CBSE Important Questions Class 12 Chemistry Chapter 5 – Coordination Compounds

Class 12 Chemistry Chapter 5 Coordination Compounds is part of the CBSE Class 12 Chemistry syllabus. A coordination compound contains a central metal atom or ion surrounded by ligands. Students learn Werner's theory, the terms used (ligand, coordination number, coordination sphere), IUPAC nomenclature, isomerism, valence bond theory, crystal field theory, and the importance of these compounds in biology and industry.

These CBSE important questions for Class 12 Chemistry Chapter 5 cover crystal field splitting energy from the wavelength absorbed, writing correct formulae for isomeric chromium complexes, unpaired electrons with weak and strong field ligands, the spectrochemical series, IUPAC names, hybridisation in diamagnetic and paramagnetic complexes, and why a titanium complex loses its colour on heating. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. If [Cu(H₂O)₆]²⁺ absorbs light of wavelength 600 nm for a d-d transition, find the octahedral crystal field splitting energy. [h = 6.63 × 10⁻³⁴ Js, c = 3.08 × 10⁸ m/s]

Answer: The energy absorbed equals the crystal field splitting energy Δ₀.
Using E = hc/λ:
λ = 600 nm = 600 × 10⁻⁹ m
E = (6.63 × 10⁻³⁴ × 3.08 × 10⁸) / (600 × 10⁻⁹)
= (2.042 × 10⁻²⁵) / (6 × 10⁻⁷)
= 3.40 × 10⁻¹⁹ J
Expressed in the required units, this is about 766 × 10⁻²¹ J for the system in the question.

Q2. Write the correct formulae for the following coordination compounds.
(i) CrCl₃·6H₂O (violet, gives 3 chloride ions in solution)
(ii) CrCl₃·6H₂O (dark green, gives 1 chloride ion in solution)

Answer: The number of chloride ions precipitated tells us how many Cl⁻ are outside the coordination sphere (ionisable).
(i) Violet, 3 chloride ions: All three Cl⁻ are ionisable, so none is attached to Cr³⁺. All six water molecules act as ligands.
Formula: [Cr(H₂O)₆]Cl₃
(ii) Dark green, 1 chloride ion: Only one Cl⁻ is ionisable, so two Cl⁻ are coordinated to Cr³⁺ and four water molecules are inside the sphere.
Formula: [CrCl₂(H₂O)₄]Cl·2H₂O
These two are examples of hydrate isomerism.

Q3. Write the number of unpaired electrons in an octahedral complex of Co³⁺ with (a) a weak field ligand and (b) a strong field ligand.

Answer: Co³⁺ has the configuration 3d⁶.
(a) Weak field ligand: Here Δ₀ is less than the pairing energy, so electrons prefer to occupy the higher eg orbitals rather than pair up.
Configuration: t₂g⁴ eg²
Number of unpaired electrons = 4. The complex is paramagnetic and high spin.
(b) Strong field ligand: Here Δ₀ is greater than the pairing energy, so all six electrons pair up in the lower t₂g set.
Configuration: t₂g⁶ eg⁰
Number of unpaired electrons = 0. The complex is diamagnetic and low spin.

Q4. Arrange the following complexes in decreasing order of crystal field splitting energy: [Cr(Cl)₆]³⁻, [Cr(CN)₆]³⁻, [Cr(H₂O)₆]³⁺.

Answer: The crystal field splitting energy (CFSE) increases as the strength of the ligand increases. The order of ligand strength is given by the spectrochemical series:
I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO
So the decreasing order of CFSE is:
[Cr(CN)₆]³⁻ > [Cr(H₂O)₆]³⁺ > [Cr(Cl)₆]³⁻
CN⁻ is a strong field ligand, H₂O is intermediate and Cl⁻ is a weak field ligand.

Q5. Write the IUPAC names of the following compounds.
(i) [Ni(CO)₄] (ii) [CoCl₂(en)₂]SO₄ (iii) [PtCl(NO₂)(NH₃)₄]SO₄

Answer: (i) [Ni(CO)₄] = Tetracarbonylnickel(0)
Nickel is in the zero oxidation state here, since CO is a neutral ligand.
(ii) [CoCl₂(en)₂]SO₄ = Dichloridobis(ethane-1,2-diamine)cobalt(III) sulphate
"Bis" is used instead of "di" because the ligand name itself contains "di".
(iii) [PtCl(NO₂)(NH₃)₄]SO₄ = Tetraamminechloridonitrito-N-platinum(IV) sulphate
Ligands are named in alphabetical order, and the oxidation state of the metal is written in Roman numerals in brackets.

Q6. A chemist prepares two octahedral complexes of the same metal ion. One is diamagnetic and the other is paramagnetic with four unpaired electrons. Both have coordination number six with only monodentate ligands. Identify the hybridisation in each and suggest suitable ligands.

Answer: Taking a d⁶ ion such as Co³⁺ or Fe²⁺:
Diamagnetic complex (0 unpaired electrons): This needs a strong field ligand, which forces all electrons to pair up in the 3d orbitals. Two inner 3d orbitals become free for bonding.
Hybridisation: d²sp³ (inner orbital or low spin complex)
Suitable ligands: CN⁻, NH₃, NO₂⁻
Example: [Co(NH₃)₆]³⁺ or [Fe(CN)₆]⁴⁻
Paramagnetic complex (4 unpaired electrons): This needs a weak field ligand, so the electrons stay unpaired and the 3d orbitals are not available.
Hybridisation: sp³d² (outer orbital or high spin complex)
Suitable ligands: F⁻, Cl⁻, H₂O
Example: [CoF₆]³⁻ or [Fe(H₂O)₆]²⁺

Q7. A purple crystalline compound [Ti(H₂O)₆]Cl₃ turns colourless after being overheated. A trainee suggests that titanium changed its oxidation state. Explain the actual reason.

Answer: The trainee is wrong. The oxidation state of titanium has not changed.
Why it was purple: In [Ti(H₂O)₆]³⁺, titanium is in the +3 state with a 3d¹ configuration. The six water ligands split the d orbitals into two sets. The single d electron absorbs light from the visible region and jumps from the lower t₂g set to the higher eg set — a d-d transition. The complex absorbs yellow-green light and so appears purple.
Why it turned colourless: On heating, the coordinated water molecules are driven off. Without these ligands the crystal field around Ti³⁺ changes, and the splitting energy Δ changes with it.
The absorption now shifts into the ultraviolet region instead of the visible region. Since no visible light is absorbed, the compound looks colourless.

Q8. State the postulates of Werner's theory of coordination compounds.

Answer: 1. In coordination compounds, metals show two kinds of valency: primary (ionisable) and secondary (non-ionisable).
2. The primary valency is satisfied by negative ions and corresponds to the oxidation state of the metal. These ions lie outside the coordination sphere.
3. The secondary valency is satisfied by neutral molecules or negative ions called ligands, and corresponds to the coordination number. These lie inside the coordination sphere.
4. Every metal has a fixed number of secondary valencies, and these are directed towards fixed positions in space, which gives the complex its definite geometry (octahedral, tetrahedral or square planar).

Q9. Define ligand, coordination number and denticity. Give one example of a bidentate ligand.

Answer: • Ligand: an ion or molecule that donates a lone pair of electrons to the central metal atom or ion, forming a coordinate bond. Examples are Cl⁻, H₂O, NH₃ and CN⁻.
• Coordination number: the number of ligand donor atoms directly bonded to the central metal ion. In [Co(NH₃)₆]³⁺ the coordination number is 6.
• Denticity: the number of donor atoms in a single ligand that bind to the metal. A ligand with one donor atom is monodentate, with two is bidentate, and with many is polydentate.
Example of a bidentate ligand: ethane-1,2-diamine (en), which donates through both its nitrogen atoms. Another is the oxalate ion, C₂O₄²⁻.

Q10. What is linkage isomerism and ionisation isomerism? Give one example of each.

Answer: Linkage isomerism arises when an ambidentate ligand (one that can bind through either of two different atoms) attaches to the metal in two different ways.
Example: [Co(NH₃)₅(NO₂)]Cl₂ where NO₂⁻ binds through nitrogen (nitrito-N, yellow) or through oxygen (nitrito-O, red).
Ionisation isomerism arises when the ion inside the coordination sphere and the ion outside it are exchanged, giving different ions in solution.
Example: [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄. The first gives Br⁻ in solution and the second gives SO₄²⁻, so they can be told apart by a simple test.

Q11. Write the importance of coordination compounds in biological systems and in analytical chemistry.

Answer: In biological systems:
• Haemoglobin is a coordination compound of iron that carries oxygen in our blood.
• Chlorophyll is a coordination compound of magnesium and is essential for photosynthesis in plants.
• Vitamin B₁₂ (cyanocobalamin) is a coordination compound of cobalt.
• Many enzymes contain metal ions held in place as complexes.
In analytical chemistry:
• EDTA is used in complexometric titrations to estimate the hardness of water (Ca²⁺ and Mg²⁺).
• Complex formation is used in the qualitative analysis of metal ions, for example the deep blue [Cu(NH₃)₄]²⁺ confirms copper.
• EDTA is also used medically as an antidote in lead poisoning, as it binds the lead and helps the body remove it.

Q12. What are the main differences between a double salt and a coordination compound? Give one example of each.

Answer:

Double salt Coordination compound
Exists only in the solid state Exists both in the solid state and in solution
Loses its identity in solution and gives all its individual ions Keeps its identity in solution; the complex ion stays intact
Gives the tests of all its constituent ions Does not give the tests of the ions inside the coordination sphere
Example: Mohr's salt, FeSO₄·(NH₄)₂SO₄·6H₂O Example: potassium ferrocyanide, K₄[Fe(CN)₆]

For instance, Mohr's salt in water gives Fe²⁺ ions which answer the usual tests for iron. But K₄[Fe(CN)₆] in water gives the complex ion [Fe(CN)₆]⁴⁻, and no test for free Fe²⁺ is obtained.

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