Coordination Compounds NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Coordination Compounds question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Coordination Compounds previous year questions
45 questions. Tap an option to check your answer.
Q1NEET 2021
Ethylene diaminetetraacetate (EDTA) ion is :
Not quite — the answer is A.
EDTA⁴⁻ donates through its two amine nitrogen atoms and four carboxylate oxygen atoms, so it is a hexadentate ligand. Ethylenediamine (en), not EDTA, is the bidentate ligand with two N donors.
Q2NEET 2021
Match List - I with List - II. List - I | List - II (a) [Fe(CN)₆]³⁻ | (i) 5.92 BM (b) [Fe(H₂O)₆]³⁺ | (ii) 0 BM (c) [Fe(CN)₆]⁴⁻ | (iii) 4.90 BM (d) [Fe(H₂O)₆]²⁺ | (iv) 1.73 BM Choose the correct answer from the options given below.
Not quite — the answer is D.
[Fe(CN)₆]³⁻: Fe³⁺ (d⁵) low spin, 1 unpaired electron, 1.73 BM. [Fe(H₂O)₆]³⁺: high spin d⁵, 5 unpaired, 5.92 BM. [Fe(CN)₆]⁴⁻: Fe²⁺ (d⁶) low spin, 0 unpaired, 0 BM. [Fe(H₂O)₆]²⁺: high spin d⁶, 4 unpaired, 4.90 BM.
Q3NEET 2020 · Q116
Which of the following is the correct order of increasing field strength of ligands to form coordination compounds ?
Not quite — the answer is A.
In the spectrochemical series SCN⁻ < F⁻ < C₂O₄²⁻ < CN⁻, with cyanide one of the strongest-field ligands. Placing oxalate above cyanide, or fluoride below thiocyanate, contradicts that series.
Q4NEET 2019 · Q75
What is the correct electronic configuration of the central atom in K₄[Fe(CN)₆] based on crystal field theory?
Not quite — the answer is B.
Fe is +2 (d⁶) in an octahedral field; CN⁻ is a strong-field ligand causing pairing, so the configuration is t₂g⁶ e_g⁰ (low spin, diamagnetic). t₂g⁴ e_g² is the high-spin arrangement with weak-field ligands, and e/t₂ labels belong to tetrahedral fields.
Q5NEET 2018 · Q71
Iron carbonyl, Fe(CO)₅ is
Not quite — the answer is B.
Fe(CO)₅ contains a single Fe atom bonded to five CO ligands, so it is mononuclear (trigonal bipyramidal). Fe₂(CO)₉ is the dinuclear and Fe₃(CO)₁₂ the trinuclear carbonyl.
Q6NEET 2018 · Q72
The type of isomerism shown by the complex [CoCl₂(en)₂] is
Not quite — the answer is C.
The octahedral [MA₂(AA)₂] complex can have the two Cl ligands cis or trans, so it shows geometrical isomerism (the cis form is also optically active). It has no counter-ion exchange, no complex cation–anion pair and no ambidentate ligand, so the other types do not apply.
Q7NEET 2018 · Q73
The geometry and magnetic behaviour of the complex [Ni(CO)₄] are
Not quite — the answer is B.
Ni is in the 0 state (3d⁸4s²); the strong-field CO ligands pair all electrons into 3d¹⁰, and sp³ hybridisation gives a tetrahedral, diamagnetic complex. Square planar dsp² geometry is for Ni²⁺ with strong-field ligands like CN⁻.
Q8NEET 2017 · Q11
Correct increasing order for the wavelengths of absorption in the visible region for the complexes of Co³⁺ is:
Not quite — the answer is A.
Field strength follows the spectrochemical series en > NH₃ > H₂O, so Δₒ is largest for the en complex. Absorbed wavelength is inversely related to Δₒ (λ = hc/Δₒ), so the order of increasing wavelength is [Co(en)₃]³⁺ < [Co(NH₃)₆]³⁺ < [Co(H₂O)₆]³⁺. Writing the reverse order confuses increasing Δₒ with increasing wavelength.
Q9NEET 2017 · Q12
The correct order of the stoichiometries of AgCl formed when AgNO₃ in excess is treated with the complexes: CoCl₃.6NH₃, CoCl₃.5NH₃, CoCl₃.4NH₃ respectively is:
Not quite — the answer is C.
The complexes are [Co(NH₃)₆]Cl₃, [Co(NH₃)₅Cl]Cl₂ and [Co(NH₃)₄Cl₂]Cl. Only chloride outside the coordination sphere is precipitated by Ag⁺, giving 3, 2 and 1 mol AgCl respectively.
Q10NEET 2017 · Q26
An example of a sigma bonded organometallic compound is:
Not quite — the answer is B.
In a Grignard reagent RMgX the alkyl carbon is joined to magnesium by a single σ bond. Ferrocene, ruthenocene and cobaltocene are sandwich compounds in which the metal is π-bonded to cyclopentadienyl rings.
Q11NEET 2017 · Q29
Pick out the correct statement with respect to [Mn(CN)₆]³⁻:
Not quite — the answer is C.
Mn³⁺ is 3d⁴. CN⁻ is a strong-field ligand, so the four electrons pair into three 3d orbitals, leaving two inner 3d orbitals free; with 4s and 4p they give d²sp³ hybridisation and an octahedral (inner-orbital) complex. sp³d² would apply to a weak-field outer-orbital complex.
Q12NEET 2017 (Gujarati paper) · Q1
The electron distribution in dⁿ coordination complexes depends on magnitude of crystal field splitting, (Δₒ) and pairing energy (P). The condition which favours formation of high spin complexes is:
Not quite — the answer is B.
When the crystal field splitting is smaller than the pairing energy, it costs less energy to put electrons in the higher eg set than to pair them in t₂g, so electrons stay unpaired and a high spin complex forms. When Δₒ exceeds P, pairing is preferred and a low spin complex results.
Q13NEET 2017 (Gujarati paper) · Q4
Which one of the following ions is not tetrahedral in shape?
Not quite — the answer is C.
[Cu(NH₃)₄]²⁺ is a d⁹ dsp² complex and is square planar. NH₄⁺ and BF₄⁻ are sp³ tetrahedral, and [NiCl₄]²⁻ with the weak-field Cl⁻ ligand is sp³ tetrahedral.
Q14NEET 2017 (Gujarati paper) · Q14
The [Co(H₂O)₆]²⁺ ion has three unpaired electrons. The hybridization of Co in [Co(H₂O)₆]²⁺ is:
Not quite — the answer is C.
Co²⁺ is 3d⁷. Three unpaired electrons means the weak-field water ligands do not force pairing, so the inner 3d orbitals are not free and the six ligands use 4s, 4p and 4d orbitals: sp³d² (outer orbital, octahedral). d²sp³ would need pairing down to one unpaired electron.
Q15NEET 2017 (Gujarati paper) · Q16
For the tetrahedral complex [MnBr₄]²⁻, the spin only magnetic moment value is: [At. no. of Mn = 25]
Not quite — the answer is B.
Mn²⁺ is 3d⁵, and in a tetrahedral field with weak Br⁻ ligands all five electrons stay unpaired. μ = √(n(n + 2)) = √35 ≈ 5.9 BM.
Q16NEET 2017 (Gujarati paper) · Q24
Which of the following complex ions is not diamagnetic?
Not quite — the answer is B.
Cr³⁺ is 3d³, with three unpaired electrons in t₂g, so [Cr(NH₃)₆]³⁺ is paramagnetic. Ti⁴⁺ and Sc³⁺ are d⁰ and Zn²⁺ is d¹⁰, so those complexes have no unpaired electrons and are diamagnetic.
Q17NEET 2016 Phase 2 · Q75
Jahn-Teller effect is not observed in high spin complexes of
Not quite — the answer is B.
High-spin d⁸ in an octahedral field is t₂g⁶e_g², with symmetrically filled degenerate sets, so there is no Jahn–Teller distortion. High-spin d⁴ (t₂g³e_g¹) and d⁹ (t₂g⁶e_g³) have unevenly filled e_g orbitals and show strong distortion; d⁷ high spin (t₂g⁵e_g²) has an uneven t₂g set.
Q18NEET 2016 Phase 1 · Q74
Which of the following has longest C–O bond length? (Free C–O bond length in CO is 1.128 Å.)
Not quite — the answer is C.
These carbonyls are isoelectronic, but the more negative the metal centre, the more electron density it back-donates into the π* orbitals of CO, weakening and lengthening the C–O bond. [Fe(CO)₄]²⁻ carries the highest negative charge, so it has the longest C–O bond, while the cationic [Mn(CO)₆]⁺ has the shortest.
Q19AIPMT 2015
Which of these statements about [Co(CN)₆]³⁻ is true?
Not quite — the answer is B.
Co³⁺ is 3d⁶. CN⁻ is a strong-field ligand, so the electrons pair up as t₂g⁶e_g⁰ (d²sp³, inner orbital): low spin with no unpaired electrons. Four unpaired electrons would occur only with a weak-field ligand such as F⁻.
Q20AIPMT 2015
Cobalt(III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at 25°C?
Not quite — the answer is B.
Octahedral Co(III) needs six ligands. With only three NH₃ the complex is [Co(NH₃)₃Cl₃], all three Cl⁻ inside the coordination sphere, so no Cl⁻ ions are free to precipitate with AgNO₃. The 6, 5 and 4 NH₃ complexes have 3, 2 and 1 ionisable Cl⁻.
Q21AIPMT 2014
Which of the following complexes is used to be as an anticancer agent?
Not quite — the answer is A.
cis-[PtCl₂(NH₃)₂] (cisplatin) binds DNA and is used in cancer chemotherapy. The trans isomer and the other complexes listed do not have this use.
Q22AIPMT 2014
Among the following complexes the one which shows Zero crystal field stabilization energy (CFSE) is:
Not quite — the answer is A.
Fe³⁺ is 3d⁵, and with weak-field H₂O it is high spin t₂g³e_g²: CFSE = 3(−0.4Δ₀) + 2(+0.6Δ₀) = 0. Co²⁺ (d⁷), Co³⁺ (d⁶) and Mn³⁺ (d⁴) all have non-zero CFSE.
Q23NEET 2013
A magnetic moment of 1.73 BM will be shown by one among the following
Not quite — the answer is A.
1.73 BM = √(n(n + 2)) with n = 1 unpaired electron. Cu²⁺ (3d⁹) in [Cu(NH₃)₄]²⁺ has one unpaired electron. [Ni(CN)₄]²⁻ is square planar and diamagnetic, TiCl₄ has Ti⁴⁺ (3d⁰), and [CoCl₆]⁴⁻ has high-spin Co²⁺ (3d⁷) with three unpaired electrons.
Q24NEET 2013
An excess of AgNO₃ is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium(III) chloride. The number of moles of AgCl precipitated would be
Not quite — the answer is A.
Dichlorotetraaquachromium(III) chloride is [Cr(H₂O)₄Cl₂]Cl, with only one chloride outside the coordination sphere. Moles of complex = 0.01 × 0.100 = 0.001 mol, giving 0.001 mol AgCl. Coordinated chlorides are not precipitated, so 0.003 is wrong.
Q25AIPMT 2012 Prelims
Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?
Not quite — the answer is B.
Ni²⁺ is d⁸; in an octahedral complex the two remaining inner d orbitals are not vacant, so it uses 4s4p³4d² (sp³d², outer orbital) and keeps two unpaired electrons, making it paramagnetic. [Co(NH₃)₆]³⁺ is inner orbital and diamagnetic, [Zn(NH₃)₆]²⁺ is outer orbital but diamagnetic (d¹⁰), and [Cr(NH₃)₆]³⁺ is inner orbital (d²sp³).
Q26AIPMT 2011 Mains
Which of the following complex compounds will exhibit highest paramagnetic behaviour? (At. No. Ti = 22, Cr = 24, Co = 27, Zn = 30)
Not quite — the answer is D.
Cr³⁺ is d³ with 3 unpaired electrons in octahedral complexes. Co³⁺ (d⁶) with NH₃ is low spin with no unpaired electrons, Zn²⁺ is d¹⁰ with none, and Ti³⁺ is d¹ with one. So [Cr(NH₃)₆]³⁺ is the most paramagnetic.
Q27AIPMT 2011 Mains
Which of the following carbonyls will have the strongest C – O bond?
Not quite — the answer is C.
Back-donation from metal d orbitals into CO π* orbitals weakens the C–O bond. The positive charge in Mn(CO)₆⁺ holds the metal electrons tightly, so back-bonding is least and the C–O bond is strongest. The anion V(CO)₆⁻ has the most back-bonding and the weakest C–O bond.
Q28AIPMT 2011 Prelims
The complex [Pt(Py)(NH₃)BrCl] will have how many geometrical isomers?
Not quite — the answer is B.
[Pt(py)(NH₃)BrCl] is square planar of type [Mabcd]. Fixing one ligand, any of the other three can be trans to it, giving 3 geometrical isomers. Two isomers is the answer only for [Ma₂b₂]-type complexes.
Q29AIPMT 2011 Prelims
The complexes [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆] are the examples of which type of isomerism?
Not quite — the answer is D.
Both compounds have the same formula, but the ligands are exchanged between the cobalt and chromium centres of the cationic and anionic complexes. This is coordination isomerism. Ionisation isomerism would involve exchange of an ion between the coordination sphere and the counter-ion.
Q30AIPMT 2011 Prelims
The d-electron configurations of Cr²⁺, Mn²⁺, Fe²⁺ and Co²⁺ are d⁴, d⁵, d⁶ and d⁷ respectively. Which one of the following will exhibit minimum paramagnetic behaviour? (At. nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)
Not quite — the answer is D.
H₂O is a weak-field ligand, so all are high-spin: d⁴ (Cr²⁺) has 4 unpaired electrons, d⁵ (Mn²⁺) 5, d⁶ (Fe²⁺) 4 and d⁷ (Co²⁺) 3. Fewest unpaired electrons means least paramagnetism, so the cobalt complex. The manganese complex is the most paramagnetic.
Q31AIPMT 2011 Prelims
Of the following complex ions, which is diamagnetic in nature?
Not quite — the answer is C.
Ni²⁺ is d⁸; strong-field CN⁻ pairs all electrons and gives a dsp² square planar complex with no unpaired electrons, so [Ni(CN)₄]²⁻ is diamagnetic. [NiCl₄]²⁻ is tetrahedral sp³ with 2 unpaired electrons, [CoF₆]³⁻ has 4 and [CuCl₄]²⁻ (d⁹) has 1.
Q32AIPMT 2010 Mains · Q111
Which one of the following complexes is not expected to exhibit isomerism?
Not quite — the answer is C.
[Ni(NH₃)₂Cl₂] is tetrahedral, and an MA₂B₂ tetrahedral complex shows no geometrical or optical isomers. [Ni(NH₃)₄(H₂O)₂]²⁺ (octahedral MA₄B₂) and square planar [Pt(NH₃)₂Cl₂] show cis-trans isomerism, and [Ni(en)₃]²⁺ is optically active.
Q33AIPMT 2010 Prelims · Q76
The existence of two different coloured complexes with the composition of [Co(NH₃)₄Cl₂]⁺ is due to
Not quite — the answer is C.
[Co(NH₃)₄Cl₂]⁺ is octahedral MA₄B₂, which exists as cis (violet) and trans (green) isomers: geometrical isomerism. Ionisation and linkage isomerism need different ions or an ambidentate ligand, which are absent here.
Q34AIPMT 2010 Prelims · Q80
Crystal field stabilization energy for high spin d⁴ octahedral complex is
Not quite — the answer is A.
High-spin d⁴ in an octahedral field is t₂g³ e_g¹. CFSE = 3(−0.4Δ₀) + 1(+0.6Δ₀) = −0.6Δ₀. −1.6Δ₀ + P is the low-spin t₂g⁴ value.
Q35AIPMT 2009 Prelims · Q181
Out of TiF₆²⁻, CoF₆³⁻, Cu₂Cl₂ and NiCl₄²⁻ (Z of Ti = 22, Co = 27, Cu = 29, Ni = 28) the colourless species are:
Not quite — the answer is B.
Ti⁴⁺ in TiF₆²⁻ is 3d⁰ and Cu⁺ in Cu₂Cl₂ is 3d¹⁰, so no d–d transition is possible and both are colourless. Co³⁺ (3d⁶) and Ni²⁺ (3d⁸) have partly filled d orbitals and are coloured.
Q36AIPMT 2009 Prelims · Q185
Which of the following does not show optical isomerism?
Not quite — the answer is A.
[Co(NH₃)₃Cl₃] is an MA₃B₃ octahedral complex; both its fac and mer forms have a mirror plane, so it is not optically active. [Co(en)₃]³⁺, cis-[Co(en)₂Cl₂]⁺ and one isomer of [Co(en)Cl₂(NH₃)₂]⁺ are chiral.
Q37AIPMT 2009 Prelims · Q196
Which of the following complex ions is expected to absorb visible light? (At. no. Zn = 30, Sc = 21, Ti = 22, Cr = 24)
Not quite — the answer is B.
Cr³⁺ is 3d³, so d–d transitions are possible and the complex absorbs visible light. Ti⁴⁺ and Sc³⁺ are 3d⁰ and Zn²⁺ is 3d¹⁰, so they have no d–d transitions and are colourless.
Q38AIPMT 2008 Prelims · Q80
In which of the following coordination entities the magnitude of Δo (CFSE in octahedral field) will be maximum? (at. no. Co = 27)
Not quite — the answer is D.
For the same metal ion (Co³⁺), Δo depends on the ligand's place in the spectrochemical series, where CN⁻ is the strongest-field ligand here, well above NH₃, H₂O and oxalate. So [Co(CN)₆]³⁻ has the largest splitting. [Co(NH₃)₆]³⁺ is a strong-field complex too, but ammonia splits less than cyanide.
Q39AIPMT 2007 Prelims · Q76
Which of the following will give a pair of enantiomorphs? (en = NH₂CH₂CH₂NH₂)
Not quite — the answer is B.
cis-[Co(en)₂Cl₂]⁺ has no plane of symmetry and exists as a pair of enantiomers. The other complexes show coordination or geometrical isomerism but no optical isomers ([Co(NH₃)₄Cl₂]⁺ cis/trans are both achiral).
Q40AIPMT 2007 Prelims · Q78
The d electron configurations of Cr²⁺, Mn²⁺, Fe²⁺ and Ni²⁺ are 3d⁴, 3d⁵, 3d⁶ and 3d⁸ respectively. Which one of the following aqua complexes will exhibit the minimum paramagnetic behaviour? (At. No. Cr = 24, Mn = 25, Fe = 26, Ni = 28)
Not quite — the answer is B.
H₂O is a weak-field ligand, so the ions stay high-spin: Cr²⁺ (d⁴) 4, Mn²⁺ (d⁵) 5, Fe²⁺ (d⁶) 4 and Ni²⁺ (d⁸) 2 unpaired electrons. [Ni(H₂O)₆]²⁺ has the fewest, so it is least paramagnetic; [Mn(H₂O)₆]²⁺ is the most.
Q41AIPMT 2006 Prelims · Q61
Copper sulphate dissolves in excess of KCN to give :
Not quite — the answer is B.
CN⁻ reduces Cu²⁺ to Cu⁺ (2Cu²⁺ + 4CN⁻ → 2CuCN + (CN)₂), and CuCN dissolves in excess cyanide to give the stable tetracyanidocuprate(I) ion, [Cu(CN)₄]³⁻. The Cu(II) complex [Cu(CN)₄]²⁻ is not the product, and CuCN is only the intermediate before excess KCN acts.
Q42AIPMT 2006 Prelims · Q79
[Co(NH₃)₄(NO₂)₂]Cl exhibits :
Not quite — the answer is B.
NO₂⁻ can bind through N (nitro) or O (nitrito), giving linkage isomers. Exchanging Cl⁻ with a coordinated NO₂⁻ gives ionisation isomers, and the two NO₂ ligands can be cis or trans, giving geometrical isomers. Neither cis nor trans [Co(NH₃)₄(NO₂)₂]⁺ is chiral, because each has mirror planes, so optical isomerism is absent.
Q43AIPMT 2006 Prelims · Q80
[Cr(H₂O)₆]Cl₃ (at. no. of Cr = 24) has a magnetic moment of 3.83 BM, the correct distribution of 3d electrons in the chromium of the complex is :
Not quite — the answer is C.
μ = √(n(n + 2)) = 3.83 BM gives n = 3 unpaired electrons, matching Cr³⁺ (3d³). In the octahedral field of six water ligands these three electrons occupy the lower t₂g set, dxy, dyz and dxz, one each. Any set that includes dx²−y² or dz² places an electron in the higher eg level.
Q44AIPMT 2005 Prelims · Q87
Which one of the following is an inner orbital complex as well as diamagnetic in behaviour ? (Atomic no. : Zn = 30, Cr = 24, Co = 27, Ni = 28)
Not quite — the answer is C.
Co³⁺ is 3d⁶; with NH₃ the electrons pair up in t₂g, leaving two 3d orbitals for d²sp³ (inner orbital) hybridisation with no unpaired electrons, so it is diamagnetic. [Cr(NH₃)₆]³⁺ is inner orbital but has 3 unpaired electrons; the Zn²⁺ and Ni²⁺ complexes are outer orbital.
Q45AIPMT 2005 Prelims · Q91
Which one of the following is expected to exhibit optical isomerism ? (en = ethylenediamine)
Not quite — the answer is D.
cis-[Co(en)₂Cl₂]⁺ has no plane of symmetry and exists as non-superimposable mirror images, so it is optically active. The trans isomer has a mirror plane through the Cl–Co–Cl axis, and square planar [Pt(NH₃)₂Cl₂] isomers are planar and achiral.
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