Practice MCQs for Class 12 Chemistry Unit 5 Coordination Compounds
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Multiple Choice Questions
(a) EDTA4−
(b) (COO)22−
(c) dien
(d) NO2−
Show Answer & Explanation
Answer: (c) dien
Explanation:
1. Dien (Diethylenetriamine) has the following structure
2. H2N̈ – CH2 – CH2 – N̈H – CH2 – CH2 – N̈H2
Teacher's Note:
1. Count donor atoms with lone pairs: dien has three N atoms; EDTA4− has six and oxalate two.
(a) [Cr(H2O)6]3+
(b) [Fe(H2O)6]2+
(c) [Cu(H2O)6]2+
(d) [Zn(H2O)6]2+
Show Answer & Explanation
Answer: (b) [Fe(H2O)6]2+
Explanation:
1. More the number of unpaired electrons, higher is its paramagnetism.
2. Cr3+ : 3d3, Fe2+ : 3d6, Cu2+ : 3d9, Zn2+ : 3d10
3. Fe2+ has four unpaired electrons hence it shows highest paramagnetism.
Teacher's Note:
1. With weak field water all these ions are high spin, so just count unpaired electrons from the d configuration: 3, 4, 1 and 0.
(a) Coordinate compounds are mainly known for transition metals.
(b) Coordination number and oxidation state of a metal are same.
(c) Tetrahedral complexes rarely form low spin complexes.
(d) A ligand donates at least one electron pair to the metal atom to form a bond.
Show Answer & Explanation
Answer: (b) Coordination number and oxidation state of a metal are same.
Explanation:
1. Coordination number is the number of ligands that are directly bound to the central metal atom by coordinate bonds. Oxidation state is the residual charge on the central metal atom left after removing all ions.
Teacher's Note:
1. In [Co(NH3)6]3+ the oxidation state is +3 but the coordination number is 6 – the two are independent quantities.
(a) On adding KF, H2O being weak field ligand is replaced by F− ions forming [CuF4]2− which is green in colour.
(b) Potassium is coordinated to [Cu(H2O)4]2+ ion present in CuSO4 and gives green colour.
(c) On adding KF, Cu2+ are replaced by K+ forming a green complex.
(d) Blue colour of CuSO4 and yellow colour of KI form green colour on mixing.
Show Answer & Explanation
Answer: (a) On adding KF, H2O being weak field ligand is replaced by F− ions forming [CuF4]2− which is green in colour.
Explanation:
1. Aqueous CuSO4 solution contains [Cu(H2O)4]2+ ions which are blue in colour. When aqueous solution of KF is added, H2O being weak field ligand can be replaced by F− ions forming [CuF4]2− which is green in colour.
2. [Cu(H2O)4]2+ + 4F− (from KF) → [CuF4]2− (green) + 4H2O
Teacher's Note:
1. A colour change on adding a new ligand signals ligand substitution in the coordination sphere.
(a) [Pt(NH3)2Cl(en)NO2]Cl2
(b) Pt[Pt(NH3)2(en)Cl2NO2]
(c) Pt[(NH3)2(en)NO2]Cl2
(d) Pt[(NH3)2(en)NO2Cl2]
Show Answer & Explanation
Answer: (a) [Pt(NH3)2Cl(en)NO2]Cl2
Explanation:
1. [Pt(NH3)2Cl(en)NO2]Cl2 : Diamminechlorido(ethylenediamine)nitroplatinum(IV) chloride
Teacher's Note:
1. Check: Pt(+4) + Cl(−1) + NO2(−1) = +2, balanced by two chloride counter-ions.
(a) sp3d2, outer orbital complex, paramagnetic
(b) dsp2, inner orbital complex, diamagnetic
(c) d2sp3, inner orbital complex, paramagnetic
(d) d2sp3, outer orbital complex, diamagnetic.
Show Answer & Explanation
Answer: (c) d2sp3, inner orbital complex, paramagnetic
Explanation:
1. In [Cr(H2O)6]3+, oxidation state of Cr = +3, Cr3+ = 3d3
2. [Cr(H2O)6]3+ :
| 3d | 4s | 4p | ||||||||
| ↑ | ↑ | ↑ | ×× | ×× | ×× | ×× | ×× | ×× | ||
| d2sp3 | ||||||||||
3. It is an inner orbital (3d – 4p) complex. Due to presence of three unpaired electrons, it is paramagnetic.
Teacher's Note:
1. For d3 ions two 3d orbitals are already empty, so they form inner orbital complexes whatever the ligand.
(a) bidentate
(b) tridentate
(c) tetradentate
(d) pentadentate.
Show Answer & Explanation
Answer: (c) tetradentate
Explanation:
1. Number of donor atoms in N(CH2CH2NH2)3 is four hence it is a tetradentate ligand.
Teacher's Note:
1. Count the central N as a donor too: one tertiary N + three NH2 groups = four donor atoms.
(a) Coordination compound containing cationic complex ion : [Fe(H2O)2(C2O4)2]2SO4
(b) Coordination compound containing anionic complex ion : [Ag(NH3)2]Cl
(c) Non-ionic coordination compound : [Co(NO2)3(NH3)3]
(d) Coordination compound containing cationic and anionic complex ion : [Pt(NH3)4] [CuCl4]
Show Answer & Explanation
Answer: (b) Coordination compound containing anionic complex ion : [Ag(NH3)2]Cl
Explanation:
1. [Ag(NH3)2]Cl is a coordination compound containing cationic complex ion.
Teacher's Note:
1. When the counter-ion is an anion like Cl−, the complex in brackets must be the cation.
(a) X = [Co(NH3)5SO4]Br, Y = [Co(NH3)5Br]SO4
(b) X = [Co(NH3)5Br]SO4, Y = [Co(NH3)5SO4]Br
(c) X = [Co(NH3)5Br(SO4)], Y = [CoBr(SO4)(NH3)5]
(d) X = [Co(Br)5NH3]SO4, Y = [CoBr(SO4)]NH3
Show Answer & Explanation
Answer: (a) X = [Co(NH3)5SO4]Br, Y = [Co(NH3)5Br]SO4
Explanation:
1. [Co(NH3)5SO4]Br (X) → [Co(NH3)5SO4]+ + Br−
2. AgNO3 + Br− → AgBr (pale yellow ppt.) + NO3−
3. [Co(NH3)5Br]SO4 (Y) → [Co(NH3)5Br]2+ + SO42−
4. BaCl2 + SO42− → BaSO4 (white ppt.) + 2Cl−
Teacher's Note:
1. These two compounds are ionisation isomers – they swap the ligand and the counter-ion.
(a) [Co(NH3)6]Cl3
(b) [Co(NH3)5Cl]Cl2
(c) [Co(NH3)4Cl2]Cl
(d) [Co(NH3)3Cl3]
Show Answer & Explanation
Answer: (a) [Co(NH3)6]Cl3
Explanation:
1. [Co(NH3)6]Cl3 gives 3 moles of AgCl.
2. [Co(NH3)5Cl]Cl2 gives 2 moles of AgCl.
3. [Co(NH3)4Cl2]Cl gives 1 mole of AgCl.
4. [Co(NH3)3Cl3] will not give AgCl.
Teacher's Note:
1. Moles of AgCl = number of Cl− written outside the square brackets.
(a) sp3d, inner orbital complex, diamagnetic
(b) sp3d2, outer orbital complex, paramagnetic
(c) d2sp3, inner orbital complex, paramagnetic
(d) d2sp3, outer orbital complex, diamagnetic
Show Answer & Explanation
Answer: (b) sp3d2, outer orbital complex, paramagnetic
Explanation:
1. In [FeCl6]4−, oxidation state of Fe = +2, Fe2+ = 3d6
2. [FeCl6]4− :
| 3d | 4s | 4p | 4d | |||||||||||||
| ↑↓ | ↑ | ↑ | ↑ | ↑ | ×× | ×× | ×× | ×× | ×× | ×× | ||||||
| sp3d2 (Outer orbital complex) | ||||||||||||||||
3. Paramagnetic due to presence of four unpaired electrons.
Teacher's Note:
1. Weak field Cl− cannot pair the 3d electrons, so the outer 4d orbitals are used.
(a) primary valency is ionisable
(b) secondary valency is ionisable
(c) primary and secondary valencies are ionisable
(d) neither primary nor secondary valency is ionisable.
Show Answer & Explanation
Answer: (a) primary valency is ionisable
Explanation:
1. Primary valency is ionisable according to Werner’s theory of coordination compounds.
Teacher's Note:
1. Primary valency = oxidation state (ionisable, satisfied by anions); secondary valency = coordination number (non-ionisable).
(a) presence of strong field ligand in CuSO4·5H2O
(b) due to absence of water (ligand), d-d transition are not possible in CuSO4
(c) anhydrous CuSO4 undergoes d-d transitions due to crystal field splitting
(d) colour is lost due to loss of unpaired electrons.
Show Answer & Explanation
Answer: (b) due to absence of water (ligand), d-d transition are not possible in CuSO4
Explanation:
1. In CuSO4·5H2O, water acts as ligand and as a result it causes crystal field splitting making d-d transition possible in CuSO4·5H2O. Hence, it is coloured. In anhydrous CuSO4 due to absence of ligand, crystal field splitting is not possible hence no colour is observed.
Teacher's Note:
1. Cu2+ is still d9 in anhydrous CuSO4 – it is the missing ligand field, not the unpaired electrons, that removes the colour.
(a) 3d4
(b) 3d5
(c) 3d6
(d) 3d7
Show Answer & Explanation
Answer: (b) 3d5
Explanation:
1. 3d5 has maximum number of unpaired electrons.
Teacher's Note:
1. In the high-spin state the number of unpaired electrons rises up to d5 and then falls.
(a) K2Cr2O7
(b) [Co(SO4)]
(c) (NH4)2[TiCl6]
(d) K3[Cu(CN)4]
Show Answer & Explanation
Answer: (b) [Co(SO4)]
Explanation:
1. In [Co(SO4)], the oxidation state of Co is +2. Configuration of Co2+ = 3d7, it has unpaired electrons in 3d-orbitals so it is paramagnetic. Because of incompletely filled d-orbitals it is coloured.
Teacher's Note:
1. K2Cr2O7 is coloured but diamagnetic (Cr6+, d0) – its colour is due to charge transfer, not unpaired electrons.
(a) Cation is named first and then anion.
(b) In coordination sphere, the ligands are named alphabetically.
(c) Positively charged ligands have suffix ‘ate’.
(d) More than one ligand of a particular type are indicated by using di, tri, tetra, etc.
Show Answer & Explanation
Answer: (c) Positively charged ligands have suffix ‘ate’.
Explanation:
1. Positively charged ligands have suffix ‘ium’.
Teacher's Note:
1. ‘ate’ is used for the metal in an anionic complex, never for a positive ligand (e.g. NO2+ nitronium).
(i) The geometry of the complex ion depends upon the coordination number.
(ii) If coordination number is 6, the complex is octahedral.
(iii) If coordination number is 4, the geometry of the complex may be tetrahedral or square planar.
(a) (i), (ii) and (iii)
(b) (i) and (ii) only
(c) (i) and (iii) only
(d) (ii) and (iii) only
Show Answer & Explanation
Answer: (a) (i), (ii) and (iii)
Explanation:
1. All the statements are correct with respect to the geometry of complex ions.
Teacher's Note:
1. For coordination number 4, decide between tetrahedral (sp3) and square planar (dsp2) from the hybridisation.
(a) H2O
(b) NH3
(c) ONO
(d) CO
Show Answer & Explanation
Answer: (c) ONO
Explanation:
1. ONO is an anionic ligand (ONO−).
Teacher's Note:
1. Nitrito (ONO−) and nitro (NO2−) are both the nitrite ion, carrying −1 charge.
Conductivity of complex I corresponds to 1 : 3 electrolyte, conductivity of complex II corresponds to 1 : 2 electrolyte while conductivity of complex 3 and 4 corresponds to 1 : 1 electrolyte. So correct option about these complexes is
(a) complex I is [Co(NH3)6]Cl3 with purple colour
(b) complex II is [Co(NH3)6]Cl3 with purple colour
(c) complex III is [Co(NH3)4Cl2]Cl with violet colour
(d) both (b) and (c)
Show Answer & Explanation
Answer: (c) complex III is [Co(NH3)4Cl2]Cl with violet colour
Explanation:
1.
| Colour | Formation | Solution conductivity corresponds to |
|---|---|---|
| Yellow | [Co(NH3)6]3+3Cl− | 1 : 3 electrolyte |
| Purple | [CoCl(NH3)5]2+2Cl− | 1 : 2 electrolyte |
| Green | [CoCl2(NH3)4]+Cl− | 1 : 1 electrolyte |
| Violet | [CoCl2(NH3)4]+Cl− | 1 : 1 electrolyte |
Teacher's Note:
1. The green and violet 1 : 1 complexes are the trans and cis isomers of [CoCl2(NH3)4]Cl.
[Cr(NH3)3(H2O)3]Cl3 – Triamminetriaquachromium(III) chloride
[Ag(NH3)2][Ag(CN)2] – Diamminesilver(I) dicyanidosilver(I)
[CoCl2(en)2]Cl – Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride
K3[Al(C2O4)3] – Potassium trioxalatoaluminium(III)
[Ni(CO)4] – Tetracarbonylnickel(0)
Few names given by her were not correct as she didn’t follow one rule while naming these compounds. That one rule is
(a) the ligands are name in an alphabetical order before the name of central atom/ion.
(b) prefixes mono, di, tri etc. are used to indicate the number of individual ligands in the coordination entity.
(c) if the complex ion is cation, the metal is named same as the element.
(d) if the complex ion is anion, the metal ends with suffix – ate.
Show Answer & Explanation
Answer: (d) if the complex ion is anion, the metal ends with suffix – ate.
Explanation:
1. [Ag(NH3)2] [Ag(CN)2] : Diamminesilver(I) dicyanidoargentate(I)
2. K3[Al(C2O4)3] : Potassium trioxalatoaluminate(III)
Teacher's Note:
1. Some metals use their Latin names in anionic complexes: argentate (Ag), ferrate (Fe), cuprate (Cu), plumbate (Pb).
Fey+ + CN− → Octahedral complex
(x and y may or may not be equal)
The difference between the spin-only magnetic moments is 4.2 B.M. approximately. The reason for this difference in magnetic moment is
(a) CN− is a strong ligand while SCN− is a weak ligand
(b) Fe is present in O.S. I in complex with SCN− while in O.S. III in complex with CN−
(c) SCN− is a strong ligand while CN− is a weak ligand
(d) x is 3 while y is 1.
Show Answer & Explanation
Answer: (a) CN− is a strong ligand while SCN− is a weak ligand
Explanation:
1. Fe (26) : 3d64s2; Fe3+ : 3d5
2. In [Fe(CN)6]3−, CN− is a strong field ligand which causes pairing of electrons.
3.
| 3d | 4s | 4p | ||||||||
| ↑↓ | ↑↓ | ↑ | ×× | ×× | ×× | ×× | ×× | ×× | ||
| d2sp3 hybridisation | ||||||||||
4. \( \mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \) = 1.732 BM
5. In [Fe(SCN)6]3−, SCN− being a weak field ligand does not cause pairing of electrons.
6.
| 3d | 4s | 4p | 4d | |||||||||||||
| ↑ | ↑ | ↑ | ↑ | ↑ | ×× | ×× | ×× | ×× | ×× | ×× | ||||||
| sp3d2 hybridisation | ||||||||||||||||
7. \( \mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \) = 5.916 BM
8. Difference = 5.916 − 1.732 = 4.184 ≈ 4.2 BM
Teacher's Note:
1. A difference of about 4.2 B.M. corresponds to 5 vs 1 unpaired electrons – the signature of high-spin vs low-spin d5.
I. [Cr(H2O)6]2+ II. [CoCl4]2− III. [Fe(H2O)6]2+ IV. [Mn(H2O)6]2+ V. [Ni(CO)4] VI. [Ni(CN)4]2−
The complexes which have zero magnetic moment are.
(a) I and V
(b) II and VI only
(c) III and IV
(d) V and VI only
Show Answer & Explanation
Answer: (d) V and VI only
Explanation:
1. [Cr(H2O)6]2+ : Cr2+ → [Ar]3d44s0 :
| 3d | 4s | |||||
| ↑ | ↑ | ↑ | ↑ | |||
2. [CoCl4]2− : Co2+ → [Ar]3d74s0 :
| 3d | 4s | |||||
| ↑↓ | ↑↓ | ↑ | ↑ | ↑ | ||
3. [Fe(H2O)6]2+ : Fe2+ → [Ar]3d64s0 :
| 3d | 4s | |||||
| ↑↓ | ↑ | ↑ | ↑ | ↑ | ||
4. [Mn(H2O)6]2+ : Mn2+ → [Ar]3d54s0 :
| 3d | 4s | |||||
| ↑ | ↑ | ↑ | ↑ | ↑ | ||
5. [Ni(CO)4] : Ni0 → [Ar]3d84s2 :
| 3d | 4s | |||||
| ↑↓ | ↑↓ | ↑↓ | ↑↓ | ↑↓ | ||
6. [Ni(CN)4]2− : Ni2+ → [Ar]3d84s0 :
| 3d | 4s | |||||
| ↑↓ | ↑↓ | ↑↓ | ↑↓ | |||
7. Hence, [Ni(CO)4] and [Ni(CN)4]2− have zero unpaired electrons i.e., zero magnetic moment.
Teacher's Note:
1. Strong field ligands (CO, CN−) pair the electrons; H2O and Cl− leave the high-spin count unchanged.
Complex I : Diamagnetic and square planar
Complex II : Paramagnetic and tetrahedral
Complex III : Diamagnetic and tetrahedral
Complex IV : Paramagnetic and Octahedral
Which is not correct option for the given complexes?
(a) The ligand in complex I is CN− and it has dsp2 hybridisation.
(b) The ligand in complex II is Cl− and it has sp3 hybridisation.
(c) The ligand in complex IV is H2O and it has d2sp3 hybridisation.
(d) The ligand in complex III is CO and it has sp3 hybridiation.
Show Answer & Explanation
Answer: (c) The ligand in complex IV is H2O and it has d2sp3 hybridisation.
Explanation:
1. [Ni(CN)4]2− : dsp2 hybridisation (square planar complex) and have zero unpaired electrons (diamagnetic)
2. [NiCl4]2− : sp3 hybridisation (tetrahedral) and have two unpaired electrons (paramagnetic)
3. [Ni(H2O)6]2+ : sp3d2 hybridisation (octahedral) and have two unpaired electrons (paramagnetic)
4. [Ni(CO)4] : sp3 hybridisation (tetrahedral) and have zero unpaired electrons (diamagnetic)
Teacher's Note:
1. Octahedral Ni2+ (d8) complexes are always outer orbital (sp3d2) since pairing frees only one 3d orbital.
(a) 0
(b) 2
(c) 3
(d) 4
Show Answer & Explanation
Answer: (a) 0
Explanation:
1. CO is a neutral ligand and its oxidation state is zero. Since the overall charge on the complex is zero too, hence oxidation state of Ni is 0.
Teacher's Note:
1. All simple metal carbonyls such as [Ni(CO)4], [Fe(CO)5] and [Cr(CO)6] have the metal in the zero oxidation state.
(a) K2[Pt(en)2Cl2]
(b) [Co(NH3)3Cl3]
(c) [Cr(H2O)6]Cl3
(d) [Fe(H2O)3Cl3]
Show Answer & Explanation
Answer: (c) [Cr(H2O)6]Cl3
Explanation:
1. AgNO3 + [Cr(H2O)6]Cl3 → AgCl ppt
2. Since Cl is outside the coordination sphere, it can react with AgNO3 forming the white AgCl precipitate.
Teacher's Note:
1. Only chloride ions outside the square bracket (counter ions) are free in solution; chlorido ligands inside the sphere do not react with AgNO3.
(a) [Co(ONO)3(NH3)3]
(b) [Co(NO2)3(NH3)3]
(c) [Co(ONO2)3(NH3)3]
(d) [Co(NO2)(NH3)3]
Show Answer & Explanation
Answer: (a) [Co(ONO)3(NH3)3]
Explanation:
1. According to IUPAC nomenclature, central metal atom with oxidation number +3, bonded with ligands :NH3, ONO− such that coordination formula is [Co(ONO)3(NH3)3].
Teacher's Note:
1. Nitrito-O means the ligand binds through oxygen and is written ONO; nitrito-N binds through nitrogen and is written NO2.
(a) 4
(b) 2
(c) 3
(d) 5
Show Answer & Explanation
Answer: (c) 3
Explanation:
1. [Co(NH3)5Cl]Cl2 → [Co(NH3)5Cl]2+(aq) + 2Cl−
Teacher's Note:
1. Count the complex ion as one ion plus every counter ion outside the bracket: 1 + 2 = 3.
(a) 3
(b) 6
(c) 4
(d) 5
Show Answer & Explanation
Answer: (b) 6
Explanation:
1. Coordination number is the number of ligands joined to the central metal ion or atom. Since ethylenediamine is a bidentate ligand, Co has coordination number of 6.
Teacher's Note:
1. Coordination number counts donor atoms, not ligands: three bidentate en ligands give 3 × 2 = 6.
(a) [Fe(CO)5]
(b) [Fe(H2O)6]3+
(c) [Fe(C2O4)3]3−
(d) [Fe(CN)6]3−
Show Answer & Explanation
Answer: (c) [Fe(C2O4)3]3−
Explanation:
1. [Fe(C2O4)3]3− acts as the chelating complex ion because C2O42− is a bidentate ligand.
Teacher's Note:
1. In such questions look for a polydentate (chelating) ligand first: ring formation (chelate effect) greatly increases the stability of a complex.
(a) unidentate
(b) didentate
(c) polydentate
(d) has variable denticity
Show Answer & Explanation
Answer: (a) unidentate
Explanation:
1. Ambidentate ligands like NO2− and SCN− are such ligands which can attach through atoms of two different elements.
2. But at a time they bind through only one donor atom (N or O in NO2−, S or N in SCN−), so they are unidentate.
Teacher's Note:
1. Do not confuse ambidentate (two possible donor atoms, but only one used at a time) with flexidentate/variable denticity, as shown by EDTA.
(a) [Co(NH3)4(H2O)Cl]Cl2
(b) [Co(NH3)4(H2O)Cl]Cl3
(c) [Co(NH3)2(H2O)Cl]Cl2
(d) [Co(NH3)4(H2O)Cl]Cl
Show Answer & Explanation
Answer: (a) [Co(NH3)4(H2O)Cl]Cl2
Explanation:
1. Tetraammineaquachloridocobalt(III) chloride is [Co(NH3)4(H2O)Cl]Cl2.
2. Central atom – Cobalt(III)
3. Coordination sphere ligands: Tetraammine – 4 NH3 groups, neutral ligand; Aqua – 1 H2O group, neutral ligand; Chlorido – 1 Cl group, negatively charged ligand, one negative charge
4. Counter ion – 2 Chloride ions
5. Since the valency of cobalt is 3+, one valency is satisfied with Cl in coordination sphere and 2 by chlorine counter ions.
6. Hence, the formula of the coordination compound is [Co(NH3)4(H2O)Cl]Cl2.
Teacher's Note:
1. Find the charge on the complex ion first: +3 − 1 = +2, so two Cl− counter ions are needed.
Assertion–Reason Questions
Assertion (A): [Fe(CN)6]3− is weakly paramagnetic while [Fe(CN)6]4− is diamagnetic.
Reason (R): [Fe(CN)6]3− has +3 oxidation state while [Fe(CN)6]4− has +2 oxidation state.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. [Fe(CN)6]3− has one unpaired electron, hence it shows paramagnetic nature while [Fe(CN)6]4− possesses no unpaired electron and thus shows diamagnetic nature.
Teacher's Note:
1. Magnetic behaviour is decided by the number of unpaired electrons (low spin d5 vs d6), not by the oxidation state as such.
Assertion (A): [Al(NH3)6]3+ does not exist in aqueous solution.
Reason (R): NH3 is a neutral ligand.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. The complex ion [Al(NH3)6]3+ undergoes the change into new complex ion [Al(H2O)6]3+ in aqueous medium due to higher heat of hydration of aluminium ion on account of its small size.
2. [Al(NH3)6]3+ + 6H2O → [Al(H2O)6]3+ + 6NH3
Teacher's Note:
1. The small, highly charged Al3+ ion prefers water molecules, so NH3 is displaced in aqueous solution.
Assertion (A): Low spin complexes have less number of unpaired electrons.
Reason (R): [FeF6]3− is a low spin complex.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. [FeF6]3− is a high spin complex since F− is a weak field ligand.
Teacher's Note:
1. [FeF6]3− has five unpaired electrons (μ = 5.92 B.M.) – the classic high spin example.
Assertion (A): [Pt(NH3)2Cl2] is square planar.
Reason (R): The oxidation state of platinum is + 2.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. The outer electronic configuration of platinum in ground state is 5d96s1. The Pt2+ ion formed by the loss of two electrons has outer electronic configuration of 5d8. In the presence of strong ligands (NH3) two unpaired electrons in the 5d-subshell pair up. This is followed by dsp2 hybridisation resulting in the formation of four hybridised vacant orbitals which accommodate four pairs of electrons from four ligands (two from ammonia and two from Cl−). As such the resulting complex is square planar.
Teacher's Note:
1. d8 ions of the 4d/5d series (Pd2+, Pt2+) are almost always square planar.
Assertion (A): Cu(OH)2 is soluble in NH4OH but not in NaOH.
Reason (R): Cu(OH)2 forms a soluble complex with NH3.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Cu(OH)2 + 4NH3 → [Cu(NH3)4]2+ (deep blue, soluble) + 2OH−; NaOH supplies no ligand that can form a soluble complex, so Cu(OH)2 remains undissolved.
Teacher's Note:
1. Precipitates that dissolve in excess ammonia (Cu2+, Ag+, Zn2+) do so by forming ammine complexes.
Assertion (A): NF3 is weaker ligand than N(CH3)3.
Reason (R): NF3 ionizes to give F− ions in aqueous solution.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. F is highly electronegative atom which will withdraw electrons from nitrogen and create deficiency to donate electrons whereas –CH3 is electron donating group that will increase electron density on nitrogen to donate. Thus, N(CH3)3 is strong field ligand than NF3.
2. NF3 is a covalent molecule and does not ionise to give F− ions in water, so the reason is false.
Teacher's Note:
1. Ligand strength here depends on how easily the N lone pair is donated: electron-withdrawing F atoms make it poor, electron-donating CH3 groups make it better.
Assertion (A): [Cr(H2O)6]Cl2 and [Fe(H2O)6]Cl2 are reducing in nature.
Reason (R): Due to presence of unpaired electrons, complexes are reducing in nature.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Due to formation of more stable complex ion after gaining electrons, these are reducing in nature. Cr+2 and Fe+2 will oxidise to Cr+3 and Fe+3 and achieve stable state.
Teacher's Note:
1. Reducing nature comes from the metal ion being easily oxidised to a more stable state (Cr3+: t2g3, Fe3+: d5), not from unpaired electrons as such.
Assertion (A): Among [Co(NH3)6]3+ and [Co(en)3]3+, coordination compound [Co(en)3]3+ is a more stable complex.
Reason (R): en is a chelating ligand/bidentate ligand.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Since, en (ethylene diamine) is a chelating ligand/bidentate ligand, [Co(en)3]3+ is a more stable complex as compared to the other one due to the formation of chelating ring.
Teacher's Note:
1. Both complexes have six N donor atoms around Co3+; the extra stability of [Co(en)3]3+ comes only from the chelate rings.
Assertion (A): [Fe(CN)6]3− ion shows magnetic moment corresponding to the two unpaired electrons.
Reason (R): [Fe(CN)6]3− has d2sp3 type of hybridisation.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. [Fe(CN)6]3− ion shows magnetic moment corresponding to one unpaired electron.
2. Fe3+ is 3d5; strong field CN− pairs the electrons (t2g5), leaving two 3d orbitals free for d2sp3 hybridisation.
Teacher's Note:
1. For a low-spin d5 ion, write t2g5eg0: two pairs and one unpaired electron, so μ ≈ 1.73 BM.
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