JEE Chemistry Coordination Compounds MCQs Set 11

Answer: (c) [Pt(trien)]²⁺

Explanation:
1. [Ag(S₂O₃)₂]³⁻ is linear and [Co(EDTA)]²⁻ is octahedral
2. [Ni(H₂O)₄]²⁺ is not square planar (water is weak field)
3. Pt²⁺ (d⁸) forms square planar complexes, and the four nitrogen donors of trien occupy the four positions.

Teacher's Note:
1. A homoleptic complex has only one kind of ligand.
2. Pt(II) is the typical square planar ion.

Answer: (d) a = H₂O, b = NH₃, c = Cl⁻

Explanation:
1. The formula Ma₂b₂c₂ means six monodentate ligands
2. en, EDTA and oxalate are polydentate, so options 1 to 3 give a wrong coordination number
3. With H₂O, NH₃ and Cl⁻ as monodentate ligands, Ma₂b₂c₂ has 5 geometrical isomers, one of which (all cis) is chiral.

Teacher's Note:
1. Ma₂b₂c₂ has 5 geometrical isomers and 6 stereoisomers in all.
2. Polydentate ligands cannot be counted as separate positions here.

Answer: (a) Grignard's reagent

Explanation:
1. In Grignard's reagent R–MgX the carbon is directly σ-bonded to Mg
2. Ferrocene, cobaltocene and ruthenocene are π-bonded sandwich compounds.

Teacher's Note:
1. Organometallic compounds have at least one metal–carbon bond.
2. Sandwich compounds bond through the π-electrons of the ring.

Answer: (b) It is d²sp³ hybridised and octahedral

Explanation:
1. Mn³⁺ is d⁴ and CN⁻ is a strong field ligand, so the four electrons pair in two t₂g orbitals
2. Two 3d orbitals are vacant, giving d²sp³ hybridisation
3. The shape is octahedral, with 2 unpaired electrons.

Teacher's Note:
1. Strong field ligand means inner orbital (d²sp³) complex.
2. Six ligands give an octahedral shape.

Answer: (d) [Co(en)₃]³⁺, [Co(NH₃)₆]³⁺, [Co(H₂O)₆]³⁺

Explanation:
1. Ligand field strength: H₂O < NH₃ < en
2. A stronger field gives larger Δ₀ and a shorter wavelength of absorption
3. So wavelength increases in the order en, NH₃, H₂O.

Teacher's Note:
1. λ ∝ 1/Δ₀.
2. Strongest ligand: absorbs at the shortest wavelength.

Answer: (a) Violet

Explanation:
1. The observed colour is the complementary colour of the absorbed colour
2. The complementary colour of yellow is violet.

Teacher's Note:
1. Complementary pairs: yellow and violet, red and green, blue and orange.
2. [CoCl(NH₃)₅]Cl₂ is a purple-violet solid.

Answer: (a) geometrical isomerism

Explanation:
1. The complex is of the type [M(AA)₂b₂]
2. It has cis and trans forms (geometrical isomerism), and the cis form is also optically active
3. There is no second metal for coordination isomerism, no counter ion for ionisation isomerism and no ambidentate ligand for linkage isomerism.

Teacher's Note:
1. Check the ligands for ambidentate character and the second metal for coordination isomerism.
2. Geometrical isomerism is the main type here.

Answer: (b) tetrahedral geometry and diamagnetic

Explanation:
1. In [Ni(CO)₄] nickel is in the zero oxidation state, 3d⁸4s²
2. CO is a strong field ligand which pairs the electrons into 3d¹⁰
3. sp³ hybridisation gives tetrahedral geometry, and there are no unpaired electrons, so it is diamagnetic.

Teacher's Note:
1. CO is a strong field ligand, but with Ni(0) the d¹⁰ arrangement leads to sp³ hybridisation.
2. Ni(CO)₄ is a volatile liquid used in Mond's process.

Answer: (b) mononuclear

Explanation:
1. Fe(CO)₅ contains a single iron atom, so it is mononuclear
2. Fe₂(CO)₉ is dinuclear and Fe₃(CO)₁₂ is trinuclear.

Teacher's Note:
1. Count the metal atoms in the formula.
2. Fe(CO)₅ has trigonal bipyramidal geometry.

Answer: (d) Ni²⁺

Explanation:
1. All are M²⁺ ions with the same ligand (H₂O), giving high spin complexes
2. CFSE (in Δ₀): Mn²⁺ d⁵ = 0, Fe²⁺ d⁶ = −0.4, Co²⁺ d⁷ = −0.8, Ni²⁺ d⁸ = −1.2
3. Ni²⁺ has the largest CFSE, so its aqua complex is the most stable (Irving–Williams series).

Teacher's Note:
1. Stability of M²⁺ complexes rises from Mn to Cu.
2. Compute CFSE for each configuration.

Answer: (a) [Co(ox)₃]³⁻ and [Fe(CN)₆]⁴⁻

Explanation:
1. [Co(ox)₃]³⁻: Co³⁺ d⁶, low spin t₂g⁶, diamagnetic
2. [Fe(CN)₆]⁴⁻: Fe²⁺ d⁶, low spin t₂g⁶, diamagnetic
3. [Fe(CN)₆]³⁻ has 1 unpaired electron and [CoF₆]³⁻ has 4, so the other pairs are paramagnetic.

Teacher's Note:
1. Low spin d⁶ is always diamagnetic.
2. Fluoride gives a high spin Co³⁺ complex.

Answer: (c) PPh₃

Explanation:
1. CO, CN⁻ and C₂H₄ have low-lying π* orbitals and are good π acceptors
2. PPh₃ accepts into its weaker σ* (P–C) orbitals and is a much weaker π acceptor, so it shows the least back bonding of the four.

Teacher's Note:
1. The better the π* acceptor, the more back bonding.
2. CO is the classic back bonding ligand.

Answer: (d) [Cr(NO)₄]

Explanation:
1. NO is a stronger π-acceptor than CO and replaces all the CO ligands
2. With excess NO, [Cr(CO)₆] gives [Cr(NO)₄]; NO donates three electrons, so Cr gets 6 + 4×3 = 18 electrons.

Teacher's Note:
1. Apply the effective atomic number (18 electron) rule.
2. NO⁺ is isoelectronic with CO but a better π acceptor.

Answer: (a) CO

Explanation:
1. A synergic bond needs a ligand with a lone pair for σ donation and empty π* orbitals for back donation
2. CO is the best π acceptor; en has no π* orbitals.

Teacher's Note:
1. Synergic bonding strengthens both the M–C and the ligand to metal donation.
2. CO is the strongest π acceptor ligand of this group.

Answer: (d) trans isomer

Explanation:
1. The two B ligands occupy opposite corners of the square plane, so they are at 180° to each other
2. B ligands opposite each other means the trans arrangement
3. Cis/trans describe the position of identical ligands; d and l refer to optical activity, and this isomer has a plane of symmetry, so it is optically inactive.

Teacher's Note:
1. In MA₄B₂ the cis isomer has B groups adjacent (90°) and the trans isomer has them opposite (180°).
2. The trans isomer has a centre of symmetry.

Answer: (d) identical

Explanation:
1. In the first drawing the two A ligands are above and below the metal, i.e. trans to each other, and the two en ligands lie in the plane
2. In the second drawing the two A ligands are at diagonally opposite corners of the square plane, which are also trans (180°); the en ligands span the remaining cis positions
3. Both are trans-[M(en)₂A₂]; one drawing is just the other rotated, so the complexes are identical.

Teacher's Note:
1. Check the angle between the two A ligands in each drawing before deciding.
2. Rotating a molecule in space does not create a new isomer.

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

Explanation:
1. Potash alum is a double salt; in water it dissociates completely into the simple ions K⁺, Al³⁺ and SO₄²⁻
2. Al³⁺ is hydrated and hydrolyses to give H⁺, so the solution is acidic: A is true
3. A double salt does not give a complex ion, so R is false.

Teacher's Note:
1. Double salts lose their identity in solution, unlike complex salts.
2. The acidity comes from hydrolysis of [Al(H₂O)₆]³⁺.

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

Explanation:
1. Six Cl⁻ ligands are bonded to Pt(IV), so the coordination number is 6: A is true
2. The coordination number is the number of coordinate bonds formed, so R is true and explains A.

Teacher's Note:
1. Coordination number equals the number of ligand donor atoms attached to the metal.
2. Here each Cl⁻ is monodentate.

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

Explanation:
1. In a tetrahedron every position is adjacent to every other position, so there is no cis or trans distinction: A is true
2. This is because all four positions are equivalent: R is true and explains A.

Teacher's Note:
1. Geometrical isomerism needs positions that differ in relative arrangement.
2. Tetrahedral complexes can still be optically active.

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

Explanation:
1. Fe(0) has 3d⁶4s²; CO is a strong field ligand and the electrons rearrange to 3d⁸, leaving a 3d orbital free for dsp³ hybridisation (inner orbital): A is true
2. The oxidation state of iron is indeed zero: R is true
3. The inner orbital character arises from the strong field of CO, not from the oxidation state, so R is not the explanation.

Teacher's Note:
1. Inner orbital means a (n−1)d orbital is used for hybridisation.
2. The oxidation state alone does not decide the type.

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

Explanation:
1. Fe³⁺ is d⁵; with strong CN⁻ the arrangement is t₂g⁵, leaving one unpaired electron: A is true
2. The complex is low spin: R is true
3. Being low spin does not itself make a complex paramagnetic (the odd d electron count does), so R is not the explanation.

Teacher's Note:
1. A d⁵ ion always has an odd number of electrons, so some stay unpaired.
2. Low spin d⁵ has 1 unpaired electron; high spin d⁵ has 5.

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

Explanation:
1. [PtCl₆]²⁻ carries a negative charge, so it is a complex anion: A is true
2. The ligands are negatively charged Cl⁻: R is true
3. The sign of the overall charge depends on the oxidation state of the metal as well as on the ligands (Pt⁴⁺ with six Cl⁻ gives 2−), so R does not fully explain A.

Teacher's Note:
1. Overall charge = metal charge + total charge of ligands.
2. Complex anions are named with the suffix -ate.