CBSE Class 12 Chemistry Coordination Compounds MCQs Set 12

Answer: (a) potassium pentacyanonitrosylferrate(II)

Explanation:
1. NO is a neutral nitrosyl ligand here, so Fe + 5(−1) + 0 = −3 gives Fe = +2.
2. Ligands are named alphabetically (cyano before nitrosyl) and the anionic complex takes the suffix ‘ate’: potassium pentacyanonitrosylferrate(II).

Teacher's Note:
1. In an anionic complex iron is named ‘ferrate’, never ‘ironate’ or ‘iron’.
2. Find the oxidation state first from the overall charge – it rules out options (b), (c) and (d) at once.

Answer: (c) NH₄⁺ ion does not have any lone pair of electrons

Explanation:
1. NH₄⁺ ion does not have any lone pair of electrons which it can donate to central metal ion hence it does not form complexes.

Teacher's Note:
1. A ligand must be a Lewis base – in NH₄⁺ the nitrogen lone pair is already used in bonding to H⁺.

Answer: (c) [Pt(NH₃)₄(NO₂)Cl]SO₄

Explanation:
1. Tetraamminechloridonitroplatinum(IV) sulphate can be written as [Pt(NH₃)₄(NO₂)Cl]SO₄.

Teacher's Note:
1. ‘Nitro’ means N-bonded NO₂; the O-bonded form is ‘nitrito’ and is written ONO.
2. Check the charge: Pt(+4) + 2(−1) = +2, balanced by one SO₄²⁻.

Answer: (a) 2.82 B.M.

Explanation:
1. In [NiCl₄]²⁻, Ni is in +2 oxidation state.
2. In [NiCl₄]²⁻ :

Teacher's Note:
1. Cl⁻ is a weak field ligand, so the two unpaired 3d electrons of Ni²⁺ stay unpaired.
2. √8 = 2.83; options may round it to 2.82 – both mean two unpaired electrons.

Answer: (a) (i) will give a pale yellow and (ii) will give a white precipitate with AgNO₃ solution.

Explanation:
1. [Pt(NH₃)₄Cl₂]Br₂ → [Pt(NH₃)₄Cl₂]²⁺ + 2Br⁻
2. Br⁻ + AgNO₃ → AgBr (pale yellow ppt.) + NO₃⁻
3. [Pt(NH₃)₄Br₂]Cl₂ → [Pt(NH₃)₄Br₂]²⁺ + 2Cl⁻
4. Cl⁻ + AgNO₃ → AgCl (white ppt.) + NO₃⁻

Teacher's Note:
1. Only the ions outside the square brackets are precipitated by AgNO₃.
2. In (iii) the counter-ion is NO₂⁻, and both Cl⁻ are inside the coordination sphere, so no AgCl forms.

Answer: (b) [Ni(NH₃)₆]₃[Co(NO₂)₆]₂

Explanation:
1. [Ni(NH₃)₆]₃[Co(NO₂)₆]₂
2. Hexaamminenickel(II) hexanitrocobaltate(III)

Teacher's Note:
1. The cation is [Ni(NH₃)₆]²⁺ and the anion is [Co(NO₂)₆]³⁻; three cations balance two anions (3 × 2 = 2 × 3).

Answer: (c) EDTA is a chelating ligand.

Explanation:
1. EDTA⁴⁻ is a hexadentate ligand that binds through two N and four O atoms, forming chelate rings.
2. VBT cannot explain colour, [NiCl₄]²⁻ has two unpaired electrons (paramagnetic) and a bidentate ligand has only two donor sites.

Teacher's Note:
1. Colour is explained by crystal field theory (d–d transitions), not by valence bond theory – a common exam trap.

Answer: (b) t_2g⁶e_g³

Explanation:
1. In [Cu(NH₃)₆]²⁺, oxidation state of Cu = +2, Cu²⁺ = 3d⁹
2. 3d⁹ = t_2g⁶e_g³

Teacher's Note:
1. For d⁹ the answer is the same for strong and weak field ligands – t_2g can hold only 6 electrons.

Answer: (d) [Cu(NH₃)₄]SO₄  p = 4, s = 4

Explanation:
1. In [Cu(NH₃)₄]SO₄ primary valency is 2 and secondary valency is 4.

Teacher's Note:
1. Primary valency = oxidation state of the metal; secondary valency = coordination number.

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

Explanation:
1. The CFSE of the ligands is in the order : H₂O < NH₃ < CN⁻
2. Hence, excitation energies is in the order: [Co(H₂O)₆]³⁺ < [Co(NH₃)₆]³⁺ < [Co(CN)₆]³⁻
3. From the relation \( E = \frac{hc}{\lambda} \Rightarrow E \propto \frac{1}{\lambda} \)
4. The order of absorption of wavelength of light in the visible region : [Co(H₂O)₆]³⁺ > [Co(NH₃)₆]³⁺ > [Co(CN)₆]³⁻

Teacher's Note:
1. Wavelength absorbed runs opposite to ligand field strength: the weakest field ligand absorbs the longest wavelength.

Answer: (b) [Co(NH₃)₅CO₃]Cl : Pentaammine carbonatochlorocobaltate(III)

Explanation:
1. [Co(NH₃)₅CO₃]Cl : Pentaamminecarbonatocobalt(III) chloride

Teacher's Note:
1. Cl outside the bracket is the counter-ion, named separately as ‘chloride’; the cationic complex keeps the plain metal name ‘cobalt’.

Answer: (d) square planar and paramagnetic.

Explanation:
1. CuSO₄ + 4NH₃ → [Cu(NH₃)₄]²⁺ (deep blue) + SO₄²⁻
2. In [Cu(NH₃)₄]²⁺, oxidation state of Cu = +2
3. Cu²⁺ : 3d⁹ :

Teacher's Note:
1. Any Cu²⁺ complex has an odd number of d electrons (d⁹), so it must be paramagnetic.

Answer: (a) (i) < (ii) < (iv) < (iii)

Explanation:
1. Higher the number of ions in the solution, higher is the conductivity.
2. No. of ions : [Co(NH₃)₃Cl₃] = 0; [Co(NH₃)₄Cl₂]Cl = 2; [Co(NH₃)₅Cl]Cl₂ = 3; [Co(NH₃)₆]Cl₃ = 4

Teacher's Note:
1. Count the complex ion itself as one ion plus every counter-ion outside the brackets.

Answer: (d) [NiCl₄]²⁻

Explanation:
1. In [NiCl₄]²⁻, oxidation state of Ni = +2
2. [NiCl₄]²⁻ :

Teacher's Note:
1. 4d and 5d metals like Pd and Pt form square planar complexes even with weak ligands such as Cl⁻, because their splitting is much larger.

Answer: (c) pentaamminenitrito-N-cobalt(III) chloride

Explanation:
1. The name of [Co(NH₃)₅NO₂]Cl₂ is pentaamminenitrito-N-cobalt(III) chloride.

Teacher's Note:
1. Co + 0 + (−1) = +2 gives Co(III); ‘nitrito-N’ (or nitro) tells you NO₂⁻ is bonded through nitrogen.

Answer: (b) Oxalate

Explanation:
1. Oxalate is a bidentate ligand hence forms a chelate.
2. ⁻O–CO–CO–O⁻ binds the metal through both O⁻ atoms, forming a ring.

Teacher's Note:
1. A chelate needs at least two donor atoms in one ligand that close a ring with the metal; acetate, CN⁻ and NH₃ bind through one atom here.

Answer: (b) [Cu(NH₃)₄]SO₄

Explanation:
1. CuSO₄ + 4NH₃ → [Cu(NH₃)₄]SO₄

Teacher's Note:
1. The deep blue colour of this solution is a standard test for Cu²⁺ ions.

Answer: (b) [Cu(CN)₄]²⁻ does not give Cu²⁺ ion in the solution

Explanation:
1. When excess of aqueous KCN solution is added to aqueous CuSO₄ solution, highly stable [Cu(CN)₄]²⁻ is formed which does not give Cu²⁺ ion in the solution, hence copper sulphide, is not formed.
2. [Cu(H₂O)₄]²⁺ + 4CN⁻ → [Cu(CN)₄]²⁻ + 4H₂O

Teacher's Note:
1. A very stable complex keeps the free metal ion concentration so low that its usual precipitation tests fail.

Answer: (c) d⁶ (low spin)

Explanation:
1. In d⁶ (low spin), electrons get paired up to make two empty d-orbitals. Hybridisation is d²sp³ (octahedral) and the complex is low spin.

Teacher's Note:
1. Inner orbital octahedral bonding needs two vacant 3d orbitals after pairing; d⁸ and d⁹ can never free two orbitals.

Answer: (c) Cl⁻ < F⁻ < H₂O < NH₃ < CN⁻ < CO

Explanation:
1. In general, the ligands can be arranged in a series in the order of increasing field strength as
2. I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < NO₂⁻ < CN⁻ < CO

Teacher's Note:
1. Remember the two ends of the spectrochemical series: halides are weakest, CN⁻ and CO are strongest.

Answer: (d) zero

Explanation:
1. Ni (Z = 28) : 3d⁸ 4s²
2. Oxidation state of Ni in [Ni(CO)₄] = 0
3. In [Ni(CO)₄] :

Teacher's Note:
1. Strong field CO pushes the 4s electrons into 3d, giving 3d¹⁰ and a diamagnetic tetrahedral complex.

Answer: (a) CN⁻ is a strong field ligand and H₂O is a weak field ligand hence magnitude of CFSE is different

Explanation:
1. CN⁻ is a strong field ligand hence pairing of electrons takes place while in case of H₂O pairing does not take place. Both ligands show different magnitude of crystal field splitting energy hence absorb different wavelengths and show different colours.

Teacher's Note:
1. Same metal ion, same oxidation state – when only the ligand changes, the colour change comes from a change in Δ_o.

Answer: (b) [Ni(CO)₄]

Explanation:
1. In metal carbonyls, metal is in zero oxidation state.

Teacher's Note:
1. CO is a neutral ligand, so in a neutral homoleptic carbonyl the metal must be 0.

Answer: (a) it absorbs light at specific wavelength in the visible part and reflects rest of the wavelengths

Explanation:
1. A substance absorbs light at specific wavelength in the visible part of the spectrum and reflects the rest of the wavelengths. Each wavelength represents a different colour hence corresponding colour is observed.

Teacher's Note:
1. The colour you see is the complementary colour of the light absorbed, not the colour absorbed.

Answer: (a) Electronic configuration = 3d⁷ → t_2g⁵e_g², no. of unpaired electrons = 3, μ = 3.87 B.M.

Explanation:
1. In [Co(H₂O)₆]²⁺, oxidation state of Co = +2, Co²⁺ = 3d⁷
2.
3. t_2g⁵e_g²,  \( \mu = \sqrt{n(n+2)} \) = \( \sqrt{3(3+2)} \) = \( \sqrt{15} \) = 3.87 B.M.  [∵ n = 3]

Teacher's Note:
1. H₂O is a weak field ligand, so Co²⁺ (d⁷) stays high spin with three unpaired electrons.

Answer: (b) hexacyano complexes absorb indigo thus appearing yellow while hexaaqua complexes absorb orange or red light thus appear blue or green

Explanation:
1. The ligand CN⁻ absorbs indigo, the high energy radiation and thus appears yellow. The aqua complexes have much smaller CFSE, they absorb orange or red light and thus appear blue or green.

Teacher's Note:
1. Strong field ligand → large Δ → high-energy (short wavelength) light absorbed; the colour seen is the complement.

Answer: (a) for tetrahedral complexes, the CFSE is lower than pairing energy

Explanation:
1. Crystal field stabilisation energy for tetrahedral complexes is less than pairing energy hence they do not pair up to form low spin complexes.

Teacher's Note:
1. Δ_t = (4/9)Δ_o, so even strong field ligands rarely make Δ_t exceed the pairing energy.

Answer: (d) Trigonal bipyramidal – [Fe(NH₃)₆]²⁺, [CuCl₄]²⁻

Explanation:
1. [CuCl₄]²⁻ – Tetrahedral
2. [Fe(NH₃)₆]²⁺ – Octahedral

Teacher's Note:
1. Trigonal bipyramidal geometry needs coordination number 5; a complex with six ligands cannot be trigonal bipyramidal.

Answer: (c) [Cr(H₂O)₄Cl₂]Cl

Explanation:
1. It gives precipitate with AgNO₃ it means it gives Cl⁻ ions in the solution. Since conductivity corresponds to two ions, it shows one Cl⁻ is outside the coordination sphere. The structure will be
2. [Cr(H₂O)₄Cl₂]Cl → [Cr(H₂O)₄Cl₂]⁺ + Cl⁻
3. AgNO₃ + Cl⁻ → AgCl (white ppt.) + NO₃⁻

Teacher's Note:
1. Two ions in solution = one complex ion + one counter-ion; the coordination number 6 of Cr(III) is completed by 4H₂O + 2Cl.

Answer: (a) (i) and (iii)

Explanation:
1. SCN⁻ and NO₂⁻ are ambidentate ligands since they have more than one donor atoms to attach to the central metal atom.

Teacher's Note:
1. Ambidentate = two possible donor atoms but only one used at a time; oxalate uses both at once, so it is bidentate, not ambidentate.

Answer: (a) [Co(CN)₆]³⁻

Explanation:
1. Electronic configuration of Co³⁺ = 3d⁶
2. [Co(CN)₆]³⁻ :

Teacher's Note:
1. With a strong ligand like CN⁻, count electrons in t_2g: Co³⁺ d⁶ fills it completely, leaving no unpaired electron.

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

Explanation:
1. [Ni(en)₃]Cl₂ is a chelating compound and chelated complexes are more stable than similar complexes with unidentate ligands as dissociation of the complex involves breaking of two bonds rather than one.
2. In [Ni(en)₃]Cl₂, Ni with d⁸ configuration shows octahedral geometry.

Teacher's Note:
1. Chelate effect: replacing unidentate ligands by a polydentate ligand of similar donor atoms always increases stability.

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

Explanation:
1. EDTA⁴⁻ is a hexadentate ligand; its two N and four O donor atoms bind the metal ion at the same time, occupying all six octahedral positions.

Teacher's Note:
1. One EDTA⁴⁻ alone satisfies coordination number 6 – that is why it is used to titrate Ca²⁺ and Mg²⁺ (hardness of water).

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

Explanation:
1. Ni²⁺ configuration :

Teacher's Note:
1. For Ni²⁺ the outer orbital nature is forced by the d⁸ configuration, whatever the ligand strength.

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

Explanation:
1. 4d and 5d elements have greater tendency to form low spin complex (allows better pairing of electrons) in comparison to 3d because the difference in energy of t_2g and e_g (CFSE, Δ_o) increases in 4d and 5d.

Teacher's Note:
1. Δ_o rises by roughly 30–50% from 3d to 4d to 5d, so heavier metals are almost always low spin.

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

Explanation:
1. [Fe(CN)₆]³⁻ ion shows magnetic moment corresponding to one unpaired electron.

Teacher's Note:
1. Fe³⁺ is d⁵; with strong field CN⁻ it becomes t_2g⁵, i.e., one unpaired electron (μ ≈ 1.73 B.M.).

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

Explanation:
1. [Sc(H₂O)₆]³⁺ has no unpaired electron in its d subshell and thus d – d transition is not possible whereas [Ti(H₂O)₆]³⁺ has one unpaired electron in its d-subshell which gives rise to d – d transition to impart colour.

Teacher's Note:
1. d⁰ and d¹⁰ ions give colourless complexes because no d–d transition can occur.

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

Explanation:
1. Thiocarbonyl (CS) has one donor atom.

Teacher's Note:
1. CS, like CO, is a neutral unidentate ligand bonding through carbon.

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

Explanation:
1. N₃⁻ is named as azido. It is derived from HN₃.

Teacher's Note:
1. Nitride is N³⁻; azide is N₃⁻ – check the subscript and charge before naming.

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

Explanation:
1. Correct IUPAC name is tetraaquadichloridochromium(III) nitrate.

Teacher's Note:
1. Ligands are named alphabetically: aqua (a) comes before chlorido (c), and anionic ligands end in ‘-ido’.