Home › Practice › Class 12 › Chemistry
Class 12 Chemistry — questions with answers and worked explanations.
These questions are drawn from the Pariksha Sutra question bank for Coordination Compounds, part of the Class 12 Chemistry syllabus. Each one shows the correct answer and, where a method helps, the working behind it.
Read the question, decide your answer before looking, then check the explanation — that is what turns practice into marks. If a question catches you out, the explanation is the part worth re-reading.
Primary valency is the oxidation state of the metal, which can be neutralised by anions that are ionisable.
These ionisable negative ions combine with the metal to give the overall charge of the complex.
Thus the primary valency is ionisable and satisfied by negative ions.
The complex ion is [Fe(CN)6]4–, so the oxidation state of Fe is +2 because each CN⁻ contributes –1 (6 × –1 = –6) and the overall charge is –4.
In the name the metal is given its oxidation number in Roman numerals, giving “hexacyanoferrate(II)”.
The counter‑cation is K⁺, four of them balance the –4 charge, so the full formula is potassium hexacyanoferrate(II).
The complex must be chiral, i.e., have no plane or centre of symmetry.
[Co(en)2Cl2]+ has a plane of symmetry when the two identical en ligands are arranged trans, so it is achiral.
[Co(NH3)6]3+ and [PtCl4]2‑ are octahedral with identical ligands all around, giving a centre of symmetry, thus achiral.
Only [Co(en)3]3+ has three bidentate en ligands arranged in a non‑superimposable left‑ and right‑handed fashion, giving optical isomerism.
In an octahedral field the ligands lie along the x, y and z axes, so orbitals that point directly at them experience greater repulsion.
The d‑orbitals with lobes along the axes are dz² (lobes on the z‑axis) and dx²‑y² (lobes on the x‑ and y‑axes).
These two orbitals are thus raised in energy to form the eg set, while the other three (dxy, dxz, dyz) form the lower t2g set.
Hence the eg set is dz² and dx²‑y².
The complex absorbs photons of green wavelength (≈500–570 nm).
The absorbed green light is removed from the white light that reaches our eyes.
The complementary colour of green in the visible spectrum is red, so the transmitted/reflected light appears red.
Hence the complex appears red.
Oxalate (C₂O₄²⁻) is a bidentate ligand, each donating two donor atoms to the metal.
Three oxalate ligands therefore provide 3 × 2 = 6 coordination sites around Fe.
Thus the coordination number of iron in [Fe(C₂O₄)₃]³⁻ is 6.
In a square‑planar d⁸ complex the four ligands occupy the corners of a plane; if two ligands are identical and the other two are different, they can be placed adjacent (cis) or opposite (trans).
[Pt(NH₃)₂Cl₂] has two NH₃ and two Cl⁻ ligands, allowing both arrangements, so it shows cis‑trans isomerism.
The other complexes have either all identical ligands or only one type of different ligand, so no geometric isomerism is possible.
Thus the square planar complex that can exhibit cis‑trans isomerism is [Pt(NH₃)₂Cl₂].
Mn2+ has d⁵ configuration. In a weak‑field (high‑spin) octahedral complex the crystal‑field splitting Δo is small, so electrons occupy all five d orbitals singly before pairing. Hence each of the five d orbitals contains one electron. Therefore the complex possesses five unpaired electrons. 5
Fe3+ is d5; strong‑field CN⁻ makes a low‑spin complex, pairing all five electrons → one unpaired electron.
Spin‑only magnetic moment μ = √[n(n+2)] BM, where n = number of unpaired electrons.
For n = 1, μ = √[1·3] = √3 ≈ 1.73 BM.
Thus the correct answer is 1.73 BM.
CO is a neutral ligand, so it does not change the oxidation state of the metal.
In Ni(CO)4 the overall charge of the complex is zero.
Therefore the oxidation state of Ni must balance the zero charge contributed by the four CO ligands, giving Ni an oxidation state of zero.
Secondary valency refers to the coordination number of the metal ion – the number of ligands it can bind.
It is a fixed spatial arrangement, not involved in ionisation of the complex.
Thus it is non‑ionisable and directional in space, which matches the given answer.
The complex ion is [Co(NH₃)₆]³⁺, so the metal is cobalt in the +3 oxidation state.
The six NH₃ ligands are neutral, therefore the charge on the metal is +3 and the counter‑ions are three Cl⁻.
In naming, the ligands are named first in alphabetical order (ammine) followed by the metal with its oxidation state in Roman numerals.
Hence the correct IUPAC name is Hexaamminecobalt(III) chloride.
In an octahedral complex MA4B2 the two identical ligands B can be either adjacent (cis) or opposite (trans).
Only these two distinct arrangements satisfy the symmetry of an octahedron; any other placement is equivalent to one of them.
Thus the complex has exactly two geometrical isomers. 2
In a tetrahedral field the ligands approach the metal along the edges of a cube, so the d‑orbitals experience a weaker electrostatic interaction than in an octahedral field where ligands lie directly opposite each other.
The crystal‑field splitting is proportional to the square of the direction cosine between the metal‑ligand axis and the orbital lobes; for tetrahedral geometry this factor is 4/9 of the octahedral value.
Hence Δt = (4/9) Δo, which matches the given answer.
Spin‑only magnetic moment μ = √[n(n+2)] BM, where n = number of unpaired electrons.
Set √[n(n+2)] ≈ 3.87 and square both sides: n(n+2) ≈ 15.
Trial values: n = 3 gives 3×5 = 15, matching the required value.
Thus the complex has 3 unpaired electrons, giving a spin‑only moment of 3.87 BM. 3
CoCl₃·6NH₃ is a Werner-type octahedral complex where all three chloride ligands are outside the coordination sphere (ionic).
When AgNO₃ is added, each free Cl⁻ combines with Ag⁺ to give AgCl precipitate.
Since there are three external Cl⁻ ions per formula unit, three moles of AgCl are formed.
Hence the number of chloride ions precipitated is 3.
Zn is in +2 oxidation state, each OH⁻ contributes –1, so four OH⁻ give –4 charge on the complex ion.
The overall compound must be neutral, therefore two K⁺ ions (each +1) are needed to balance the –4 charge of [Zn(OH)₄]²⁻.
Thus the formula is K₂[Zn(OH)₄].
Both complexes contain the same overall composition: one Co, one Cr, six NH₃ and six CN⁻ ligands.
In one compound the cobalt centre is coordinated by NH₃ and the chromium centre by CN⁻; in the other the ligands are interchanged.
Since the metal ions exchange their respective ligand sets while the overall formula remains unchanged, they are examples of coordination isomers.
Hence the correct answer is Coordination isomers.
Strong field ligands cause a large crystal field splitting (Δo) in octahedral complexes.
When Δo > pairing energy, electrons pair in the lower‑energy t2g set before occupying eg orbitals.
Thus the complex has paired electrons and exhibits low spin.
Therefore a strong field ligand favours the formation of Low-spin complexes.
Ti³⁺ has atomic number 22, so Ti³⁺ loses three electrons → configuration [Ar] 3d¹ 4s⁰.
In an octahedral aqua complex the 3d orbital is split into t₂g (lower) and e_g (higher).
Only one electron occupies the lower t₂g set, giving a d¹ configuration.
Thus the metal ion in [Ti(H₂O)₆]³⁺ is d¹.
The full chapter test is timed, gives an All-India rank and a subject-wise breakdown of where you lost marks.
See the test series →