CBSE Class 12 Chemistry Coordination Compounds MCQs Set 16

Answer: (a) [Ni(CN)₄]²⁻

Explanation:
1. In [Ni(CN)₄]²⁻, the strong field CN⁻ pairs up all the 3d electrons of Ni²⁺ (dsp², no unpaired electron), so no d–d absorption is expected in the visible region.
2. [Cr(NH₃)₆]³⁺ (3 unpaired), [Fe(H₂O)₆]²⁺ (4 unpaired) and [Ni(H₂O)₆]²⁺ (2 unpaired) absorb visible light and are coloured.

Teacher's Note:
1. Colour of a complex is linked to d–d transitions, which need partly filled d orbitals with a suitable gap.

Answer: (b) [Co(CN)₆]³⁻

Explanation:
1. Energy absorbed, Δ_o = hc/λ, so the smaller the wavelength absorbed, the larger the splitting.
2. [Co(CN)₆]³⁻ absorbs at the shortest wavelength (310 nm), so its crystal field splitting is maximum, in agreement with CN⁻ being the strongest ligand.

Teacher's Note:
1. Δ_o is inversely proportional to λ_max: strong field ligand → larger Δ_o → shorter wavelength absorbed.

Answer: (b) Co³⁺, octahedral complex with strong field ligand

Explanation:
1. ₂₇Co = 3d⁷ 4s², Co³⁺ = 3d⁶ 4s⁰
2. In a strong field ligand, electrons pair up (t_2g⁶). Hence, it is diamagnetic in nature.
3. Co²⁺ (3d⁷) has an odd number of electrons and can never be diamagnetic.

Teacher's Note:
1. An ion with an odd number of electrons is always paramagnetic.

Answer: (d) All of these

Explanation:
1. All the given ligands are weak field ligands (with respect to these metal ions). Hence, all the given complexes are high spin complexes.

Teacher's Note:
1. The same ligand can give high spin with one metal ion and low spin with another: NH₃ gives low spin with Co³⁺ but high spin with Co²⁺.

Answer: (a) I and II

Explanation:
1. In I and II, charge on metal ion = + 2 while in III, charge on metal ion = + 3.
2. Hence, I and II are less stable (labile, their ligands are replaced fast) while III is quite stable (inert), so substitution in III is slow.

Teacher's Note:
1. Kinetic (labile/inert) behaviour is about the speed of ligand replacement, not about the final stability constant.

Answer: (d) All of these

Explanation:
1. All the given complexes are formed which are soluble in water.
2. CuCl + KCl → K[CuCl₂]; with more Cl⁻, [CuCl₃]²⁻ and [CuCl₄]³⁻ are also formed.

Teacher's Note:
1. Many insoluble salts dissolve in excess of a common anion because a soluble complex ion is formed.

Answer: (a) 4

Explanation:
1. Cobalt in zero oxidation state forms Co₂(CO)₈, i.e., 8 CO groups for two Co atoms.
2. Hence, n = 8/2 = 4.

Teacher's Note:
1. Recall the common carbonyls: Ni(CO)₄, Fe(CO)₅, Cr(CO)₆, Mn₂(CO)₁₀, Co₂(CO)₈.

Answer: (b) mononuclear

Explanation:
1. As the number of metal atoms present in the carbonyl complex is one, the complex is mononuclear.

Teacher's Note:
1. Nuclearity = number of metal atoms in the complex: Fe(CO)₅ is mononuclear, Mn₂(CO)₁₀ is dinuclear.

Answer: (a) paramagnetic [Fe(H₂O)₅(NO)] SO₄

Explanation:
1. Brown ring is [Fe(H₂O)₅(NO)] SO₄.
2. Here NO is present as NO⁺ and iron as Fe⁺ (3d⁶4s¹), which has unpaired electrons, so the complex is paramagnetic.

Teacher's Note:
1. The brown ring test is for nitrate ions; the brown colour comes from the nitrosyl complex [Fe(H₂O)₅NO]²⁺.

Answer: (b) cis-[PtCl₂(NH₃)₂]

Explanation:
1. cis-[PtCl₂(NH₃)₂] (cisplatin) is used as anti-cancer agent.

Teacher's Note:
1. Only the cis isomer (cisplatin) is active; trans-[PtCl₂(NH₃)₂] is not used against cancer.

Answer: (b) H₂PtCl₆

Explanation:
1. Pt + aqua regia (conc. HCl + conc. HNO₃) → H₂[PtCl₆].

Teacher's Note:
1. Aqua regia dissolves noble metals because Cl⁻ forms a stable complex anion with the oxidised metal ([PtCl₆]²⁻, [AuCl₄]⁻).

Answer: (c) [Fe(CN)₅NOS]⁴⁻

Explanation:
1. [Fe(CN)₅NO]²⁻ (sodium nitroprusside) + S²⁻ → [Fe(CN)₅NOS]⁴⁻ (purple complex)

Teacher's Note:
1. Charge check: [Fe(CN)₅NO]²⁻ plus S²⁻ gives a complex of charge 4−.

Answer: (b) Chlorophyll

Explanation:
1. Chlorophyll is a complex compound containing Mg²⁺ ion.
2. Haemoglobin and ferrocene contain Fe and vitamin B₁₂ contains Co, which are transition metals.

Teacher's Note:
1. Remember the metals: chlorophyll – Mg, haemoglobin – Fe, vitamin B₁₂ – Co.

Answer: (a) 287 × 10⁻³

Explanation:
1. [CrCl(H₂O)₅] Cl₂ → [CrCl(H₂O)₅]²⁺ + 2 Cl⁻
2. Moles of complex = \( \frac{0.01}{1000} \) × 100 = 0·001 mol, so Cl⁻ = 0·001 × 2 mol = 0·002 mol
3. Ag⁺ + Cl⁻ (0·002 mol) → AgCl (0·002 mol)
4. ∴ Mass of silver chloride = 143·5 × 0·002 = 0·287 g = 287 × 10⁻³ g

Teacher's Note:
1. The Cl inside the coordination sphere is not precipitated; only the two counter-ion Cl⁻ form AgCl.

Answer: (c) 6 and 2

Explanation:
1. Five NH₃ and one SO₄²⁻ (monodentate here) give coordination number 6.
2. x + (−2) + 5(0) = 0 ⇒ x = +2. So [X(SO₄)(NH₃)₅] contains X in + 2 oxidation state and C.N. is 6.

Teacher's Note:
1. Coordination number counts donor atoms bonded to the metal, not the total number of atoms in the ligands.

Answer: (a) NO⁺

Explanation:
1. In brown ring complex, NO behaves as NO⁺ and Fe has + 1 charge.

Teacher's Note:
1. Check: Fe(+1) + NO⁺(+1) + 5H₂O(0) = +2, which balances SO₄²⁻.

Answer: (d) I and V

Explanation:
1. The ligands which have two coordination positions but can coordinate to the central metal ion through only one coordination position at a time, are called ambidentate ligands, e.g. CN⁻, SCN⁻, NO₂⁻.
2. EDTA⁴⁻, C₂O₄²⁻ and en bind through two or more atoms at the same time (chelating), not ambidentate.

Teacher's Note:
1. Ambidentate: two possible donor atoms, but one at a time. Chelating: two or more donor atoms at the same time.

Answer: (b) potassium trioxalatoaluminate(III)

Explanation:
1. Al: 3(+1) + x + 3(−2) = 0, x = +3. The complex is an anion, so ‘aluminium’ becomes ‘aluminate’.
2. IUPAC name is potassium trioxalatoaluminate (III).

Teacher's Note:
1. Ligand names come before the metal, and the oxidation state is written in Roman numerals after the metal name.

Answer: (b) 3

Explanation:
1. Three geometrical isomers are possible: (I) trans w.r.t. Cl, (II) trans w.r.t. NH₃, (III) cis w.r.t. Cl and NH₃ both (optically active).

Teacher's Note:
1. For [M(AA)B₂C₂]: 3 geometrical isomers, but 4 stereoisomers in all because the all-cis form exists as a d/l pair.

Answer: (d) fac

Explanation:
1. When the three ligands (with same donor atoms) are on the same triangular face of the octahedron, the isomer is called facial or fac isomer.
2. In this complex the three ligands a are on the same triangular face of the octahedron (and so are the three b).

Teacher's Note:
1. In a fac isomer the three identical ligands are all cis to one another; in a mer isomer two of them are trans.

Answer: (d) [Cr (C₂O₄)₃]³⁻

Explanation:
1. [Cr (C₂O₄)₃]³⁻ (type [M(AA)₃]) has no plane of symmetry and shows optical isomerism.
2. The other three contain only one kind of monodentate ligand and are symmetrical.

Teacher's Note:
1. Every [M(AA)₃] octahedral complex, like [Co(en)₃]³⁺ or [Cr(ox)₃]³⁻, exists as d and l forms.

Answer: (c) both ionization and linkage isomerisms

Explanation:
1. Ionization isomers are : [Co(NH₃)₅(NO₂)] (NO₃)₂ and [Co(NH₃)₅(NO₃)] NO₂ NO₃
2. Linkage isomers are : [Co(NH₃)₅(NO₂)] (NO₃)₂ and [Co(NH₃)₅(ONO)] (NO₃)₂.

Teacher's Note:
1. An ambidentate ligand (NO₂⁻) gives linkage isomers; a different anion outside the sphere gives ionisation isomers.

Answer: (c) [Co(NH₃)₆] Cl₃

Explanation:
1. 4·305 g AgCl = \( \frac{4.305}{143.5} \) mole = 0·03 mole
2. As 0·01 mole of the complex gives 0·03 mole of AgCl, this shows that there are three ionizable Cl⁻ ions. Hence, formula is (c).

Teacher's Note:
1. Moles of AgCl ÷ moles of complex = number of ionisable chloride ions per formula.

Answer: (b) dsp², square planar, 1.73 B.M.

Explanation:
1. In [Cu(CN)₄]²⁻, Cu undergoes dsp²-hybridization giving a square planar complex having Cu²⁺ (d⁹ configuration) with one unpaired electron (n = 1), similar to [Cu(NH₃)₄]²⁺.
2. The magnetic moment (μ) = \( \sqrt{n(n+2)} \) B.M. = \( \sqrt{1(1+2)} \) = \( \sqrt{3} \) = 1·73 B.M.

Teacher's Note:
1. For Cu²⁺ (d⁹) there is always one unpaired electron, so μ = 1.73 B.M. whatever the geometry.

Answer: (c) The complex is an outer orbital complex

Explanation:
1. The complex is an inner orbital octahedral complex with d²sp³ hybridization.
2. Cr³⁺ (3d³) has three unpaired electrons (paramagnetic), and the three Cl⁻ outside the sphere give white AgCl.

Teacher's Note:
1. A d³ ion always has two empty 3d orbitals, so its octahedral complexes are inner orbital (d²sp³).

Answer: (a) − 0·4 and − 2·4

Explanation:
1. High spin d⁶ (t_2g⁴e_g²): CFSE = (− 0·4x + 0·6y) Δ_o = (− 0·4 × 4 + 0·6 × 2) Δ_o = − 0·40 Δ_o
2. Low spin d⁶ (t_2g⁶e_g⁰): CFSE = (− 0·4x + 0·6y) Δ_o = (− 0·4 × 6 + 0·6 × 0) Δ_o = − 2·4 Δ_o
3. (x = electrons in t_2g, y = electrons in e_g)

Teacher's Note:
1. Each t_2g electron contributes −0.4Δ_o and each e_g electron +0.6Δ_o; pairing energy is ignored here.

Answer: (b) Low spin t_2g⁶ e_g⁰

Explanation:
1. The complex [RhCl₆]³⁻ contains Rh³⁺ (4d⁶) which belongs to 4d transition series. The magnitude of Δ_o increases by about 30% between adjacent members down a group of transition elements, i.e., from 3d to 4d and 4d to 5d series.
2. Therefore, elements of 4d and 5d series are more prone to spin pairing and thus, form low spin complexes (even with Cl⁻).

Teacher's Note:
1. 4d and 5d metal complexes are almost always low spin, because Δ_o is large and pairing energy is smaller.

Answer: (a) SCN⁻ < F⁻ < CN⁻ < CO

Explanation:
1. As per spectrochemical series, the correct increasing order of field strength of ligands is SCN⁻ < F⁻ < CN⁻ < CO

Teacher's Note:
1. Learn the series: I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO.

Answer: (b) [Fe(CO)₅]

Explanation:
1. Stability of a complex depends upon the strength of ligand and charge of the central metal ion: stability ∝ strength of ligand ∝ charge of central metal ion.
2. Among the given complexes, in Fe(CO)₅, the charge of central metal ion is zero while CO is the strongest field ligand. Also, it obeys EAN rule (26 + 10 = 36). Hence, [Fe(CO)₅] is the most stable complex.

Teacher's Note:
1. EAN = atomic number − electrons lost + electrons gained from ligands; an EAN equal to the next noble gas number indicates extra stability.

Answer: (c) One

Explanation:
1. Ca²⁺ forms an octahedral complex with just one molecule of EDTA as it has four donor oxygen atoms and two donor nitrogen atoms in each molecule.

Teacher's Note:
1. EDTA⁴⁻ is hexadentate: one EDTA alone fills all six positions of an octahedron.

Answer: (c) Solvate isomerism

Explanation:
1. The given pair differ in the number of coordinated H₂O molecules as ligand.
2. Such isomers are called solvate (hydrate) isomers.

Teacher's Note:
1. In hydrate isomers water moves between the coordination sphere and the lattice, so the number of ionisable Cl⁻ changes too.

Answer: (c) MABCD complexes : form three geometrical isomers :: M(AA)₂BC complexes : form total three isomers

Explanation:
1. (a) is incorrect because EDTA⁴⁻ is hexadentate.
2. (b) is incorrect because [Cr(H₂O)₅(SCN)]²⁺ shows linkage isomerism.
3. (d) is incorrect because [Fe(H₂O)₆]²⁺ forms outer orbital complex whereas [Cr(H₂O)₆]³⁺ forms inner orbital complex.
4. Only (c) is correct: MABCD gives three geometrical isomers, and M(AA)₂BC gives trans + d-cis + l-cis = three isomers.

Teacher's Note:
1. In analogy questions, check both halves of each option; one wrong half makes the whole option wrong.

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

Explanation:
1. Assertion is true.
2. Reason is false: Zeise’s salt is ionic complex with the formula, K[PtCl₃(η²–C₂H₄)].

Teacher's Note:
1. Oxidation state of Pt in K[PtCl₃(C₂H₄)]: +1 + x + 3(−1) + 0 = 0, so x = +2.

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

Explanation:
1. Assertion is false: [Ni(CO)₄] is tetrahedral (sp³) while [Ni(CN)₄]²⁻ is square planar (dsp²).
2. Reason is true: both have four monodentate ligands.

Teacher's Note:
1. Coordination number 4 can mean tetrahedral or square planar; the hybridisation decides which.

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: CN⁻ is a strong field ligand while H₂O is a weak field ligand.

Teacher's Note:
1. Whether a complex is low spin depends on field strength, not on the charge of the ligand (Cl⁻ is negative but weak field).

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: t_2g contain three electrons while e_g contains two electrons (though all are unpaired). Hence, net CFSE = 3(−0·4Δ_o) + 2(+0·6Δ_o) = 0.

Teacher's Note:
1. CFSE depends on how many electrons are in t_2g and e_g, not on whether they are paired.

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

Explanation:
1. Assertion is false: coordination compounds are also formed by certain non-transition elements, e.g. chlorophyll is a coordination compound of Mg.
2. Reason is true.

Teacher's Note:
1. Examples of non-transition metal complexes: chlorophyll (Mg), [AlF₆]³⁻, Ca-EDTA.

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

Explanation:
1. Both A and R are correct and R is the correct explanation of A.

Teacher's Note:
1. Geometrical isomerism is common in square planar (CN = 4) and octahedral (CN = 6) complexes, never in tetrahedral ones.

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

Explanation:
1. Assertion is true. Reason is false.
2. In [Ni(CO)₄], CO is a strong field ligand and results into the pairing of the two half-filled 3d orbitals by electrons of 4s orbitals of Ni atom. In [NiCl₄]²⁻, Cl⁻ is a weak field ligand and is unable to pair up the unpaired electrons. Hence, hybridisation in both cases is sp³.

Teacher's Note:
1. Both are sp³, but for different reasons: Ni(0) 3d¹⁰ with CO, and Ni²⁺ 3d⁸ unpaired with Cl⁻.

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: complexes with configuration d⁵ can have either t_2g³e_g² (high spin state) configuration in the presence of a weak field ligand or t_2g⁵e_g⁰ (low spin state) configuration in the presence of strong field ligand.

Teacher's Note:
1. High or low spin for d⁴–d⁷ ions is decided by comparing Δ_o with the pairing energy P.

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: effective atomic number of Fe in [Fe(CN)₆]⁴⁻ (26 − 2 + 12 = 36) is equal to that of next noble gas (viz. 36) whereas that of [Fe(CN)₆]³⁻ is not equal to that of noble gas (as it is 35).

Teacher's Note:
1. By the EAN rule, the extra stability comes from reaching the noble gas number (36 for Kr), not just from having a larger EAN.

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

Explanation:
1. Both A and R are correct and R is the correct explanation of A.
2. In [Cr(CO)₅NH₃], NH₃ is a pure σ-donor; it increases the electron density on Cr, which is passed on to the CO groups by back bonding. More electrons in the antibonding π* orbitals of CO lower the C–O bond order, so the C–O bond is longer than in [Ni(CO)₄].

Teacher's Note:
1. Greater back bonding means a stronger M–C bond but a weaker and longer C–O bond (lower C–O stretching frequency).