25 questions in 25 minutes on the NTA computer-based test interface, all from Coordination Compounds. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Coordination Compounds, of which 14 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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NEET CBT mock test: common questions
How many questions are in this Coordination Compounds mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 14 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Coordination Compounds?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
All 25 questions with answers and solutions
The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
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Chemistry
- Q1The coordination number of the central metal in [Co(en)₂Cl₂]⁺ is:
- 4
- 5
- 8
- 6
Answer: (D) 6
en (ethylenediamine) is bidentate, so 2 en donate through 4 atoms; plus 2 Cl⁻ = 2 more donors. Total donor atoms = 4 + 2 = 6.
- Q2The IUPAC name of K₃[Fe(CN)₆] is:
- Potassium hexacyanidoferrate(II)
- Tripotassium hexacyanoiron
- Potassium hexacyanidoferrate(III)
- Potassium ferricyanide(III)
Answer: (C) Potassium hexacyanidoferrate(III)
3 K⁺ = +3, so the complex anion is −3; 6 CN⁻ = −6, hence Fe = +3. The complex is anionic so the metal becomes ferrate: potassium hexacyanidoferrate(III).
- Q3Which one of the following is an inner orbital complex as well as diamagnetic in behaviour? (Zn = 30, Cr = 24, Co = 27, Ni = 28)
- [Zn(NH₃)₆]²⁺
- [Ni(NH₃)₆]²⁺
- [Cr(NH₃)₆]³⁺
- [Co(NH₃)₆]³⁺
Answer: (D) [Co(NH₃)₆]³⁺
[Co(NH₃)₆]³⁺: Co³⁺ is d⁶; NH₃ pairs all six electrons into the t₂g set, freeing two inner 3d orbitals → d²sp³ (inner orbital) and zero unpaired electrons (diamagnetic). [Cr(NH₃)₆]³⁺ is inner but has 3 unpaired (paramagnetic); the Ni and Zn complexes are outer-orbital.
- Q4Which statement about Werner’s theory is INCORRECT?
- Primary valencies are ionisable
- Secondary valencies are non-directional and variable
- Secondary valency equals the coordination number
- Primary valency equals the oxidation state of the metal
Answer: (B) Secondary valencies are non-directional and variable
Secondary valencies are DIRECTIONAL and FIXED (they determine geometry), not non-directional or variable. All other statements are correct postulates of Werner’s theory.
- Q5Which of the following ligands produces the largest crystal field splitting (Δ₀)?
- NH₃
- Cl⁻
- H₂O
- CO
Answer: (D) CO
Per the spectrochemical series I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO, CO is the strongest field ligand and gives the largest Δ₀.
- Q6The coordination number and oxidation state of Cr in K₃[Cr(C₂O₄)₃] are respectively:
- 6 and +3
- 3 and +3
- 3 and 0
- 4 and +2
Answer: (A) 6 and +3
Oxalate (C₂O₄²⁻) is bidentate, so three oxalates give CN = 3 × 2 = 6. Charge: 3(+1) + x + 3(−2) = 0 ⟹ x = +3. So CN = 6 and oxidation state = +3.
- Q7Which of the following is an ambidentate ligand?
- Oxalate
- Ammonia
- Ethylenediamine
- Thiocyanate (SCN⁻)
Answer: (D) Thiocyanate (SCN⁻)
SCN⁻ can coordinate through either sulphur (thiocyanato) or nitrogen (isothiocyanato), making it ambidentate. en and oxalate are bidentate (chelating); NH₃ is monodentate.
- Q8Which ligand, despite having multiple atoms with lone pairs, can only act as a monodentate ligand in a normal complex due to its rigid linear geometry?
- EDTA
- Ethylenediamine
- Oxalate
- Carbon monoxide (CO)
Answer: (D) Carbon monoxide (CO)
CO binds the metal through carbon as a monodentate ligand (carbonyl); it cannot chelate because its two atoms cannot simultaneously reach the same metal. The others are chelating bidentate/hexadentate ligands.
- Q9Which of the following complex ions is expected to absorb visible light? (Zn = 30, Sc = 21, Ti = 22, Cr = 24)
- [Sc(H₂O)₃(NH₃)₃]³⁺
- [Ti(en)₂(NH₃)₂]⁴⁺
- [Cr(NH₃)₆]³⁺
- [Zn(NH₃)₆]²⁺
Answer: (C) [Cr(NH₃)₆]³⁺
Absorption of visible light needs unpaired d-electrons that can undergo d-d transitions. [Cr(NH₃)₆]³⁺ has Cr³⁺ (d³, 3 unpaired) → absorbs visible light. Sc³⁺ (d⁰), Ti⁴⁺ (d⁰) and Zn²⁺ (d¹⁰) have no d-d transition and do not absorb in the visible.
- Q10The correct order of the number of moles of AgCl formed when excess AgNO₃ is treated with the complexes CoCl₃·6NH₃, CoCl₃·5NH₃ and CoCl₃·4NH₃ respectively is:
- 3 AgCl, 2 AgCl, 1 AgCl
- 3 AgCl, 1 AgCl, 2 AgCl
- 1 AgCl, 3 AgCl, 2 AgCl
- 2 AgCl, 3 AgCl, 1 AgCl
Answer: (A) 3 AgCl, 2 AgCl, 1 AgCl
These are [Co(NH₃)₆]Cl₃ (3 free Cl⁻ → 3 AgCl), [Co(NH₃)₅Cl]Cl₂ (2 free Cl⁻ → 2 AgCl) and [Co(NH₃)₄Cl₂]Cl (1 free Cl⁻ → 1 AgCl). Only the counter ions outside the sphere precipitate. Order: 3, 2, 1.
- Q11What is the correct formula for the complex potassium trioxalatoaluminate(III)?
- K₃[Al(C₂O₄)₃]
- K[Al(C₂O₄)₃]
- K₃[Al(C₂O₄)₂]
- Al₃[K(C₂O₄)₃]
Answer: (A) K₃[Al(C₂O₄)₃]
Al is +3, oxalate (C₂O₄²⁻) is −2, three oxalates give −6, so complex charge = +3 − 6 = −3, needing 3 K⁺. Formula: K₃[Al(C₂O₄)₃].
- Q12Which of the following complexes exhibits the HIGHEST paramagnetic behaviour? (Ti = 22, V = 23, Fe = 26, Co = 27; gly = glycinate, en = ethylenediamine, bpy = bipyridyl)
- [Co(ox)₂(OH)₂]⁻
- [Ti(NH₃)₆]³⁺
- [Fe(en)(bpy)(NH₃)₂]²⁺
- [V(gly)₂(OH)₂(NH₃)₂]⁺
Answer: (A) [Co(ox)₂(OH)₂]⁻
Paramagnetism grows with the number of unpaired electrons. In [Co(ox)₂(OH)₂]⁻, oxalate and OH⁻ are weak field, and Co is +5 (x + 2(−2) + 2(−1) = −1 ⟹ x = +5), giving Co⁵⁺ = d⁴ high spin with 4 unpaired electrons – the most among the options (V⁵⁺ d⁰: 0; Fe²⁺ with strong en/bpy: 0; Ti³⁺ d¹: 1).
- Q13The oxidation state of platinum in [Pt(NH₃)₂Cl₂] is:
- 0
- +2
- +4
- +1
Answer: (B) +2
NH₃ is neutral (0), each Cl⁻ is −1, complex is neutral overall. So Pt + 2(0) + 2(−1) = 0 ⟹ Pt = +2.
- Q14The extra stability of a complex due to ring formation by a polydentate ligand is called the:
- Trans effect
- Chelate effect
- Jahn-Teller effect
- Inert pair effect
Answer: (B) Chelate effect
The chelate effect is the enhanced stability when a bidentate or polydentate ligand forms a ring with the metal (e.g., en, EDTA). More rings give greater stability.
- Q15Which one of the following ligands is expected to be bidentate?
- CH₃NH₂
- CH₃C≡N
- C₂O₄²⁻ (oxalate)
- Br⁻
Answer: (C) C₂O₄²⁻ (oxalate)
A bidentate ligand has two donor atoms that can bind the same metal. Oxalate, C₂O₄²⁻, coordinates through two of its oxygen atoms. CH₃NH₂ (one N), CH₃CN (one N) and Br⁻ are all monodentate.
- Q16The shape of Fe(CO)₅ is:
- square planar
- square pyramidal
- trigonal bipyramidal
- octahedral
Answer: (C) trigonal bipyramidal
In Fe(CO)₅ the Fe atom is dsp³ hybridised (strong field CO pairs the 4s electrons into 3d, then one 3d + s + three p mix). Five dsp³ hybrid orbitals give a trigonal bipyramidal geometry.
- Q17The hypothetical complex chlorodiaquatriamminecobalt(III) chloride is represented by the formula:
- [Co(NH₃)₂(H₂O)₂Cl]
- [CoCl(NH₃)₃(H₂O)₂]Cl₂
- [Co(NH₃)₃(H₂O)Cl₃]
- [Co(NH₃)₃(H₂O)₃]Cl₃
Answer: (B) [CoCl(NH₃)₃(H₂O)₂]Cl₂
Decode the name: one chlorido + two aqua + three ammine on Co(III). Inner charge = +3 + (−1) + 0 + 0 = +2, so two Cl⁻ counter ions: [CoCl(NH₃)₃(H₂O)₂]Cl₂.
- Q18[Co(NH₃)₄(NO₂)₂]Cl can exhibit which combination of isomerism?
- linkage, ionisation and geometrical isomerism
- linkage, ionisation and optical isomerism
- ionisation, geometrical and optical isomerism
- linkage, geometrical and optical isomerism
Answer: (A) linkage, ionisation and geometrical isomerism
NO₂⁻ is ambidentate → linkage isomerism. The Cl⁻ counter ion can swap with NO₂⁻ inside the sphere → ionisation isomerism. The MA₄B₂ octahedral arrangement gives cis/trans → geometrical isomerism. It is not optically active (the cis/trans forms have symmetry).
- Q19Which one of the following statements about [Co(CN)₆]³⁻ is TRUE? (Co = 27)
- It has no unpaired electrons and is high spin
- It has four unpaired electrons and is low spin
- It has four unpaired electrons and is high spin
- It has no unpaired electrons and is low spin
Answer: (D) It has no unpaired electrons and is low spin
Co³⁺ is d⁶. CN⁻ is a strong field ligand, so all six electrons pair into the t₂g set (t₂g⁶ eg⁰): no unpaired electrons → diamagnetic, low spin.
- Q20The hybridisation involved in the complex [Ni(CN)₄]²⁻ is (atomic number of Ni = 28):
- dsp²
- sp³
- d²sp³
- d²sp²
Answer: (A) dsp²
Ni²⁺ is 3d⁸. CN⁻ is a strong field ligand, so the d-electrons pair up, vacating one 3d orbital. With one (n−1)d + s + two p orbitals the hybridisation is dsp² → square planar, diamagnetic.
- Q21The d-electron configuration of the central metal in K₄[Fe(CN)₆], based on crystal field theory, is:
- e⁴ t₂²
- t₂g⁴ eg²
- e³ t₂³
- t₂g⁶ eg⁰
Answer: (D) t₂g⁶ eg⁰
In K₄[Fe(CN)₆], Fe is +2 (3d⁶). CN⁻ is a strong field ligand, so all six d-electrons occupy the lower t₂g set: t₂g⁶ eg⁰ (low spin, diamagnetic).
- Q22Assertion (A): Tetrahedral complexes of the type [MA₂B₂] do not show cis-trans (geometrical) isomerism. Reason (R): In a tetrahedron all four positions are adjacent and equivalent, so no two ligands are uniquely ‘cis’ or ‘trans’.
- A is true but R is false
- Both A and R are true but R is not the correct explanation of A
- A is false but R is true
- Both A and R are true and R is the correct explanation of A
Answer: (D) Both A and R are true and R is the correct explanation of A
In a tetrahedron every vertex is adjacent (equidistant) to every other, so there is no distinct cis/trans relationship; hence no geometrical isomerism. R correctly explains A.
- Q23Urea reacts with water to form A, which on heating gives gas B; B passed through Cu²⁺(aq) gives a deep blue solution C. The formula of C is:
- CuCO₃·Cu(OH)₂
- CuSO₄
- Cu(OH)₂
- [Cu(NH₃)₄]²⁺
Answer: (D) [Cu(NH₃)₄]²⁺
Urea + H₂O → ammonium carbamate/ammonium carbonate, which on heating releases NH₃ (gas B). Excess NH₃ passed into a Cu²⁺ solution forms the deep blue tetraamminecopper(II) ion, [Cu(NH₃)₄]²⁺ (C).
- Q24An ‘inner orbital’ (low spin) octahedral complex uses which set of d-orbitals for hybridisation?
- Inner 3d orbitals (d²sp³)
- f-orbitals
- Outer 4d orbitals (sp³d²)
- Only 4s and 4p
Answer: (A) Inner 3d orbitals (d²sp³)
An inner orbital complex uses the inner (n−1)d orbitals, giving d²sp³ hybridisation, which occurs with strong field ligands that cause pairing.
- Q25Which of the following organometallic compound is σ (sigma) bonded?
- ferrocene
- ruthenocene
- Grignard’s reagent (RMgX)
- cobaltocene
Answer: (C) Grignard’s reagent (RMgX)
A σ-bonded organometallic has a direct metal-carbon σ bond. Grignard reagent R-Mg-X has an Mg-C σ bond. Ferrocene, ruthenocene and cobaltocene are π-bonded (sandwich) compounds.