JEE Chemistry Coordination Compounds MCQs Set 27

Answer: (c) \( \ce{[Ag(NH3)2]^{+}} \)

Explanation:
1. AgCl + 2NH₃ → \( \ce{[Ag(NH3)2]^{+}} \) + Cl⁻.
2. The soluble complex ion removes Ag⁺ from the precipitate.
3. Answer: (c)

Teacher's Note:
1. Diamminesilver(I) is linear.
2. Tollens' reagent is a related complex.

Answer: (d) \( \ce{[Cr(SCN)2(NH3)4]^{+}} \)

Explanation:
1. \( \ce{[Cr(SCN)2(NH3)4]^{+}} \) shows both geometrical and linkage isomerism: the two ambidentate ligands can be SCN/SCN, NCS/NCS or SCN/NCS, each with cis and trans forms, giving 6 isomers.
2. \( \ce{[Co(NH3)4Cl2]^{+}} \) gives 2 and the other two give 1.
3. Answer: (d)

Teacher's Note:
1. Ambidentate ligands add linkage isomers.
2. Count geometric forms for each linkage set.

Answer: (a) K₄[Fe(CN)₆]

Explanation:
1. Conductivity depends on the number of ions produced per formula unit.
2. K₄[Fe(CN)₆] gives 5 ions, [Co(NH₃)₆]Cl₃ gives 4 ions, [Cu(NH₃)₄]Cl₂ gives 3 ions and [Ni(CO)₄] gives none.
3. Answer: (a)

Teacher's Note:
1. More ions means higher conductance.
2. Neutral complexes do not conduct.

Answer: (d) \( \ce{[CuBr4]^{2-}} \)

Explanation:
1. \( \ce{[CuBr4]^{2-}} \): Cu²⁺ is d⁹ with 1 unpaired electron.
2. [Ni(CO)₄] (Ni 0, d¹⁰), \( \ce{[Ag(CN)2]^{-}} \) (Ag⁺, d¹⁰) and the low-spin Co³⁺ complex (d⁶) have none.
3. Answer: (d)

Teacher's Note:
1. Count \( d \) electrons of the metal ion.
2. Strong field ligands pair electrons.

Answer: (c) \( \ce{[MnBr4]^{2-}} \), tetrahedral

Explanation:
1. \( \mu = 5.92 \) BM means 5 unpaired electrons, high spin d⁵.
2. Weak field bromide with Mn²⁺ gives tetrahedral \( \ce{[MnBr4]^{2-}} \) (sp³), and the charge works out as \( +2 - 4 = -2 \).
3. Answer: (c)

Teacher's Note:
1. \( \mu = \sqrt{35} = 5.92 \) BM for five unpaired electrons.
2. Halides give tetrahedral complexes with \( 3d \) ions.

Answer: (a) linkage isomerism, ionization isomerism and geometrical isomerism

Explanation:
1. NO₂ is ambidentate, so linkage isomers (nitro/nitrito) are possible.
2. The counter ion Cl⁻ can exchange with NO₂⁻ to give ionization isomers.
3. cis and trans forms give geometrical isomerism; neither has optical activity, since the cis form has a mirror plane.
4. Answer: (a)

Teacher's Note:
1. Check each type of isomerism.
2. Optical isomerism needs no mirror plane.

Answer: (b) \( \ce{[Cu(NH3)4]^{2+}} \)

Explanation:
1. Fe(CO)₅ is dsp³ and uses d_z², and \( \ce{[Co(NH3)6]^{2+}} \) and \( \ce{[IrF6]^{3-}} \) are d²sp³ (d_z² and d_x²−y²).
2. \( \ce{[Cu(NH3)4]^{2+}} \) is square planar dsp², which uses d_x²−y² only.
3. Answer: (b)

Teacher's Note:
1. Square planar dsp² uses d_x²−y².
2. Octahedral inner complexes use both e_g orbitals.

Answer: (c) Magnetic behaviour

Explanation:
1. Both complexes are square planar with dsp² hybridisation and have one stereoisomer each.
2. \( \ce{[Cu(en)2]^{2+}} \) has one unpaired electron (Cu²⁺, d⁹) and is paramagnetic, while [Ni(dmg)₂] (Ni²⁺, d⁸) is diamagnetic.
3. Answer: (c)

Teacher's Note:
1. Count unpaired electrons to compare magnetism.
2. Both ligands are bidentate.

Answer: (b) [Ti(H₂O)₆]Cl₄

Explanation:
1. Zero magnetic moment means no unpaired electrons, so Ti has d⁰, Ti⁴⁺.
2. [Ti(H₂O)₆]Cl₄ contains Ti⁴⁺, while the other formulas have Ti²⁺ or Ti³⁺, which have unpaired electrons.
3. Answer: (b)

Teacher's Note:
1. Ti⁴⁺ is d⁰.
2. Match the oxidation state to zero moment.

Answer: (d) \( \ce{[Cu(NH3)4]^{2+}} \) has tetrahedral structure and is paramagnetic

Explanation:
1. \( \ce{[Cu(NH3)4]^{2+}} \) is square planar (dsp²), not tetrahedral, and has one unpaired electron.
2. Statements \( (a), (b) \) and \( (c) \) are true.
3. Answer: (d)

Teacher's Note:
1. Cu²⁺ with strong field ammonia: square planar.
2. Colour change confirms complex formation.

Answer: (a) \( \ce{[Co(H2O)6]^{2+}} \)

Explanation:
1. Water is a weak field ligand, so all are high spin.
2. Unpaired electrons: Cr²⁺ 4, Mn²⁺ 5, Fe²⁺ 4, Co²⁺ 3.
3. The lowest paramagnetism belongs to Co²⁺.
4. Answer: (a)

Teacher's Note:
1. Fewer unpaired electrons means lower paramagnetism.
2. High spin d⁷ has 3 unpaired electrons.

Answer: (b) Ionisation isomerism

Explanation:
1. A gives AgCl, so Cl⁻ is outside: [M(NH₃)₅SO₄]Cl.
2. \( B \) gives BaSO₄, so sulphate is outside: [M(NH₃)₅Cl]SO₄.
3. Exchange of ions between coordination sphere and outside is ionisation isomerism.
4. Answer: (b)

Teacher's Note:
1. Test with AgNO₃ for Cl⁻ and BaCl₂ for sulphate.
2. Ionisation isomers give different ions in solution.

Answer: (a) Linkage isomerism

Explanation:
1. SCN⁻ is an ambidentate ligand that can bind through \( S \) or \( N \).
2. The two compounds differ only in the donor atom, so they are linkage isomers.
3. Answer: (a)

Teacher's Note:
1. Ambidentate ligands: SCN⁻, NO₂⁻, CN⁻.
2. Linkage isomers have different donor atoms.

Answer: (b) 3 isomers (Geometrical)

Explanation:
1. The complex is square planar of the type [Mabcd].
2. For [Mabcd] there are 3 geometrical isomers.
3. Answer: (b)

Teacher's Note:
1. [Mabcd] square planar: 3.
2. [Ma2bc]: 2, [Ma2b2]: 2.

Answer: (a) zero

Explanation:
1. Ni in Ni(CO)₄ is in the zero oxidation state with configuration 3d¹⁰, sp³ hybridised and tetrahedral.
2. All electrons are paired, so there are zero unpaired electrons.
3. Answer: (a)

Teacher's Note:
1. CO is a strong field ligand.
2. Ni(0) is d¹⁰.

Answer: (b) square planar

Explanation:
1. Cu²⁺ is d⁹; one electron is promoted to \( 4p \) and dsp² hybridisation gives a square planar shape.
2. Answer: (b)

Teacher's Note:
1. Cu²⁺ ammine complex is dark blue.
2. Ni(CO)₄ is tetrahedral instead.

Answer: (a) 6, 3, 6, 0

Explanation:
1. Five NH₃ and one monodentate CO₃²⁻ give coordination number 6.
2. Oxidation state: \( x + 0 - 2 = +1 \), so \( x = +3 \) (Co³⁺, d⁶).
3. With the strong field ammonia, d⁶ is low spin with 0 unpaired electrons.
4. Answer: (a)

Teacher's Note:
1. Coordination number counts donor atoms.
2. Co(III) with NH₃ is low spin.

Answer: (d) In the presence of CN⁻ as a ligand Δ_o < P

Explanation:
1. CN⁻ is a strong field ligand, so Δ_o > \( P \) (pairing energy) and electrons pair up.
2. The other statements are correct, so \( (d) \) is incorrect.
3. Answer: (d)

Teacher's Note:
1. Strong field means Δ_o > \( P \).
2. Weak field means Δ_o < \( P \).

Answer: (c) d²sp³

Explanation:
1. Diamagnetic means low spin d⁶ (t_2g⁶), with two \( 3d \) orbitals vacant.
2. These two \( 3d \) orbitals take part with \( 4s \) and \( 4p \), so the hybridisation is d²sp³ (inner orbital).
3. Answer: (c)

Teacher's Note:
1. Inner orbital complexes use \( (n-1)d \) orbitals.
2. Co(III) with strong field ligands is low spin.