CBSE Class 12 Chemistry Coordination Compounds MCQs Set 15

Answer: (d) ambidentate ligand

Explanation:
1. NO₂⁻ can bind the metal either through N (nitro, –NO₂) or through O (nitrito, –ONO).
2. A ligand that can coordinate through two different donor atoms (but only one at a time) is an ambidentate ligand.

Teacher's Note:
1. Ambidentate ligands (NO₂⁻, SCN⁻, CN⁻) give rise to linkage isomerism.

Answer: (b) Monodentate ligand

Explanation:
1. In 2,4-dinitrophenylhydrazine, O₂N–C₆H₃(NO₂)–NH–NH₂, the lone pair of the terminal –NH₂ group is donated to the metal.
2. Hence it is a monodentate ligand.

Teacher's Note:
1. Having many N atoms does not make a ligand polydentate; the donor atoms must be able to bind the same metal ion together.

Answer: (c) Ferrous ammonium sulphate

Explanation:
1. Ferrous ammonium sulphate, FeSO₄.(NH₄)₂SO₄.6H₂O (Mohr’s salt), dissociates completely into Fe²⁺, NH₄⁺ and SO₄²⁻ ions in water, so it is a double salt.
2. Potassium ferrocyanide and ferricyanide are complexes; they give the complex ions [Fe(CN)₆]⁴⁻ / [Fe(CN)₆]³⁻, not Fe²⁺/Fe³⁺ and CN⁻.

Teacher's Note:
1. Test of a double salt: its solution gives the tests of ALL its constituent simple ions.

Answer: (a) + 3

Explanation:
1. Let the oxidation state of Fe be x: x + 6(−1) = −3
2. x = +3

Teacher's Note:
1. Always use the charge of the complex ion (here 3−), not the number of counter ions, when finding the oxidation state.

Answer: (b) Potassium hexacyanoferrate (III)

Explanation:
1. 3(+1) + x + 6(−1) = 0 gives x = +3 for Fe.
2. In an anionic complex the metal name ends in –ate (iron → ferrate), so the name is potassium hexacyanoferrate(III).

Teacher's Note:
1. For anionic complexes use the Latin names: ferrate, cuprate, argentate, aurate, plumbate, stannate.

Answer: (c) Both are included in stereoisomerism

Explanation:
1. Both geometrical and optical isomers have the same ligands bonded through the same atoms; they differ only in the spatial arrangement of the ligands.
2. Hence both are types of stereoisomerism. (a) and (b) are not true in general, e.g. [Pt(NH₃)₂Cl₂] has geometrical but no optical isomers.

Teacher's Note:
1. Structural isomerism = different bonds (ionisation, linkage, hydrate, coordination); stereoisomerism = same bonds, different arrangement in space (geometrical, optical).

Answer: (d) [Co(en)₂(NH₃)₂]³⁺

Explanation:
1. The cis isomer of the octahedral complex [Co(en)₂(NH₃)₂]³⁺ (type [M(AA)₂B₂]) has no plane of symmetry and exists as d and l forms.
2. The other complexes have a plane of symmetry in every arrangement, so they are optically inactive.

Teacher's Note:
1. cis-[M(AA)₂B₂] is optically active, trans-[M(AA)₂B₂] is not – a favourite exam point.

Answer: (b) 3

Explanation:
1. It is a square planar complex of the type MABCD.
2. Fixing A, it can be trans to B, C or D, giving three geometrical isomers.

Teacher's Note:
1. Square planar MABCD → 3 geometrical isomers; tetrahedral complexes show no geometrical isomerism at all.

Answer: (c) [Co(NH₃)₂(en)₂]³⁺

Explanation:
1. Geometrical isomerism is shown by disubstituted octahedral complexes such as [Co(NH₃)₂(en)₂]³⁺.
2. The two NH₃ groups can be cis (adjacent) or trans (opposite) to each other.

Teacher's Note:
1. [M(AA)₃], [MA₅B] and [MA₄(AA)] have no geometrical isomers; [M(AA)₂B₂] has cis and trans forms.

Answer: (c) Geometrical isomerism

Explanation:
1. The cis and trans forms of the given complex have different colours (cis is violet, trans is green).

Teacher's Note:
1. Same formula, same ions, different colours – think of cis/trans isomers of [MA₄B₂].

Answer: (a) two cis and one trans isomer

Explanation:
1. Fixing A, the ligand trans to it can be B, X or L, giving three isomers.
2. Taking A and B as reference, one isomer has A trans to B and two have A cis to B (two cis and one trans).

Teacher's Note:
1. For square planar MABCD the total is always three geometrical isomers.

Answer: (a) [Co(en) (NH₃)₂Cl₂] Cl

Explanation:
1. In [Co(en)(NH₃)₂Cl₂]⁺ (type [M(AA)B₂C₂]) there are three geometrical arrangements: (i) the two NH₃ trans, (ii) the two Cl trans, (iii) both pairs cis.
2. Arrangement (iii) has no plane of symmetry and exists as d and l forms, so the total is 2 + 2 = 4 isomers.
3. [Co(en)₂Cl₂]⁺ has only three stereoisomers (trans, d-cis and l-cis) and [Co(en)₃]³⁺ has two (d and l).

Teacher's Note:
1. Count isomers systematically: first list the geometrical arrangements, then check each one for a plane of symmetry.

Answer: (b) Pauling

Explanation:
1. Valence bond theory (hybridisation of metal orbitals) was applied to coordination compounds by Linus Pauling.
2. Crystal field theory was developed by Bethe and Van Vleck; Werner gave the theory of primary and secondary valencies.

Teacher's Note:
1. Match the theory with its scientist: Werner – valencies, Pauling – VBT, Bethe & Van Vleck – CFT.

Answer: (a) 4

Explanation:
1. [Co(en)₃]Cl₃ → [Co(en)₃]³⁺ + 3Cl⁻
2. 1 complex ion + 3 chloride ions = 4 ions.

Teacher's Note:
1. Ligands inside the square bracket do not ionise; only the counter ions outside the bracket are released.

Answer: (c) [Fe(C₂O₄)₃]³⁻

Explanation:
1. Oxalate (C₂O₄²⁻) is a chelating ligand. Hence, it forms most stable complex.

Teacher's Note:
1. Chelate effect: a ring-forming (polydentate) ligand gives a more stable complex than comparable monodentate ligands.

Answer: (b) [Fe(H₂O)₆]²⁺

Explanation:
1. Unpaired electrons (H₂O is a weak field ligand): Cr³⁺ (3d³) = 3, Fe²⁺ (3d⁶) = 4, Cu²⁺ (3d⁹) = 1, Zn²⁺ (3d¹⁰) = 0.
2. Fe²⁺ has the maximum number of unpaired electrons, so [Fe(H₂O)₆]²⁺ is the most paramagnetic.

Teacher's Note:
1. More unpaired electrons = more paramagnetism; with weak field ligands count unpaired electrons as in the free ion.

Answer: (c) 2·82 BM

Explanation:
1. Ni²⁺ is 3d⁸; Cl⁻ is a weak field ligand, so the two unpaired electrons remain unpaired (sp³, tetrahedral).
2. μ = \( \sqrt{n(n+2)} \) = \( \sqrt{2(2+2)} \) = \( \sqrt{8} \) ≈ 2·82 BM

Teacher's Note:
1. Remember the spin-only values: 1 → 1.73, 2 → 2.83, 3 → 3.87, 4 → 4.90, 5 → 5.92 BM.

Answer: (c) [Cr(NH₃)₆]³⁺

Explanation:
1. Unpaired electrons: Zn²⁺ (3d¹⁰) = 0, Ti³⁺ (3d¹) = 1, Cr³⁺ (3d³) = 3, Co³⁺ (3d⁶, low spin with NH₃) = 0.
2. Cr³⁺ has the most unpaired electrons, so [Cr(NH₃)₆]³⁺ is the most paramagnetic.

Teacher's Note:
1. With NH₃, Co³⁺ pairs up all six d electrons (t_2g⁶) and becomes diamagnetic.

Answer: (b) Tetrahedral, square planar

Explanation:
1. [NiCl₄]²⁻: Cl⁻ is a weak ligand, no pairing, sp³ → tetrahedral.
2. [Ni(CN)₄]²⁻: CN⁻ is a strong ligand, the two unpaired 3d electrons pair up, dsp² → square planar.

Teacher's Note:
1. For four-coordinate Ni²⁺: weak field ligand → tetrahedral and paramagnetic; strong field ligand → square planar and diamagnetic.

Answer: (d) [Co(H₂O)₃Cl₃].3 H₂O

Explanation:
1. Complex (d) is non-ionic complex. All other produce 2 or more ions in the solution.
2. Hence, (d) has least depression in freezing point or highest freezing point.

Teacher's Note:
1. Depression in freezing point is a colligative property: fewer particles → smaller ΔT_f → higher freezing point.

Answer: (b) [Co(NH₃)₅Cl] Cl₂

Explanation:
1. Only 1 mole of (b) produces two moles of Cl⁻ ions in the solution.
2. Coordination number of Co(III) is 6: 5 NH₃ + 1 Cl inside the sphere, 2 Cl⁻ outside.

Teacher's Note:
1. Moles of AgCl precipitated = number of Cl⁻ outside the coordination sphere.

Answer: (b) Tetrahedral geometry and diamagnetic

Explanation:
1. In Ni(CO)₄, Ni is in zero oxidation state (3d⁸4s²). CO is a strong ligand, so the 4s electrons move into 3d, giving 3d¹⁰.
2. The empty 4s and three 4p orbitals give sp³ hybridisation: tetrahedral and diamagnetic.

Teacher's Note:
1. Do not confuse Ni(CO)₄ (Ni⁰, sp³, tetrahedral) with [Ni(CN)₄]²⁻ (Ni²⁺, dsp², square planar); both are diamagnetic.

Answer: (b) [Co(H₂O)₅Cl] Cl₂.H₂O

Explanation:
1. 100 ml of 0·1 M CoCl₃.6H₂O solution contains complex = 100 × 0·1 millimole = 10 millimole = 0·01 mole
2. As the ions produced are precipitated by AgNO₃, the ions produced must be Cl⁻ ions.
3. 0·01 mole of the complex give Cl⁻ ions = 1·2 × 10²² = \( \frac{1.2 \times 10^{22}}{6.0 \times 10^{23}} \) mole = 0·02 mole
4. ∴ 1 mole of the complex gives Cl⁻ ions = 2 moles (which are precipitated as AgCl)
5. Remembering that coordination number of Co is 6 and two Cl⁻ ions are present outside the coordination sphere, the formula of the complex will be [Co(H₂O)₅Cl] Cl₂.H₂O.

Teacher's Note:
1. Convert number of ions to moles first, then find moles of Cl⁻ per mole of complex.

Answer: (c) L, M, O, P

Explanation:
1. K : K₃[Fe(CN)₆], Fe³⁺ = 3d⁵, d²sp³ hybridization, number of unpaired electrons = 1, paramagnetic.
2. L : [Co(NH₃)₆]Cl₃, Co³⁺ = 3d⁶, d²sp³ hybridization, number of unpaired electrons = 0, diamagnetic.
3. M : Na₃[Co (oxalate)₃], Co³⁺ = 3d⁶, d²sp³ hybridization, number of unpaired electrons = 0, diamagnetic.
4. N : [Ni(H₂O)₆] Cl₂, Ni²⁺ = 3d⁸, sp³d² hybridization, number of unpaired electrons = 2, paramagnetic.
5. O : K₂[Pt(CN)₄], Pt²⁺ = 5d⁸, dsp² hybridization, number of unpaired electrons = 0, diamagnetic.
6. P : [Zn(H₂O)₆](NO₃)₂, Zn²⁺ = 3d¹⁰, sp³d² hybridization, number of unpaired electrons = 0, diamagnetic.
7. Hence, diamagnetic complexes are L, M, O and P.

Teacher's Note:
1. A d¹⁰ ion (Zn²⁺) is always diamagnetic, whatever the ligand; a d⁵ ion can never be diamagnetic.

Answer: (c) CoCl₃.3 NH₃

Explanation:
1. As cobalt forms octahedral complexes for which coordination number = 6, the formulae of the given complexes will be [Co(NH₃)₅Cl] Cl₂, [Co(NH₃)₆] Cl₃, [Co(NH₃)₃Cl₃] and [Co(NH₃)₄Cl₂] Cl
2. Thus, only (c) does not have any ionizable chloride ion.

Teacher's Note:
1. Fill the six coordination positions with NH₃ first and then with Cl; whatever Cl is left over is ionisable.

Answer: (c) CoCl₃ . 6 NH₃ and PtCl₄ . 5 NH₃

Explanation:
1. CoCl₃ . 6 NH₃ = [Co(NH₃)₆] Cl₃ → [Co(NH₃)₆]³⁺ + 3 Cl⁻ (4 ions)
2. PtCl₄ . 5 NH₃ = [Pt(NH₃)₅ Cl] Cl₃ → [Pt(NH₃)₅ Cl]³⁺ + 3 Cl⁻ (4 ions)
3. Hence, they will show same electrical conductance.

Teacher's Note:
1. Equal number of ions per formula unit (at the same concentration) means nearly equal conductance.

Answer: (c) Ni(CO)₄ and [Ni(CN)₄]²⁻ are diamagnetic and [NiCl₄]²⁻ is paramagnetic

Explanation:
1. [Ni(CO)₄] and [Ni(CN)₄]²⁻ do not contain any unpaired electron while [NiCl₄]²⁻ contains two unpaired electrons.

Teacher's Note:
1. CO and CN⁻ are strong field ligands and force pairing; Cl⁻ is weak and leaves the two electrons of Ni²⁺ unpaired.

Answer: (a) [Ni(NH₃)₆]²⁺

Explanation:
1. In [Ni(NH₃)₆]²⁺, Ni²⁺ is 3d⁸. Its electronic configuration is:
2.

Teacher's Note:
1. An octahedral d⁸ ion has no two empty 3d orbitals even after pairing, so it must use 4d orbitals (sp³d²).

Answer: (a) 0

Explanation:
1. In [Cr(CO)₆], Cr is in zero oxidation state: ₂₄Cr = [Ar] 3d⁵ 4s¹
2. As CO is a strong ligand, all the six unpaired electrons will pair up, i.e., there will be no unpaired electron. Hence, μ = 0.

Teacher's Note:
1. In metal carbonyls the metal is usually in the zero oxidation state; count all valence electrons (3d + 4s) before pairing.

Answer: (c) [Co(CN)₆]³⁻ has no unpaired electrons and will be in a low-spin configuration

Explanation:
1. [Co(CN)₆]³⁻ has cobalt in +3 oxidation state: Co³⁺ = 3d⁶
2. As CN⁻ is a strong field ligand, all electrons will be paired up and complex will be a low-spin complex (t_2g⁶e_g⁰).

Teacher's Note:
1. “Four unpaired electrons” and “low spin” can never go together for a d⁶ ion; low spin d⁶ always has zero unpaired electrons.

Answer: (c) [Fe(H₂O)₆]³⁺

Explanation:
1. Fe³⁺ is 3d⁵ and H₂O is a weak field ligand, so the configuration is t_2g³e_g².
2. CFSE = 3(−0·4Δ_o) + 2(+0·6Δ_o) = 0. Hence, it has zero crystal field stabilization energy.

Teacher's Note:
1. High spin d⁵ (and d⁰, d¹⁰) always have zero CFSE in an octahedral field.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Chelating ligands such as EDTA form very stable chelate complexes with toxic metal ions (e.g., Pb²⁺), which are then removed from the body.

Teacher's Note:
1. This is the principle of chelate therapy – e.g., EDTA for lead poisoning, D-penicillamine for excess copper.

Answer: (d) A is false, but R is true.

Explanation:
1. Assertion is false: ambidentate ligands do not form chelates, because they bind through only one donor atom at a time.
2. Reason is true: they contain two donor atoms (e.g. N and O in NO₂⁻).

Teacher's Note:
1. Chelation needs two donor atoms of the same ligand bonded to the metal together; an ambidentate ligand uses only one.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Reason R is the correct explanation of Assertion A.
2. In a tetrahedron every position is adjacent to every other position, so cis/trans arrangements are not possible.

Teacher's Note:
1. All bond angles in a tetrahedron are equal (109°28′), so no ligand can be ‘trans’ to another.

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion is true: NO₂⁻ and SCN⁻ are ambidentate.
2. Reason is false: linkage isomers have the same ionization sphere. They differ in the donor atom of a ligand.

Teacher's Note:
1. Different ionisation spheres describe ionisation isomers, not linkage isomers.

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion is true: trans, d-cis and l-cis forms exist.
2. Reason is false: cis has two optical isomers but trans is optically inactive.

Teacher's Note:
1. Trans-[M(AA)₂B₂] has a plane of symmetry; the cis form does not and is chiral.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is the correct explanation of A.
2. Glycinate (H₂N–CH₂–COO⁻) binds Cu²⁺ through N and O, forming a five-membered ring.

Teacher's Note:
1. A chelate needs a polydentate ligand that closes a ring with the metal; five- and six-membered rings are the most stable.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true, but R is not the correct explanation.
2. Correct explanation: trans [CoCl₂(en)₂]⁺ has a plane of symmetry.

Teacher's Note:
1. Being octahedral does not decide optical activity; the cis isomer is also octahedral but is optically active.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is the correct explanation of A (The complex is non-ionic and does not contain ionizable Cl⁻ ions).

Teacher's Note:
1. AgNO₃ precipitates only free Cl⁻ ions; Cl bonded to the metal inside the sphere is not precipitated.

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion is true: Fe²⁺ (3d⁶, t_2g⁶) has no unpaired electron, Fe³⁺ (3d⁵, t_2g⁵) has one.
2. Reason is false: in ferrocyanide, Fe is in oxidation state + 2 while in ferricyanide, Fe is in oxidation state + 3. Higher the oxidation state of the metal, greater is the crystal field splitting.

Teacher's Note:
1. For the same ligand, Δ_o increases with the charge on the metal ion (M³⁺ > M²⁺).