Chemistry Objective Questions and Answers: Coordination Compounds
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Multiple Choice Questions
(a) \( \ce{[Pt(NH3)4]Cl2} \)
(b) \( \ce{[Co(NH3)5Cl](NO2)2} \)
(c) \( \ce{[Pt(NH3)2Cl2]} \)
(d) \( \ce{[Pt(en)Cl2]} \)
Show Answer & Explanation
Answer: (a) \( \ce{[Pt(NH3)4]Cl2} \)
Explanation:
1. Only the complex with ionic chloride, \( \ce{[Pt(NH3)4]Cl2} \), gives white AgCl.
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.
(a) 3 and 6
(b) 4 and 2
(c) 2 and 4
(d) 4 and 4
Show Answer & Explanation
Answer: (c) 2 and 4
Explanation:
1. Ni is +2 (the complex ion balances sulphate), which is the primary valency.
2. Four NH3 ligands give a coordination number of 4.
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.
(a) carboxypeptidase-A is an enzyme and contains zinc
(b) chlorophylls are green pigments in plants and contain calcium
(c) cyanocobalamin is vitamin \( \ce{B12} \) and contains cobalt
(d) haemoglobin is the red pigment of blood and contains iron
Show Answer & Explanation
Answer: (b) chlorophylls are green pigments in plants and contain calcium
Explanation:
1. Chlorophyll contains magnesium, not calcium.
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.
(a) it is a small molecule
(b) it is a negatively charged ion
(c) it has an unshared electron pair
(d) it is a positively charged ion
Show Answer & Explanation
Answer: (c) it has an unshared electron pair
Explanation:
1. A ligand must donate a lone pair to the metal to form a coordinate bond.
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.
(a) hydrogenation
(b) polymerisation
(c) halogenation
(d) condensation
Show Answer & Explanation
Answer: (a) hydrogenation
Explanation:
1. Wilkinson's catalyst, \( \ce{[RhCl(PPh3)3]} \), is a homogeneous hydrogenation catalyst.
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.
(a) \( \ce{[NiCl4]^{2-}} \)
(b) \( \ce{[Cu(NH3)4]^{2+}} \)
(c) \( \ce{[Ni(H2O)6]^{2+}} \)
(d) \( \ce{Ni(CO)4} \)
Show Answer & Explanation
Answer: (b) \( \ce{[Cu(NH3)4]^{2+}} \)
Explanation:
1. Cu(II) with NH3 forms the square planar tetraamminecopper(II) ion.
2. The nickel species are tetrahedral, tetrahedral and octahedral.
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.
(a) for creating knocking
(b) for reducing knocking
(c) as a catalyst in polymerisation reaction of alkenes
(d) as a catalyst in addition reaction of alkenes
Show Answer & Explanation
Answer: (b) for reducing knocking
Explanation:
1. Tetraethyllead is added to petrol as an anti-knock 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.
(a) arsenic
(b) nitrogen
(c) oxygen
(d) both nitrogen and oxygen
Show Answer & Explanation
Answer: (d) both nitrogen and oxygen
Explanation:
1. N (as in \( \ce{NH3} \)) and O (as in \( \ce{H2O} \)) are the most common donor atoms.
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.
(a) ionisation isomers
(b) geometrical isomers
(c) coordination isomers
(d) linkage isomers
Show Answer & Explanation
Answer: (c) coordination isomers
Explanation:
1. The ligands NH3 and oxalate are exchanged between the cationic and anionic complexes.
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.
(a) 4
(b) 8
(c) 2
(d) 6
Show Answer & Explanation
Answer: (d) 6
Explanation:
1. Four ammonia and two chloride ligands are bonded to platinum: 4 + 2 = 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.
(a) \( \ce{[Pt(en)2]Cl2} \)
(b) \( \ce{K2[Ni(EDTA)]} \)
(c) \( \ce{Ag(NH3)2Cl} \)
(d) \( \ce{K3[Al(C2O4)3]} \)
Show Answer & Explanation
Answer: (b) \( \ce{K2[Ni(EDTA)]} \)
Explanation:
1. EDTA is hexadentate and forms several five-membered chelate rings, which gives a very large stability constant (chelate effect).
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.
(a) ferrous ammonium sulphate reacts with \( \ce{FeCl3} \)
(b) ferrous sulphate reacts with \( \ce{FeCl3} \)
(c) ferric sulphate reacts with \( \ce{K4[Fe(CN)6]} \)
(d) ammonium sulphate reacts with \( \ce{FeCl3} \)
Show Answer & Explanation
Answer: (c) ferric sulphate reacts with \( \ce{K4[Fe(CN)6]} \)
Explanation:
1. Ferric ions combine with ferrocyanide to give Prussian blue.
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.
(a) iron(III) hexacyanoferrate(II)
(b) hexacyanoferrate(III-II)
(c) ferroso-ferric cyanide
(d) ferric-ferrous hexacyanate
Show Answer & Explanation
Answer: (a) iron(III) hexacyanoferrate(II)
Explanation:
1. Four Fe3+ balance three \( \ce{[Fe(CN)6]^{4-}} \) (+12 and -12).
2. The name is iron(III) hexacyanoferrate(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.
(a) \( sp^3 \)
(b) \( sp^3d^2 \)
(c) \( d^3sp^2 \)
(d) \( dsp^2 \)
Show Answer & Explanation
Answer: (b) \( sp^3d^2 \)
Explanation:
1. Water is a weak-field ligand, so Ni(II) uses outer 4d orbitals: sp³d² (octahedral, outer orbital).
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.
(a) \( \ce{[Cu{O=C(NH2)2}Cl2]} \)
(b) \( \ce{[CuCl2][O=C(NH2)2H2]} \)
(c) \( \ce{[Cu{O=C(NH2)2}Cl]Cl} \)
(d) \( \ce{[CuCl2{O=C(NH2)2}2]} \)
Show Answer & Explanation
Answer: (d) \( \ce{[CuCl2{O=C(NH2)2}2]} \)
Explanation:
1. 'Dichloro' gives two Cl and 'bis(urea)' gives two urea molecules bonded through oxygen, all inside the bracket.
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.
(a) less than 1
(b) 9-11
(c) 3-4
(d) 2-3
Show Answer & Explanation
Answer: (b) 9-11
Explanation:
1. The precipitation needs the ligand to lose a proton, so slightly basic (ammoniacal) conditions of pH 9-11 are best.
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.
(a) \( d^2sp^3 \)
(b) \( sp^3d^2 \)
(c) \( d^3sp^2 \)
(d) \( dsp^2 \)
Show Answer & Explanation
Answer: (a) \( d^2sp^3 \)
Explanation:
1. Cr(0) is 3d⁵ 4s¹; with CO strong field the six electrons pair in 3d, leaving two 3d orbitals for d²sp³.
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.
(a) two types of ions
(b) three types of ions
(c) only one type of ion
(d) four types of ions
Show Answer & Explanation
Answer: (b) three types of ions
Explanation:
1. Potash alum is a double salt that gives \( \ce{K^+} \), \( \ce{Al^{3+}} \) and \( \ce{SO4^{2-}} \) in water.
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.
(a) 3
(b) 5
(c) 2
(d) 4
Show Answer & Explanation
Answer: (a) 3
Explanation:
1. Co(II) is d⁷. In the tetrahedral weak-field complex the configuration is e⁴ t₂³, which has 3 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.
(a) 4
(b) 3
(c) 1
(d) 2
Show Answer & Explanation
Answer: (b) 3
Explanation:
1. \( \ce{[Co(NH3)5Cl]Cl2} \) gives one complex cation and two chloride ions: 3 ions.
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.
(a) \( sp^3, sp^3, dsp^2 \)
(b) \( dsp^2, sp^3, sp^3 \)
(c) \( sp^3, dsp^2, sp^3 \)
(d) \( sp^3, dsp^2, dsp^2 \)
Show Answer & Explanation
Answer: (c) \( sp^3, dsp^2, sp^3 \)
Explanation:
1. Ni(CO)4 is tetrahedral (sp³).
2. [Ni(CN)4]2- is square planar (dsp²).
3. [NiCl4]2- is tetrahedral (sp³).
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.
(a) a substance with no special name
(b) recipient
(c) coordinate ion
(d) ligand
Show Answer & Explanation
Answer: (d) ligand
Explanation:
1. A species that donates a lone pair to the central metal is a ligand.
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.
(a) \( \ce{[Co(NH3)3(H2O)3]Cl2} \)
(b) \( \ce{[Co(NH2)3(H2O)2]Cl2} \)
(c) \( \ce{[CoCl(NH3)3(H2O)2]Cl2} \)
(d) \( \ce{[CoCl(NH3)3(H2O)2]Cl3} \)
Show Answer & Explanation
Answer: (c) \( \ce{[CoCl(NH3)3(H2O)2]Cl2} \)
Explanation:
1. The complex ion has Co(III), one Cl-, three NH3 and two H2O: charge +3 - 1 = +2, balanced by two 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.
(a) \( \ce{K^+, Cl^-, SO4^{2-}, Br^-} \)
(b) \( \ce{K^+, Mg^{2+}, Cl^-, Br^-} \)
(c) \( \ce{K^+, Mg^{2+}, CO3^{2-}} \)
(d) \( \ce{K^+, Mg^{2+}, Cl^-} \)
Show Answer & Explanation
Answer: (d) \( \ce{K^+, Mg^{2+}, Cl^-} \)
Explanation:
1. Carnallite is \( \ce{KCl.MgCl2.6H2O} \), a double salt.
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.
(a) \( \ce{Ag^+} \)
(b) \( \ce{Fe^{2+}} \)
(c) \( \ce{Cu^+} \)
(d) \( \ce{Ni^{2+}} \)
Show Answer & Explanation
Answer: (a) \( \ce{Ag^+} \)
Explanation:
1. \( \ce{Ag^+} \) forms the linear dicyanoargentate complex \( \ce{[Ag(CN)2]^-} \) used in silver extraction.
2. The others give four- or six-coordinate cyanide 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.
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