JEE Chemistry Coordination Compounds MCQs Set 08

Multiple Choice Questions (MCQs) for JEE Chemistry: Coordination Compounds

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Multiple Choice Questions

Question 1: Select the correct statement:
(a) potassium ferrocyanide and potassium ferricyanide can be differentiated by measuring the solid state magnetic moment
(b) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br can be differentiated by adding aqueous BaCl2
(c) [Co(NH3)5Cl]Br and [Co(NH3)5Br]Cl can be differentiated by adding aqueous AgNO3
(d) all of these
Show Answer & Explanation

Answer: (d) all of these

Explanation:
1. (1) Ferrocyanide is diamagnetic, ferricyanide paramagnetic
2. (2) Only the first gives BaSO4
3. (3) The first gives cream AgBr, the second white AgCl.

Teacher's Note:
1. Each statement is true.
2. Use the ionisable ion to distinguish ionisation isomers.

Question 2: Which of the following complexes does not exhibit geometrical isomerism?
(a) [PtCl2(NH3)2]
(b) [PdCl2BrI]
(c) [Pt(NH3)(py)(Cl)(Br)]
(d) [Pt(NH3)3Br]+
Show Answer & Explanation

Answer: (d) [Pt(NH3)3Br]+

Explanation:
1. Square planar Ma3b has only one arrangement.

Teacher's Note:
1. Ma2b2, Ma2bc and Mabcd show geometrical isomerism.
2. The last has only one distinct ligand different from three.

Question 3: The pair of complexes which exhibit geometrical isomerism is
(a) [Cr(NH3)4Cl2]+ and [Co(NH3)4Cl2]+
(b) [Cr(NH3)5Cl]2+ and [Co(NH3)5Cl]2+
(c) [Cr(NH3)6]3+ and [Co(NH3)6]3+
(d) all of the above show geometrical isomerism
Show Answer & Explanation

Answer: (a) [Cr(NH3)4Cl2]+ and [Co(NH3)4Cl2]+

Explanation:
1. Octahedral M a4 b2 complexes exist as cis and trans
2. M a5 b and M a6 have no geometrical isomers.

Teacher's Note:
1. Two different ligands in 4 : 2 ratio give isomers.
2. The others are single arrangements.

Question 4: Which of the following is a π-acid ligand?
(a) NH3
(b) CO
(c) glycinate
(d) ethylenediamine
Show Answer & Explanation

Answer: (b) CO

Explanation:
1. CO accepts electrons from filled metal d orbitals into its empty π* orbitals, so it is a π-acid ligand.

Teacher's Note:
1. NH3 and en are only σ donors.
2. π-acid ligands stabilise low oxidation states.

Question 5: Which of the following statements is/are wrong? (a) Al4C3 is an organometallic compound, (b) metal carbonyls are organometallic compounds, (c) TEL is a π-bonded organometallic compound, (d) Frankland reagent is a σ-bonded organometallic compound
(a) (c) and (d)
(b) (a) and (c)
(c) (a) and (b)
(d) all are correct
Show Answer & Explanation

Answer: (b) (a) and (c)

Explanation:
1. (a) is wrong: Al4C3 is an ionic carbide
2. (c) is wrong: tetraethyl lead is σ-bonded
3. (b) and (d) are correct.

Teacher's Note:
1. Carbides are not organometallic.
2. σ-bonded: R–M with simple alkyl groups.

Question 6: Compounds containing one or more metal–carbon bonds are called
(a) organic compounds
(b) complex compounds
(c) metal carbides
(d) organometallic compounds
Show Answer & Explanation

Answer: (d) organometallic compounds

Explanation:
1. By definition, an organometallic compound has at least one metal–carbon bond.

Teacher's Note:
1. Carbides are ionic or interstitial compounds, not organometallic.
2. Examples: Grignard reagents, ferrocene.

Question 7: Which of the following is used as a heterogeneous catalyst?
(a) Wilkinson's catalyst
(b) tetraethyl lead
(c) Ziegler–Natta catalyst
(d) Grignard reagent
Show Answer & Explanation

Answer: (c) Ziegler–Natta catalyst

Explanation:
1. The Ziegler–Natta catalyst (TiCl4 + Al(C2H5)3) is a solid in the hydrocarbon medium, so it is heterogeneous
2. Wilkinson's catalyst is homogeneous.

Teacher's Note:
1. Homogeneous catalysts dissolve in the reaction medium.
2. Grignard reagent and TEL are not catalysts.

Question 8: Which of the following statements is true?
(a) FeCO3 and Fe3C are organometallic compounds
(b) in ferrocene the ligand is cyclopentadienyl
(c) Pb(C2H5)4 is π-bonded organometallic
(d) in Zeise's salt the central atom is sp3 hybridised
Show Answer & Explanation

Answer: (b) in ferrocene the ligand is cyclopentadienyl

Explanation:
1. Ferrocene is bis(cyclopentadienyl)iron(II)
2. Carbonate and carbide have no metal–carbon σ or π organometallic bond, TEL is σ-bonded, and Pt in Zeise's salt is dsp2.

Teacher's Note:
1. The cyclopentadienyl ligand is π-bonded to iron.
2. Zeise's salt is square planar.

Question 9: Synergic bonding involves
(a) transfer of electrons from ligands to the metal
(b) transfer of electrons from filled metal orbitals to anti-bonding orbitals of the ligands
(c) both of the above
(d) none of these
Show Answer & Explanation

Answer: (c) both of the above

Explanation:
1. In synergic bonding, the ligand donates a σ pair to the metal and the metal donates electrons from filled d orbitals into the ligand's π* orbitals
2. So both transfers occur.

Teacher's Note:
1. The two effects strengthen each other.
2. This is seen in metal carbonyls.

Question 10: An organometallic compound formed during the purification of a metal is
(a) Ni(CO)4
(b) Pb(C2H5)4
(c) Li–C4H9
(d) Na2[Ni(CN)4]
Show Answer & Explanation

Answer: (a) Ni(CO)4

Explanation:
1. In the Mond process, impure nickel reacts with CO to give volatile Ni(CO)4, which decomposes to pure nickel on heating.

Teacher's Note:
1. Ni(CO)4 is a metal carbonyl.
2. Pure Ni is regenerated at about 460 K.

Question 11: Which of the following is not an organometallic compound?
(a) (C2H5)2Zn
(b) CH3B(OCH3)2
(c) B(OCH3)3
(d) Ni(CO)4
Show Answer & Explanation

Answer: (c) B(OCH3)3

Explanation:
1. B(OCH3)3 has B–O–C bonds only, so it is not organometallic
2. The others contain Zn–C, B–C and Ni–C bonds.

Teacher's Note:
1. Alkoxides and esters have M–O–C bonds.
2. Look for a direct metal–carbon bond.

Question 12: The brown ring test for nitrites and nitrates is due to the formation of a complex ion with the formula
(a) [Fe(H2O)5NO+]2+
(b) [Fe(H2O)6]2+
(c) [Fe(H2O)(NO)5]2−
(d) [Fe(NO)(CN)5]2+
Show Answer & Explanation

Answer: (a) [Fe(H2O)5NO+]2+

Explanation:
1. Fe2+ reduces nitrate to NO, which replaces a water ligand: [Fe(H2O)5NO]2+, the brown ring.

Teacher's Note:
1. The test uses freshly prepared FeSO4 and conc. H2SO4.
2. The ring forms at the junction of the two liquids.

Question 13: When potassium ferrocyanide is added to a ferric salt, a Prussian blue colour is obtained. The compound formed is
(a) K3[Fe(CN)6]
(b) Fe3[Fe(CN)6]4
(c) FeSO4[Fe(CN)6]
(d) Fe4[Fe(CN)6]3
Show Answer & Explanation

Answer: (d) Fe4[Fe(CN)6]3

Explanation:
1. Fe3+ with [Fe(CN)6]4− gives ferric ferrocyanide, Fe4[Fe(CN)6]3, Prussian blue.

Teacher's Note:
1. Charge balance: 4(+3) = 3(+4).
2. This is a test for Fe3+.

Question 14: Hypo is used in photography because it is
(a) a strong reducing agent
(b) a strong oxidising agent
(c) a strong complexing agent
(d) a photosensitive compound
Show Answer & Explanation

Answer: (c) a strong complexing agent

Explanation:
1. Thiosulphate dissolves unexposed AgBr by forming the soluble complex [Ag(S2O3)2]3−.

Teacher's Note:
1. This process is called fixing.
2. Hypo is sodium thiosulphate.

Question 15: The solubility of AgBr in hypo solution is due to the formation of
(a) Ag2SO3
(b) Ag2S2O3
(c) [Ag(S2O3)]−
(d) [Ag(S2O3)2]3−
Show Answer & Explanation

Answer: (d) [Ag(S2O3)2]3−

Explanation:
1. AgBr + 2S2O32− → [Ag(S2O3)2]3− + Br−, a soluble complex.

Teacher's Note:
1. Ag+ has coordination number 2.
2. Thiosulphate binds through sulphur.

Question 16: Which of the following is related to Nessler's reagent?
(a) PtCl4 + KCl →
(b) AgCl + NH3 →
(c) AgBr + Na2S2O3 →
(d) HgI2 + KI →
Show Answer & Explanation

Answer: (d) HgI2 + KI →

Explanation:
1. Nessler's reagent is K2[HgI4], formed from HgI2 and excess KI.

Teacher's Note:
1. It gives a brown precipitate with NH4+.
2. Hg(II) is tetrahedral here.

Question 17: K4[Fe(CN)6] reacts with FeCl3 to form
(a) K3Fe(CN)6
(b) K4[Fe(CN)3Cl3]
(c) K3[Fe(CN)5Cl]
(d) KFe[Fe(CN)6]
Show Answer & Explanation

Answer: (d) KFe[Fe(CN)6]

Explanation:
1. Ferrocyanide plus Fe3+ gives the soluble Prussian blue, KFe[Fe(CN)6].

Teacher's Note:
1. The insoluble form is Fe4[Fe(CN)6]3.
2. The reaction is a test for ferric ion.

Question 18: A reagent used for identifying the nickel ion is
(a) potassium ferrocyanide
(b) phenolphthalein
(c) dimethylglyoxime
(d) EDTA
Show Answer & Explanation

Answer: (c) dimethylglyoxime

Explanation:
1. Ni2+ in ammoniacal solution with dimethylglyoxime gives a bright red precipitate.

Teacher's Note:
1. The red complex is [Ni(dmg)2].
2. It is a square planar complex.

Question 19: Which of the following statements is false for the nickel–dimethylglyoximate complex?
(a) the stability of the complex is only due to the presence of intramolecular hydrogen bonding
(b) the complex is stable only because dimethylglyoxime is a stronger ligand
(c) the complex is stable as it has five-membered chelate rings as well as intramolecular hydrogen bonding
(d) (1) and (2) both
Show Answer & Explanation

Answer: (d) (1) and (2) both

Explanation:
1. Stability arises from both chelate rings and hydrogen bonding together
2. Statements 1 and 2 each claim a single cause, so they are false and (4) is the answer.

Teacher's Note:
1. The word only makes statements 1 and 2 false.
2. Statement 3 gives the complete reason.

Question 20: CuCl2 + K4[Fe(CN)6] gives a chocolate brown precipitate (X). Which statement about X is correct?
(a) its IUPAC name is copper(I) hexacyanoferrate(II)
(b) it reacts with excess potassium cyanide forming another soluble complex which has tetrahedral geometry
(c) it has a spin-only magnetic moment equal to √36 BM
(d) both 2 and 3
Show Answer & Explanation

Answer: (b) it reacts with excess potassium cyanide forming another soluble complex which has tetrahedral geometry

Explanation:
1. X is Cu2[Fe(CN)6], copper(II) hexacyanoferrate(II)
2. With excess KCN it forms soluble K3[Cu(CN)4] (after reduction), which is tetrahedral
3. It has only two unpaired electrons in total (one per Cu2+), not √36 BM.

Teacher's Note:
1. Statement 1 has the wrong oxidation state.
2. Spin-only moments come from √[n(n + 2)].

Question 21: DMG + Ni(II) salt + NH4OH gives a red precipitate. Which statement about this red precipitate is wrong?
(a) it is a non-ionic complex
(b) it involves intramolecular hydrogen bonding
(c) Ni(II) is sp3 hybridised
(d) it is a diamagnetic complex
Show Answer & Explanation

Answer: (c) Ni(II) is sp3 hybridised

Explanation:
1. The red complex is square planar with dsp2 hybridisation, so the statement that Ni(II) is sp3 hybridised is wrong.

Teacher's Note:
1. Square planar d8 low spin is diamagnetic.
2. The complex is neutral.

Question 22: Wilkinson's catalyst reacts with H2 to give an octahedral complex in which Rh (Z = 45) has the configuration t2g2,2,2 eg0,0. Which statement about this new complex is correct?
(a) it is paramagnetic
(b) its IUPAC name is chlorodihydridotris(triphenylphosphine)rhodium(III)
(c) the hybridisation of Rh(I) is d2sp3
(d) none of these
Show Answer & Explanation

Answer: (b) its IUPAC name is chlorodihydridotris(triphenylphosphine)rhodium(III)

Explanation:
1. The product is [RhH2Cl(PPh3)3]: ligands alphabetical: chlorido, dihydrido, tris(triphenylphosphine), with Rh(III)
2. All six t2g electrons are paired: diamagnetic.

Teacher's Note:
1. The configuration t2g6 has no unpaired electron.
2. Oxidative addition of H2 raises Rh from +1 to +3.

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Multiple Choice Questions (MCQs) for JEE Chemistry Coordination Compounds

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