Class 12 Chemistry Chapter 5 Important Questions – Coordination Compounds

Coordination compounds contain a central metal atom or ion bonded to surrounding ions or molecules called ligands. These questions cover terminology, nomenclature, isomerism, bonding, magnetic properties, colour and applications.

Class 12 Chemistry Chapter 5 Important Questions help students revise the structure and behaviour of coordination compounds. The chapter explains how ligands bond with central metal ions and affect their shape, colour and magnetic properties.

Students should practise calculating oxidation states and coordination numbers before studying nomenclature. They should also understand how ligand strength affects electron pairing, hybridisation and crystal field splitting.

Key Takeaways

• A coordination entity contains a central metal atom or ion and attached ligands.
• Coordination number counts the ligand donor atoms bonded directly to the metal.
• Ligands may be unidentate, didentate, polydentate or ambidentate.
• Complexes may show structural and stereoisomerism.
• VBT explains geometry and magnetic behaviour qualitatively.
• CFT explains d-orbital splitting, colour and high-spin or low-spin configurations.
• Strong-field ligands can cause electron pairing.
• Weak-field ligands generally produce high-spin complexes.
• Chelating ligands form stable ring structures with metal ions.

Important Terms in Coordination Compounds

Common Ligands and Their Names

Spectrochemical Series

Increasing ligand field strength:

I⁻ < Br⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO

Ligands towards the left generally act as weak-field ligands.

Ligands towards the right generally act as strong-field ligands.

Access Class 12 Chemistry Chapter 5 Important Questions in 30 Minutes

First 10 minutes: Revise coordination terms, oxidation states and IUPAC rules.

Next 10 minutes: Review isomerism, VBT, hybridisation and magnetic behaviour.

Final 10 minutes: Practise CFT, colour, metal carbonyls and applications.

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Important Questions for Class 12 Chemistry Chapter 5

Q1. What is a coordination compound? Explain its main components.

Answer:

A coordination compound contains a central metal atom or ion surrounded by ions or molecules called ligands.

For example:

[Co(NH₃)₆]Cl₃

Its main components are:

• Central metal ion: Co³⁺
• Ligands: Six NH₃ molecules
• Coordination entity: [Co(NH₃)₆]³⁺
• Counter ions: Three Cl⁻ ions
• Coordination number: 6

The metal-ligand unit remains together when the compound dissolves in water.

Q2. State the main postulates of Werner’s theory.

Answer:

Werner proposed the following ideas:

1. A metal shows primary and secondary valences.
2. Primary valences are ionisable.
3. Primary valences are normally satisfied by negative ions.
4. Secondary valences are non-ionisable.
5. Secondary valence equals the coordination number.
6. Secondary valences have definite spatial arrangements.
7. Groups satisfying secondary valences remain inside the coordination sphere.

For example:

[Co(NH₃)₆]Cl₃

The three chloride ions satisfy the primary valence. Six ammonia molecules satisfy the secondary valence.

Q3. Distinguish between a double salt and a coordination compound.

Answer:

Mohr’s salt gives tests for Fe²⁺, NH₄⁺ and SO₄²⁻ ions.

K₄[Fe(CN)₆] does not dissociate into free Fe²⁺ and CN⁻ ions.

Q4. Explain unidentate, didentate, polydentate and ambidentate ligands.

Answer:

Unidentate ligands bond through one donor atom.

Examples:

• NH₃
• H₂O
• Cl⁻

Didentate ligands bond through two donor atoms.

Examples:

• Ethane-1,2-diamine
• Oxalate ion

Polydentate ligands bond through several donor atoms.

Example:

EDTA⁴⁻ is hexadentate because it can bond through six donor atoms.

Ambidentate ligands can bond through either of two different atoms.

Examples:

• NO₂⁻ can bond through N or O.
• SCN⁻ can bond through S or N.

Q5. Find the oxidation state and coordination number of the central metal in the following complexes.

(i) [Co(NH₃)₅Cl]Cl₂
(ii) K₄[Fe(CN)₆]
(iii) [Cr(C₂O₄)₃]³⁻
(iv) [PtCl₄]²⁻

Answer:

(i) [Co(NH₃)₅Cl]Cl₂

Let the oxidation state of Co be x.

x + 5(0) − 1 = +2

x = +3

Coordination number:

5 from NH₃ + 1 from Cl⁻ = 6

(ii) K₄[Fe(CN)₆]

x + 6(−1) = −4

x = +2

Coordination number = 6

(iii) [Cr(C₂O₄)₃]³⁻

x + 3(−2) = −3

x = +3

Each oxalate ligand is didentate.

Coordination number = 3 × 2 = 6

(iv) [PtCl₄]²⁻

x + 4(−1) = −2

x = +2

Coordination number = 4

Q6. Write the IUPAC names of the following coordination compounds.

(i) [Co(NH₃)₆]Cl₃
(ii) K₃[Fe(CN)₆]
(iii) [CoCl₂(en)₂]Cl
(iv) [Ni(CO)₄

Answer:

(i) Hexaamminecobalt(III) chloride

(ii) Potassium hexacyanidoferrate(III)

(iii) Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride

(iv) Tetracarbonylnickel(0)

Q7. Write the formulas of the following coordination compounds.

(i) Tetraamminecopper(II) sulphate
(ii) Potassium tetracyanidonickelate(II)
(iii) Tris(ethane-1,2-diamine)cobalt(III) chloride
(iv) Potassium trioxalatochromate(III)

Answer:

(i) [Cu(NH₃)₄]SO₄

(ii) K₂[Ni(CN)₄]

(iii) [Co(en)₃]Cl₃

(iv) K₃[Cr(C₂O₄)₃]

Q8. What types of isomerism are shown by coordination compounds?

Answer:

Coordination compounds show two main classes of isomerism.

Structural isomerism

• Linkage isomerism
• Coordination isomerism
• Ionisation isomerism
• Solvate isomerism

Stereoisomerism

• Geometrical isomerism
• Optical isomerism

Structural isomers differ in the bonds or groups attached to the metal.

Stereoisomers have the same bonds but different spatial arrangements.

Q9. Explain geometrical isomerism with suitable examples.

Answer:

Geometrical isomerism arises when ligands occupy different relative positions around the central metal ion.

For a square planar complex of the form [MA₂B₂]:

• Similar ligands placed next to each other form the cis isomer.
• Similar ligands placed opposite each other form the trans isomer.

Example:

[Pt(NH₃)₂Cl₂]

It exists as cisplatin and transplatin.

Octahedral complexes may also show cis-trans isomerism.

Example:

[Co(NH₃)₄Cl₂]⁺

Complexes of the type [MA₃B₃] may show facial and meridional isomerism.

Example:

[Co(NH₃)₃(NO₂)₃]

Q10. Explain linkage, ionisation and coordination isomerism.

Answer:

Linkage isomerism

It occurs when an ambidentate ligand bonds through different donor atoms.

Example:

[Co(NH₃)₅(NO₂)]Cl₂

[Co(NH₃)₅(ONO)]Cl₂

The nitrite ligand bonds through nitrogen in one form and oxygen in the other.

Ionisation isomerism

It occurs when a ligand inside the coordination sphere exchanges with a counter ion.

Example:

[Co(NH₃)₅Br]SO₄

[Co(NH₃)₅SO₄]Br

The two compounds produce different ions in solution.

Coordination isomerism

It occurs when ligands are exchanged between cationic and anionic complex ions.

Example:

[Co(NH₃)₆][Cr(CN)₆]

[Cr(NH₃)₆][Co(CN)₆]

Q11. What is optical isomerism? Which form of [CoCl₂(en)₂]⁺ can be optically active?

Answer:

Optical isomers are non-superimposable mirror images called enantiomers.

They rotate plane-polarised light in opposite directions.

The cis form of [CoCl₂(en)₂]⁺ is optically active because it lacks a plane of symmetry.

The trans form is optically inactive because it has a symmetrical arrangement.

Q12. Explain the geometry and magnetic behaviour of [Ni(CN)₄]²⁻.

Answer:

Oxidation state of nickel:

x + 4(−1) = −2

x = +2

Ni²⁺ has the configuration:

3d⁸

CN⁻ is a strong-field ligand. It causes the 3d electrons to pair.

One 3d orbital, one 4s orbital and two 4p orbitals undergo dsp² hybridisation.

Therefore:

• Hybridisation: dsp²
• Shape: Square planar
• Unpaired electrons: 0
• Magnetic behaviour: Diamagnetic

Q13. Explain the geometry and magnetic behaviour of [NiCl₄]²⁻.

Answer:

The oxidation state of nickel is +2.

Ni²⁺ has the configuration:

3d⁸

Cl⁻ is a weak-field ligand and does not cause pairing of the 3d electrons.

One 4s and three 4p orbitals undergo sp³ hybridisation.

Therefore:

• Hybridisation: sp³
• Shape: Tetrahedral
• Unpaired electrons: 2
• Magnetic behaviour: Paramagnetic

Q14. Why is [NiCl₄]²⁻ paramagnetic while [Ni(CO)₄] is diamagnetic, although both are tetrahedral?

Answer:

In [NiCl₄]²⁻, nickel is in the +2 oxidation state.

Ni²⁺ has a 3d⁸ configuration. Chloride is a weak-field ligand and does not pair the two unpaired electrons.

Therefore, [NiCl₄]²⁻ is paramagnetic.

In [Ni(CO)₄], nickel is in the zero oxidation state.

CO is a strong-field ligand. The electrons become paired, producing a 3d¹⁰ arrangement.

Therefore, [Ni(CO)₄] has no unpaired electrons and is diamagnetic.

Both complexes use sp³ hybridisation and have tetrahedral geometry.

Q15. Why is [Co(NH₃)₆]³⁺ an inner-orbital complex while [CoF₆]³⁻ is an outer-orbital complex?

Answer:

In both complexes, cobalt is in the +3 oxidation state.

Co³⁺ has a 3d⁶ configuration.

In [Co(NH₃)₆]³⁺, NH₃ causes pairing of the 3d electrons. Two inner 3d orbitals become available.

The complex uses d²sp³ hybridisation.

Therefore, it is an inner-orbital octahedral complex and is diamagnetic.

In [CoF₆]³⁻, F⁻ is a weak-field ligand and does not cause pairing.

The complex uses the outer 4d orbitals and undergoes sp³d² hybridisation.

Therefore, it is an outer-orbital octahedral complex and is paramagnetic.

Q16. Explain crystal field splitting in an octahedral complex.

Answer:

In a free metal ion, all five d-orbitals have the same energy.

In an octahedral complex, six ligands approach the metal along the coordinate axes.

The dₓ²₋ᵧ² and d𝓏² orbitals point directly towards the ligands. They experience greater repulsion and form the higher-energy e𝓰 set.

The dₓᵧ, dᵧ𝓏 and d𝓏ₓ orbitals lie between the axes. They experience less repulsion and form the lower-energy t₂𝓰 set.

The energy difference between the two sets is called octahedral crystal field splitting energy:

Δ₀

If Δ₀ is smaller than the pairing energy, a high-spin complex forms.

If Δ₀ is greater than the pairing energy, a low-spin complex forms.

Q17. What is the spectrochemical series? Distinguish between weak-field and strong-field ligands.

Answer:

The spectrochemical series arranges ligands according to increasing crystal field splitting strength.

A part of the series is:

I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO

Weak-field ligands

• Produce small crystal field splitting
• Usually do not cause electron pairing
• Generally form high-spin complexes
• Examples: I⁻, Br⁻, Cl⁻ and F⁻

Strong-field ligands

• Produce large crystal field splitting
• Can cause electron pairing
• Generally form low-spin complexes
• Examples: CN⁻ and CO

Q18. Why are many coordination compounds coloured?

Answer:

Ligands split the metal d-orbitals into groups of different energies.

An electron can absorb visible light and move from a lower d-energy level to a higher d-energy level. This is called a d–d transition.

The colour observed is complementary to the colour absorbed.

For example, [Ti(H₂O)₆]³⁺ absorbs light from the blue-green region and appears violet.

The colour can change when the ligand changes because different ligands produce different crystal field splitting energies.

Q19. Explain the bonding in metal carbonyls.

Answer:

The metal-carbon bond in a metal carbonyl contains sigma and pi components.

Sigma bonding

The carbon atom of CO donates a lone pair into an empty metal orbital.

CO → Metal

Pi back-bonding

A filled metal d-orbital donates electron density into an empty antibonding pi orbital of CO.

Metal → CO

These two processes strengthen each other. This is called synergic bonding.

Examples of metal carbonyl structures include:

• [Ni(CO)₄]: Tetrahedral
• [Fe(CO)₅]: Trigonal bipyramidal
• [Cr(CO)₆]: Octahedral

Q20. Explain four applications of coordination compounds.

Answer:

Biological systems

Haemoglobin is an iron coordination compound that transports oxygen.

Chlorophyll contains magnesium and supports photosynthesis.

Vitamin B₁₂ contains cobalt.

Analytical chemistry

EDTA forms complexes with Ca²⁺ and Mg²⁺. It is used to determine water hardness.

Metallurgy

Gold and silver are extracted through soluble cyanide complexes such as [Au(CN)₂]⁻.

Medicine

Cisplatin is used in cancer treatment.

EDTA may be used in the treatment of lead poisoning.

Electroplating

Silver and gold can be deposited evenly from solutions containing [Ag(CN)₂]⁻ and [Au(CN)₂]⁻.

Photography

Undecomposed AgBr is removed using thiosulphate because a soluble complex forms:

[Ag(S₂O₃)₂]³⁻

How to Prepare Coordination Compounds

Begin with common ligand names and rules for finding oxidation states. Practise naming and writing formulas every day because small spelling or charge errors can change the answer.

Next, prepare a chart showing which complexes have tetrahedral, square planar or octahedral geometry. Add their hybridisation and magnetic behaviour.

For CFT questions, first determine the metal oxidation state and d-electron count. Then identify whether the ligand is weak-field or strong-field before filling the split orbitals.

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