Exams › JEE Main › Chemistry › Coordination Compounds
301 questions with worked solutions.
Answer: anionic form
The correct option is right because when excess NaOH is added to the solution, it forms a soluble zincate complex, indicating that zinc is present in anionic form as zincate ions (Zn(OH)4²-), rather than remaining solely as cations.
Answer: Cu(NH3)4²+
Excess NH3 with CuSO4 forms the deep blue tetraamminecopper(II) ion [Cu(NH3)4]^2+. Cu(II) in aqueous ammonia takes 4 NH3 ligands in the square-planar/equatorial plane giving the characteristic intense blue colour.
Answer: measuring the magnetic moment in the solid state
Fe2+(NO+) and Fe3+(NO) have different numbers of unpaired electrons, giving different magnetic moments. Measuring the magnetic moment in the solid state distinguishes the two formulations.
Answer: [Fe(H2O)5NO]2+
In the brown ring test the NO produced replaces one water ligand of [Fe(H2O)6]2+ to give the brown nitrosyl complex [Fe(H2O)5NO]2+ (iron in +1, NO+). Hence the formula is [Fe(H2O)5NO]2+.
Answer: 2.82
Ni2+ = [Ar]3d8 has 2 unpaired electrons, so spin-only moment = sqrt(n(n+2)) = sqrt(2*4) = sqrt(8) = 2.83 BM (~2.82).
Answer: [Ag(NH3)2]+
AgCl dissolves in aqueous ammonia by forming the soluble diammine silver(I) complex ion [Ag(NH3)2]+, which shifts the dissolution equilibrium and dissolves the precipitate.
Answer: I > II > III > IV
[FeF6]3- is d5 high-spin (5 unpaired), [CrF6]3- is d3 (3), [V(H2O)6]3+ is d2 (2), [Ti(H2O)6]3+ is d1 (1). Spin-only moment rises with unpaired electrons, so I > II > III > IV.
Answer: In acidic medium, H+ ions combine with ammonia to form NH4+, so free NH3 is not available
In acidic medium, H+ reacts with ammonia to form NH4+, leaving no free NH3 available to coordinate to Cu2+, so the complex [Cu(NH3)4]2+ cannot form. In alkaline medium free NH3 is present and complexation occurs.
Answer: 0.001
In [CrCl2(H2O)4]Cl only the one chloride outside the coordination sphere is precipitated. Moles of complex = 0.01 M x 0.100 L = 0.001 mol, so 0.001 mol AgCl forms.
Q10. Which of the following complexes is an example of dsp2 hybridisation?
Answer: [Ni(CN)4]2-
dsp2 hybridisation gives square-planar geometry. [Ni(CN)4]2- has Ni(II) (d8) with the strong-field CN- causing pairing and one (n-1)d orbital becoming available, giving dsp2 square planar. [Fe(CN)6]3-, [Co(CN)6]3- are d2sp3 octahedral and [Ag(CN)2]- is sp linear.
Q11. Which of the following coordination compounds can exist in the greatest number of isomeric forms?
Answer: [Co(en)2Cl2]+
[Co(en)2Cl2]+ shows geometrical isomerism (cis and trans) and the cis form is optically active (d and l), giving three isomeric forms total. The MA4B2-type [Ru(NH3)4Cl2]+ gives only cis/trans (two), and [Co(NH3)5Cl]2+ gives one.
Q12. In the complex [Fe(CO)5], the Fe–C bond is best described as having which type of bonding?
Answer: both sigma and pi bonding
In Fe(CO)5 the M-C bond is synergic: CO donates its lone pair into a metal orbital (sigma bond) while filled metal d-orbitals back-donate into the CO pi* orbital (pi bond). Hence the Fe-C bond has both sigma and pi character.
Q13. Which of the following chloro complexes can be explained by dsp2 hybridization?
Answer: [PdCl4]2−
The complex [PdCl4]2− involves palladium in a +2 oxidation state, which allows for dsp2 hybridization, resulting in a square planar geometry. This hybridization occurs because palladium can utilize its d orbitals along with s and p orbitals to accommodate the four chloride ligands.
Answer: L1 < L3 < L2 < L4
The correct option indicates that as the ligand field strength increases, the absorption shifts from red (lower energy) to blue (higher energy), which corresponds to the order of ligands L1, L3, L2, and L4. This sequence reflects the increasing ability of the ligands to split the d-orbitals of the metal ion, thus affecting the energy of the absorbed light.
Q15. Which of the following coordination compounds is employed to suppress tumour growth?
Answer: Cis-platin
Cis-platin, cis-[Pt(NH3)2Cl2], is used in chemotherapy to suppress tumour growth by binding DNA. Trans-platin is inactive, and [Ph3P]3RhCl is Wilkinson's hydrogenation catalyst.
Answer: tetrahedral and square planar
[NiCl2(PEt2Ph)2] shows a tetrahedral <-> square planar equilibrium. The tetrahedral form has two unpaired electrons (paramagnetic), while the square planar form is spin-paired (diamagnetic).
Q17. Which of the following complexes will produce a white precipitate with aqueous AgNO3?
Answer: [Pt(NH3)6]Cl4
A white precipitate (AgCl) forms only from chloride present as a counter-ion. In [Pt(NH3)6]Cl4 all four Cl- are ionizable and precipitate as AgCl; in the others Cl is coordinated (or there is no Cl), so no white precipitate.
Answer: dsp2, zero
The complex [NiL4]2− involves nickel in a +2 oxidation state, leading to a d8 electron configuration. In this case, the dsp2 hybridization occurs due to the square planar geometry, and since all electrons are paired, there are zero unpaired electrons.
Q19. Which of the following species would not be expected to act as a ligand?
Answer: NH4+
A ligand must have a donatable lone pair. In NH4+ all four bonds use nitrogen's electrons, leaving no lone pair, so it cannot act as a ligand. NO and ethylenediamine both have donor lone pairs, so only NH4+ fails.
Q20. Which of the following organometallic compounds contains both σ-bonding and π-bonding interactions?
Answer: [PtCl3(η2-C2H4)]
In [PtCl3(eta2-C2H4)]- (Zeise's anion) ethylene bonds via sigma donation from its pi orbital to Pt and pi back-donation from filled Pt d orbitals into the alkene pi* orbital, so both sigma and pi interactions are present. Ferrocene's Cp rings bond only through pi interactions.
Q21. Which of the following complexes is the most stable?
Answer: K2[Ni(EDTA)]
K2[Ni(EDTA)] is the most stable complex due to the chelating effect of the EDTA ligand, which forms multiple bonds with the nickel ion, enhancing stability through a more favorable enthalpic and entropic interaction compared to the other complexes.
Q22. Which of the following species would contain the greatest number of unpaired electrons?
Answer: A coordination compound having a magnetic moment of 5.92 B.M.
The magnetic moment of 5.92 B.M. indicates the presence of five unpaired electrons, which is the maximum for a transition metal complex, confirming that this option has the greatest number of unpaired electrons compared to the others.
Answer: 10
The complex can form multiple stereoisomers due to the presence of two unsymmetrical bidentate ligands and two monodentate ligands, leading to a total of 10 distinct arrangements. Each arrangement can exist as a pair of enantiomers, resulting in 10 enantiomeric pairs.
Answer: +3, 0, and +6
The oxidation number of Cr in [Cr(H2O)6]Cl3 is +3 because the complex is neutral and the water molecules are neutral ligands. In [Cr(C6H6)2], Cr has an oxidation state of 0 since benzene is a neutral ligand and does not contribute to the charge. Finally, in K2[Cr(CN)2(O2)(O2)(NH3)], Cr is +6 as the overall charge of the complex must balance with the contributions from the ligands and the potassium ions.
Q25. In CuSO4·5H2O, how many water molecules are not directly bonded to the copper ion?
Answer: 1
In CuSO4·5H2O, one water molecule is directly coordinated to the copper ion, while the remaining four water molecules are not bonded directly to the copper but are part of the crystal structure, making the correct answer 1.
Answer: c, a
The CN− ion acts as a complexing agent by forming stable complexes with metal ions, and it also exhibits reducing properties by donating electrons. However, it does not typically function as an oxidizing agent.
Q27. The correct order of magnetic moments (spin only values in B.M.) among the following is
Answer: [MnCl4]2− > [CoCl4]2− > [Fe(CN)6]4−
The correct order reflects the number of unpaired electrons in each complex, with [MnCl4]2− having the highest spin state due to its d5 configuration, followed by [CoCl4]2− with d7 and unpaired electrons, and [Fe(CN)6]4− being low-spin d6 with no unpaired electrons.
Q28. The equation which is balanced and represents correct product(s) is:
Answer: [Mg(H2O)6]2+ + (EDTA)4− —excess NaOH→ [Mg(EDTA)]2− + 6H2O
[Mg(H2O)6]^2+ + (EDTA)^4- -> [Mg(EDTA)]^2- + 6H2O is correctly balanced in mass and charge (-2 = -2) and is a genuine complexometric reaction. The Li2O/KCl swap does not occur, the [CoCl(NH3)5] equation is charge-unbalanced, and CuSO4 + KCN gives Cu(I) as K3[Cu(CN)4], not K2[Cu(CN)4].
Q29. Identify the complex that is formed as an outer-orbital complex.
Answer: [Ni(NH₃)₆]²⁺
The complex [Ni(NH₃)₆]²⁺ is an outer-orbital complex because nickel in this oxidation state utilizes its 3d and 4s orbitals for bonding, allowing for the formation of a six-coordinate complex with a higher coordination number.
Answer: Chlorophyll is the green pigment found in plants and contains calcium
Chlorophyll is actually a complex molecule that contains magnesium, not calcium, which is essential for its role in photosynthesis. This makes the statement incorrect.
Answer: [Co(en)2Cl2]+
The compound [Co(en)2Cl2]+ can exhibit both geometric and optical isomerism due to the presence of the bidentate ligand ethylenediamine (en), which allows for multiple arrangements of the ligands around the cobalt center, leading to a greater variety of isomeric forms compared to the other options.
Q32. What is the oxidation number of chromium in the complex ion [Cr(NH3)4Cl2]+ ?
Answer: +3
In the complex ion [Cr(NH3)4Cl2]+, ammonia (NH3) is a neutral ligand contributing no charge, while each chloride (Cl) has a charge of -1. To balance the overall charge of +1 for the complex, chromium must have an oxidation state of +3.
Q33. What is the IUPAC name of the coordination compound K3[Fe(CN)6]?
Answer: Potassium hexacyanoferrate (III)
The correct option is right because the compound contains iron in the +3 oxidation state, indicated by the 'III' in the name, and 'hexacyanoferrate' correctly describes the complex ion with six cyanide ligands coordinated to the iron.
Q34. Which one of the following complexes exhibits optical isomerism?
Answer: [Cr(C2O4)3]3−
[Cr(C2O4)3]3− exhibits optical isomerism because it has a chiral arrangement of ligands around the chromium center, allowing for non-superimposable mirror images. In contrast, the other complexes either have symmetrical arrangements or do not possess chirality.
Q35. Among the following cyanide complexes, which one would show the least paramagnetic character?
Answer: [Co(CN)6]3−
[Co(CN)6]3− has a low-spin configuration due to the strong field strength of cyanide ligands, resulting in all electrons being paired and thus exhibiting the least paramagnetic character compared to the other complexes.
Q36. What is the IUPAC name of the coordination compound [Co(NO2)(NH3)5]Cl2?
Answer: nitrito-N-pentaamminecobalt(III) chloride
In [Co(NO2)(NH3)5]Cl2 the complex ion is +2 (balances 2 Cl-); with NO2 as -1 and NH3 neutral, Co is +3. Bound through N it is the nitrito-N (nitro) ligand, giving pentaamminenitrito-N-cobalt(III) chloride. Only the cobalt(III) option is consistent with the oxidation state.
Answer: two, tetrahedral
Nickel in the +2 oxidation state has the electron configuration of [Ar] 3d8, which results in two unpaired electrons in its d-orbitals. The tetrahedral geometry of the complex arises from the arrangement of the monodentate ligands around the central nickel ion, minimizing electron pair repulsion.
Q38. In Fe(CO)5, the bond between iron and carbon has which type of bonding character?
Answer: both sigma and pi bonding
In Fe(CO)5, the bonding between iron and carbon involves both sigma and pi interactions. The sigma bond is formed by the overlap of the iron's d-orbitals with the carbon's sp-hybridized orbitals, while the pi bonds arise from the donation of electron density from the filled p-orbitals of carbon to the empty d-orbitals of iron, allowing for a stable complex.
Answer: One
An octahedral complex with a Ca2+ ion can be formed by a single EDTA molecule, as EDTA acts as a hexadentate ligand, coordinating through its six donor atoms to the metal ion.
Q40. Which one of the following complexes has square planar geometry?
Answer: [PtCl4]2−
[PtCl4]2− has a square planar geometry due to the presence of a d8 electron configuration in platinum, which allows for this arrangement to minimize electron pair repulsion and achieve stability.
Answer: 6 and 3
The coordination number of 6 indicates that element E is surrounded by six donor atoms from the ligands, which include two ethylenediamine (en) molecules and one oxalate (C2O4) ion. The oxidation state of 3 for element E is determined by balancing the overall charge of the complex with the charges contributed by the ligands and the nitrate counterion.
Answer: [Co(CN)6]3−
The complex [Co(CN)6]3− has the largest value of Δo because cyanide (CN−) is a strong field ligand that causes greater splitting of the d-orbitals compared to the other ligands listed, leading to a higher energy difference between the split d-orbitals.
Q43. Which of the following shows optical isomerism ?
Answer: [Co(en)2(NH3)2]3+
The complex [Co(en)2(NH3)2]3+ exhibits optical isomerism because it contains two bidentate ethylenediamine (en) ligands that create a chiral environment, allowing for non-superimposable mirror images. This characteristic is essential for optical isomerism, which is not present in the other complexes listed.
Q44. Which of the following pairs represent linkage isomers?
Answer: [Pd(PPh3)2(NCS)2] and [Pd(PPh3)2(SCN)2]
The correct option represents linkage isomers because the two compounds differ in the way the thiocyanate ligand is bonded to the metal center; one is bonded through the sulfur atom (NCS) while the other is bonded through the nitrogen atom (SCN), resulting in distinct chemical properties.
Answer: [Co(NH3)6]Cl3
The correct option, [Co(NH3)6]Cl3, indicates that all six ammonia ligands are coordinated to the cobalt ion, which is consistent with the stoichiometry of the reaction and the amount of chloride ions released, as evidenced by the formation of AgCl from the chloride ions.
Q46. Which one of the following has an optical isomer? (en = ethylenediamine)
Answer: [Co(en)3]3+
The complex [Co(en)3]3+ has three bidentate ethylenediamine ligands, creating a chiral environment around the cobalt center, which results in optical isomerism. In contrast, the other options either have a symmetrical arrangement or do not contain enough bidentate ligands to create chirality.
Q47. Which one of the following complex ions has geometrical isomers ? (en = ethylenediamine)
Answer: [Co(NH3)2(en)2]3+
The complex ion [Co(NH3)2(en)2]3+ can exhibit geometrical isomerism due to the presence of two different types of ligands (NH3 and en) and the square planar geometry around the cobalt center, allowing for different spatial arrangements of the ligands.
Q48. Which among the following will be named as dibromido bis(ethylenediamine) chromium (III) bromide?
Answer: [Cr(en)2Br2]Br
The correct option, [Cr(en)2Br2]Br, indicates that there are two ethylenediamine (en) ligands and two bromide (Br) ligands coordinated to the chromium ion, which aligns with the naming convention of dibromido bis(ethylenediamine) chromium (III) bromide.
Q49. Which of the following complex species is not expected to exhibit optical isomerism ?
Answer: [Co(NH3)3Cl3]
[Co(NH3)3Cl3] does not exhibit optical isomerism because it has a symmetrical arrangement of ligands, leading to a lack of non-superimposable mirror images, which is a requirement for optical isomerism.
Answer: L1 < L3 < L2 < L4
Crystal-field splitting (and hence ligand strength) increases as the energy of absorbed light increases, i.e. as wavelength decreases. Wavelengths run red > yellow > green > blue, so field strength runs L4(blue) > L2(green) > L3(yellow) > L1(red). Increasing order of ligand strength: L1 < L3 < L2 < L4.