CBSE Class 12 Chemistry Coordination Compounds NEET JEE Questions Set 01

Single Correct Answer

Question 1. The ligand N(CH₂CH₂NH₂)₃ is (Kerala PMT 2011)
(a) tridentate
(b) pentadentate
(c) tetradentate
(d) bidentate
(e) hexadentate
Answer:
Ans. (c)
Given ligand has four donor N-atoms and hence is tetradentate.

Teacher's Note:
1. Count the donor atoms: one tertiary N and three terminal NH₂ groups give four donor sites.

Question 2. The correct structure of ethylenediaminetetraacetic acid (EDTA) is (IIT 2010)
(a) (HOOC–CH₂)₂N–CH=CH–N(CH₂–COOH)₂
(b) (HOOC)₂N–CH₂–CH₂–N(COOH)₂
(c) (HOOC–CH₂)₂N–CH₂–CH₂–N(CH₂–COOH)₂
(d) HOOC–CH₂–NH–CH(CH₂–COOH)–CH(CH₂–COOH)–NH–CH₂–COOH
Answer:
Ans. (c)
(c) is the correct structure of EDTA: an ethylene (–CH₂–CH₂–) bridge joins two N atoms, each carrying two –CH₂COOH groups.

Teacher's Note:
1. The name tells the structure: ethylenediamine (H₂N–CH₂CH₂–NH₂) with all four N–H hydrogens replaced by acetic acid (–CH₂COOH) groups.

Question 3. Oxidation number and coordination number of silver in Tollen's reagent respectively are
(a) 1, 1
(b) 2, 1
(c) 2, 2
(d) 1, 2
Answer:
Ans. (d)
Tollen's reagent contains silver-ammonia complex ion, [Ag(NH₃)₂]⁺. Thus, oxidation number is +1 and coordination number is 2.

Teacher's Note:
1. Coordination number counts donor atoms around the metal (two NH₃), not the charge on the ion.

Question 4. Oxidation number of Cr in the following complex is :
[(H₂O)₆Cr(O₂)(O–O)Cr(H₂O)₆]³⁺
(the two Cr atoms are bridged by an O₂ group and an O–O group)
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
Ans. (b)
2x + 12(0) + (−2) + (−1) = +3
(for Cr) (for H₂O) (for O–O, peroxide ion) (for O₂, superoxide ion)
or x = +3

Teacher's Note:
1. The bridging O–O is a peroxide (O₂²⁻, −2) and the bridging O₂ is a superoxide (O₂⁻, −1); remember there are two Cr atoms.

Question 5. Which one of the following is a homoleptic complex ? (Kerala PMT 2015)
(a) tris(ethane-1, 2-diamine) cobalt (III) chloride
(b) triamminetriaquachromium (III) chloride
(c) diamminechloridonitro-N-platinum (II)
(d) dichlorido bis(ethane-1, 2-diamine) cobalt (III) chloride
(e) pentaamminecarbonato cobalt (III) chloride
Answer:
Ans. (a)
Homoleptic complexes are those in which metal atom or ion is linked to only one type of ligands. In [Co(en)₃]Cl₃, only en is attached to Co.

Teacher's Note:
1. Ions outside the bracket (counter ions) do not count; look only at the ligands inside the coordination sphere.

Question 6. Ethylene diaminetetraacetate (EDTA) ion is : (NEET 2021)
(a) hexadentate ligand with four “O” and two “N” donor atoms
(b) unidentate ligand
(c) bidentate ligand with two “N” donor atoms
(d) tridentate ligand with three “N” donor atoms
Answer:
Ans. (a)
Ethylene diaminetetraacetate (EDTA) ion is a hexadentate ligand having two donor nitrogen atoms and four donor oxygen atoms.
(−OOCCH₂)₂N–CH₂–CH₂–N(CH₂COO⁻)₂ (EDTA⁴⁻)

Teacher's Note:
1. EDTA⁴⁻ = 2 N + 4 carboxylate O = 6 donor atoms; it wraps around one metal ion forming five chelate rings.

Question 7. The formula, dichlorobis(urea)copper (II) is
(a) [Cu{O = C(NH₂)₂}]Cl₂
(b) [CuCl₂{O = C(NH₂)₂}₂]
(c) [Cu{O = C(NH₂)₂}Cl] Cl
(d) [CuCl₂{O = C(NH₂)₂}₂]
Answer:
Ans. (b)
Two chloro ligands and two (bis) urea ligands are all inside the coordination sphere and no counter ion is named, so the formula is [CuCl₂{O=C(NH₂)₂}₂].

Teacher's Note:
1. A name with no separate anion or cation word describes a neutral complex, so everything goes inside one bracket.

Question 8. IUPAC name of sodium cobaltinitrite is
(a) Sodium cobaltinitrite
(b) Sodium hexanitritocobaltate (III)
(c) Sodium hexanitritocobalt (III)
(d) Sodium hexanitritocobaltate (II).
Answer:
Ans. (b)
Sodium cobaltinitrite is Na₃[Co(ONO)₆].
Its IUPAC name is Sodium hexanitritocobaltate (III).

Teacher's Note:
1. In an anionic complex the metal name ends in “-ate”; Co here is +3 because 3Na⁺ balance six NO₂⁻.

Question 9. According to IUPAC nomenclature, sodium nitroprusside is named as
(a) Sodium nitroferricyanide
(b) Sodium nitroferrocyanide
(c) Sodium pentacyanonitrosylferrate (II)
(d) Sodium pentacyano nitrosylferrate (III)
Answer:
Ans. (d)
Sodium nitroprusside, Na₂[Fe(CN)₅NO], is named as sodium pentacyano nitrosyl ferrate (III).
In this compound, NO is present as NO⁺ and Fe has + 2 charge. On the basis of this, the actual name should be sodium pentacyanonitrosonium ferrate (II).

Teacher's Note:
1. Taking NO as a neutral ligand gives Fe +3 (the name used in this question); taking it as NO⁺ gives Fe +2.

Question 10. Which of the following has largest number of isomers ? (AMU Engg. 2011)
(a) [Ru(NH₃)₄Cl₂]⁺
(b) [Co(en)₂Cl₂]⁺
(c) [Ir(PR₃)₂H(CO)]²⁺
(d) [Co(NH₃)₅Cl]²⁺
Answer:
Ans. (b)
All given compounds have cis and trans isomers but only cis-isomer of (b) will also have optical isomers (d and l) due to presence of symmetrical didentate, en. So (b) has three isomers.

Teacher's Note:
1. [M(AA)₂X₂] gives three stereoisomers: trans plus a pair of cis enantiomers.

Question 11. Which one of the following complexes is not expected to exhibit isomerism ? (AIPMT Main 2010)
(a) [Ni(NH₃)₄(H₂O)₂]²⁺
(b) [Pt(NH₃)₂Cl₂]
(c) [Ni(NH₃)₂Cl₂]
(d) [Ni(en)₃]²⁺
Answer:
Ans. (c)
[Ni(NH₃)₂Cl₂] does not exhibit isomerism because it is a tetrahedral complex.

Teacher's Note:
1. Pt(II) complexes are square planar (cis/trans possible), but Ni(II) with weak-field Cl⁻ is tetrahedral, where all positions are adjacent.

Question 12. For the given complex [CoCl₂ (en) (NH₃)₂]⁺, the number of geometrical isomers, the number of optical isomers and total number of isomers of all type possible respectively are (DCE 2009)
(a) 2, 2 and 4
(b) 2, 2 and 3
(c) 2, 0 and 2
(d) 0, 2 and 2
Answer:
Ans. (b)
The octahedral complex of the type [M(AA)X₂Y₂], i.e., [CoCl₂ (en) (NH₃)₂]⁺ will show two cis- and trans-geometrical isomers. Out of these, cis-isomer will be optically active and shows two d- and l-forms. So the total number of possible isomers will be three.

Teacher's Note:
1. Do not add the geometrical and optical counts blindly: the cis form itself is one of the two optical isomers.

Question 13. The number of geometrical isomers that can exist for square planar [Pt(Cl) (py) (NH₃) (NH₂OH)]⁺ is (py = pyridine) (JEE Main 2015)
(a) 4
(b) 6
(c) 2
(d) 3
Answer:
Ans. (d)
A square planar complex of the type MABCD forms three isomers (A trans to B, A trans to C, A trans to D).
Two of these are cis and one is trans (with respect to any chosen pair).

Teacher's Note:
1. For [MABCD], fix ligand A and ask which of B, C or D sits opposite it: three choices give three isomers.

Question 14. The total number of possible isomers for the complex compound [Cu^II (NH₃)₄] [Pt^IICl₄] are
(a) 3
(b) 6
(c) 5
(d) 4
Answer:
Ans. (d)
The possible isomers of the given complex are [Cu (NH₃)₄] [PtCl₄], [Cu (NH₃)₃Cl] [PtCl₃ (NH₃)], [Cu (NH₃) Cl₃] [PtCl (NH₃)₃], [CuCl₄] [Pt (NH₃)₄]
Note. [Cu (NH₃)₂Cl₂] [PtCl₂ (NH₃)₂] is not possible because in this case, there will be no cation and anion.

Teacher's Note:
1. Coordination isomers exchange ligands between the cationic and anionic spheres, but each sphere must keep an ionic charge.

Question 15. Number of possible isomers for the complex [Co(en)₂Cl₂] Cl will be : (en = ethylenediamine) (RE-AIPMT 2015)
(a) 3
(b) 4
(c) 2
(d) 1
Answer:
Ans. (a)
[Co(en)₂Cl₂]⁺ exists as a cis-isomer (d and l-form) and a trans-isomer (optically inactive).
Hence, total isomers = 3.

Teacher's Note:
1. The trans isomer has a mirror plane through Co and both en rings, so it cannot be optically active.

Question 16. [Co(NH₃)₄(NO₂)₂] Cl exhibits
(a) ionization isomerism, geometrical isomerism and optical isomerism
(b) linkage isomerism, geometrical isomerism and optical isomerism
(c) linkage isomerism, ionization isomerism and optical isomerism
(d) linkage isomerism, ionization isomerism and geometrical isomerism
Answer:
Ans. (d)
[Co(NH₃)₄(NO₂)₂] Cl and [Co(NH₃)₄(Cl) (NO₂)] NO₂ → ionization isomers
[Co(NH₃)₄(NO₂)₂] Cl and [Co(NH₃)₄ (ONO)₂] Cl → linkage isomers
The two NO₂ groups can be cis (adjacent) or trans (opposite) → geometrical isomers. Both forms of [MA₄B₂] have a plane of symmetry, so there is no optical isomerism.

Teacher's Note:
1. NO₂⁻ is ambidentate (N or O donor), so any complex containing it can show linkage isomerism.

Question 17. Fac-Mer isomerism is associated with which one of the following complexes ? (M = central metal) (Kerala PMT 2007)
(a) [M(AA)₂]
(b) [MA₃B₃]
(c) [M(AA)₃]
(d) [MABCD]
(e) [MA₄B₂]
Answer:
Ans. (b)
In octahedral [MA₃B₃], the three A ligands can occupy one triangular face (fac) or a meridian (mer) of the octahedron.

Teacher's Note:
1. fac/mer needs two sets of three identical ligands on an octahedron, e.g. [Co(NH₃)₃(NO₂)₃].

Question 18. Which of the following can exhibit geometrical isomerism ? (Odisha JEE 2008)
(a) [MnBr₄]²⁻
(b) [Pt(NH₃)₃Cl]⁺
(c) [PtCl₂P(C₂H₅)₃]₂
(d) [Fe(H₂O)₅NO]²⁺
Answer:
Ans. (c)
(c) is a bridged complex, [(C₂H₅)₃P(Cl)Pt(μ-Cl)₂Pt(Cl)P(C₂H₅)₃], and shows geometrical isomerism:
in the trans-form the two P(C₂H₅)₃ groups lie on opposite sides of the Pt(μ-Cl)₂Pt unit; in the cis-form they lie on the same side.

Teacher's Note:
1. Tetrahedral [MA₄] and octahedral [MA₅B] or square planar [MA₃B] never show geometrical isomerism.

Question 19. Which of the following complex species is not expected to exhibit optical isomerism ? (JEE Main 2013)
(a) [Co(en)(NH₃)₂Cl₂]⁺
(b) [Co(en)₃]³⁺
(c) [Co(en)₂Cl₂]⁺
(d) [Co(NH₃)₃Cl₃]
Answer:
Ans. (d)
[Co(NH₃)₃Cl₃] shows cis or facial (fac) as well as trans or meridional (mer) isomers. But both these isomers contain plane of symmetry. So they do not exhibit optical isomerism.

Teacher's Note:
1. Complexes with only monodentate ligands of the MA₃B₃ type always have a mirror plane; chirality usually needs chelating ligands.

One or More Correct Answers

Question 20. The compound K₃[Co(ONO)₆] is called
(a) potassium cobaltinitrite
(b) Fischer's salt
(c) potassium hexanitritocobaltate (III)
(d) potassium cobalt nitrate
Answer:
Ans. (a, b, c)
K₃[Co(ONO)₆] is the yellow compound commonly called potassium cobaltinitrite or Fischer's salt; its IUPAC name is potassium hexanitritocobaltate (III) (Co is +3).
It is not a nitrate: the ligand is the nitrite ion (ONO⁻), so (d) is wrong.

Teacher's Note:
1. “Nitrito” means O-bonded ONO⁻, “nitro” means N-bonded NO₂⁻; neither is nitrate (NO₃⁻).

Question 21. Which of the following are correct about [Cu(NH₃)₄]SO₄ ?
(a) It is a square planar complex.
(b) It is paramagnetic with one unpaired electron in the d-subshell.
(c) It gives white precipitate with BaCl₂ solution.
(d) Its aqueous solution does not conduct electricity.
Answer:
Ans. (a, c)
(b) is wrong because the unpaired electron is present in 4p subshell.
(d) is wrong because it gives ions in the solution and hence can conduct electricity.
(a) is correct (dsp², square planar) and (c) is correct because the SO₄²⁻ outside the coordination sphere gives white BaSO₄.

Teacher's Note:
1. In the valence bond picture of [Cu(NH₃)₄]²⁺, the odd 3d electron is promoted to 4p to make room for dsp².

Passage-Based Questions

Question 22. Comprehension 1: A, B and C are three complexes of chromium (III) with the empirical formula H₁₂O₆Cl₃Cr. All the three complexes have water and chloride ions as ligands. Complex A does not react with conc. H₂SO₄ whereas complexes B and C lose 6.75% and 13.5% of their original weight respectively, on treatment with concentrated H₂SO₄.
The complex A is
(a) [Cr(H₂O)₄(H)₄(O)₂] Cl₃
(b) [Cr(H₂O)₅ (H)₂(O)] Cl₃
(c) [Cr(H₂O)₆] Cl₃
(d) [Cr(H₂O)₅Cl] Cl₂ . H₂O
Answer:
Ans. (c)
As A does not react with conc. H₂SO₄, no H₂O is present as water of hydration. Hence, the formula is [Cr(H₂O)₆]Cl₃.

Teacher's Note:
1. No mass loss with conc. H₂SO₄ means no water of crystallisation: all six H₂O are coordinated.

Matrix Match Questions

Teacher's Note:
1. Low spin CFSE peaks at d⁶ (−2.4Δ₀) and falls once electrons enter e_g.

Teacher's Note:
1. Decide magnetism from the d-count and ligand field, then decide isomerism from the formula; match only when both parts agree.

Integer and Numerical Answer Questions

Question 25. The number of chelate rings present in the complex K₂ [Ni(EDTA)] is
Answer:
Ans. 5
EDTA forms 5 chelate rings with metal ion (M²⁺): two Ni–O–C–C–N rings on each N atom (four glycinate-type rings) and one Ni–N–C–C–N ring through the ethylene bridge.

Teacher's Note:
1. A ligand with n donor atoms bound to one metal forms n − 1 chelate rings: 6 − 1 = 5 for EDTA.

Question 26. Number of CO groups that can be attached to chromium is
Answer:
Ans. 6
Applying Effective Atomic Number rules, number of groups that can be attached to Cr = 6: EAN = 24 + 2 × 6 = 36.

Teacher's Note:
1. Cr(0) needs 12 more electrons to reach the krypton count of 36, and each CO donates 2: hence Cr(CO)₆.

Question 27. Taking CFSE value of octahedral complex (Δ₀) as 10 Dq, the e_g orbitals are ...... Dq above the bari centre.
Answer:
Ans. 6
The e_g orbitals are 6 Dq above the bari centre (and t_2g are 4 Dq below it).

Teacher's Note:
1. Barycentre rule: 2 × (+6 Dq) + 3 × (−4 Dq) = 0, so the average energy is unchanged.

Question 28. The number of rings present in the structure of chlorophyll is
Answer:
Ans. 9
There are 9 rings present in chlorophyll: four pyrrole rings, one five-membered carbocyclic ring (carrying the C=O and CO₂CH₃ groups) and four chelate rings formed by Mg with the four pyrrole N atoms.

Teacher's Note:
1. Chlorophyll is a magnesium complex of a porphyrin-type ligand; the Mg–N bonds create the extra rings.

Question 29. The volume (in mL) of 0.1 M AgNO₃ required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H₂O)₅Cl]Cl₂, as silver chloride is close to (IIT 2011)
Answer:
Ans. 6
[Cr(H₂O)₅Cl] Cl₂ + 2 AgNO₃ → 2 AgCl + [Cr(H₂O)₅Cl] (NO₃)₂
Number of ionizable Cl⁻ ions in the complex = 2
Millimoles of Cl⁻ ions = Molarity × volume × 2 = 0.01 × 30 × 2 = 0.6
Therefore, Ag⁺ required for complete precipitation of Cl⁻ ions = 0.6 millimoles
As Millimoles = Molarity × V ml
0.6 = 0.1 × V
V = 6 ml.

Teacher's Note:
1. The coordinated Cl does not react with Ag⁺; only the two counter-ion chlorides are precipitated.

Assertion–Reason Questions

Question 30. Statement-1: The number of unpaired electrons present in [CuCl₂]⁻ complex is zero.
Statement-2: The complex is linear in the solid state with sp-hybridization.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
Correct Statement-2. The complex, [CuCl₂]⁻ (as well as [CuCl₃]²⁻ and [CuCl₄]³⁻) are tetrahedral in the solid state. In KCuCl₂, there are chains in which Cu is tetrahedrally surrounded by Cl with simple units joined into a chain by shared halide ions.

Teacher's Note:
1. Cu⁺ is 3d¹⁰, so all its complexes have zero unpaired electrons regardless of geometry.

Question 31. Statement-1: Coordination entities with d⁴ to d⁷ ions are more stable for high spin state.
Statement-2: If Δ_o < P, high spin state is more stable.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (d)
Correct Statement-1. Coordination entities with d⁴ to d⁷ ions are more stable for low spin state.

Teacher's Note:
1. Only d⁴–d⁷ ions have a high spin/low spin choice; the stable state depends on whether Δ_o or P is larger.

Question 32. Statement-1: K₂ [Ni(EDTA)] is more stable than K₃ [Al (C₂O₄)₃]
Statement-2: Nickel (Ni) is a transition element while aluminium (Al) is non-transition element
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (b)
Both Statement-1 and Statement-2 are True but the greater stability is due to greater number of chelate rings (five) formed by EDTA than by oxalate.

Teacher's Note:
1. Stability of chelates rises with the number of chelate rings (chelate effect); EDTA forms five rings with one metal ion.

Question 33. Statement-1: [Fe(H₂O)₅NO]SO₄ is paramagnetic.
Statement-2: The Fe in [Fe(H₂O)₅NO]SO₄ has three unpaired electrons. (IIT 2008)
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
In the compound [Fe(H₂O)₅NO]SO₄, NO is present as NO⁺. Hence, oxidation state of Fe is x + 5 × 0 + 1 = +2 or x = + 1. (In fact, odd e⁻ on N is transferred from NO to Fe, changing NO to NO⁺ and Fe²⁺ to Fe⁺).
E.C. of Fe⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s¹
The electron present in 4s shifts to 3d giving 3d⁷ configuration, with 3 unpaired electrons; 4s, 4p and 4d orbitals give sp³d².
H₂O and NO⁺ both are weak field ligands and cannot cause pairing of 3d electrons. Thus, there are 3 unpaired electrons and hybridisation is sp³d² (an outer orbital complex).

Teacher's Note:
1. The brown ring complex has μ ≈ 3.9 BM, matching three unpaired electrons on Fe(I).