JEE Chemistry Coordination Compounds MCQs Set 19

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

Explanation:
1. Ni(II) (d⁸) in an octahedral aqua complex has two unpaired electrons.
2. The others are diamagnetic: d¹⁰, d¹⁰ and low spin d⁶.

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) potassium hexachloroplatinate(IV)

Explanation:
1. Pt = +4 (2 + x - 6 = 0), and an anionic complex takes -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: (a) \( \ce{Na2[Fe(CN)5NO]} \)

Explanation:
1. Sodium nitroprusside is sodium pentacyanonitrosylferrate(II), \( \ce{Na2[Fe(CN)5NO]} \).

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) paramagnetism

Explanation:
1. Mn(II) is d⁵ and water is a weak-field ligand, so the ion has five unpaired electrons.

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: (b) \( \ce{[NiCl4]^{2-}} \)

Explanation:
1. Tetrahedral Ni(II) in \( \ce{[NiCl4]^{2-}} \) is d⁸ with two unpaired electrons.
2. The other three are diamagnetic.

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) \( \ce{CN^-} \)

Explanation:
1. Potassium ferricyanide is \( \ce{K3[Fe(CN)6]} \); the ligand is cyanide.

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) \( \ce{[(C5H5)2Fe]} \)

Explanation:
1. Ferrocene is bis(cyclopentadienyl)iron(II), a sandwich compound.

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) \( \ce{Na[Ag(CN)2]} \)

Explanation:
1. \( \ce{Ag2S + 4NaCN <=> 2Na[Ag(CN)2] + Na2S} \).
2. Ag(I) forms the linear dicyanoargentate(I) ion.

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) 2

Explanation:
1. The cis form is chiral and exists as a pair of enantiomers.
2. The trans form is achiral, so 2 optical isomers exist.

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: (b) one co-ordinate site

Explanation:
1. Mono means one: it binds through one donor atom.

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) five

Explanation:
1. Mn(II) is d⁵. With a weak-field ligand the electrons occupy t₂g³ eg² singly, so there are 5 unpaired electrons.

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) +3

Explanation:
1. The complex ion carries +3 (three Cl- balance it); NH3 is neutral, so Co = +3.

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) \( \ce{[Ag(NH3)2]Cl} \)

Explanation:
1. Ag(I) is 4d¹⁰, so \( \ce{[Ag(NH3)2]Cl} \) is diamagnetic.
2. Cu(II) is d⁹, and NO and \( \ce{NO2} \) have an odd number of electrons.

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) \( \ce{Al(OC2H5)3} \)

Explanation:
1. In aluminium ethoxide aluminium is bonded to oxygen 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) \( \ce{[CoCl4]^{2-}} \)

Explanation:
1. Excess chloride replaces the water ligands, giving the blue tetrahedral tetrachlorocobaltate(II) ion.

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) \( \ce{[Cr(C2O4)3]^{3-}} \)

Explanation:
1. Tris(oxalato)chromate(III) is a propeller-shaped tris-chelate with non-superimposable mirror images.
2. The others are symmetric.

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

Explanation:
1. \( \ce{AgCl + 2NH3 -> [Ag(NH3)2]^+ + Cl^-} \).

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) one unpaired electron

Explanation:
1. Ti(III) is 3d¹, which has one unpaired electron.

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) \( \ce{cis-[Co(en)2Cl2]} \)

Explanation:
1. cis-[Co(en)2Cl2] has no plane of symmetry, so it exists as two enantiomers.
2. The others have a mirror plane.

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) \( \ce{[Cr(CO)6]} \)

Explanation:
1. CO is neutral and the complex is neutral, so Cr is 0.
2. The others have chloride ligands that make the metal +2 or +3.

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) hexaamminechromium(III) ion

Explanation:
1. Ammonia is neutral, so the oxidation state equals the charge, +3.

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) Zeise's salt

Explanation:
1. Zeise's salt is \( \ce{K[PtCl3(C2H4)]} \); ethene binds to Pt through its pi electrons.

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) tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)

Explanation:
1. The cation is named first: tetraamminedichloroplatinum(IV).
2. The anion is tetrachloroplatinate(II).

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) zero

Explanation:
1. CO is a strong-field ligand, so the 4s electrons of Ni move into 3d, giving 3d¹⁰ with sp³ hybridisation.
2. All electrons are paired, so there are no unpaired electrons.

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) the number of ligands around a metal ion bonded by sigma bonds

Explanation:
1. Coordination number counts the donor atoms joined to the metal by sigma bonds.
2. Pi interaction with the metal does not add to the count.

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.