JEE Chemistry Coordination Compounds MCQs Set 20

Answer: (c) the same as secondary valency

Explanation:
1. Werner's secondary valency is what is now called the 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: (a) ionisation

Explanation:
1. The anion inside the coordination sphere (Br or sulphate) is exchanged with the counter ion: ionisation isomerism.

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) triamminediaquachlorocobalt(III) chloride

Explanation:
1. Alphabetical order: ammine, aqua, chloro.
2. Co: x + 0 + 0 - 1 = +2 so x = +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: (a) two

Explanation:
1. Square planar \( \ce{[MA2B2]} \) has a cis and a trans isomer.

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) \( \ce{K4[Ni(CN)2(Ox)2]} \)

Explanation:
1. Charge on the anion = +2 + 2(-1) + 2(-2) = -4, so four K+ are needed.

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) \( d^4 \) (in strong ligand field)

Explanation:
1. A moment of 2.84 BM corresponds to n = 2 (\( \sqrt{2\times4}=2.83 \)).
2. Strong-field d⁴ is t₂g⁴ with 2 unpaired electrons; weak-field d⁴ has 4, d³ has 3 and strong-field d⁵ has 1.

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

Explanation:
1. Both fac and mer isomers of \( \ce{[Co(NH3)3Cl3]} \) have a plane of symmetry, so neither is chiral.

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: (d) 5.91 BM

Explanation:
1. Fe(III) in \( \ce{[FeF6]^{3-}} \) is high spin d⁵ with n = 5.
2. \( \mu=\sqrt{5\times7}=5.92 \) BM, closest to 5.91 BM.

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) \( \ce{triethylaluminium + TiCl4} \)

Explanation:
1. The Ziegler-Natta catalyst is made from \( \ce{Al(C2H5)3} \) and \( \ce{TiCl4} \) and polymerises alkenes.

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) \( \ce{[Co(NH3)5(NO2)]Cl2} \)

Explanation:
1. Two free chlorides precipitate as AgCl, and three ions in total means one complex cation and two Cl-: \( \ce{[Co(NH3)5(NO2)]Cl2} \).

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) \( sp^3d^2 \)

Explanation:
1. F- is weak field, so Co(III) keeps four unpaired electrons and uses outer 4d orbitals: sp³d².

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) \( sp^3d^2 \)

Explanation:
1. Six hybrid orbitals directed to the corners of an octahedron are needed.
2. Of these options, only \( sp^3d^2 \) provides six.

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) \( \ce{C2H5ONa} \)

Explanation:
1. Sodium ethoxide has a Na-O-C link and no metal-carbon bond.

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) potassium hexacyanoferrate(III)

Explanation:
1. Fe = +3 (three K+ balance six CN-: 3 + x - 6 = 0).

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) zero

Explanation:
1. Magnetic moment \( \sqrt{n(n+2)}=0 \) means n = 0.

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) trigonal bipyramidal

Explanation:
1. Iron(0) uses dsp³ hybrid orbitals for five CO ligands, giving a trigonal bipyramid.

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) hydrate isomerism

Explanation:
1. They differ in how many water molecules are inside the coordination sphere versus outside as water of crystallisation.

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: (d) \( \ce{C2O4^{2-}} \)

Explanation:
1. Oxalate has two donor oxygens; the others have one donor atom each.

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) low spin

Explanation:
1. Strong-field ligands cause a large splitting that forces electrons to pair, giving low-spin complexes.

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) geometrical isomerism

Explanation:
1. The octahedral complex \( \ce{[MA4B2]} \) has cis and trans 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) potassium dicyanoargentate(I)

Explanation:
1. Ag is +1 and the anionic complex takes the Latin stem argent- with -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) \( \ce{[Co(NH3)5Br]SO4} \)

Explanation:
1. Only \( \ce{[Co(NH3)5Br]SO4} \) has a ligand (Br-) that can exchange with the counter ion (sulphate) to give \( \ce{[Co(NH3)5SO4]Br} \).

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) oxalate

Explanation:
1. A chelate needs a ligand that binds through two or more donor atoms to form a ring.
2. Oxalate is the only polydentate ligand listed.

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) Both A and R are true and R is the correct explanation of A

Explanation:
1. Ti(III) is d¹, so a d-d transition is possible and it is coloured.
2. Sc(III) is d⁰, so no d-d transition is possible and it is colourless; R explains A.

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) A is true but R is false

Explanation:
1. Geometrical isomers are cis-trans isomers, so A is true.
2. In a tetrahedron all positions are adjacent, so tetrahedral complexes do not show it; 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.