar (Cu²⁺ typically square planar)
Number of NH₃ ligands: 4
Charge: +2
Q2. Explain the crystal field theory and its application to octahedral complexes.
Crystal Field Theory (CFT):
Theory that explains the structure, color, and magnetic properties of coordination complexes based on electrostatic interactions between the central metal ion and the surrounding ligands.
Basic Assumption:
Ligands are treated as point charges or dipoles that create an electric field.
This field causes splitting of d-orbitals of the central metal ion.
Origin of d-orbital Splitting:
In free metal ion: All five d-orbitals are degenerate (same energy)
d_xy, d_xz, d_yz, d_x²-y², d_z²
In octahedral complex:
Six ligands approach along ±x, ±y, ±z axes
Two sets of d-orbitals formed based on orientation:
1. Lower energy (t₂g): d_xy, d_xz, d_yz
Orbitals point between axes, minimum repulsion
Energy: -0.4Δ (relative to average)
2. Higher energy (e_g): d_x²-y², d_z²
Orbitals point directly toward axes, maximum repulsion
Energy: +0.6Δ (relative to average)
Energy difference = Δ (crystal field splitting parameter)
Δ depends on:
- Nature of central metal ion (charge, size)
- Nature of ligands (ligand field strength)
- Geometry (octahedral, tetrahedral, etc.)
Ligand Field Strength Series (Spectrochemical series):
Increasing ligand field strength:
I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO
Weaker ligands → smaller Δ
Stronger ligands → larger Δ
Applications to Octahedral Complexes:
1. Color of complexes:
Color arises from d-d electronic transitions between t₂g and e_g levels.
Δ corresponds to visible light wavelengths
Example: [Fe(CN)₆]⁴⁻ is yellow (CN⁻ strong field, large Δ)
[FeCl₆]³⁻ is pale yellow (Cl⁻ weak field, small Δ)
2. Magnetic Properties:
Weak field ligands (small Δ):
Electrons enter higher e_g orbitals even with unpaired electrons
High-spin complex (paramagnetic)
Example: [FeF₆]³⁻ (weak field F⁻, high-spin, 5 unpaired electrons)
Strong field ligands (large Δ):
Electrons pair up in lower t₂g orbitals first
Low-spin complex (diamagnetic or less paramagnetic)
Example: [Fe(CN)₆]⁴⁻ (strong field CN⁻, low-spin, 0 unpaired electrons)
3. Electron configuration in octahedral [M(H₂O)₆]ⁿ⁺:
For d⁶ metal (Fe²⁺):
Weak field: t₂g⁴ e_g² (high-spin, 4 unpaired)
Strong field: t₂g⁶ (low-spin, 0 unpaired)
For d⁵ metal (Mn²⁺, high-spin):
t₂g³ e_g²
4. Pairing energy:
Δ > P: Low-spin (electrons pair up)
Δ < P: High-spin (electrons remain unpaired)
P = pairing energy (cost of pairing electrons)
5. Stability:
More negative value of Δ provides greater stabilization energy
Can predict complex stability from ligand field strength
Limitations of CFT:
1. Does not account for covalent bonding
2. Treats ligands as point charges
3. Explains colors and magnetism well, but less effective for bonding
4. Mod: Molecular orbital theory provides better bonding picture
Q3. Write the IUPAC nomenclature rules for coordination compounds with examples.
IUPAC Nomenclature of Coordination Compounds:
General Rules (in order):
1. NAME THE CATION FIRST, then the anion
2. In the coordination sphere:
- Name ligands in alphabetical order
- Prefix numbers: mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octa-
- Use Greek prefixes regardless of ligand name
- For complicated names, use bis-, tris-, tetrakis-, etc. with parentheses
3. LIGAND NAMING:
Neutral ligands retain name:
- H₂O = aqua
- NH₃ = ammine (note double m)
- CO = carbonyl
- NO = nitrosyl
- C₂H₄ = ethene
Anionic ligands: -ido, -ito, or -ate:
- Cl⁻ = chloro
- Br⁻ = bromo
- F⁻ = fluoro
- OH⁻ = hydroxo
- CN⁻ = cyano
- SCN⁻ = thiocyanato
- NO₂⁻ = nitrito
- NO₃⁻ = nitrato
- SO₄²⁻ = sulfato
- C₂O₄²⁻ = oxalato
Cationic ligands: -ium:
- NH₄⁺ = ammonium
- NO⁺ = nitrosonium
4. COMPLEX NAME:
- If complex is cation: add -um suffix to metal name
- If complex is anion: change metal name to -ate form
- If complex is neutral: use metal name as is
Examples:
1. [Cu(NH₃)₄]²⁺
Alphabetically: NH₃ (ammine)
Ligand: tetraammine
Metal: Cu is copper, complex is cation → copper
Full name: tetraammine copper(II) ion
or: [tetraammine copper(II)]
2. K₃[Fe(CN)₆]
Cation: 3 K⁺
Anion: [Fe(CN)₆]³⁻
Ligand: CN⁻ is cyano, 6 ligands = hexacyano
Metal: Fe, complex is anion → ferrate
Oxidation state of Fe: +3
Full name: potassium hexacyanoferrate(III)
3. [Cr(H₂O)₄(OH)₂]⁺
Alphabetically: aqua before hydroxo
Ligands: tetraaquadihydroxo
Metal: Cr is chromium, cation
Oxidation state: +3
Full name: tetraaquadihydroxochromium(III) ion
4. [PtCl₂(NH₃)₂]
Alphabetically: ammine before chloro
Ligands: diamminedichlorido
Metal: Pt is platinum, neutral
Oxidation state: +2
Full name: diamminedichloridoplatinum(II)
5. [Co(en)₃]³⁺
Ligand: ethylenediamine = en (bidentate)
Ligands: tris(ethylenediamine) [use parentheses for complicated names]
Metal: cobalt, cation
Oxidation state: +3
Full name: tris(ethylenediamine)cobalt(III) ion
6. [Fe(CN)₆]⁴⁻
Ligand: cyano, 6 ligands = hexacyano
Metal: iron, anion → ferrate
Oxidation state: +2
Full name: hexacyanoferrate(II) ion
Additional Rules:
- When same ligand name starts with vowel, drop final vowel of Greek prefix
Example: tetra + ammine = tetraammine (not tetrarammine)
- Oxidation number in Roman numerals in parentheses after metal name
- For polynuclear complexes, name bridging ligands with μ- prefix
Example: μ-hydroxo
- Common monodentate ligands and their names:
NH₃ (ammine), H₂O (aqua), CO (carbonyl), NO (nitrosyl)
Cl⁻ (chloro), Br⁻ (bromo), I⁻ (iodo), CN⁻ (cyano)
- Common bidentate and polydentate:
Ethylenediamine (en), oxalate, EDTA
Q4. Calculate the percentage composition of [Cu(NH₃)₄]SO₄.
Given complex: [Cu(NH₃)₄]SO₄
Step 1: Calculate molar mass
Atomic masses: Cu = 63.5, N = 14, H = 1, S = 32, O = 16
Central complex [Cu(NH₃)₄]²⁺:
- Cu: 1 × 63.5 = 63.5
- N: 4 × 14 = 56
- H: 4 × 3 = 12 (in 4 NH₃ groups)
- Subtotal: 63.5 + 56 + 12 = 131.5
Counter ion SO₄²⁻:
- S: 1 × 32 = 32
- O: 4 × 16 = 64
- Subtotal: 32 + 64 = 96
Total molar mass = 131.5 + 96 = 227.5 g/mol
Step 2: Calculate percentage of each element
% Cu = (63.5 / 227.5) × 100 = 27.92%
% N = (56 / 227.5) × 100 = 24.62%
% H = (12 / 227.5) × 100 = 5.27%
% S = (32 / 227.5) × 100 = 14.07%
% O = (64 / 227.5) × 100 = 28.13%
Verification:
27.92 + 24.62 + 5.27 + 14.07 + 28.13 = 100.01% ✓ (rounding)
Answer:
% Cu = 27.92%
% N = 24.62%
% H = 5.27%
% S = 14.07%
% O = 28.13%
Q5. Explain the stability of complexes and factors affecting complex stability.
Complex Stability:
The tendency of a complex to remain intact and resist dissociation into its constituent ions and molecules.
[M(L)_n]ⁿ⁺ ⇌ Mⁿ⁺ + nL
Stability Constant (K_formation or K_stability):
Quantifies tendency of complex formation.
For [M(L)_n]ⁿ⁺ ⇌ Mⁿ⁺ + nL
K_stability = 1 / K_dissociation = [Mⁿ⁺][L]ⁿ / [M(L)_n]ⁿ⁺
Larger K_stability → more stable complex
Factors Affecting Complex Stability:
1. Nature of the Central Metal Ion:
Size: Smaller metal → higher charge density → greater attraction to ligands
Example: Cr³⁺ forms more stable complexes than Fe³⁺ (smaller Cr³⁺)
Charge: Higher charge → stronger electrostatic attraction
Example: Al³⁺ complexes > Al²⁺ complexes
Electronic configuration: d⁶ and d¹⁰ configurations (filled/half-filled) form stable complexes
2. Nature of the Ligand:
Basic strength: Stronger Lewis bases form stronger complexes
Example: NH₃ > H₂O (NH₃ is stronger base)
Ligand field strength: Stronger field ligands → more stable complexes
CN⁻ > NH₃ > H₂O > Cl⁻
Chelating effect:
Bidentate and polydentate ligands form more stable complexes than monodentate
Example: [Cr(en)₃]³⁺ > [Cr(NH₃)₆]³⁺
Reason: Entropy effect (one bidentate = two monodentates)
3. Geometry and Crystal Field Effects:
Octahedral geometry more stable than tetrahedral
Square planar preferred for d⁸ metals (Ni²⁺, Pd²⁺, Pt²⁺)
4. Steric Effects:
Bulky ligands may reduce stability due to steric hindrance
Example: Triphenylphosphine complexes less stable due to size
5. pH Effect:
For complexes with OH⁻ or H₂O ligands
Lower pH (more H⁺) destabilizes complex
Higher pH favors complex formation
6. Temperature:
Complexity varies with temperature
Usually decreases with temperature (exothermic formation)
Quantitative Expression:
Stability constant = ∏ K_i
[M(L)_n]ⁿ⁺ = K_overall = K₁ × K₂ × K₃ × ... × K_n
where:
K₁ = [M(L)]ⁿ⁺ / ([Mⁿ⁺][L])
K₂ = [M(L)₂]ⁿ⁺ / ([M(L)]ⁿ⁺[L])
K_n is successive stability constant
Example values (at 25°C):
[Cu(NH₃)₄]²⁺: K = 4.8 × 10¹² (very stable)
[Cu(en)₂]²⁺: K = 10²⁰ (extremely stable, chelate effect)
[Fe(CN)₆]⁴⁻: K = 10⁴³ (extremely stable)
[FeCl₃]: K = 10² (less stable)
Chelate Effect Example:
For Cu²⁺:
- [Cu(NH₃)₄]²⁺: K = 4.8 × 10¹²
- [Cu(en)₂]²⁺: K = 10²⁰
Ethylenediamine complex is ~10⁸ times more stable
Despite same denticity (1:1 ligand ratio)
Reason: Entropy gain from formation of fewer particles
Applications:
1. Qualitative analysis: Using specific complexes for ion identification
2. Analytical chemistry: Complexometric titrations
3. Medicine: Chelation therapy (EDTA removes heavy metal ions)
4. Industry: Extraction and separation of metals
5. Environmental remediation: Complex formation removes toxic ions
Q6. Explain the concept of isomerism in coordination compounds with examples.
Isomerism in Coordination Compounds:
Two or more coordination compounds with same molecular formula but different structural arrangements or spatial orientations.
Types of Isomerism:
A. STRUCTURAL ISOMERISM:
1. Ionization Isomerism:
Difference in which ion remains outside coordination sphere
Different counter-ions or different complex ion charge
Example:
[Pt(NH₃)₃(NO₂)]Cl and [Pt(NH₃)₃Cl]NO₂
Both have formula Pt(NH₃)₃(NO₂)Cl
First: coordination sphere = [Pt(NH₃)₃(NO₂)]⁺, counter-ion = Cl⁻
Second: coordination sphere = [Pt(NH₃)₃Cl]⁺, counter-ion = NO₂⁻
Electrochemical properties differ
2. Coordination Isomerism:
Exchange of ligands between two metal centers in polynuclear complex
Both metals exchange coordinated ligands
Example:
[Cu(NH₃)₄][PtCl₄] and [Cu(Cl)₄][Pt(NH₃)₄]
First: Cu²⁺ coordinates NH₃, Pt²⁺ coordinates Cl⁻
Second: Cu²⁺ coordinates Cl⁻, Pt²⁺ coordinates NH₃
3. Linkage Isomerism:
Ligand with two different donor atoms (ambidentate ligand)
Ligand bonds to metal through different atoms
Example with NO₂⁻ (nitrite ion):
- Nitro complex: M-ONO (bonding through O)
- Nitrito complex: M-NO₂ (bonding through N)
Example: [Co(NH₃)₅(NO₂)]²⁺
Cobalt-bonded through O: [Co(NH₃)₅(ONO)]²⁺ (orange color)
Cobalt-bonded through N: [Co(NH₃)₅(NO₂)]²⁺ (red color)
Other ambidentate ligands:
- SCN⁻: bonding through S (thiocyanato) or N (isothiocyanato)
- CN⁻: bonding through C (normal) or N (rare)
4. Hydration Isomerism:
Difference in number of water molecules inside/outside coordination sphere
Example:
[Fe(H₂O)₆]Cl₃ (ferric chloride)
[Fe(H₂O)₅Cl]Cl₂·H₂O (one water outside, one Cl⁻ inside)
[Fe(H₂O)₄Cl₂]Cl·2H₂O (two waters outside, two Cl⁻ inside)
Same molecular formula: Fe(H₂O)₆Cl₃
Different properties, different reactions
B. STEREOISOMERISM:
1. Geometric Isomerism:
Difference in spatial arrangement of ligands
Occurs in square planar (CN=4) and octahedral (CN=6) complexes
Square planar (CN=4):
Complex [MA₂B₂] (A and B are different monodentate ligands)
cis-isomer: A ligands adjacent (90°)
Example: [Pt(NH₃)₂Cl₂] cis
Properties: polar, insoluble in ether
trans-isomer: A ligands opposite (180°)
Example: [Pt(NH₃)₂Cl₂] trans
Properties: non-polar, soluble in ether
Octahedral (CN=6):
Complex [MA₃B₃] or [MA₄B₂]
Example: [Co(NH₃)₃(NO₂)₃]
fac-isomer: Three NH₃ at faces (120° apart)
mer-isomer: Three NH₃ at meridian (linear arrangement)
Example: [Cr(H₂O)₄Cl₂]⁺
cis: Cl⁻ at 90° to each other
trans: Cl⁻ at 180° to each other
2. Optical Isomerism (Enantiomerism):
Chiral complexes that are non-superimposable mirror images
Show optical activity (rotate plane of polarized light)
Example: [Co(en)₃]³⁺ (tris(ethylenediamine)cobalt(III))
No plane of symmetry
Exists as d-form (dextrorotatory) and l-form (levorotatory)
Identical chemical/physical properties except optical rotation
Conditions for optical isomerism:
1. Complex must be chiral (no plane of symmetry, no center of inversion)
2. Usually with chelating ligands (en, C₂O₄²⁻)
3. Octahedral complexes common
Example: [Cr(C₂O₄)₃]³⁻
Three bidentate C₂O₄²⁻ chelate
Chiral complex
Exists as (+) and (-) enantiomers
Summary Table:
Isomerism Type | Difference | Example
---|---|---
Ionization | Counter-ion | [Pt(NH₃)₃Cl]NO₂ vs [Pt(NH₃)₃NO₂]Cl
Coordination | Metal coordination | [Cu(NH₃)₄][PtCl₄] vs [CuCl₄][Pt(NH₃)₄]
Linkage | Donor atom | [Co(NO₂)(NH₃)₅]²⁺ vs [Co(ONO)(NH₃)₅]²⁺
Hydration | Water position | [Fe(H₂O)₆]Cl₃ vs [Fe(H₂O)₅Cl]Cl₂·H₂O
Geometric | Spatial arrangement | cis-[Pt(NH₃)₂Cl₂] vs trans-[Pt(NH₃)₂Cl₂]
Optical | Mirror image | d-[Co(en)₃]³⁺ vs l-[Co(en)₃]³⁺
Coordination Compounds — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the coordination number of the central metal ion in [Co(en)₂(Cl)₂]Cl?
- A. 3
- B. 4
- C. 5
- D. 6
Answer: D. 6 — 'en' (ethylenediamine) is a bidentate ligand, contributing 2 coordination sites each. There are two 'en' ligands and two 'Cl' ligands, making the total coordination number (2 * 2) + (2 * 1) = 6.
MCQ 2. The IUPAC name for [Cr(NH₃)₃Cl₃] is:
- A. Triamminetrichlorochromium(III)
- B. Triamminetrichloridochromium(III)
- C. Triamminetrichlorochromate(III)
- D. Triamminetrichloridochromiumate(III)
Answer: B. Triamminetrichloridochromium(III) — In a neutral complex, the metal name is unchanged. Anionic ligands (like chloro/chlorido) are named with an 'o' suffix. The oxidation state of Cr is +3 (3*0 + 3*(-1) = -3, so Cr is +3).
MCQ 3. Which of the following ligands can exhibit linkage isomerism?
- A. NH₃
- B. H₂O
- C. NO₂⁻
- D. Cl⁻
Answer: C. NO₂⁻ — Linkage isomerism arises when a ligand can coordinate to the central metal atom through two different donor atoms. NO₂⁻ is an ambidentate ligand that can coordinate through nitrogen (nitro) or oxygen (nitrito).
MCQ 4. A complex [NiCl₄]²⁻ is paramagnetic. What is its geometry and hybridization?
- A. Square planar, dsp²
- B. Tetrahedral, sp³
- C. Square planar, sp³
- D. Tetrahedral, dsp²
Answer: B. Tetrahedral, sp³ — Ni²⁺ has d⁸ configuration. For [NiCl₄]²⁻ to be paramagnetic, it must have unpaired electrons. Cl⁻ is a weak field ligand, so it does not cause pairing. In a tetrahedral geometry (sp³), there will be two unpaired electrons, leading to paramagnetism.
MCQ 5. According to Werner's theory, the primary valency of a metal ion in a complex is its:
- A. Coordination number
- B. Oxidation state
- C. Number of ligands
- D. Number of counter ions
Answer: B. Oxidation state — Werner's theory proposed two types of valencies: primary valency, which is ionizable and corresponds to the oxidation state of the central metal ion.
MCQ 6. For an octahedral complex, which of the following statements about Crystal Field Splitting Energy (CFSE) is correct?
- A. Strong field ligands cause greater splitting (Δo).
- B. Weak field ligands cause greater splitting (Δo).
- C. CFSE is always positive.
- D. CFSE depends only on the metal ion.
Answer: A. Strong field ligands cause greater splitting (Δo). — Strong field ligands cause a larger separation between the t₂g and eg orbitals (greater Δo) because they create a stronger electrostatic field that repels the d-electrons more effectively.
MCQ 7. Which of the following complexes is expected to show optical isomerism?
- A. [Co(NH₃)₆]³⁺
- B. [Co(en)₃]³⁺
- C. [Pt(NH₃)₂Cl₂]
- D. [Ni(CN)₄]²⁻
Answer: B. [Co(en)₃]³⁺ — [Co(en)₃]³⁺ is a chelate complex of the type [M(AA)₃]ⁿ⁺. Such complexes are chiral and exist as non-superimposable mirror images (enantiomers), thus exhibiting optical isomerism.
MCQ 8. What is the oxidation state of Nickel in [Ni(CO)₄]?
- A. +4
- B. +2
- C. +1
- D. 0
Answer: D. 0 — Carbonyl (CO) is a neutral ligand, so it does not contribute to the charge of the complex. Since the overall charge of the complex [Ni(CO)₄] is zero, the oxidation state of Nickel must also be zero.
MCQ 9. A bidentate ligand has how many donor atoms?
- A. One
- B. Two
- C. Three
- D. Four
Answer: B. Two — The term 'bidentate' indicates that the ligand has two donor atoms that can simultaneously coordinate to the central metal ion, forming a chelate ring.
MCQ 10. If a complex ion [ML₆]ⁿ⁺ forms an outer orbital complex, what type of hybridization is involved?
- A. d²sp³
- B. sp³d²
- C. dsp²
- D. sp³
Answer: B. sp³d² — Outer orbital complexes are formed when the metal ion uses its outer d-orbitals (nd) for hybridization. For an octahedral geometry (coordination number 6), this corresponds to sp³d² hybridization.
MCQ 11. The color of a coordination compound is primarily due to:
- A. s-p transitions
- B. d-d transitions
- C. p-p transitions
- D. Charge transfer transitions
Answer: B. d-d transitions — The color in most transition metal coordination compounds arises from d-d transitions. Electrons in lower energy d-orbitals absorb specific wavelengths and are promoted to higher energy d-orbitals, leading to the observed complementary color.
MCQ 12. Which of the following biological molecules contains a coordination compound of iron?
- A. Chlorophyll
- B. Vitamin B12
- C. Hemoglobin
- D. Insulin
Answer: C. Hemoglobin — Hemoglobin, the oxygen-carrying protein in red blood cells, contains a heme group, which is a porphyrin ring coordinated to an iron(II) ion.
MCQ 13. Which of the following pairs represents ionization isomers?
- A. [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br
- B. [Co(NH₃)₅NO₂]Cl₂ and [Co(NH₃)₅ONO]Cl₂
- C. [Cr(H₂O)₆]Cl₃ and [Cr(H₂O)₅Cl]Cl₂·H₂O
- D. [Pt(NH₃)₄Cl₂]Br₂ and [Pt(NH₃)₄Br₂]Cl₂
Answer: A. [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br — Ionization isomers have the same molecular formula but differ in the ions produced in solution. In this pair, the sulfate ion and bromide ion exchange positions between the coordination sphere and the counter ion.
MCQ 14. Calculate the spin-only magnetic moment for a d³ metal ion in an octahedral field.
- A. 0 BM
- B. 1.73 BM
- C. 2.83 BM
- D. 3.87 BM
Answer: D. 3.87 BM — A d³ metal ion in an octahedral field will have 3 unpaired electrons (t₂g³ eg⁰). Using the spin-only magnetic moment formula μ = √[n(n+2)], where n=3, μ = √[3(3+2)] = √15 ≈ 3.87 BM.
MCQ 15. In a d⁶ octahedral complex, if Δo > P (P = pairing energy), what will be the number of unpaired electrons?
- A. 0
- B. 2
- C. 4
- D. 5
Answer: A. 0 — When Δo > P, it indicates a strong field ligand. For a d⁶ octahedral complex, electrons will pair up in the t₂g orbitals (t₂g⁶ eg⁰) rather than occupying the higher energy eg orbitals, resulting in 0 unpaired electrons.
MCQ 16. Which of the following ligands is a chelating ligand?
- A. CN⁻
- B. H₂O
- C. NO₃⁻
- D. C₂O₄²⁻ (oxalate)
Answer: D. C₂O₄²⁻ (oxalate) — A chelating ligand is a polydentate ligand that can bind to the central metal atom through more than one donor atom, forming a ring structure. Oxalate (C₂O₄²⁻) is a bidentate ligand and thus a chelating ligand.
MCQ 17. The term 'ligand' in coordination chemistry refers to:
- A. The central metal atom
- B. An electron pair donor
- C. An electron pair acceptor
- D. A counter ion
Answer: B. An electron pair donor — Ligands are species (ions or molecules) that act as Lewis bases, donating a pair of electrons to the central metal atom or ion to form a coordinate bond.
MCQ 18. What is the hybridization of the central metal ion in [Fe(CN)₆]³⁻? (Assume it is a low spin complex)
- A. sp³d²
- B. d²sp³
- C. dsp²
- D. sp³
Answer: B. d²sp³ — Fe³⁺ is d⁵. CN⁻ is a strong field ligand, forcing electrons to pair up. In a low spin d⁵ octahedral complex, the inner (n-1)d orbitals are used for hybridization, leading to d²sp³ hybridization.
MCQ 19. Which of the following complexes can exist as cis and trans isomers?
- A. [Ma₄b₂]
- B. [Ma₃b₃]
- C. [Ma₅b]
- D. [Ma₆]
Answer: A. [Ma₄b₂] — Geometric isomerism (cis-trans isomerism) is possible in octahedral complexes of the type [Ma₄b₂]. The two 'b' ligands can be adjacent (cis) or opposite (trans) to each other.
MCQ 20. The IUPAC name of K₃[Fe(C₂O₄)₃] is:
- A. Potassium trioxalateferrate(III)
- B. Tripotassium trioxalatoiron(III)
- C. Potassium trioxalatoferrate(III)
- D. Potassium trioxalateiron(III)
Answer: C. Potassium trioxalatoferrate(III) — The counter ion is potassium. Oxalate (C₂O₄²⁻) is an anionic ligand, named 'oxalato'. Since the complex ion is anionic, the metal name ends in '-ate'. The oxidation state of Fe is +3.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.