ChemCA.in
CBSE Class 12 Chapter Test: COORDINATION COMPOUNDS
General Instructions:
- This question paper consists of 16 questions divided into 5 sections: A, B, C, D, and E.
- Section A comprises 6 MCQs and 2 Assertion-Reasoning questions of 1 mark each.
- Section B comprises 2 Very Short Answer questions of 2 marks each.
- Section C comprises 3 Short Answer questions of 3 marks each.
- Section D comprises 1 Case-Based question of 4 marks.
- Section E comprises 2 Long Answer questions of 5 marks each.
(Atomic number of Co = 27)
Directions for Q7 & Q8: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Reason (R): The nitrite ion (NO2-) is an ambidentate ligand that can bind through either nitrogen or oxygen.
Reason (R): CN- is a weak field ligand, whereas Cl- is a strong field ligand.
(i) Chelate ligand
(ii) Homoleptic complex
(iii) Coordination polyhedron
(Atomic Number of Fe = 26)
Crystal Field Theory (CFT) considers the metal-ligand bond to be entirely ionic arising from electrostatic interactions between the metal ion and the ligands. Ligands are treated as point charges in the case of anions or as point dipoles in the case of neutral molecules. When ligands approach the central metal ion, the five degenerate d-orbitals split into two sets of different energies (t2g and eg). The magnitude of this splitting (Δo) depends on the nature of the ligand. Ligands can be arranged in a series in the order of increasing field strength, known as the spectrochemical series. The distribution of electrons in d-orbitals is heavily influenced by the relationship between Δo and the pairing energy (P).
(a) What is the spectrochemical series? [1 Mark]
(b) Why do the d-orbitals split into two different energy levels in an octahedral field? [1 Mark]
(c) On the basis of CFT, write the electronic configuration for a d4 ion if:
(i) Δo < P
(ii) Δo > P [2 Marks]
(b) Predict the number of unpaired electrons in the square planar complex [Pt(CN)4]2- and explain your answer.
(Atomic Number of Pt = 78) [2 Marks]
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SOLUTIONS & MARKING SCHEME
1. Answer: (c) SCN-
Reasoning: An ambidentate ligand has two different donor atoms but binds to the metal through only one at a time. The thiocyanate ion (SCN-) can bind through the Sulphur atom (forming thiocyanato) or through the Nitrogen atom (forming isothiocyanato).
2. Answer: (b) Diamminechloridonitrito-N-platinum(II)
Reasoning: Ligands are named alphabetically: ammine (NH3), chlorido (Cl), nitrito-N (NO2 attached via N). The oxidation state of Pt is x + 2(0) - 1 - 1 = 0 ⇒ x = +2. Since the complex is neutral, the metal is named 'platinum'.
3. Answer: (c) 0
Reasoning: Carbon monoxide (CO) is a neutral ligand with a charge of 0. Therefore, x + 4(0) = 0 ⇒ x = 0. The oxidation state of nickel is zero.
4. Answer: (c) trans-[Co(en)2Cl2]+
Reasoning: In the trans-isomer, the identical ligands (Cl atoms) are placed opposite to each other (180° apart). This gives the molecule a plane of symmetry, making it superimposable on its mirror image and therefore optically inactive.
5. Answer: (a) CO
Reasoning: According to the spectrochemical series, Carbon monoxide (CO) is the strongest field ligand due to synergistic back-bonding, resulting in the maximum crystal field splitting energy (Δo).
6. Answer: (c) 4.90 BM
Reasoning: In [CoF6]3-, Co is in a +3 oxidation state. Co (Z=27) is [Ar] 3d7 4s2. Co3+ is 3d6. F- is a weak field ligand, so electrons do not pair up (high spin). Configuration is t2g4 eg2, which gives 4 unpaired electrons.
μ = √[n(n+2)] = √[4(6)] = √24 ≈ 4.90 BM.
7. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: Linkage isomerism arises in coordination compounds containing an ambidentate ligand. Since NO2- can coordinate through N (nitrito-N) or through O (nitrito-O), the given compounds are perfectly explained by the reason.
8. Answer: (c) A is true but R is false.
Reasoning: The assertion is true: [Ni(CN)4]2- is dsp2 hybridized and diamagnetic, while [NiCl4]2- is sp3 hybridized and paramagnetic. However, the reason states CN- is a weak field ligand, which is false. CN- is a strong field ligand.
9. Solution:
IUPAC Name: Pentaamminecarbonatocobalt(III) chloride.
Isomerism: If the positions of CO32- and Cl- are interchanged, the formula becomes [Co(NH3)5Cl]CO3. This type of isomerism, where the counter ion in the complex acts as a ligand and vice-versa, is called Ionization Isomerism.
10. Solution:
In [Ni(H2O)6]2+, H2O is a weak field ligand. Ni2+ (3d8) has 2 unpaired electrons. These electrons can absorb red light from the visible spectrum and undergo a d-d transition, emitting the complementary green colour.
In [Ni(CN)4]2-, CN- is a strong field ligand causing the pairing of 3d electrons (dsp2 hybridization). Since there are no unpaired electrons left in the d-orbitals, d-d transitions cannot occur in the visible region, rendering it colourless.
11. Solution:
(i) Chelate Ligand: When a di- or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it forms a ring structure. Such a ligand is called a chelate ligand. Example: Ethylenediamine (en).
(ii) Homoleptic Complex: Complexes in which a metal is bound to only one kind of donor group (ligand). Example: [Co(NH3)6]3+.
(iii) Coordination Polyhedron: The spatial arrangement of the ligand atoms directly attached to the central atom/ion. Example: Octahedral shape for [Co(NH3)6]3+.
12. Solution:
Complex: [Fe(CN)6]3-.
Oxidation state of Fe = +3. Electronic configuration of Fe3+ = [Ar] 3d5.
CN- is a strong field ligand, which causes forced pairing of electrons against Hund's rule.
The 5 electrons pair up in the inner 3d orbitals: ↑↓ | ↑↓ | ↑ | _ | _ .
Two 3d orbitals become empty. Along with one 4s and three 4p orbitals, they undergo d2sp3 hybridization.
Geometry: Octahedral.
Magnetic Behaviour: Since there is 1 unpaired electron left, the complex is Paramagnetic.
13. Solution:
1. Medicine: Cis-platin ([Pt(NH3)2Cl2]) is a coordination compound widely used as an anti-tumor agent in the treatment of cancer.
2. Analytical Chemistry: EDTA is used in the complexometric titration to estimate the hardness of water (due to Ca2+ and Mg2+). EDTA is also used in the treatment of lead poisoning.
3. Metallurgy: Gold and silver are extracted via the Macarthur-Forrest cyanide process, where they form water-soluble coordination complexes like [Au(CN)2]- to separate them from the ore.
14. Solution (Case-Based):
(a) The spectrochemical series is an experimentally determined arrangement of ligands in the increasing order of their crystal field splitting energy (Δo) field strength.
(b) In an octahedral field, ligands approach the central metal ion along the Cartesian axes (x, y, z). The d-orbitals lying along the axes (dx²-y², dz²) experience greater electrostatic repulsion from the ligands and their energy is raised (eg set). The d-orbitals lying between the axes (dxy, dyz, dzx) experience less repulsion and their energy is lowered (t2g set).
(c) (i) If Δo < P: The splitting energy is weak (weak field ligand), so the 4th electron prefers to enter the higher eg orbital rather than pair up. Configuration: t2g3 eg1 (High Spin).
(ii) If Δo > P: The splitting energy is strong (strong field ligand), so the 4th electron prefers to pair up in the lower t2g orbital. Configuration: t2g4 eg0 (Low Spin).
15. Solution:
Geometrical Isomers: [Co(en)2Cl2]+ exists in two geometrical isomeric forms: cis and trans.
- In the cis-isomer, the two Cl- ligands are adjacent to each other (at 90°).
- In the trans-isomer, the two Cl- ligands are opposite to each other (at 180°).
Optical Activity: The cis-isomer is optically active because it lacks any plane or centre of symmetry. Its mirror images are non-superimposable (enantiomers). The trans-isomer is optically inactive due to the presence of a plane of symmetry.
*Note for Students: Draw the octahedral structures showing the 'en' (ethylenediamine) rings bridging adjacent positions for the cis form, and draw its non-superimposable mirror image. For trans, place Cl atoms at top and bottom (axial positions) and the 'en' rings in the equatorial plane.*
16. Solution:
(a) Synergic Bonding in Metal Carbonyls:
The Metal-Carbon bond in metal carbonyls (e.g., [Ni(CO)4]) possesses both σ and π character.
1. σ-bond: Formed by the donation of lone pair of electrons from the carbon atom of CO into an empty orbital of the metal.
2. π-back bond: Formed by the donation of a pair of electrons from a filled d-orbital of the metal into the empty antibonding π* orbital of CO.
The mutual donation strengthens the bond. The σ bond drift increases electron density on the metal, which the metal relieves via the π back-bond. This creates a "synergic effect" which heavily strengthens the M-CO bond.
*Note for Students: Draw the orbital diagram showing the forward donation (C→M) and backward donation (M→C π*).*
(b) Number of unpaired electrons in [Pt(CN)4]2-:
Platinum (Z=78) is in group 10 (like Ni). Its oxidation state here is +2.
The configuration of Pt2+ is 5d8.
For 4d and 5d series elements (like Pt), the crystal field splitting (Δ) is very large regardless of the ligand. Therefore, the electrons are always forced to pair up.
In a square planar geometry (dsp2 hybridization), the 8 electrons completely pair up in the lower energy d-orbitals.
Hence, the number of unpaired electrons is Zero (0), and the complex is completely diamagnetic.
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