Class 12 Chemistry Notes
Chapter 5 – Coordination Compounds
Question Bank (Part 1)
A. Multiple Choice Questions (MCQs)
1. The species present inside the square brackets in a coordination compound is called:
A. Counter ion
B. Ligand
C. Coordination sphere
D. Ion pair
Answer: C
2. The coordination number of cobalt in [Co(NH₃)₆]³⁺ is:
A. 3
B. 4
C. 6
D. 9
Answer: C
3. Which of the following is a bidentate ligand?
A. NH₃
B. H₂O
C. en
D. Cl⁻
Answer: C
4. EDTA acts as:
A. Monodentate ligand
B. Bidentate ligand
C. Hexadentate ligand
D. Ambidentate ligand
Answer: C
5. The oxidation state of Fe in K₄[Fe(CN)₆] is:
A. +1
B. +2
C. +3
D. +4
Answer: B
6. Which ligand is an ambidentate ligand?
A. NH₃
B. H₂O
C. NO₂⁻
D. en
Answer: C
7. In naming coordination compounds, ligands are arranged:
A. According to charge
B. According to size
C. Alphabetically
D. According to atomic number
Answer: C
8. The oxidation state of cobalt in [Co(NH₃)₆]Cl₃ is:
A. +1
B. +2
C. +3
D. 0
Answer: C
9. The suffix used for metals in anionic complexes is:
A. -ium
B. -ate
C. -ide
D. -ous
Answer: B
10. Which complex shows geometrical isomerism?
A. [Co(NH₃)₆]³⁺
B. [Pt(NH₃)₂Cl₂]
C. [Zn(NH₃)₄]²⁺
D. [Ni(CO)₄]
Answer: B
B. Concept-Based MCQs
11. Primary valency of a metal ion represents:
A. Coordination number
B. Oxidation state
C. Number of ligands
D. Geometry
Answer: B
12. Secondary valency in Werner theory represents:
A. Oxidation number
B. Coordination number
C. Charge
D. Atomic number
Answer: B
13. Which complex is expected to be diamagnetic?
A. [NiCl₄]²⁻
B. [CoF₆]³⁻
C. [Ni(CN)₄]²⁻
D. [FeF₆]³⁻
Answer: C
14. Strong field ligand among the following is:
A. F⁻
B. Cl⁻
C. CN⁻
D. Br⁻
Answer: C
15. Crystal field splitting in an octahedral complex is represented by:
A. Δt
B. Δ₀
C. P
D. λ
Answer: B
16. In an octahedral crystal field, the lower energy orbitals are:
A. eg
B. t₂g
C. dz²
D. dx²-y²
Answer: B
17. A low spin complex is formed when:
A. Weak field ligand is present
B. Strong field ligand is present
C. No ligand is present
D. Metal is absent
Answer: B
18. Magnetic moment depends on:
A. Number of ligands
B. Number of unpaired electrons
C. Coordination number only
D. Atomic mass
Answer: B
19. The geometry of Ni(CO)₄ is:
A. Square planar
B. Octahedral
C. Tetrahedral
D. Linear
Answer: C
20. CO acts as:
A. Weak field ligand
B. Strong field ligand
C. Neutral atom only
D. Counter ion
Answer: B
C. Fill in the Blanks
1. The atom or ion to which ligands are attached is called the ______.
Answer: Central metal atom/ion
2. Ligands donate ______ pair of electrons to metal ions.
Answer: Lone
3. Coordination number of [Co(en)₃]³⁺ is ______.
Answer: 6
4. EDTA is a ______ ligand.
Answer: Hexadentate
5. Complex compounds were explained by ______ theory.
Answer: Werner’s theory
6. NH₃ ligand is called ______.
Answer: Ammine
7. H₂O ligand is called ______.
Answer: Aqua
8. The magnetic moment formula is ______.
Answer: √n(n+2) BM
9. Strong field ligands produce ______ spin complexes.
Answer: Low
10. Colour in coordination compounds is mainly due to ______ transition.
Answer: d–d
D. True or False
1. Secondary valency is ionisable.
False
2. EDTA can form six coordinate bonds.
True
3. [Ni(CN)₄]²⁻ is tetrahedral.
False
4. CO is a strong field ligand.
True
5. Weak field ligands produce low-spin complexes.
False
6. Optical isomers are mirror images.
True
7. Coordination compounds have no biological importance.
False
8. d⁰ and d¹⁰ complexes are generally colourless.
True
E. Match the Following
| Column A | Column B |
|---|---|
| NH₃ | Ammine |
| H₂O | Aqua |
| CN⁻ | Cyanido |
| CO | Carbonyl |
| Cl⁻ | Chlorido |
Answer:
- NH₃ → Ammine
- H₂O → Aqua
- CN⁻ → Cyanido
- CO → Carbonyl
- Cl⁻ → Chlorido
Question Bank (Part 2)
Nomenclature + Isomerism (Selected Important Questions)
A. Multiple Choice Questions (MCQs)
1. The correct IUPAC name of K₄[Fe(CN)₆] is:
A. Potassium ferrocyanide
B. Potassium hexacyanidoferrate(II)
C. Potassium hexacyanidoferrate(III)
D. Potassium iron cyanide
Answer: B
2. The correct name of [Co(NH₃)₆]Cl₃ is:
A. Hexaamminecobalt(III) chloride
B. Hexammine cobalt chloride
C. Cobalt ammine chloride
D. Cobalt(III) chloride ammine
Answer: A
3. In the name “tetraammine dichloridocobalt(III) chloride”, the oxidation state of cobalt is:
A. +1
B. +2
C. +3
D. +4
Answer: C
4. The IUPAC name of [Pt(NH₃)₂Cl₂] is:
A. Diamminedichloridoplatinum(II)
B. Dichlorodiammineplatinum(II)
C. Platinum diamine chloride
D. Diammineplatinum chloride
Answer: A
5. Which complex has an anionic coordination entity?
A. [Co(NH₃)₆]Cl₃
B. K₄[Fe(CN)₆]
C. [Cu(NH₃)₄]SO₄
D. [Ag(NH₃)₂]Cl
Answer: B
6. In anionic complexes, the metal name ends with:
A. -ide
B. -ous
C. -ate
D. -ium
Answer: C
7. The ligand name for Cl⁻ in IUPAC nomenclature is:
A. Chlorine
B. Chloro
C. Chlorido
D. Chlorate
Answer: C
8. Which complex can show cis-trans isomerism?
A. [Pt(NH₃)₂Cl₂]
B. [Co(NH₃)₆]³⁺
C. [Ni(CO)₄]
D. [Zn(NH₃)₄]²⁺
Answer: A
9. The type of isomerism shown by [Co(NH₃)₅(NO₂)]²⁺ and [Co(NH₃)₅(ONO)]²⁺ is:
A. Optical isomerism
B. Linkage isomerism
C. Coordination isomerism
D. Ionisation isomerism
Answer: B
10. Which pair represents ionisation isomers?
A. [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄
B. cis and trans forms
C. d and l forms
D. fac and mer forms
Answer: A
B. Assertion–Reason Questions
Choose:
A. Both Assertion and Reason are true and Reason correctly explains Assertion.
B. Both are true but Reason is not the correct explanation.
C. Assertion is true but Reason is false.
D. Assertion is false but Reason is true.
1.
Assertion: [Pt(NH₃)₂Cl₂] shows geometrical isomerism.
Reason: Square planar complexes can have different ligand arrangements.
Answer: A
2.
Assertion: EDTA forms stable complexes.
Reason: EDTA is a hexadentate ligand.
Answer: A
3.
Assertion: Tetrahedral complexes generally do not show geometrical isomerism.
Reason: All positions in a tetrahedral structure are equivalent.
Answer: A
4.
Assertion: Linkage isomerism is shown by NO₂⁻ ligand.
Reason: NO₂⁻ can attach through nitrogen or oxygen.
Answer: A
5.
Assertion: Optical isomers are non-superimposable mirror images.
Reason: They rotate plane polarised light in opposite directions.
Answer: A
C. Fill in the Blanks
1. The arrangement of ligands around a metal ion is called ______.
Answer: Coordination geometry
2. The prefix used for six ligands is ______.
Answer: Hexa
3. NH₃ is named as ______ in coordination compounds.
Answer: Ammine
4. CO ligand is called ______.
Answer: Carbonyl
5. Complexes having the same formula but different spatial arrangements show ______ isomerism.
Answer: Stereoisomerism
6. Cis-trans isomerism is a type of ______ isomerism.
Answer: Geometrical
7. Mirror image isomers are called ______.
Answer: Enantiomers
8. Ligands capable of binding through two different atoms show ______ isomerism.
Answer: Linkage
9. EDTA forms ring structures known as ______.
Answer: Chelates
10. fac-mer isomerism is observed in ______ complexes.
Answer: Octahedral
D. True or False
1. Ligands are named after the metal in IUPAC nomenclature.
False
2. Oxidation state of metal is written in Roman numerals.
True
3. Optical isomers have identical physical properties.
False
4. Ambidentate ligands can attach through different donor atoms.
True
5. [Co(NH₃)₆]³⁺ shows geometrical isomerism.
False
6. Square planar complexes can show cis-trans isomerism.
True
7. Coordination isomerism occurs due to exchange of ligands between complex ions.
True
8. Hydrate isomerism involves different positions of water molecules.
True
E. Short Answer Questions
1. What is linkage isomerism?
Answer:
Linkage isomerism occurs when an ambidentate ligand attaches to the metal through different donor atoms.
Example:
- M–NO₂
- M–ONO
2. What is optical isomerism?
Answer:
Optical isomerism occurs when coordination compounds exist as non-superimposable mirror images called enantiomers.
3. Why does [Pt(NH₃)₂Cl₂] show cis-trans isomerism?
Answer:
Because the square planar structure allows ligands to occupy adjacent (cis) or opposite (trans) positions.
4. Differentiate between homoleptic and heteroleptic complexes.
Answer:
Homoleptic: Contains only one type of ligand.
Example: [Co(NH₃)₆]³⁺
Heteroleptic: Contains more than one type of ligand.
Example: [Co(NH₃)₄Cl₂]⁺
5. What is chelation?
Answer:
The formation of ring structures when a multidentate ligand attaches to a metal ion is called chelation.
F. Important Exam-Based Questions
1. Explain types of structural isomerism in coordination compounds.
Answer Points:
Structural isomerism includes:
- Ionisation isomerism
- Exchange between ligand and counter ion.
- Hydrate isomerism
- Different positions of water molecules.
- Linkage isomerism
- Different attachment through ambidentate ligands.
- Coordination isomerism
- Exchange of ligands between complex ions.
2. Explain geometrical isomerism with examples.
Answer Points:
- Different arrangement of ligands around metal.
- Cis form: Similar ligands adjacent.
- Trans form: Similar ligands opposite.
Example:
[Pt(NH₃)₂Cl₂]
3. What are fac and mer isomers?
Answer:
Found in octahedral complexes of type MA₃B₃.
- Fac: Three similar ligands occupy one face.
- Mer: Three similar ligands lie in the same plane.
Question Bank (Part 3)
Valence Bond Theory (VBT), Hybridisation, Magnetic Properties & Numericals
A. Multiple Choice Questions (MCQs)
1. According to Valence Bond Theory, coordination compounds are formed due to:
A. Transfer of electrons
B. Overlap of orbitals containing lone pairs
C. Nuclear reactions
D. Ionic interaction only
Answer: B
2. The hybridisation of [Ni(CN)₄]²⁻ is:
A. sp³
B. dsp²
C. d²sp³
D. sp³d²
Answer: B
3. The geometry of [Ni(CN)₄]²⁻ is:
A. Tetrahedral
B. Square planar
C. Octahedral
D. Linear
Answer: B
4. The hybridisation present in [NiCl₄]²⁻ is:
A. dsp²
B. sp³
C. d²sp³
D. sp³d²
Answer: B
5. [NiCl₄]²⁻ is:
A. Diamagnetic
B. Paramagnetic
C. Colourless
D. Non-coordination compound
Answer: B
6. Which ligand causes maximum electron pairing?
A. F⁻
B. Cl⁻
C. CN⁻
D. Br⁻
Answer: C
7. The hybridisation of [Co(NH₃)₆]³⁺ is:
A. sp³
B. sp³d²
C. d²sp³
D. dsp²
Answer: C
8. The hybridisation of [CoF₆]³⁻ is:
A. d²sp³
B. sp³d²
C. dsp²
D. sp³
Answer: B
9. An inner orbital complex contains:
A. Outer d-orbitals only
B. Inner d-orbitals in hybridisation
C. No d-orbitals
D. Only p-orbitals
Answer: B
10. A high-spin complex generally contains:
A. No unpaired electrons
B. Maximum possible unpaired electrons
C. Only paired electrons
D. No metal ion
Answer: B
B. Concept-Based MCQs
11. Which complex is diamagnetic?
A. [CoF₆]³⁻
B. [NiCl₄]²⁻
C. [Ni(CN)₄]²⁻
D. [FeF₆]³⁻
Answer: C
12. Strong field ligands generally produce:
A. High-spin complexes
B. Low-spin complexes
C. No splitting
D. Tetrahedral complexes only
Answer: B
13. Weak field ligands generally:
A. Cause electron pairing
B. Increase pairing energy only
C. Do not cause electron pairing
D. Remove electrons from metal
Answer: C
14. The shape of Ni(CO)₄ is:
A. Square planar
B. Linear
C. Tetrahedral
D. Octahedral
Answer: C
15. The magnetic moment formula is:
A. √n(n+2) BM
B. n² BM
C. n+2 BM
D. √n BM
Answer: A
16. A complex with all electrons paired is:
A. Paramagnetic
B. Diamagnetic
C. Ferromagnetic
D. Magnetic only
Answer: B
17. Which complex has sp³d² hybridisation?
A. [CoF₆]³⁻
B. [Ni(CN)₄]²⁻
C. [Co(NH₃)₆]³⁺
D. [NiCl₄]²⁻
Answer: A
18. Square planar complexes are commonly formed by:
A. d⁸ metal ions
B. d¹ ions
C. s-block ions
D. Noble gases
Answer: A
19. Inner orbital complexes are usually formed by:
A. Weak field ligands
B. Strong field ligands
C. No ligands
D. Halogens only
Answer: B
20. The number of unpaired electrons in a complex determines its:
A. Colour only
B. Magnetic behaviour
C. Molecular mass
D. Oxidation state
Answer: B
C. Fill in the Blanks
1. VBT was used to explain bonding in ______ compounds.
Answer: Coordination
2. sp³ hybridisation gives ______ geometry.
Answer: Tetrahedral
3. dsp² hybridisation gives ______ geometry.
Answer: Square planar
4. d²sp³ hybridisation forms ______ orbital complexes.
Answer: Inner
5. sp³d² hybridisation forms ______ orbital complexes.
Answer: Outer
6. Strong field ligands cause ______ of electrons.
Answer: Pairing
7. Complexes having unpaired electrons are called ______.
Answer: Paramagnetic
8. Complexes having all paired electrons are called ______.
Answer: Diamagnetic
9. CN⁻ is a ______ field ligand.
Answer: Strong
10. Cl⁻ is a ______ field ligand.
Answer: Weak
D. True or False
1. [Ni(CN)₄]²⁻ is paramagnetic.
False
2. [NiCl₄]²⁻ contains unpaired electrons.
True
3. NH₃ is generally a stronger ligand than F⁻.
True
4. High-spin complexes have fewer unpaired electrons.
False
5. Low-spin complexes are produced by strong field ligands.
True
6. Tetrahedral complexes usually show strong crystal field splitting.
False
7. Magnetic moment depends on unpaired electrons.
True
8. VBT can explain the colour of coordination compounds completely.
False
E. Numerical Type Questions
1. Calculate magnetic moment of a complex containing 3 unpaired electrons.
Formula:μ=n(n+2)
n = 3μ=3(3+2) μ=15
Answer: 3.87 BM
2. Calculate magnetic moment when no unpaired electrons are present.
n = 0μ=0(0+2)
Answer: 0 BM (Diamagnetic)
3. A complex has two unpaired electrons. Find magnetic moment.
n = 2μ=2(2+2) μ=8
Answer: 2.83 BM
F. Short Answer Questions
1. Why is [Ni(CN)₄]²⁻ square planar?
Answer:
CN⁻ is a strong field ligand. It causes pairing of electrons and allows dsp² hybridisation, producing a square planar structure.
2. Why is [NiCl₄]²⁻ tetrahedral?
Answer:
Cl⁻ is a weak field ligand. It does not cause electron pairing, so sp³ hybridisation occurs and the structure becomes tetrahedral.
3. Differentiate between high-spin and low-spin complexes.
| High Spin | Low Spin |
|---|---|
| Weak field ligand | Strong field ligand |
| More unpaired electrons | Fewer unpaired electrons |
| Paramagnetic | Less magnetic/diamagnetic |
4. What are inner orbital complexes?
Answer:
Complexes in which inner (n−1)d orbitals participate in hybridisation are called inner orbital complexes.
Example: [Co(NH₃)₆]³⁺
5. What are outer orbital complexes?
Answer:
Complexes in which outer d orbitals participate in hybridisation are called outer orbital complexes.
Example: [CoF₆]³⁻
G. Important Board Questions
1. Explain the bonding in [Co(NH₃)₆]³⁺ using VBT.
Answer Points:
- Co³⁺ has d⁶ configuration.
- NH₃ is a strong field ligand.
- Electrons pair in d-orbitals.
- Inner d-orbitals participate.
- Hybridisation = d²sp³.
- Geometry = octahedral.
- Complex is diamagnetic.
2. Explain the bonding in [CoF₆]³⁻ using VBT.
Answer Points:
- F⁻ is a weak field ligand.
- Pairing does not occur.
- Outer d-orbitals are used.
- Hybridisation = sp³d².
- Geometry = octahedral.
- Complex is paramagnetic.
Question Bank (Part 4)
Crystal Field Theory (CFT), Spectrochemical Series, Colour & Stability
A. Multiple Choice Questions (MCQs)
1. Crystal Field Theory explains:
A. Only bonding
B. Colour and magnetic properties of complexes
C. Only nomenclature
D. Molecular mass
Answer: B
2. In an octahedral complex, the d-orbitals split into:
A. Two groups
B. Three groups
C. Four groups
D. Five groups
Answer: A
3. In an octahedral field, the lower energy orbitals are:
A. eg
B. t₂g
C. dz²
D. dx²−y²
Answer: B
4. In an octahedral complex, the higher energy orbitals are:
A. t₂g
B. eg
C. dxy
D. dxz
Answer: B
5. The crystal field splitting energy for octahedral complexes is represented by:
A. Δt
B. Δ₀
C. P
D. E
Answer: B
6. The ligands causing maximum splitting are:
A. Weak field ligands
B. Strong field ligands
C. Neutral ligands only
D. Negative ligands only
Answer: B
7. Which of the following is a strong field ligand?
A. F⁻
B. Cl⁻
C. CN⁻
D. Br⁻
Answer: C
8. Which of the following is a weak field ligand?
A. CO
B. CN⁻
C. NH₃
D. F⁻
Answer: D
9. The correct spectrochemical series is:
A. CN⁻ < CO < I⁻
B. I⁻ < Br⁻ < Cl⁻ < F⁻ < CN⁻ < CO
C. CO < CN⁻ < F⁻
D. NH₃ < H₂O < I⁻
Answer: B
10. A strong field ligand generally forms:
A. High spin complexes
B. Low spin complexes
C. Tetrahedral complexes only
D. Colourless complexes only
Answer: B
B. Concept-Based MCQs
11. High-spin complexes are formed when:
A. Δ₀ > P
B. Δ₀ < P
C. Δ₀ = 0
D. No splitting occurs
Answer: B
12. Low-spin complexes contain:
A. Maximum unpaired electrons
B. More paired electrons
C. No electrons
D. Only s-electrons
Answer: B
13. The colour of coordination compounds is mainly due to:
A. Nuclear transition
B. d–d transition
C. Proton transfer
D. Bond breaking
Answer: B
14. A complex with d⁰ configuration is usually:
A. Coloured
B. Colourless
C. Magnetic only
D. Unstable always
Answer: B
15. A complex with d¹⁰ configuration is generally:
A. Colourless
B. Highly coloured
C. Always paramagnetic
D. Always unstable
Answer: A
16. Absorption of visible light in complexes causes:
A. Electron movement between split d-orbitals
B. Loss of ligand
C. Metal ion removal
D. Ionisation
Answer: A
17. The value of Δ₀ increases when:
A. Ligand field strength increases
B. Ligand field strength decreases
C. Metal ion disappears
D. Coordination number becomes zero
Answer: A
18. The colour observed in a complex is generally:
A. The absorbed colour
B. Complementary to absorbed colour
C. Always violet
D. Always red
Answer: B
19. CFT assumes interaction between metal and ligand is mainly:
A. Covalent
B. Electrostatic
C. Nuclear
D. Metallic
Answer: B
20. Which theory explains the splitting of d-orbitals?
A. VBT
B. CFT
C. Werner theory only
D. Lewis theory only
Answer: B
C. Assertion–Reason Questions
Choose:
A. Both Assertion and Reason are true and Reason correctly explains Assertion.
B. Both are true but Reason is not the correct explanation.
C. Assertion is true but Reason is false.
D. Assertion is false but Reason is true.
1.
Assertion: Coordination compounds are often coloured.
Reason: They absorb certain wavelengths due to d–d transitions.
Answer: A
2.
Assertion: CN⁻ forms low-spin complexes.
Reason: CN⁻ is a strong field ligand.
Answer: A
3.
Assertion: F⁻ forms high-spin complexes.
Reason: F⁻ is a weak field ligand.
Answer: A
4.
Assertion: d⁰ complexes are colourless.
Reason: No d–d transition is possible.
Answer: A
5.
Assertion: CFT completely explains all bonding in coordination compounds.
Reason: It considers only electrostatic interactions.
Answer: D
D. Fill in the Blanks
1. Crystal field splitting in octahedral complexes is represented by ______.
Answer: Δ₀
2. The lower energy set of orbitals in octahedral complexes is ______.
Answer: t₂g
3. The higher energy set of orbitals is ______.
Answer: eg
4. Strong field ligands produce ______ spin complexes.
Answer: Low
5. Weak field ligands produce ______ spin complexes.
Answer: High
6. Colour in complexes is due to ______ transition.
Answer: d–d
7. The arrangement of ligands according to splitting ability is called the ______.
Answer: Spectrochemical series
8. CO is a ______ field ligand.
Answer: Strong
9. F⁻ is a ______ field ligand.
Answer: Weak
10. Complexes having d¹⁰ configuration are generally ______.
Answer: Colourless
E. True or False
1. CFT explains the colour of coordination compounds.
True
2. t₂g orbitals have higher energy than eg orbitals in octahedral complexes.
False
3. CN⁻ produces large crystal field splitting.
True
4. Weak field ligands produce low-spin complexes.
False
5. d–d transition requires partially filled d-orbitals.
True
6. CFT assumes metal-ligand bonding is purely covalent.
False
7. The spectrochemical series arranges ligands according to field strength.
True
8. High-spin complexes contain fewer unpaired electrons.
False
F. Short Answer Questions
1. What is crystal field splitting?
Answer:
The separation of degenerate d-orbitals into groups of different energies due to the approach of ligands is called crystal field splitting.
2. Why are coordination compounds coloured?
Answer:
Ligands split the d-orbitals. Electrons absorb visible light and move between these split levels, producing colour.
3. What is the difference between high-spin and low-spin complexes?
Answer:
| High Spin | Low Spin |
|---|---|
| Weak field ligands | Strong field ligands |
| More unpaired electrons | More paired electrons |
| Paramagnetic | Less magnetic/diamagnetic |
4. Write the spectrochemical series.
Answer:
I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO
5. Why are Zn²⁺ complexes generally colourless?
Answer:
Zn²⁺ has a d¹⁰ configuration. Since no d–d transition is possible, the complexes are colourless.
G. Long Answer Questions
1. Explain crystal field splitting in an octahedral complex.
Answer Points:
- Six ligands approach the metal ion along axes.
- d-orbitals pointing directly toward ligands experience more repulsion.
- dz² and dx²−y² become higher energy (eg).
- dxy, dxz, dyz remain lower energy (t₂g).
- The energy difference is Δ₀.
2. Explain the origin of colour in coordination compounds.
Answer Points:
Remaining light gives the observed colour.
d-orbitals split into different energy levels.
Electrons absorb visible light.
Absorption causes d–d transition.
Metal Carbonyls + Applications + Case-Based Questions + Full Chapter Mixed Practice
A. Multiple Choice Questions (MCQs)
Metal Carbonyls
1. In metal carbonyls, the ligand is:
A. NH₃
B. CO
C. Cl⁻
D. CN⁻
Answer: B
2. The general formula of metal carbonyls is:
A. Mx(CO)y
B. MxClᵧ
C. M(NH₃)ₙ
D. MX₂
Answer: A
3. The geometry of Ni(CO)₄ is:
A. Square planar
B. Octahedral
C. Tetrahedral
D. Linear
Answer: C
4. The geometry of Fe(CO)₅ is:
A. Trigonal bipyramidal
B. Octahedral
C. Tetrahedral
D. Square planar
Answer: A
5. CO forms a bond with metals through:
A. Only ionic attraction
B. Sigma donation and pi back bonding
C. Hydrogen bonding
D. Van der Waals force
Answer: B
6. CO is considered a strong field ligand because:
A. It has no lone pair
B. It causes strong d-orbital splitting
C. It removes metal electrons
D. It forms ionic bonds only
Answer: B
B. Applications of Coordination Compounds
7. Haemoglobin contains which metal ion?
A. Mg
B. Fe
C. Co
D. Cu
Answer: B
8. Chlorophyll contains:
A. Iron complex
B. Magnesium complex
C. Cobalt complex
D. Platinum complex
Answer: B
9. Vitamin B₁₂ contains:
A. Nickel
B. Iron
C. Cobalt
D. Zinc
Answer: C
10. Cisplatin is used in:
A. Water purification
B. Cancer treatment
C. Fertiliser production
D. Extraction of gold
Answer: B
11. EDTA is commonly used for:
A. Polymer formation
B. Metal ion estimation
C. Making acids
D. Producing gases
Answer: B
12. Coordination compounds are used in metallurgy for:
A. Making plastics
B. Extraction and purification of metals
C. Making fuels
D. Producing salts only
Answer: B
C. Fill in the Blanks
1. CO acts as a ______ ligand in metal carbonyls.
Answer: Strong field
2. Ni(CO)₄ has ______ geometry.
Answer: Tetrahedral
3. Fe(CO)₅ has ______ geometry.
Answer: Trigonal bipyramidal
4. Haemoglobin contains ______ metal.
Answer: Iron
5. Chlorophyll contains ______ metal.
Answer: Magnesium
6. Vitamin B₁₂ contains ______.
Answer: Cobalt
7. EDTA is a ______ ligand.
Answer: Hexadentate
8. Ring formation by multidentate ligands is called ______.
Answer: Chelation
9. Metal carbonyl bonding involves ______ donation.
Answer: Sigma
10. Electron donation from metal to CO is called ______ bonding.
Answer: Pi back bonding
D. True or False
1. CO is a weak field ligand.
False
2. Ni(CO)₄ is a coordination compound.
True
3. EDTA forms stable complexes because it is multidentate.
True
4. Haemoglobin contains magnesium.
False
5. Chlorophyll contains magnesium.
True
6. Chelate complexes are generally more stable.
True
7. Metal carbonyls contain CO as ligand.
True
8. Cisplatin is a platinum coordination compound.
True
E. Assertion–Reason Questions
Choose:
A. Both Assertion and Reason are true and Reason correctly explains Assertion.
B. Both are true but Reason is not the correct explanation.
C. Assertion is true but Reason is false.
D. Assertion is false but Reason is true.
1.
Assertion: EDTA forms highly stable complexes.
Reason: EDTA forms ring structures with metal ions.
Answer: A
2.
Assertion: CO forms strong bonds with transition metals.
Reason: CO shows sigma donation and pi back bonding.
Answer: A
3.
Assertion: Metal carbonyls are coordination compounds.
Reason: CO acts as a ligand.
Answer: A
4.
Assertion: Vitamin B₁₂ is biologically important.
Reason: It contains a cobalt coordination complex.
Answer: A
F. Case-Based Questions
Case Study 1
A student studies a complex [Ni(CN)₄]²⁻ and observes that it is diamagnetic and square planar.
Questions:
1. Which type of ligand is CN⁻?
A. Weak field
B. Strong field
C. Neutral ligand
D. Ambidentate only
Answer: B
2. The hybridisation of this complex is:
A. sp³
B. dsp²
C. d²sp³
D. sp³d²
Answer: B
3. The geometry is:
Answer: Square planar
Case Study 2
A coordination compound absorbs visible light and shows colour due to electronic transition.
Questions:
1. The phenomenon responsible for colour is:
A. Ionisation
B. d–d transition
C. Ligand loss
D. Oxidation
Answer: B
2. Complexes with d¹⁰ configuration are generally:
Answer: Colourless
3. Which theory explains this phenomenon?
Answer: Crystal Field Theory
G. Full Chapter Mixed MCQs
1. Coordination number of Fe in [Fe(CN)₆]⁴⁻ is:
A. 4
B. 5
C. 6
D. 8
Answer: C
2. The ligand that can attach through two different atoms is:
A. NH₃
B. H₂O
C. NO₂⁻
D. CO
Answer: C
3. A compound with formula [Co(NH₃)₆]Cl₃ gives:
A. [Co(NH₃)₆]³⁺ ions
B. Free NH₃ only
C. Co metal
D. No ions
Answer: A
4. Which is an example of a chelating ligand?
A. Cl⁻
B. NH₃
C. EDTA
D. F⁻
Answer: C
5. The number of donor atoms in EDTA is:
A. 1
B. 2
C. 4
D. 6
Answer: D
6. The complex [Pt(NH₃)₂Cl₂] shows:
A. Optical isomerism only
B. Cis-trans isomerism
C. Linkage isomerism
D. Ionisation isomerism
Answer: B
7. The formula for magnetic moment is:
A. √n(n+2) BM
B. n² BM
C. n+1 BM
D. √n BM
Answer: A
8. A ligand which donates one pair of electrons is:
A. Monodentate
B. Bidentate
C. Polydentate
D. Chelating only
Answer: A
9. Werner theory was proposed by:
A. Lewis
B. Werner
C. Arrhenius
D. Bohr
Answer: B
10. The stability of chelate complexes is due to:
A. Chelate effect
B. Oxidation
C. Ionisation
D. Hydrolysis
Answer: A
H. Important Long Answer Practice Questions
1. Explain the applications of coordination compounds.
Answer Points:
- Haemoglobin → oxygen transport.
- Chlorophyll → photosynthesis.
- Vitamin B₁₂ → cobalt complex.
- Cisplatin → anticancer drug.
- EDTA → analytical chemistry.
- Metallurgy → extraction of metals.
2. Explain bonding in metal carbonyls.
Answer Points:
- CO donates lone pair from carbon to metal.
- Metal donates electrons back to CO through pi back bonding.
- Metal–carbon bond becomes stronger.
3. Explain chelate effect.
Answer Points:
- Multidentate ligands form ring structures.
- Ring formation increases stability.
- EDTA and en show chelation.