JEE Chemistry Coordination Compounds MCQs Set 22

Answer: (d) 6, 6, 4

Explanation:
1. Count the ligands: 6 CN-, 6 CN- and 4 Cl-.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (a) \( \ce{Al^{3+}} \)

Explanation:
1. Al(OH)3 precipitates with ammonia and does not dissolve in excess, because Al(III) has no tendency to form ammine complexes in water.
2. Cd, Cu and Ag all do.

Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.

Answer: (a) +2

Explanation:
1. Ethene is neutral and the three chlorides give -3; the overall charge is -1, so Pt = +2.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (b) 3

Explanation:
1. One complex cation and two chloride ions: 3 ions.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (a) 6

Explanation:
1. Each ethylenediamine (en) is bidentate, giving 2 x 2 = 4 donor atoms.
2. Two bromide ligands add 2 more, so the coordination number is 6.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (b) blue

Explanation:
1. The deep blue colour arises from d-d transitions of the \( \ce{[Cu(NH3)4]^{2+}} \) ion.

Teacher's Note:
1. Identify the type of isomerism from what changes: the ligands' arrangement, their donor atom, or which ions are inside the coordination sphere.
2. For magnetism, find the metal's oxidation state and d-electron count, then decide high or low spin from the ligand field.

Answer: (a) \( \ce{[Co(NH3)3Cl3]} \)

Explanation:
1. Only free (ionic) chloride precipitates AgCl.
2. In \( \ce{[Co(NH3)3Cl3]} \) all chlorides are coordinated, so no precipitate forms.

Teacher's Note:
1. Identify the type of isomerism from what changes: the ligands' arrangement, their donor atom, or which ions are inside the coordination sphere.
2. For magnetism, find the metal's oxidation state and d-electron count, then decide high or low spin from the ligand field.

Answer: (c) \( \ce{[Ni(NH3)6]^{2+}} \)

Explanation:
1. Ni(II) is d⁸, so no inner d orbitals are free for d²sp³; it must use outer 4d orbitals (sp³d²).
2. The other three are low-spin inner-orbital d²sp³ complexes.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (d) \( \ce{Li[AlH4]} \)

Explanation:
1. The complex anion is \( \ce{[AlH4]^-} \) and lithium is the cation.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (d) coordination number

Explanation:
1. This is the definition of coordination number.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (b) \( \ce{[Co(en)3]Cl3} \), \( \ce{cis-[Co(en)2Cl2]Cl} \)

Explanation:
1. Tris-chelate and cis-bis-chelate octahedral complexes lack mirror planes and are chiral.
2. \( \ce{cis-[Co(NH3)4Cl2]} \) is achiral, and so are the square planar and tetrahedral complexes.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (d) coagulating agent

Explanation:
1. Alum furnishes \( \ce{Al^{3+}} \) ions, which coagulate the blood proteins and seal the wound.

Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.

Answer: (d) A is false

Explanation:
1. Ferrocyanide (Fe(II), low spin d⁶) is diamagnetic, but ferricyanide (Fe(III), low spin d⁵) has one unpaired electron and is paramagnetic.
2. So the assertion is false, and the reason is also false because ferrocyanide has no unpaired electrons.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (c) tris(ethylenediamine)cobalt(III) bromide

Explanation:
1. Tris(chelate) octahedral complexes are chiral (propeller shape) and exist as enantiomers.
2. The other complexes have a mirror plane.

Teacher's Note:
1. Identify the type of isomerism from what changes: the ligands' arrangement, their donor atom, or which ions are inside the coordination sphere.
2. For magnetism, find the metal's oxidation state and d-electron count, then decide high or low spin from the ligand field.

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

Explanation:
1. Fe(II) low spin d⁶ has no unpaired electrons, while Fe(III) low spin d⁵ has one, so A is true.
2. In fact the splitting is slightly larger for the higher charged Fe(III) complex, so R is false.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (c) \( \ce{CN^-} \)

Explanation:
1. In the spectrochemical series \( \ce{CN^-} \) produces by far the largest splitting of these.

Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.

Answer: (b) chloropentaamminecobalt(III) chloride

Explanation:
1. Co: x - 1 = +2 so x = +3. The complex cation is named chloropentaamminecobalt(III), followed by the counter ion chloride.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (b) tetracarbonylnickel(0)

Explanation:
1. The complex is neutral and CO is neutral, so the oxidation state of Ni is 0 and the name has no -ate.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (b) ethylenediamine (en)

Explanation:
1. Ethylenediamine has two nitrogen donor atoms.
2. Cyanide and thiocyanate are monodentate; EDTA is hexadentate.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (a) \( \ce{[(C5H5)2Fe]} \)

Explanation:
1. Ferrocene has a metal-to-ring back-donation of d electrons into empty ring orbitals.
2. The others have bridging or ordinary sigma bonds only.

Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.

Answer: (d) carbonyls

Explanation:
1. Compounds of CO with metals, such as \( \ce{Ni(CO)4} \), are called metal carbonyls.

Teacher's Note:
1. Draw the complex out: identify the central metal, count its ligands, and add up the charges.
2. Remember the naming order: ligands alphabetically, then metal, with the oxidation state in Roman numerals.

Answer: (a) \( \ce{(CH3)4Sn} \)

Explanation:
1. Tetramethyltin has only sigma Sn-C bonds.
2. The others bind through pi systems (alkene or arene rings).

Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.

Answer: (c) \( \ce{Fe4[Fe(CN)6]3} \)

Explanation:
1. \( \ce{4Fe^{3+} + 3[Fe(CN)6]^{4-} -> Fe4[Fe(CN)6]3} \), ferric ferrocyanide.

Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.

Answer: (d) \( \ce{[Ni(CN)4]^{2-}} \)

Explanation:
1. Ni(II) (d⁸) with strong-field CN- pairs its electrons, leaving one 3d orbital for dsp² (square planar).

Teacher's Note:
1. Decide the oxidation state and d-count of the metal, then use the ligand strength (strong field pairs electrons, weak field does not) to pick the hybridisation.
2. Link geometry and hybridisation: sp³ is tetrahedral, dsp² is square planar, d²sp³ and sp³d² are octahedral.