CBSE Class 12 Chemistry Coordination Compounds MCQs Set 14

Answer: (b) 1 mole of CoCl₃.5NH₃

Explanation:
1. In complex structure of 1 mole of CoCl₃.xNH₃, coordination number should be 6 in coordination sphere such that:
2. [Co(NH₃)₆]Cl₃ gives 3 moles of AgCl. Similarly, [Co(NH₃)₅Cl]Cl₂ gives 2 moles of AgCl, [Co(NH₃)₄Cl₂]Cl gives 1 mole of AgCl.

Teacher's Note:
1. Fill the six coordination positions with NH₃ first and then with Cl⁻; the chlorides left over are ionisable and precipitate as AgCl.

Answer: (a) [Co(NH₃)₆]³⁺

Explanation:
1. Electronic configuration of Co³⁺ in [Co(NH₃)₆]³⁺ is
2.

Teacher's Note:
1. Co³⁺ is d⁶: in a low-spin octahedral field all six electrons pair in t_2g. Mn³⁺ (d⁴) and Fe³⁺ (d⁵) are left with 2 and 1 unpaired electrons.

Answer: (c) [Co(H₂O)₆]³⁺ > [Co(NH₃)₆]³⁺ > [Co(CN)₆]³⁻

Explanation:
1. Δ_o values follow the order: [Co(H₂O)₆]³⁺ < [Co(NH₃)₆]³⁺ < [Co(CN)₆]³⁻
2. and Δ_o ∝ \( \frac{1}{\lambda} \), therefore, absorption wavelength follows the order:
3. [Co(H₂O)₆]³⁺ > [Co(NH₃)₆]³⁺ > [Co(CN)₆]³⁻

Teacher's Note:
1. Stronger field ligand means larger Δ_o, higher energy absorbed and therefore a shorter wavelength – the order of wavelength is the reverse of the spectrochemical series.

Answer: (c) [Fe(C₂O₄)₃]³⁻

Explanation:
1. We know that the stability of a complex increases by chelation. Therefore, the most stable complex is [Fe(C₂O₄)₃]³⁻.

Teacher's Note:
1. All four complexes have Fe³⁺; the only difference is the ligand, and the only chelating ligand here is oxalate.

Answer: (a) potassium hexacyanoferrate(II)

Explanation:
1. K₄[Fe(CN)₆] ionises to give 4K⁺ + [Fe(CN)₆]⁴⁻, i.e., 5 ions per formula unit, the largest number among the given compounds.
2. [Cr(H₂O)₆]Cl₃ gives 4 ions, [Co(NH₃)₄Cl₂]Cl gives 2 ions and [Pt(NH₃)₂Cl₂] is a non-electrolyte.

Teacher's Note:
1. Molar conductivity of complexes rises with the number of ions produced – first write the formula and count the ions outside and inside the bracket.

Answer: (b) 1:2 electrolyte

Explanation:
1. 0.2 mol AgCl from 0.1 mol complex means 2 Cl⁻ ions are outside the coordination sphere: [Co(NH₃)₅Cl]Cl₂.
2. [Co(NH₃)₅Cl]Cl₂ → [Co(NH₃)₅Cl]²⁺ + 2Cl⁻, a 1:2 electrolyte.

Teacher's Note:
1. Only ionisable (counter-ion) chloride is precipitated by AgNO₃; chloride bound to the metal is not.

Answer: (d) [Cr(H₂O)₆]Cl₃

Explanation:
1. 3 mol AgCl per mol means all three Cl⁻ ions are ionisable, so all six H₂O molecules are ligands: [Cr(H₂O)₆]Cl₃.

Teacher's Note:
1. Moles of AgCl precipitated = number of chloride ions outside the square bracket.

Answer: (b) NH₄⁺

Explanation:
1. In NH₄⁺, the lone pair on nitrogen is already used in bonding with H⁺, so it has no lone pair to donate to a metal.

Teacher's Note:
1. A ligand must have at least one lone pair (or π-electrons) free to donate; positive ions with no free lone pair cannot act as ligands.

Answer: (b) CO

Explanation:
1. CO donates a lone pair through carbon (σ-bond) and accepts electron density from filled metal d-orbitals into its empty π* orbitals (back bonding), so it is a π-acid ligand.

Teacher's Note:
1. π-acid ligands (CO, CN⁻, NO) have empty π* orbitals; NH₃, F⁻ and en are only σ-donors (F⁻ is a π-donor).

Answer: (c) [Fe(C₂O₄)₃]³⁻

Explanation:
1. Oxalate is a bidentate chelating ligand; three oxalate ions form three chelate rings with Fe³⁺, so [Fe(C₂O₄)₃]³⁻ is the most stable.

Teacher's Note:
1. When the question mentions the chelate effect, look for the complex with a polydentate ligand.

Answer: (a) diamminedichloridoplatinum(II)

Explanation:
1. Oxidation state of Pt: x + 2(0) + 2(−1) = 0, x = +2. Ligands are named alphabetically (ammine before chlorido).
2. Hence, the name is diamminedichloridoplatinum(II).

Teacher's Note:
1. Alphabetical order of ligands ignores the multiplying prefixes di-, tri- etc.; “ammine” comes before “chlorido”.

Answer: (a) linkage isomers

Explanation:
1. SCN⁻ is an ambidentate ligand; it binds through S in one complex and through N in the other. Hence they are linkage isomers.

Teacher's Note:
1. Linkage isomerism needs an ambidentate ligand such as SCN⁻/NCS⁻ or NO₂⁻/ONO⁻.

Answer: (a) [Cr(H₂O)₄Cl₂]⁺

Explanation:
1. [Cr(H₂O)₄Cl₂]⁺ is an octahedral MA₄B₂ type complex; the two Cl⁻ can be adjacent (cis) or opposite (trans).

Teacher's Note:
1. MA₄B₂ octahedral and MA₂B₂ square planar complexes show cis-trans isomerism; MA₃B square planar and MA₆ complexes do not.

Answer: (c) both tetrahedral

Explanation:
1. In Ni(CO)₄, Ni(0) is 3d¹⁰ and is sp³ hybridised – tetrahedral.
2. In [Ni(PPh₃)₂Cl₂], Ni²⁺ (3d⁸) with weak Cl⁻ and bulky PPh₃ ligands is sp³ hybridised – tetrahedral.

Teacher's Note:
1. Four-coordinate Ni²⁺ is square planar only with strong-field ligands like CN⁻; with Cl⁻ it is tetrahedral.

Answer: (d) Both (a) and (c)

Explanation:
1. Mn³⁺ (3d⁴) in [MnCl₆]³⁻: 4 unpaired electrons; Co³⁺ (3d⁶) in [CoF₆]³⁻: 4 unpaired electrons.
2. Fe³⁺ (3d⁵) in [FeF₆]³⁻ has 5 unpaired electrons.

Teacher's Note:
1. In outer orbital (high spin) complexes the d-electrons stay unpaired as in the free ion, so count unpaired electrons from the free-ion configuration.

Answer: (d) −0.6 Δ₀

Explanation:
1. High spin d⁴: t_2g³e_g¹.
2. CFSE = 3(−0.4Δ₀) + 1(+0.6Δ₀) = −0.6Δ₀

Teacher's Note:
1. Each t_2g electron contributes −0.4Δ₀ and each e_g electron +0.6Δ₀; no pairing energy term arises for high spin d⁴.

Answer: (d) Good π-donor properties of Cl

Explanation:
1. Cl⁻ has filled p-orbitals that donate electron density to the metal by π-donation; this reduces the splitting, so Cl⁻ is a weaker field ligand than H₂O and lies on the weak-field side of water in the series.

Teacher's Note:
1. π-donor ligands (halides) decrease Δ₀, while π-acceptor ligands (CO, CN⁻) increase it.

Answer: (c) t_2g orbital are three fold degenerate whereas e_g orbitals are two fold degenerate.

Explanation:
1. In an octahedral field, d-orbitals split into a three-fold degenerate t_2g set (d_xy, d_yz, d_zx) and a two-fold degenerate e_g set (d_x²−y², d_z²).
2. High CFSE favours low spin complexes, and the same ion (e.g., Co³⁺) can form both diamagnetic and paramagnetic complexes, so (a), (b) and (d) are wrong.

Teacher's Note:
1. Large Δ₀ (> P) forces pairing – low spin; small Δ₀ (< P) gives high spin.

Answer: (a) Ti(C₂H₄)₄

Explanation:
1. In Ti(C₂H₄)₄, the metal is bonded directly to carbon atoms of ethene; in the others Ti is bonded to oxygen.

Teacher's Note:
1. An organometallic compound must contain at least one metal–carbon bond; metal–oxygen–carbon linkages do not count.

Answer: (a) triethylaluminium

Explanation:
1. Ziegler-Natta catalyst is a combination of TiCl₄ with triethylaluminium, (C₂H₅)₃Al.

Teacher's Note:
1. Ziegler-Natta catalyst is the standard example of an organometallic catalyst (used in polymerisation of alkenes).

Answer: (d) A(4), B(1), C(2), D(3)

Explanation:
1. Isomerism in coordination compound is decided by type of ligands and geometry of coordination and arrangements of ligands.
2. (A) [Co(NH₃)₄Cl₂]⁺ shows geometrical isomerism due to presence of two types of ligands whose arrangement around the central metal ion can be cis or trans.
3. (B) cis-[Co(en)₂Cl₂]⁺ shows optical isomerism due to its non-superimposable mirror-image relationship.
4. (C) [Co(NH₃)₅(NO₂)]Cl₂ shows ionisation isomerism due to its inter-changing ligand from outside the ionisation sphere.
5. (D) [Co(NH₃)₆][Cr(CN)₆] shows coordination isomerism due to inter-changing of ligand in between two metal ions from one coordination sphere to another coordination sphere.

Teacher's Note:
1. [Co(NH₃)₅(NO₂)]Cl₂ can also show linkage isomerism (NO₂ vs ONO); here choose the code that fits all four rows, which settles C as ionisation.

Answer: (a) A(3), B(1), C(5), D(2)

Explanation:
1. Formation of inner-orbital and outer complex determines hybridisation of molecule which in turn depends upon field strength of ligand and number of vacant d orbitals.
2. (i) Strong-field ligand forms inner orbital complex with hybridisation d²sp³. (ii) Weak-field ligand forms outer-orbital complex with hybridisation sp³d².
3. (A) Cr³⁺ in [Cr(H₂O)₆]³⁺ is 3d³: hybridisation = d²sp³, n (number of unpaired electrons) = 3
4. (B) Co²⁺ in [Co(CN)₄]²⁻ is 3d⁷: hybridisation = dsp², n = 1
5. (C) Ni²⁺ in [Ni(NH₃)₆]²⁺ is 3d⁸: hybridisation = sp³d², n = 2
6. (D) Mn²⁺ in [MnF₆]⁴⁻ is 3d⁵: hybridisation = sp³d², n = 5

Teacher's Note:
1. For d⁸ Ni²⁺, only one 3d orbital could ever be freed, so a six-coordinate Ni(II) complex must use outer 4d orbitals (sp³d²) even with NH₃.

Answer: (b) Cu²⁺ + 4CN⁻ → [Cu(CN)₄]²⁻; log K = 27.3

Explanation:
1. Greater the value of the stability constant (log K), more stable is the complex.
2. [Cu(CN)₄]²⁻ has the highest log K (27.3), so it is the most stable.

Teacher's Note:
1. Stability constant K is the equilibrium constant for formation of the complex – compare log K values directly.

Answer: (d) No isomerism

Explanation:
1. Isomers must have the same molecular formula.
2. One compound contains Br and the other Cl, so their formulae are different and they are not isomers.

Teacher's Note:
1. Before naming a type of isomerism, first check that the two formulae contain exactly the same atoms.

Answer: (b) Solvate isomerism

Explanation:
1. The two compounds differ in whether a water molecule is directly bonded to the metal ion or present as free water of crystallisation.
2. This is solvate (hydrate) isomerism.

Teacher's Note:
1. Hydrate isomers often differ in colour and in the number of Cl⁻ ions precipitated by AgNO₃.

Answer: (c) Diamminechloridonitrito-N-platinum(II)

Explanation:
1. Ligands in alphabetical order: ammine, chlorido, nitrito-N (NO₂ bonded through N).
2. The complex is neutral, so the metal is named “platinum” (not platinate); Pt is +2.

Teacher's Note:
1. Use “-ate” with the metal name only when the complex ion is an anion.

Answer: (a) d²sp³ hybridisation, paramagnetic

Explanation:
1. Fe³⁺ is 3d⁵. CN⁻ is a strong field ligand, so the electrons pair up as t_2g⁵, leaving two 3d orbitals free for d²sp³ hybridisation.
2. One electron remains unpaired, so the complex is paramagnetic.

Teacher's Note:
1. Low spin d⁵ still has one unpaired electron – inner orbital does not automatically mean diamagnetic.

Answer: (a) [Co(en)₃]³⁺ and cis-[Co(en)₂Cl₂]⁺

Explanation:
1. [Co(en)₃]³⁺ and cis-[Co(en)₂Cl₂]⁺ have no plane of symmetry, so their mirror images are non-superimposable.
2. trans-[Co(en)₂Cl₂]⁺ has a plane of symmetry and [Cr(NH₃)₅Cl]²⁺ has only unidentate ligands; both are optically inactive.

Teacher's Note:
1. For [M(AA)₂B₂] only the cis form is optically active – the trans form has a mirror plane.

Answer: (a) A–3, B–1, C–5, D–2

Explanation:
1. (A) Cr³⁺, 3d³: d²sp³, 3 unpaired electrons. (B) Co²⁺, 3d⁷ with CN⁻: square planar dsp², 1 unpaired electron.
2. (C) Ni²⁺, 3d⁸: inner d orbitals not available, sp³d², 2 unpaired electrons. (D) Mn²⁺, 3d⁵ with weak F⁻: sp³d², 5 unpaired electrons.

Teacher's Note:
1. Write the d-electron count of each metal ion first; Ni²⁺ (d⁸) octahedral complexes are always outer orbital with 2 unpaired electrons.

Answer: (c) A–5, B–1, C–4, D–2

Explanation:
1. (A) x − 1 + 0 − 2 = 0, x = +3. (B) x + 0 − 2 − 2 = 0, x = +4.
2. (C) 4(+1) + x + 3(−2) = 0, x = +2. (D) CO is neutral, so Co is 0.

Teacher's Note:
1. Count the charge of the counter ion first (SO₃²⁻, SO₄²⁻, Na⁺), then the ligand charges.

Answer: (c) 6

Explanation:
1. en and C₂O₄²⁻ are didentate (2 + 2), Cl⁻ and Br⁻ are unidentate (1 + 1).
2. Coordination number = 2 + 2 + 1 + 1 = 6.

Teacher's Note:
1. Coordination number counts donor atoms, not ligands – four ligands here give six bonds.

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

Explanation:
1. In the complexes, Cr exists as Cr²⁺ and Fe as Fe²⁺. Both of the complexes become stable by oxidation of metal ion to Cr³⁺ and Fe³⁺.
2. It is due to the presence of weak field ligand, so pairing of electrons does not take place and are easily removed because of excitation in e_g orbital.

Teacher's Note:
1. Both statements are true, but having unpaired electrons only explains paramagnetism; reducing nature comes from the ease of oxidation to the +3 state.

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

Explanation:
1. In tetrahedral complexes, the splitting of the d-orbitals is inverted and is smaller in comparison to octahedral complexes. The crystal field splitting energy is not large enough to force pairing and hence, low spin complexes are rarely observed.

Teacher's Note:
1. Remember Δ_t = (4/9)Δ_o; such a small splitting is almost always less than the pairing energy P.

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

Explanation:
1. Oxalate ion, C₂O₄²⁻, binds the metal through two oxygen atoms, which is why it is bidentate.

Teacher's Note:
1. Denticity is the number of donor atoms through which one ligand binds the same metal ion.

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

Explanation:
1. An ambidentate ligand (e.g., NO₂⁻, SCN⁻) can bind through either of its two different donor atoms, giving linkage isomers.

Teacher's Note:
1. Common ambidentate pairs: nitrito-N/nitrito-O, thiocyanato-S/thiocyanato-N (isothiocyanato).

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

Explanation:
1. In MX₆ and MX₅L, all arrangements are identical, so no geometrical isomers exist. Assertion is correct.
2. Octahedral complexes of coordination number 6 such as MX₄L₂ and MX₃L₃ do show geometrical isomerism, so Reason is incorrect.

Teacher's Note:
1. Octahedral complexes need at least two ligands of a second kind (MX₄L₂) before cis/trans forms become possible.

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

Explanation:
1. Strong field CN⁻ pairs the two unpaired 3d electrons of Ni²⁺; one 3d orbital becomes vacant, giving dsp² hybridisation (square planar) with no unpaired electron (diamagnetic).

Teacher's Note:
1. Compare with [NiCl₄]²⁻: weak Cl⁻ does not cause pairing, so it is sp³, tetrahedral and paramagnetic.

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

Explanation:
1. [Ni(CO)₄] is sp³ hybridised (tetrahedral) and [Ni(CN)₄]²⁻ is dsp² hybridised (square planar), so Assertion is incorrect.
2. The nature of ligands (strong or weak field) does decide the hybridisation and geometry, so Reason is correct.

Teacher's Note:
1. Ni(0) in Ni(CO)₄ is d¹⁰, so no inner d-orbital is free for dsp² – it must be tetrahedral.

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

Explanation:
1. F⁻ is a weak field ligand; [FeF₆]³⁻ is sp³d² hybridised, a high spin complex with 5 unpaired electrons. Assertion is incorrect.
2. Low spin complexes do have fewer unpaired electrons, so Reason is correct.

Teacher's Note:
1. Fluoride complexes of 3d metals are the classic outer orbital (high spin) examples.

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

Explanation:
1. Since F⁻ cannot cause pairing, the inner d-orbitals are not vacated and the outer nd orbitals are used: sp³d² (outer orbital) complex.

Teacher's Note:
1. Outer orbital = high spin = sp³d² for octahedral complexes of weak field ligands.