JEE Chemistry Coordination Compounds MCQs Set 05

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Chapter-wise Objective Questions: Coordination Compounds

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Multiple Choice Questions

Question 1: Ziegler–Natta catalyst is
(a) [RhCl(PPh3)3]
(b) Al(C2H5)3 + TiCl4
(c) K[PtCl3(η2–C2H4)]
(d) Pt/Rh
Show Answer & Explanation

Answer: (b) Al(C2H5)3 + TiCl4

Explanation:
1. The Ziegler–Natta catalyst for polymerisation of alkenes is triethylaluminium with titanium tetrachloride.

Teacher's Note:
1. [RhCl(PPh3)3] is Wilkinson's catalyst (hydrogenation).
2. The third is Zeise's salt.

Question 2: The number of coordination isomers of [Cu(NH3)4][PtCl4] other than the compound itself (ignoring geometrical isomers) is
(a) 3
(b) 6
(c) 5
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. Distribute 4 NH3 and 4 Cl− between Cu and Pt (each with CN = 4)
2. Cu can carry 0, 1, 2 or 3 NH3 (remaining Cl− on Pt), giving 4 other combinations: [Pt(NH3)4][CuCl4], [Cu(NH3)Cl3][Pt(NH3)3Cl], [Cu(NH3)2Cl2][Pt(NH3)2Cl2], [Cu(NH3)3Cl][Pt(NH3)Cl3].

Teacher's Note:
1. Total five compounds including the original.
2. Coordination isomerism needs two metal centres.

Question 3: Which of the following is/are organometallic compounds? (I) Grignard reagent, (II) sodium methoxide, (III) trimethylboron, (IV) tetraethyl lead (TEL)
(a) only I
(b) I and II
(c) I, II and IV
(d) I, III and IV
Show Answer & Explanation

Answer: (d) I, III and IV

Explanation:
1. Organometallic compounds have a metal–carbon bond: Grignard reagent (Mg–C), B(CH3)3 (B–C) and Pb(C2H5)4 (Pb–C)
2. Sodium methoxide has Na–O–C, no metal–carbon bond.

Teacher's Note:
1. Alkoxides are not organometallic.
2. Boron is treated as a metalloid in this definition.

Question 4: Which set is an example of coordination isomerism?
(a) [Co(NH3)6][Cr(CN)6] and [Co(CN)6][Cr(NH3)6]
(b) [Cr(H2O)5Cl]Cl2·H2O and [Cr(H2O)4Cl2]Cl·2H2O
(c) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br
(d) [Pt(NH3)2Cl2] and [Pt(NH3)4][PtCl4]
Show Answer & Explanation

Answer: (a) [Co(NH3)6][Cr(CN)6] and [Co(CN)6][Cr(NH3)6]

Explanation:
1. Coordination isomers arise by exchange of ligands between the cationic and anionic complex parts
2. Set 1 shows this; set 2 is hydrate isomerism, set 3 ionisation isomerism, and set 4 polymerisation isomerism.

Teacher's Note:
1. Coordination isomerism needs both a complex cation and a complex anion.
2. Look at where the ligands sit.

Question 5: According to the rule of effective atomic number, the central atom acquires
(a) inert gas configuration
(b) duplet
(c) pseudo inert electronic configuration
(d) quartet
Show Answer & Explanation

Answer: (a) inert gas configuration

Explanation:
1. The EAN rule says the metal gains electrons from ligands until its total electrons equal those of the next noble gas.

Teacher's Note:
1. There are exceptions, e.g. [Fe(CN)6]3− has 35 electrons.
2. The rule works best for carbonyls.

Question 6: S1: [Cr(NH3)6]3+ is an inner orbital complex with CFSE of −1.2Δo. S2: the complex formed by CN− with Fe3+ has a spin-only magnetic moment of 1.73 BM. S3: in Na2S + Na2[Fe(CN)5NO] → Na4[Fe(CN)5NOS], the oxidation state of iron is the same in reactant and product. The statements S1, S2, S3 are respectively
(a) F, T, F
(b) T, T, F
(c) T, T, T
(d) F, F, F
Show Answer & Explanation

Answer: (c) T, T, T

Explanation:
1. S1: Cr3+ d3 gives t2g3, CFSE = −1.2Δo and d2sp3 hybridisation: true
2. S2: [Fe(CN)6]3− is low spin d5 with one unpaired electron, μ = √3 = 1.73 BM: true
3. S3: iron is +2 in both: true.

Teacher's Note:
1. NO+ plus S2− gives the NOS− ligand, so iron stays +2.
2. μ = √[n(n + 2)].

Question 7: A reagent used for identifying nickel ion is
(a) potassium ferrocyanide
(b) phenolphthalein
(c) dimethylglyoxime
(d) EDTA
Show Answer & Explanation

Answer: (c) dimethylglyoxime

Explanation:
1. Dimethylglyoxime gives a bright red precipitate of [Ni(dmg)2] in ammoniacal solution.

Teacher's Note:
1. The complex has five-membered chelate rings and hydrogen bonding.
2. It is also used to estimate Ni gravimetrically.

Question 8: CN− is called an ambidentate ligand because
(a) it has multiple bonds
(b) carbon has a negative charge
(c) it forms a chelate
(d) both C and N can donate a lone pair
Show Answer & Explanation

Answer: (d) both C and N can donate a lone pair

Explanation:
1. An ambidentate ligand has two different donor atoms, either of which can bind the metal
2. In CN− both carbon and nitrogen carry a lone pair (M–CN or M–NC).

Teacher's Note:
1. Other ambidentate ligands: NO2−, SCN−.
2. Ambidentate ligands cause linkage isomerism.

Question 9: In which process is an organometallic compound formed with the help of another organometallic compound?
(a) SnCl4 + butyl lithium → Bu4Sn + 4LiCl
(b) C2H5I + 4Pb/Na alloy → (C2H5)4Pb + 4NaI + 3Pb
(c) Ni + 4CO → [Ni(CO)4]
(d) K2[PtCl4] + CH2=CH2 → K[PtCl3(C2H4)] + KCl
Show Answer & Explanation

Answer: (a) SnCl4 + butyl lithium → Bu4Sn + 4LiCl

Explanation:
1. Butyllithium is itself organometallic (Li–C bond) and transfers butyl groups to tin to make tetrabutyltin
2. The other reactions start from non-organometallic reagents.

Teacher's Note:
1. Organolithium and Grignard reagents are common organometallic reagents.
2. Check which reactant has a metal–carbon bond.

Question 10: Which of the following gives the maximum number of isomers?
(a) [Co(py)3(NH3)3]3+
(b) [Ni(en)(NH3)4]2+
(c) [Fe(C2O4)(en)2]−
(d) [Cr(NO2)2(NH3)4]+
Show Answer & Explanation

Answer: (d) [Cr(NO2)2(NH3)4]+

Explanation:
1. (1) 2 geometrical isomers; (2) 1; (3) 3 stereoisomers (cis pair + trans)
2. (4) cis and trans, each with nitro/nitrito linkage combinations: 6 isomers, the maximum.

Teacher's Note:
1. Count both geometrical and linkage isomers for ambidentate NO2−.
2. The complex with an ambidentate ligand has the most isomers.

Question 11: Which of the compounds (a) PtCl4·4NH3, (b) CoCl3·3NH3, (c) impure chloroform, (d) vinyl chloride, (e) 4-chlorobenzyl chloride (ring Cl and CH2Cl), (f) carnallite are expected to yield a white precipitate with AgNO3 solution?
(a) a, c, f, e
(b) a, d, e
(c) a, b, c, f
(d) a, b, e, f
Show Answer & Explanation

Answer: (a) a, c, f, e

Explanation:
1. (a) [Pt(NH3)4Cl2]Cl2 has ionic chloride; (f) carnallite has Cl−; (c) impure chloroform contains HCl; (e) the benzylic chloride reacts with silver ion
2. (b) [Co(NH3)3Cl3] has no ionic Cl and (d) vinyl chloride has an unreactive C–Cl bond.

Teacher's Note:
1. Only free chloride (or a very reactive C–Cl bond) gives AgCl.
2. Aryl and vinyl halides do not react.

Question 12: The coordination number of Co in [CoF6]3− is
(a) 4
(b) 5
(c) 6
(d) 8
Show Answer & Explanation

Answer: (c) 6

Explanation:
1. Six F− ions are bonded directly to cobalt.

Teacher's Note:
1. Coordination number = number of donor atoms attached to the metal.
2. Each fluoride is monodentate.

Question 13: [Rh(en)2Cl2][Rh(en)Cl4] and [Rh(en)3][RhCl6] show which type of isomerism?
(a) linkage
(b) coordination
(c) ligand
(d) ionisation
Show Answer & Explanation

Answer: (b) coordination

Explanation:
1. The same ligands are distributed differently between a complex cation and a complex anion: coordination isomerism.

Teacher's Note:
1. Both ions contain Rh.
2. Ionisation isomers differ in the counter ion.

Question 14: The number of geometrical isomers of [Co(NH3)3(NO2)3] is
(a) 0
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. M a3 b3 octahedral complexes have two geometrical isomers: facial (fac) and meridional (mer).

Teacher's Note:
1. fac: three identical ligands on one face; mer: on a meridian.
2. Neither isomer is optically active.

Question 15: In [Co(H2O)6]2+, the number of electrons in the t2g and eg orbitals (weak field) respectively is
(a) 6 and 0
(b) 5 and 2
(c) 4 and 3
(d) 3 and 4
Show Answer & Explanation

Answer: (b) 5 and 2

Explanation:
1. Co2+ is d7 and H2O is a weak field ligand
2. High spin d7: t2g5 eg2.

Teacher's Note:
1. High spin: fill eg singly before pairing in t2g.
2. Three unpaired electrons, μ = √15 = 3.87 BM.

Question 16: Which of the following entities forms a racemic mixture when mixed in equimolar concentration? [Cr(ox)3]3−, cis-[Co(en)2Cl2]+, [Co(en)Cl4]−
(a) [Cr(ox)3]3− only
(b) [Cr(ox)3]3− and cis-[Co(en)2Cl2]+ only
(c) cis-[Co(en)2Cl2]+ and [Co(en)Cl4]− only
(d) all of them
Show Answer & Explanation

Answer: (b) [Cr(ox)3]3− and cis-[Co(en)2Cl2]+ only

Explanation:
1. Racemic mixtures form from optically active (chiral) complexes
2. [Cr(ox)3]3− and the cis form of [Co(en)2Cl2]+ are chiral
3. [Co(en)Cl4]− has a plane of symmetry and is achiral.

Teacher's Note:
1. A racemic mixture has equal amounts of the d and l forms.
2. The trans form of [Co(en)2Cl2]+ is achiral.

Question 17: A platinum complex gives a precipitate with AgNO3 equal to 1/4 of the total chlorine of the compound. The most likely formula is
(a) PtCl4·6H2O
(b) PtCl4·5H2O
(c) PtCl2·2H2O
(d) PtCl4·3H2O
Show Answer & Explanation

Answer: (d) PtCl4·3H2O

Explanation:
1. Pt(IV) has coordination number 6
2. PtCl4·3H2O is [Pt(H2O)3Cl3]Cl: one of the four chlorines is ionic = 1/4.

Teacher's Note:
1. PtCl4·5H2O would be [Pt(H2O)5Cl]Cl3 with 3/4 ionic.
2. Fill the coordination sphere first, the rest are ionic.

Question 18: Match the complexes (A) [Pt(NH3)2Cl2], (B) K3[Cu(CN)4], (C) [Ca(EDTA)]2−, (D) K2[Cd(CN)4] with the properties (P) reacts with H2S gas, (Q) estimation of hardness, (R) removal of lead poisoning, (S) anticancer drug, (T) coordination number of the metal is 4. The correct pairing is
(a) A–S,T; B–T; C–Q,R; D–P,T
(b) A–S; B–T; C–Q,R; D–P,T
(c) A–P; B–T; C–Q,R; D–P,T
(d) A–R; B–T; C–Q,R; D–P,T
Show Answer & Explanation

Answer: (a) A–S,T; B–T; C–Q,R; D–P,T

Explanation:
1. Cisplatin is an anticancer drug and has CN 4
2. [Cu(CN)4]3− is very stable: CN 4, no CuS with H2S
3. [Ca(EDTA)]2− is used for hardness estimation and lead poisoning
4. [Cd(CN)4]2− releases enough Cd2+ to give yellow CdS with H2S: CN 4.

Teacher's Note:
1. This difference separates Cu2+ from Cd2+ in qualitative analysis.
2. EDTA binds lead and other heavy metals.

Question 19: For K3[CoF6], the incorrect statement is
(a) it is a high spin complex
(b) its magnetic moment is √24 BM
(c) the primary valency of Co is six
(d) the hybridisation of [CoF6]3− is sp3d2
Show Answer & Explanation

Answer: (c) the primary valency of Co is six

Explanation:
1. Co3+ d6 with weak F−: four unpaired electrons, μ = √24 BM, sp3d2 high spin
2. The primary valency of cobalt is its oxidation state, three, and six is the secondary valency, so the statement is wrong.

Teacher's Note:
1. Primary valency = oxidation state; secondary valency = coordination number.
2. Werner's theory uses both terms.

Question 20: Which of (a) [Cu(NH3)4]2+, (b) XeF4, (c) [Ni(CO)4], (d) [NiCl4]2−, (e) [Ni(CN)4]2−, (f) CuSO4·5H2O is square planar and has no unpaired electron?
(a) a, e, f
(b) a, b, c
(c) b, e
(d) b, c, f
Show Answer & Explanation

Answer: (c) b, e

Explanation:
1. XeF4 (sp3d2, two lone pairs) and [Ni(CN)4]2− (dsp2) are square planar and diamagnetic
2. [Cu(NH3)4]2+ has one unpaired electron; [Ni(CO)4] and [NiCl4]2− are tetrahedral.

Teacher's Note:
1. Square planar d8 low spin complexes are diamagnetic.
2. Cu(II) always has one unpaired electron.

Question 21: Match the complexes (A) [Ni(dmg)2], (B) [Ca(EDTA)]2−, (C) [Fe(en)3]Cl3, (D) [Ag(S2O3)2]3− with the properties: (P) two 5-membered rings, (Q) two 6-membered rings (through H-bonds), (R) coordination number 2, (S) coordination number 6, (T) intramolecular H-bonding. The correct pairing is
(a) A–P,Q,T; B–S; C–S; D–R
(b) A–P,T; B–R; C–S; D–S
(c) A–S; B–P,Q; C–R; D–T
(d) A–P,Q,T; B–R; C–S; D–S
Show Answer & Explanation

Answer: (a) A–P,Q,T; B–S; C–S; D–R

Explanation:
1. [Ni(dmg)2]: two five-membered chelate rings from the dmg ligands, two six-membered rings closed by O–H···O hydrogen bonds
2. [Ca(EDTA)]2− and [Fe(en)3]3+ are six-coordinate
3. In [Ag(S2O3)2]3− two thiosulphate ions are bonded through one S each, so the coordination number is 2.

Teacher's Note:
1. Hydrogen bonding in Ni(dmg)2 adds extra stability and gives the red colour.
2. Coordination number counts donor atoms, not ligands.

Question 22: The longest C–O bond is present in
(a) [Mn(CO)6]+
(b) [Cr(CO)6]
(c) [V(CO)6]−
(d) the C–O bond length is the same in all
Show Answer & Explanation

Answer: (c) [V(CO)6]−

Explanation:
1. Back donation from metal d orbitals into π* of CO weakens and lengthens the C–O bond
2. More negative charge on the metal means stronger back donation
3. [V(CO)6]− has the most electron-rich metal, so its C–O bond is longest.

Teacher's Note:
1. In a cationic carbonyl back donation is least and the C–O bond is strongest.
2. Order of C–O bond length: [V(CO)6]− > [Cr(CO)6] > [Mn(CO)6]+.

Question 23: Which compound does not give the test for sulphate ion in aqueous solution?
(a) K2SO4·Al2(SO4)3·24H2O
(b) [Cu(H2O)4]SO4·H2O
(c) [CoSO4(NH3)5]Br
(d) FeSO4·(NH4)2SO4·6H2O
Show Answer & Explanation

Answer: (c) [CoSO4(NH3)5]Br

Explanation:
1. Alum, tetraaquacopper(II) sulphate and Mohr's salt release free SO42− ions
2. In [Co(SO4)(NH3)5]Br the sulphate is inside the coordination sphere and does not ionise, so no BaSO4 forms.

Teacher's Note:
1. Species inside the square bracket do not ionise.
2. Here the Br− is ionisable instead.

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Multiple Choice Questions (MCQs) for JEE Chemistry Coordination Compounds

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