Previous year questions (2013–2025) grouped by marks — 2, 3, 4 & 5 mark questions with answers. 160 questions total.
Chapter-wise Question Distribution
Chapter
2M
3M
4M
5M
Total
d&f Block
20
29
–
20
69
Coordination Compounds
31
24
3
3
61
p-Block Elements
13
4
–
5
22
General Principles of Metallurgy
–
5
–
–
5
General Principles of Isolation of Elements
2
1
–
–
3
Total
66
63
3
28
160
3 Mark Questions ★
IUPAC Naming of Coordination CompoundsAsked 11 times
2013201420172018201920242025
2013 · 3 marks
Write the IUPAC names of the following coordination compounds:
(i) [Cr(NH3)3Cl3]
(ii) K3[Fe(CN)6]
(iii) [CoBr2(en)2]+, (en = ethylenediamine)
Answer(i) Triamminetrichloridochromium(III)
(ii) Potassium hexacyanoferrate(III)
(iii) Dibromobis(ethylenediamine)cobalt(III) ion
2014 (i) Write the IUPAC name of the complex [Cr(NH 3 ) 4 Cl 2 ]Cl. (ii) What type of isomerism is exhibited by the complex [Co(en) 3 ] 3+ ? (en = ethane-1,2-diamine) (iii) Why is [NiCl...
2017 · 3 marks
(i) What type of isomerism is shown by the complex [Co(NH3)6][Cr(CN)6]?
(ii) Why a solution of [Ni(H2O)6]2+ is green while a solution of [Ni(CN)4]2- is colourless? (At. no. of Ni = 28)
(iii) Write IUPAC name of the following complex: [Co(NH3)5(CO3)]Cl.
Answer(i) Co-ordination isomerism.
(ii) In [Ni(H2O)6]2+, H2O is a weak field ligand. There are unpaired electrons in Ni2+. The d electrons can be excited to higher energy level (d-d transition), so it is coloured. In [Ni(CN)4]2-, CN- is a strong ligand, electrons are all paired, d-d transition not possible, hence colourless.
(iii) Pentaamminecarbonatocobalt(III) chloride.
Transition Metal PropertiesAsked 10 times
201320162019
2013 How would you account for the following? (i) Transition metals exhibit variable oxidation states. (ii) Zr(Z = 40) and Hf(Z = 72) have almost identical radii. (iii) Transition metal...
2016 · 3 marks
Give reasons:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Transition metals show variable oxidation states.
(iii) Actinoids show irregularities in their electronic configurations.
Answer(i) Oxygen can form multiple bonds (p-pi - d-pi) with Mn enabling Mn2O7. Fluorine has no d-orbitals and cannot form multiple bonds, so maximum oxidation state with F is +4 (MnF4).
(ii) Due to comparable energies of ns and (n-1)d orbitals, electrons of both can participate in bonding, giving variable oxidation states.
(iii) Due to comparable energies of 5f, 6d, and 7s orbitals, electrons can rearrange for more stable configurations (f0, f7, f14 stability).
2019 · 3 marks
Give the reasons for following:
(i) Transition elements and their compounds act as catalysts.
(ii) E° value for (Mn2+/Mn) is negative whereas for (Cu2+/Cu) is positive.
(iii) Actinoids show irregularities in their electronic configuration.
Answer(i) Due to variable oxidation state and ability to form complexes/provide surface for adsorption.
(ii) Mn2+ is stable due to exactly half filled 3d5 configuration. Due to high Delta_a H and low Delta_hyd H for Cu, E° is positive.
(iii) Due to comparable energies of 5f, 6d and 7s orbitals.
Other (d&f Block)Asked 8 times
201320162018201920202025
2013 Complete the following chemical equations: (i) Cr 2 O 7 2 + 6Fe 2+ + 14H + (ii) 2CrO 4 2 + 2H + (iii) 2MnO 4 + 5C 2 O 4 2 + 16H +
2016 · 3 marks
(i) Name the method of refining of metals such as germanium.
(ii) In the extraction of Al, impure Al2O3 is dissolved in conc. NaOH to form sodium aluminate and leaving impurities behind. What is the name of this process?
(iii) What is the role of coke in the extraction of iron from its oxides?
Answer(i) Zone refining is used for refining of germanium.
(ii) The process is called Baeyer's process (or leaching).
(iii) Coke acts as a reducing agent: Fe2O3 + 3CO -> 2Fe + 3CO2. Coke first converts to CO which then reduces iron oxide.
2016 · 3 marks
(i) Name the method of refining of nickel.
(ii) What is the role of cryolite in the extraction of aluminium?
(iii) What is the role of limestone in the extraction of iron from its oxides?
Answer(i) Mond process (vapour phase refining): Ni + 4CO -> Ni(CO)4 -> Ni + 4CO.
(ii) Cryolite (Na3AlF6) lowers the melting point of alumina from ~2050°C to ~950°C and increases conductivity of electrolyte.
(iii) Limestone (CaCO3) acts as a flux; it decomposes to CaO which combines with SiO2 (gangue) to form slag CaSiO3.
Solid State / Crystal Structure / DefectsAsked 6 times
201520232024
2015 · 3 marks
(i) Draw the geometrical isomers of complex [Pt(NH3)2Cl2].
(ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 < P.
(iii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)4].
(At.no. of Ni = 28)
Answer(i) Cis-isomer: Both NH3 on same side, both Cl on same side. Trans-isomer: NH3 on opposite sides, Cl on opposite sides.
(ii) For d4 ion with Δ0 < P (weak field/high spin): t2g3 eg1
(iii) [Ni(CO)4]: sp3 hybridization, diamagnetic (CO is strong field ligand causing pairing, no unpaired electrons).
2022-II · 3 marks
(a)
(i) Write the IUPAC name of the following complex: K2[PdCl4] (ii) Using crystal field theory, write the electronic configuration of d5 ion, if Delta_0 > P. (iii) What are Homoleptic complexes?
OR (b)
(i) Why chelate complexes are more stable than complexes with unidentate ligands? (ii) What is 'spectrochemical series'? What is the difference between a weak field ligand and a strong field ligand?
Answer(a)
(i) IUPAC name of K2[PdCl4] is Potassium tetrachloropalladate(II).
(ii) If Delta_0 > P, the electronic configuration of d5 ion will be t2g5 eg0 as it is associated with strong field and low spin situation. Thus, no electron will enter into eg orbital.
(iii) The complex compounds in which all the ligand which are connected with central atom are same or identical are called homoleptic complexes. For example: [Ni(CO)4], [Co(NH3)6]3+.
OR (b)
(i) Chelate complexes are more stable than unidentate ligand because chelate ligand forms a ring with the central metal ion and are held by strong force of attraction and are less likely to dissociate.
(ii) A series in which ligands are arranged in the order of increasing magnitude of crystal field splitting energy (CFSE), is called spectrochemical series. I- < Br- < Cl- < SCN- < F- < OH- < C2O42- < H2O < NCS- < EDTA4- < NH3 < en < CN- < CO. Weak field ligands have CFSE less than pairing energy; strong field ligands have CFSE greater than pairing energy.
2022-II · 3 marks
(a)
(i) Write the IUPAC name of the following complex: K2[PdCl4] (ii) Using crystal field theory, write the electronic configuration of d5 ion, if Delta_0 > P. (iii) What are Homoleptic complexes?
OR (b)
(i) Why chelate complexes are more stable than complexes with unidentate ligands? (ii) What is 'spectrochemical series'? What is the difference between a weak field ligand and a strong field ligand?
Answer(a)
(i) IUPAC name of K2[PdCl4] is Potassium tetrachloropalladate(II).
(ii) If Delta_0 > P, the electronic configuration of d5 ion will be t2g5 eg0 as it is associated with strong field and low spin situation. Thus, no electron will enter into eg orbital.
(iii) The complex compounds in which all the ligand which are connected with central atom are same or identical are called homoleptic complexes. For example: [Ni(CO)4], [Co(NH3)6]3+.
OR (b)
(i) Chelate complexes are more stable than unidentate ligand because chelate ligand forms a ring with the central metal ion and are held by strong force of attraction and are less likely to dissociate.
(ii) A series in which ligands are arranged in the order of increasing magnitude of crystal field splitting energy (CFSE), is called spectrochemical series. I- < Br- < Cl- < SCN- < F- < OH- < C2O42- < H2O < NCS- < EDTA4- < NH3 < en < CN- < CO. Weak field ligands have CFSE less than pairing energy; strong field ligands have CFSE greater than pairing energy.
Isomerism in Coordination CompoundsAsked 6 times
2016201720232024
2016 · 3 marks
(a) For the complex [Fe(CN)6]3-, write the hybridization, magnetic character and spin nature of the complex. (At. number: Fe = 26).
(b) Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically active.
Answer(a) Fe3+ has [Ar]3d5. CN- is a strong field ligand causing pairing. Hybridization: d2sp3 (inner orbital complex, octahedral). Magnetic character: Paramagnetic (1 unpaired electron). Spin nature: Low spin complex.
(b) The cis form of [Pt(en)2Cl2]2+ is optically active as it is non-superimposable on its mirror image.
2016 · 3 marks
(a) For the complex [Fe(H2O)6]3+, write the hybridization, magnetic character and spin of the complex. (At. number: Fe = 26).
(b) Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically inactive.
Answer(a) Fe3+: [Ar]3d5. H2O is weak field ligand (no pairing). Hybridization: sp3d2 (outer orbital, octahedral). Magnetic character: Paramagnetic (5 unpaired electrons, mu = sqrt(35) = 5.92 BM). Spin: High spin complex.
(b) The trans form of [Pt(en)2Cl2]2+ is optically inactive as it has a plane of symmetry.
2017 · 3 marks
(a) What type of isomerism is shown by the complex [Co(NH3)5(SCN)]2+?
(b) Why is [NiCl4]2- paramagnetic while [Ni(CN)4]2- is diamagnetic? (Atomic number of Ni = 28)
(c) Why are low spin tetrahedral complexes rarely observed?
Answer(a) Linkage isomerism (SCN- can bind through S or N).
(b) [NiCl4]2-: Cl- is weak field ligand, sp3 hybridization, 2 unpaired electrons (paramagnetic). [Ni(CN)4]2-: CN- is strong field ligand, dsp2 hybridization, no unpaired electrons (diamagnetic).
(c) Because of very low CFSE (Crystal Field Splitting Energy) in tetrahedral complexes which is not able to pair up the electrons.
Other (General Principles of Metallurgy)Asked 5 times
20172018
2017 · 3 marks
(a) Write the principle of method used for the refining of germanium.
(b) Out of PbS and PbCO3 (ores of lead), which one is concentrated by froth flotation process?
(c) What is significance of leaching in the extraction of aluminium?
Answer(a) Zone refining: Based on the principle that impurities are more soluble in the melt than in the solid state.
(b) PbS (sulphide ore is concentrated by froth flotation).
(c) Leaching is used to concentrate bauxite ore (Al2O3) by removing impurities like SiO2, Fe2O3, TiO2 using Bayer's process with NaOH.
2017 Write the principles of the following methods: (i) Vapour phase refining (ii) Zone refining (iii) Chromatography
2017 Write the principle of the following: (a) Zone refining (b) Froth flotation process (c) Chromatography
Lanthanoid/Actinoid ContractionAsked 4 times
20152025
2015 (a) How would you account for the following: (i) Actinoid contraction is greater than lanthanoid contraction. (ii) Transition metals form coloured compounds. (b) Complete the follo...
2022-II · 3 marks
(a) Write any two consequences of Lanthanoid Contraction.
(b) Name the element of 3d series which exhibits the largest number of oxidation states. Give reason.
Answer(a) Two consequences of lanthanide contraction: (i) Due to close similarity in electronic configuration and ionic radii, the lanthanides have identical chemical properties which makes their separation difficult. (ii) Due to lanthanide contraction, the size of Ln3+ ions decrease regularly with increase in atomic number. According to Fajan's rule, decrease in size of Ln3+ ions increase the covalent character and decrease the basic character between Ln3+ and OH- ion in Ln(OH)3. Order of size: La3+ > Ce3+ > ... > Lu3+.
(b) The element which shows largest number of oxidation states in 3d series is Manganese (Mn). It shows oxidation states +2, +3, +4, +5, +6 and +7. It is because it contains the maximum number of unpaired electrons in the outermost shell, i.e., 3d5 4s2.
2025 · 3 marks
(a) Of the d4 species, Cr2+ is strongly reducing while Mn3+ is strongly oxidising. Why?
(b) Write two consequences of lanthanoid contraction.
(c) Which element of 3d series has lowest enthalpy of atomisation and why?
Answer(a) Cr2+ is a strong reducing agent, it has 3d4 electronic configuration. Hence, it loses electron to form 3d3 configuration, which has half filled t2g and is more stable. But in case of Mn3+, the electronic configuration 3d4 gains electron to become 3d5 (Mn2+). It attains extra stability due to 3d5 electronic configuration. Hence it is a strong oxidising agent. (b)(1) It causes the radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. (2) The basic strength decreases from La(OH)3 to Lu(OH)3.
(c) In 3d elements, d-orbitals are involved in metallic bonding but not in case of zinc. Hence it shows lowest enthalpy of atomisation.
Other (p-Block Elements)Asked 3 times
20142017
2014 · 3 marks
(a) Draw the structures of the following molecules:
(i) XeOF4
(ii) H3SO4 (should be H2SO4)
(b) Write the structural difference between white phosphorus and red phosphorus.
Answer(a)
(i) XeOF4: Square pyramidal structure with one O at apex.
(ii) H2SO4: Tetrahedral structure around S with 2 OH groups and 2 =O.
(b) White phosphorus exists as discrete P4 tetrahedral molecules. Red phosphorus has a polymeric chain structure where P4 units are linked together.
2014 Give reasons for the following: (i) (CH 3 ) 3 P = O exists but (CH 3 ) 3 N = O does not. (ii) Oxygen has less electron gain enthalpy with negative sign than sulphur. (iii) H 3 PO 2...
2017 Give reasons: (i) Thermal stability decreases from H 2 O to H 2 Te. (ii) Fluoride ion has higher hydration enthalpy than chloride ion. (iii) Nitrogen does not form pentahalide.
VBT / Hybridization / Magnetic MomentAsked 3 times
20192025
2019 · 3 marks
Give reasons:
(a) Transition metals have high enthalpies of atomization.
(b) Manganese has lower melting point even though it has higher number of unpaired electrons.
(c) Ce4+ is a strong oxidizing agent.
Answer(a) Due to strong interatomic interactions/strong metallic bonding.
(b) Due to stable 3d5 configuration, interatomic interaction is poor between unpaired electrons.
(c) Ce is more stable in +3 oxidation state; Ce4+ readily gains electron to achieve +3.
2025 · 3 marks
Using valence bond theory, explain the hybridisation and magnetic character of the following:
(a) [Co(NH3)6]3+
(b) [Ni(CO)4]. [At. no.: Co = 27, Ni = 28]
Answer(a) [Co(NH3)6]3+: Co3+ has [Ar]3d6 configuration. In the presence of ammonia (strong field ligand), electrons in 3d level get paired up leaving two of the 3d orbitals for hybridisation. Hence, d2sp3 hybridisation. The complex is diamagnetic as there are no unpaired electrons.
(b) [Ni(CO)4]: The oxidation state of Ni is 0. The valence shell electronic configuration is 3d8 4s2, but all 10 electrons are pushed into 3d orbital due to the strong field CO ligands, forming sp3 hybridisation. The complex is tetrahedral and diamagnetic.
2025 · 3 marks
(a) Using valence bond theory, explain the hybridisation and magnetic behaviour of the following:
(i) [Co(NH3)6]Cl3
(ii) K2[NiCl4]. (At. no.: Co = 27, Ni = 28)
(b) Write the electronic configuration of d5 ion when delta_o > P.
Answer(a)
(i) [Co(NH3)6]Cl3: Co3+ has [Ar]3d6 configuration. NH3 is strong field ligand. Electrons in 3d get paired up. d2sp3 hybridisation (inner orbital octahedral complex). Diamagnetic due to absence of unpaired electrons.
(ii) K2[NiCl4]: Ni2+ has [Ar]3d8 4s0 configuration. Cl- is a weak field ligand. sp3 hybridisation with tetrahedral geometry. Paramagnetic due to two unpaired electrons.
(b) When CFSE (delta_o) is more than pairing energy, it indicates a low spin complex. The electron will pair in lower energy level before filling in higher energy level. Electronic configuration will be t2g5 eg0.
5 Mark Questions ★★★
Lanthanoid/Actinoid ContractionAsked 9 times
20142017202020232024
2014 · 5 marks
(i) Name the element of 3d transition series which shows maximum number of oxidation states. Why does it show so?
(ii) Which transition metal of 3d series has positive E°(M2+/M) value and why?
(iii) Out of Cr3+ and Mn3+, which is a stronger oxidizing agent and why?
(iv) Name a member of the Lanthanoid series which is well known to exhibit +2 oxidation state.
(v) Complete the following equation: MnO4− + 8H+ + 5e− →
Answer(i) Manganese ([Ar] 3d5 4s2) shows maximum number of oxidation states as its atoms have five unpaired electrons in 3d orbitals. It shows all the oxidation states from +2 to +7.
(ii) Cu has positive E°(M2+/M) value because the sum of enthalpies of sublimation and ionization is not balanced by hydration enthalpy.
(iii) Mn3+ is stronger oxidizing agent as the charge from Mn3+ to Mn2+ results in half filled d5 configuration which has extra stability.
(iv) Europium (Eu2+) is formed by losing the two 5s electrons and its electronic configuration becomes [Xe]4f7 which is quite stable configuration.
(v) MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
2014 · 5 marks
(i) With reference to structural variability and chemical reactivity, write the difference between lanthanoids and actinoids.
(ii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
(iii) Complete the following equation: MnO4− + 8H+ + 5e− →
(iv) Out of Mn3+ and Cr3+, which is more paramagnetic and why? (Mn = 25, Cr = 24)
Answer(i)(a) Actinoids are radioactive; lanthanoids are not.
(b) Ionisation enthalpies of early actinoids are lower. Actinoids show oxidation states from +3 to +7; lanthanoids show +3 (most common) up to +4.
(c) Actinoids are more reactive and have more complex magnetic properties.
(ii) Cerium (Ce) is known to exhibit +4 oxidation state.
(iii) MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
(iv) Mn2+ is more paramagnetic because it has more number of unpaired electrons (5) than Cr3+ (3).
2017 (a) Account for the following: (i) Transition metals form large number of complex compounds. (ii) The lowest oxide of transition metal is basic whereas the highest oxide is amphote...
Other (p-Block Elements)Asked 4 times
201320152018
2013 (a) Give reasons for the following: (i) Bond enthalpy of F 2 is lower than that of Cl 2 . (ii) PH 3 has lower boiling point than NH 3 . (b) Draw the structures of the following mol...
2013 (a) Account for the following: (i) Helium is used in diving apparatus. (ii) Fluorine does not exhibit positive oxidation state. (iii) Oxygen shows catenation behaviour less than su...
2015 (a) Account for the following: (i) Acidic character increases from HF to HI. (ii) There is large difference between the melting and boiling points of oxygen and sulphur. (iii) Nitr...
p-Block Element Properties/ReactionsAsked 4 times
201520162019
2015 · 5 marks
(i) Which allotrope of phosphorus is more reactive and why?
(ii) How the supersonic jet aeroplanes are responsible for the depletion of ozone layers?
(iii) F2 has lower bond dissociation enthalpy than Cl2. Why?
Answer(i) White phosphorus is more reactive because it is a discrete P4 tetrahedron with angular strain (60° bond angle). It is unstable and highly reactive.
(ii) Supersonic jet aeroplanes release NO (nitric oxide) in the stratosphere which acts as a catalyst for ozone decomposition: NO + O3 → NO2 + O2; NO2 + O → NO + O2
(iii) F2 has lower bond dissociation enthalpy than Cl2 because F atoms are very small, leading to high lone pair-lone pair repulsion between the two F atoms, weakening the F–F bond.
2016 · 5 marks
(a) Account for the following:
(i) Ozone is thermodynamically unstable.
(ii) Solid PCl5 is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
(b) Draw the structure of
(i) BrF3
(ii) XeF4.
OR
(i) Compare the oxidising action of F2 and Cl2 by considering bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
(ii) Write conditions to maximize yield of H2SO4 by contact process.
(iii) Arrange in increasing order:
(a) H3PO3, H3PO4, H3PO2 (Reducing character).
(b) NH3, PH3, AsH3, SbH3, BiH3 (Base strength).
Answer(a)
(i) Ozone decomposes to oxygen as delta-G is negative for decomposition.
(ii) Solid PCl5 exists as [PCl4]+[PCl6]- (tetrahedral cation and octahedral anion).
(iii) Due to high electronegativity and small size of fluorine. (b)
(i) BrF3: T-shaped (sp3d, 2 lone pairs).
(ii) XeF4: Square planar (sp3d2, 2 lone pairs).
OR
(i) F2 is stronger oxidizing agent due to low bond dissociation enthalpy and very high hydration enthalpy of F-.
(ii) Low temperature (~720 K), high pressure, V2O5 catalyst, excess O2. (iii)(a) H3PO4 < H3PO3 < H3PO2.
(b) BiH3 < SbH3 < AsH3 < PH3 < NH3.
2019 · 5 marks
(a)
(i) Write the disproportionation reaction of H3PO3.
(ii) Draw the structure of XeF4.
(b) Account for:
(i) Although fluorine has less negative electron gain enthalpy yet F2 is a strong oxidizing agent.
(ii) Acidic character decreases from N2O3 to Bi2O3 in group 15.
(c) Write a chemical reaction to test sulphur dioxide gas.
Answer(a)
(i) 4H3PO3 -> 3H3PO4 + PH3.
(ii) XeF4 has square planar geometry. (b)
(i) F2 is strong oxidizing agent due to low F-F bond dissociation energy, high hydration enthalpy of F-.
(ii) Metallic character increases down group 15; N2O3 is acidic (non-metallic) while Bi2O3 is basic (metallic).
(c) SO2 turns acidified K2Cr2O7 from orange to green: K2Cr2O7 + 3SO2 + H2SO4 -> K2SO4 + Cr2(SO4)3 + H2O.
Transition Metal PropertiesAsked 3 times
20142024
2014 (a) How do you prepare: (i) KMnO 4 from MnO 2 (ii) Na 2 Cr 2 O 7 from Na 2 CrO 4 (b) Account for the following: (i) Mn 2+ is more stable than Fe 2+ towards oxidation to +3 state. (...
2014 (a) Complete the following equations: (i) Cr 2 O 7 2 + 2OH (ii) MnO 4 + 4H + + 3e (b) Account for the following: (i) Zn is not considered as a transition element. (ii) Transition m...
2024 Attempt any five of the following: (a) Ce(III) is easily oxidised to Ce(IV). Comment. (b) E°(Mn2+/Mn) is -1.18 V. Why is this value highly negative in comparison to neighbouring d ...
IUPAC Naming of Coordination CompoundsAsked 3 times
20192025
2019 · 5 marks
(a) Give reasons:
(i) Sulphur in vapour state shows paramagnetic behaviour.
(ii) N-N bond is weaker than P-P bond.
(iii) Ozone is thermodynamically less stable than oxygen.
(b) Write the name of the gas released when Cu is added to
(i) Dilute HNO3
(ii) Conc. HNO3.
Answer(a)
(i) S2 molecules in vapour state have two unpaired electrons in antibonding molecular orbitals.
(ii) N-N single bond is weaker due to high interelectronic repulsion of lone pairs on small nitrogen atoms.
(iii) Ozone has higher energy and tendency to decompose to more stable O2. (b)
(i) Dilute HNO3: NO (nitric oxide).
(ii) Conc. HNO3: NO2 (nitrogen dioxide).
2025 · 5 marks
(A) Answer the following:
(a) Low spin tetrahedral complexes are not known.
(b) Co2+ is easily oxidised to Co3+ in the presence of a strong ligand [At. No. of Co = 27].
(c) What type of isomerism is shown by the complex [Co(NH3)6][Cr(CN)6]?
(d) Why a solution of [Ni(H2O)6]2+ is green while a solutions of [Ni(CN)4]2- is colourless. (At. No. of Ni = 28)
(e) Write the IUPAC name of the following complex: [Co(NH3)5(CO3)]Cl.
OR (B)(a) What is meant by 'Chelate effect'? Give an example.
(b) Write the hybridisation and magnetic behaviour of [Fe(CN)6]4-. (Atomic number: Fe = 26)
(c) If PtCl2.2NH3 does not react with AgNO3, what will be its formula?
Answer(A)(a) A tetrahedral complex has sp3 hybridisation. During crystal field splitting, the splitting energy of d orbitals is much lower than that of the pairing energy, and thus the electrons jump to the higher energy level d-orbitals rather than pairing, resulting in high-spin complexes being formed.
(b) Co2+ can be oxidised to Co3+ in the presence of strong field ligands. Strong field ligands create a strong ligand field that stabilizes the higher oxidation state of the metal ion, making the +3 oxidation state more favourable.
(c) Coordination isomerism - both positive and negative ions of a salt are complex ions and the two isomers differ in the distribution of ligands between the cation and anion.
(d) In both complexes, Ni is in +2 oxidation state with 3d8 configuration. In the presence of weak field water ligands, two unpaired electrons do not pair up. The [Ni(H2O)6]2+ has two unpaired electrons which result in green colour due to d-d transition. In presence of strong field cyanide ligand, the unpaired electrons in 3d orbital pair up. Due to absence of unpaired electrons, no d-d transitions are possible and [Ni(CN)4]2- is colourless.
(e) Pentaamminecarbonatocobalt(III)chloride.
OR (B)(a) When a ligand uses its two or more donor atoms simultaneously a ring-like structure is formed. The formation of such ring-like structures is called a chelate effect. Compounds having chelate rings are more stable.
(b) The hybridisation is d2sp3 and it forms an inner orbital complex. Since there are no unpaired electrons, the complex is diamagnetic.
(c) [Pt(NH3)2Cl2] - since it does not react with AgNO3, the chloride ions are inside the coordination sphere.
2025 · 5 marks
(A)(a) In the following complex ions, explain the type of hybridisation, shape and magnetic property:
(a) [Fe(H2O)6]2+
(b) [NiCl4]2-. (At. Nos.: Fe = 26, Ni = 28) OR (B)(a) Write IUPAC names of the following:
(i) [Co(H2O)(CN)(en)2]2+
(ii) [PtCl4]2-
(iii) [Cr(NH3)4Cl(ONO)]+.
(b) What is spectrochemical series? Write the difference between a strong field ligand and a weak field ligand.
Answer(A)(a) [Fe(H2O)6]2+: Fe2+ has [Ar]3d6 configuration. H2O is a weak field ligand, so no pairing occurs. Hybridisation is sp3d2, shape is octahedral, and the complex is paramagnetic with 4 unpaired electrons.
(b) [NiCl4]2-: Ni2+ has [Ar]3d8 configuration. Cl- is a weak field ligand. Hybridisation is sp3, shape is tetrahedral, and the complex is paramagnetic with 2 unpaired electrons.
(b) The spectrochemical series is the arrangement of ligands in increasing order of their crystal field splitting energy. Weak field ligands: I- < Br- < SCN- < Cl- < S2- < F- < OH- < C2O42- < H2O. Strong field ligands: NCS- < edta4- < NH3 < en < CN- < CO. Strong field ligands cause high crystal field splitting and often lead to low-spin complexes. Weak field ligands cause low crystal field splitting and often lead to high-spin complexes.
2014 · 2 marks
Write the IUPAC name of the complex [Cr(NH3)4Cl2]+. What type of isomerism does it exhibit?
AnswerIUPAC name: Tetraamminedichlorochromium(III) ion. It exhibits geometrical isomerism, i.e., cis and trans isomers.
2015 · 2 marks
(i) Write down the IUPAC name of the following complex: [Cr(NH3)2Cl2(en)]Cl (en = ethylenediamine)
Write IUPAC name of the complex [Pt(en)2Cl2]. Draw structures of geometrical isomers for this complex.
AnswerBis(ethane-1,2-diamine)dichloridoplatinum(II). Two geometrical isomers: cis-isomer (both Cl on same side) and trans-isomer (Cl on opposite sides).
2019 · 2 marks
Using IUPAC norms write the formulae for the following:
(i) Hexaamminecobalt(III) sulphate
(ii) Potassium trioxalatochromate(III)
Answer(i) [Co(NH3)6]2(SO4)3.
(ii) K3[Cr(C2O4)3].
2019 · 2 marks
Out of [CoF6]3- and [Co(en)3]3+, which one complex is:
(i) Paramagnetic,
(ii) More stable,
(iii) Inner orbital complex and
(iv) High spin complex? (Atomic number of Co = 27)
Answer(i) [CoF6]3- is paramagnetic.
(ii) [Co(en)3]3+ is more stable.
(iii) [Co(en)3]3+ is inner orbital complex.
(iv) [CoF6]3- is high spin complex.
2019 · 2 marks
Write IUPAC name of the complex [Cr(NH3)4Cl2]+. Draw structures of geometrical isomers for this complex.
AnswerTetraamminedichloridochromium(III) ion. Two geometrical isomers: cis (both Cl adjacent) and trans (Cl opposite).
2019 · 2 marks
Using IUPAC norms write the formulae for:
(i) Pentaamine nitrito-O-cobalt(III) Chloride
(ii) Potassium tetracyanonickelate(II)
Answer(i) [Co(NH3)5ONO]Cl2.
(ii) K2[Ni(CN)4].
2019 · 2 marks
Out of [CoF6]3- and [Co(C2O4)3]3-, which one complex is:
(i) Diamagnetic,
(ii) More stable,
(iii) Outer orbital complex and
(iv) Low spin complex? (Atomic number of Co = 27)
Answer(i) [Co(C2O4)3]3-.
(ii) [Co(C2O4)3]3-.
(iii) [CoF6]3-.
(iv) [Co(C2O4)3]3-.
2019 · 2 marks
Write the hybridization and magnetic character of the following complexes:
(i) [Fe(H2O)6]2+
(ii) [Fe(CO)5]
Answer(i) sp3d2 hybridization, Paramagnetic (4 unpaired electrons; H2O is weak field ligand).
(ii) dsp3/trigonal bipyramidal, Diamagnetic (CO is strong field ligand causing pairing).
2019 · 2 marks
Write IUPAC name of the complex [Co(en)2(NO2)Cl]+. What type of structural isomerism is shown by this complex?
AnswerChloridobis(ethane-1,2-diamine)nitrito-N-cobalt(III) ion. It shows linkage isomerism.
2019 · 2 marks
Using IUPAC norms, write formulae for:
(a) Hexaaquachromium(III) chloride
(b) Sodium trioxalatoferrate(III)
Answer(a) [Cr(H2O)6]Cl3.
(b) Na3[Fe(C2O4)3].
2019 · 2 marks
Define the following terms with a suitable example:
(a) Chelate complex
(b) Ambidentate ligand
Answer(a) Chelate complex: Complex formed by bi/polydentate ligands with metal forming ring. Example: [Co(en)3]3+.
(b) Ambidentate ligand: Ligand that can ligate through two different donor atoms. Example: SCN- (through S or N), NO2- (through N or O).
(a) Using valence bond theory, write the hybridization and magnetic character of [Fe(CN)6]4-. (Atomic no. of Fe = 26).
(b) Write the electronic configuration of d6 on the basis of crystal field theory when:
(i) Delta_0 < P and
(ii) Delta_0 > P.
Answer(a) d2sp3 hybridization, Diamagnetic. CN- is strong field ligand causing pairing. (b)
(i) Delta_0 < P: t2g4 eg2 (high spin).
(ii) Delta_0 > P: t2g6 eg0 (low spin).
2019 · 2 marks
Define with suitable example:
(a) Polydentate ligand
(b) Homoleptic complex
Answer(a) Polydentate ligand: Ligand with several donor atoms. Example: EDTA (hexadentate).
(b) Homoleptic complex: Metal bound to only one kind of ligand. Example: [Co(NH3)6]3+.
2019 · 2 marks
Using IUPAC norms, write formulae for:
(a) Potassium tri(oxalate)chromate(III)
(b) Hexaaquamanganese(II) sulphate
Answer(a) K3[Cr(C2O4)3].
(b) [Mn(H2O)6]SO4.
2019 · 2 marks
(a) Although both [NiCl4]2- and [Ni(CO)4] have sp3 hybridisation yet [NiCl4]2- is paramagnetic and [Ni(CO)4] is diamagnetic. Give reason.
(b) Write electronic configuration of d5 by crystal field theory when:
(i) Delta_0 < P and
(ii) Delta_0 > P.
Answer(a) In [NiCl4]2-, Ni is +2, Cl- is weak field (no pairing, 2 unpaired electrons, paramagnetic). In [Ni(CO)4], Ni is 0, CO is strong field (forced pairing, no unpaired electrons, diamagnetic). (b)
(i) t2g3 eg2 (high spin).
(ii) t2g5 eg0 (low spin).
2020 · 2 marks
(a) Write the IUPAC name and hybridisation of the complex [Fe(CN)6]3-. (Given: Atomic number of Fe = 26)
(b) What is the difference between an ambidentate ligand and a chelating ligand?
(b) Ambidentate ligand can bond through different atoms to form different coordination compounds, e.g. NO2- can bind at either the nitrogen atom or one of the oxygen atoms. Chelating ligand: If the ligands with two or more electron donor groups positioned in such a way that they form five or six membered ring with central metal ion are called chelating ligands, e.g. ethane-1,2-diamine (en).
2020 · 2 marks
Discuss the nature of bonding in metal carbonyls.
AnswerIn metal carbonyls, the metal-carbon bond has both sigma and pi character. The M-C sigma bond is formed by donation of lone pair of electrons from C of CO to empty orbital of metal. The M-C pi bond is formed by back donation of electrons from filled d-orbital of metal into the vacant antibonding pi* orbital of CO (synergic bonding).
2020 · 2 marks
Write the IUPAC name and hybridization of the following complexes:
(i) [Ni(CN)4]2-
(ii) [Fe(H2O)6]2+. (Given: Atomic number of Ni = 28, Fe = 26)
(ii) [NiCl4]2- - Tetrachloronickelate(II), Hybridization - sp3.
2023 · 2 marks
(a) Which of the following species cannot act as a ligand? Give reason. OH-, NH4+, CH3, NH3, H2O
(b) The complex [Co(NH3)5(NO2)] Cl2 is red in colour. Give IUPAC name of its linkage isomer.
Answer(a) NH4+ (Ammonium ion) can not act as ligand. Ligand donate electron to central atom or they can have lone pair of electron to donate and form bond between ligand and central atom. But NH4+ do not have lone pair of electron to donate.
(b) Complex [Co(NH3)5(NO2)]Cl2 is Red colour when it is present in hydrated form. It absorb moisture and turn into Red colour. It is showing linkage isomerism: [Co(NH3)5(NO2)]Cl2 (Pentaamine N-nitro cobalt(III) Chloride) ⇌ [Co(NH3)5(ONO)]Cl2 (Pentaamine o-nitro cobalt(III) Chloride)
2023 · 2 marks
(a) Write the IUPAC names of the following:
(i) [Co(NH3)5(ONO)]2+
(ii) K2[NiCl4]
Answer(a)
(i) [Co(NH3)5(ONO)]2+: IUPAC: Pentaammine onitrito cobalt (III) ion
(ii) K2[NiCl4]: potassiumtetra chlornickelate (II)
2023 · 2 marks
(b)
(i) What is a chelate complex? Give one example.
(ii) What are heteroleptic complexes? Give one example. (OR)
Answer(b)
(i) Chelate complex: A class of coordination or complex compounds consisting of a central metal atom attached through two or more co-ordinate bonds with ligands in a cyclic or ring structure. Example: EDIA
(ii) Heteroleptic complex: Coordination complexes which contain more than one type of ligands. Example: [Fe(NH3)4Cl2]+
2025 · 2 marks
Name the following coordination compounds according to IUPAC norms:
Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex. [At. No. Co = 27, Ni = 28]
Answer[Co(NH3)6]3+: Co3+ = [Ar]3d6 4s0. Co3+ has d6 configuration. Ammonia acts as strong field ligand due to +3 oxidation state. Electrons get paired up forming d2sp3 hybridisation. Inner 3d orbital is involved, so it is called inner orbital complex. [Ni(NH3)6]2+: Ni2+ = [Ar]3d8 4s0. Ammonia in this case acts as weak field ligand due to +2 oxidation state. It undergoes sp3d2 hybridisation. As outer d-orbital is involved it is outer orbital complex.
2013 · 2 marks
(a) Which solution is used for the leaching of silver metal in the extraction of silver?
(b) Out of C and CO, which is a better reducing agent at the lower temperature range in the blast furnace to extract iron from the oxide ore?
Answer(a) NaCN solution (sodium cyanide) is used for leaching of silver metal.
(b) CO is a better reducing agent than C at lower temperature range in the blast furnace.
2014 · 2 marks
Outline the principles behind the refining of metals by the following methods:
(i) Zone refining method
(ii) Chromatographic method
Answer(i) Zone refining: Based on the principle that impurities are more soluble in the melt than in the solid state. A moving heater melts a narrow zone which moves along the rod, carrying impurities to one end.
(ii) Chromatographic method: Based on the principle that different components of a mixture are adsorbed to different extents on an adsorbent. The more strongly adsorbed component moves slowly through the column.
2013 · 2 marks
(a) Which metal in the first transition series (3d series) exhibits +1 oxidation state most frequently and why?
(b) Which of the following cations are coloured in aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+ (At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
Answer(a) Copper metal shows +1 oxidation state because its electronic configuration is [Ar]3d104s1. It can easily donate one electron from 4s1 and exhibit +1 oxidation state.
(b) V3+ and Mn2+ are coloured transition metal ions because of presence of unpaired electrons and they show d-d transition. Sc3+ and Ti4+ are colourless because their d-orbitals are empty and no d-d transition takes place.
2015 · 2 marks
What are the transition elements? Write two characteristics of the transition elements.
AnswerTransition elements are elements which have partially filled d-orbital in its ground states or any one of its oxidation states. Characteristics: (i) They exhibit variable oxidation states. (ii) They form coloured ions/compounds.
2016 · 2 marks
When chromite ore FeCr2O4 is fused with NaOH in presence of air, a yellow coloured compound
(A) is obtained which on acidification with dilute sulphuric acid gives a compound (B). Compound
(B) on reaction with KCl forms an orange coloured crystalline compound (C).
(i) Write the formulae of the compounds (A),
(B) and (C).
(ii) Write one use of compound (C).
OR Complete the following chemical equations:
(i) MnO4- + 3S2O32- + H2O -> ?
(ii) Cr2O72- + 3Sn2+ + 14H+ -> ?
Answer(i) A = Na2CrO4 (sodium chromate, yellow), B = Na2Cr2O7 (sodium dichromate), C = K2Cr2O7 (potassium dichromate, orange).
(ii) K2Cr2O7 is used as a strong oxidizing agent / as a primary standard in volumetric analysis.
(a) Transition elements generally form coloured compounds.
(b) Zinc is not regarded as a transition element.
Answer(a) Transition metals form coloured solutions and compounds because these elements have unfilled d orbitals. Which allow for transition between different energy level by absorption of specific wavelength and results a complementary colour.
(b) Zinc is not a transition metal since its d subshells are not partly filled (n-1). As it does not have a vacant d-orbital it is not a transition element.
2025 · 2 marks
When FeCr2O4 is fused with Na2CO3 in free access of air it gives a yellow solution of compound (A). Compound
(A) on acidification gives compound (B). Compound
(B) on reaction with KCl forms an orange coloured (C). An acidified solution of compound
(ii) Optical isomerism (the complex has non-superimposable mirror images)
(iii) In [NiCl4]2−, Ni is +2, Cl is a weak field ligand and does not cause pairing, so Ni has 2 unpaired electrons in d-orbitals → paramagnetic. In [Ni(CO)4], CO is a strong field ligand causing electron pairing, resulting in no unpaired electrons → diamagnetic.
2015 · 3 marks
(i) Draw the geometrical isomers of complex [Pt(NH3)2Cl2].
(ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 < P.
(iii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)4].
(At.no. of Ni = 28)
Answer(i) Cis-isomer: Both NH3 on same side, both Cl on same side. Trans-isomer: NH3 on opposite sides, Cl on opposite sides.
(ii) For d4 ion with Δ0 < P (weak field/high spin): t2g3 eg1
(iii) [Ni(CO)4]: sp3 hybridization, diamagnetic (CO is strong field ligand causing pairing, no unpaired electrons).
2016 · 3 marks
(a) For the complex [Fe(CN)6]3-, write the hybridization, magnetic character and spin nature of the complex. (At. number: Fe = 26).
(b) Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically active.
Answer(a) Fe3+ has [Ar]3d5. CN- is a strong field ligand causing pairing. Hybridization: d2sp3 (inner orbital complex, octahedral). Magnetic character: Paramagnetic (1 unpaired electron). Spin nature: Low spin complex.
(b) The cis form of [Pt(en)2Cl2]2+ is optically active as it is non-superimposable on its mirror image.
2016 · 3 marks
(a) For the complex [Fe(H2O)6]3+, write the hybridization, magnetic character and spin of the complex. (At. number: Fe = 26).
(b) Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically inactive.
Answer(a) Fe3+: [Ar]3d5. H2O is weak field ligand (no pairing). Hybridization: sp3d2 (outer orbital, octahedral). Magnetic character: Paramagnetic (5 unpaired electrons, mu = sqrt(35) = 5.92 BM). Spin: High spin complex.
(b) The trans form of [Pt(en)2Cl2]2+ is optically inactive as it has a plane of symmetry.
2017 · 3 marks
(i) What type of isomerism is shown by the complex [Co(NH3)6][Cr(CN)6]?
(ii) Why a solution of [Ni(H2O)6]2+ is green while a solution of [Ni(CN)4]2- is colourless? (At. no. of Ni = 28)
(iii) Write IUPAC name of the following complex: [Co(NH3)5(CO3)]Cl.
Answer(i) Co-ordination isomerism.
(ii) In [Ni(H2O)6]2+, H2O is a weak field ligand. There are unpaired electrons in Ni2+. The d electrons can be excited to higher energy level (d-d transition), so it is coloured. In [Ni(CN)4]2-, CN- is a strong ligand, electrons are all paired, d-d transition not possible, hence colourless.
(iii) Pentaamminecarbonatocobalt(III) chloride.
2017 · 3 marks
(i) What type of isomerism is shown by the complex [Co(en)3]Cl3?
(ii) Write the hybridisation and magnetic character of [Co(C2O4)3]3-. (At. no. of Co = 27)
(iii) Write IUPAC name of the following complex [Cr(NH3)3Cl3].
Answer(i) Optical isomerism (dextrorotatory and laevorotatory).
(ii) d2sp3 hybridization, diamagnetic.
(iii) Triamminetrichloridochromium(III).
2017 · 3 marks
(a) What type of isomerism is shown by the complex [Co(NH3)5(SCN)]2+?
(b) Why is [NiCl4]2- paramagnetic while [Ni(CN)4]2- is diamagnetic? (Atomic number of Ni = 28)
(c) Why are low spin tetrahedral complexes rarely observed?
Answer(a) Linkage isomerism (SCN- can bind through S or N).
(b) [NiCl4]2-: Cl- is weak field ligand, sp3 hybridization, 2 unpaired electrons (paramagnetic). [Ni(CN)4]2-: CN- is strong field ligand, dsp2 hybridization, no unpaired electrons (diamagnetic).
(c) Because of very low CFSE (Crystal Field Splitting Energy) in tetrahedral complexes which is not able to pair up the electrons.
2018 · 3 marks
(a) Write the formula of the following co-ordination compound: Iron(III) hexacyanoferrate(II)
(b) What type of isomerism is exhibited by the complex [Co(NH3)5Cl]SO4?
(c) Write the hybridisation and number of unpaired electrons in the complex [CoF6]3-. (Atomic No. of Co = 27)
Answer(a) Fe4[Fe(CN)6]3
(b) Ionisation isomerism. [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl.
(c) sp3d2 hybridisation, 4 unpaired electrons. F- is a weak field ligand, forms outer orbital complex (high spin).
2022-II · 3 marks
(a)
(i) Write the IUPAC name of the following complex: K2[PdCl4] (ii) Using crystal field theory, write the electronic configuration of d5 ion, if Delta_0 > P. (iii) What are Homoleptic complexes?
OR (b)
(i) Why chelate complexes are more stable than complexes with unidentate ligands? (ii) What is 'spectrochemical series'? What is the difference between a weak field ligand and a strong field ligand?
Answer(a)
(i) IUPAC name of K2[PdCl4] is Potassium tetrachloropalladate(II).
(ii) If Delta_0 > P, the electronic configuration of d5 ion will be t2g5 eg0 as it is associated with strong field and low spin situation. Thus, no electron will enter into eg orbital.
(iii) The complex compounds in which all the ligand which are connected with central atom are same or identical are called homoleptic complexes. For example: [Ni(CO)4], [Co(NH3)6]3+.
OR (b)
(i) Chelate complexes are more stable than unidentate ligand because chelate ligand forms a ring with the central metal ion and are held by strong force of attraction and are less likely to dissociate.
(ii) A series in which ligands are arranged in the order of increasing magnitude of crystal field splitting energy (CFSE), is called spectrochemical series. I- < Br- < Cl- < SCN- < F- < OH- < C2O42- < H2O < NCS- < EDTA4- < NH3 < en < CN- < CO. Weak field ligands have CFSE less than pairing energy; strong field ligands have CFSE greater than pairing energy.
2022-II · 3 marks
(a)
(i) Write the IUPAC name of the following complex: [Pt(NH3)6]Cl4 (ii) On the basis of crystal field theory, write the electronic configuration of d4 ion, if Delta_0 < P. (iii) What are Heteroleptic complexes?
OR (b)
(i) Using IUPAC norms write the formulas for: (a) Pentaamminenitrito-N-Cobalt(III) (b) Tetrahydroxidozincate(II) (ii) What is crystal field splitting energy?
Answer(a)
(i) The IUPAC name of [Pt(NH3)6]Cl4 is Hexaammineplatinum(IV) chloride.
(ii) If Delta_0 < P, the electronic configuration of d4 ion will be t2g3 eg1 as it is associated with weak field and high spin situation. The fourth electron will enter into eg orbital.
(iii) The complexes in which central atom or metal ion is surrounded by more than one type of ligands are called heteroleptic complexes. For example: [Co(NH3)4Cl2]+.
OR (b)
(i) (a) Pentaamminenitrito-N-Cobalt(III) is [Co(NO2)(NH3)5]2+. (b) Tetrahydroxidozincate(II) is [Zn(OH)4]2-.
(ii) The difference of energy between the two sets of degenerated orbitals (t2g and eg) after crystal field splitting is known as crystal field splitting energy. It is denoted by Delta_o.
2022-II · 3 marks
(a) Using Valence Bond theory, predict the hybridization and magnetic character of following: [CoF6]3- [Atomic number of Co = 27]
(b) Write IUPAC name of the following complex: [CoBr2(en)2]+
(c) How many ions are produced from the complex [Co(NH3)6]Cl2 in solution?
Answer(a) [CoF6]3-: Co2+ has configuration [Ar]3d7. F- is a weak field ligand so no pairing occurs. Hybridization = sp3d2. Magnetic Nature = Paramagnetic (4 unpaired electrons).
(b) [CoBr2(en)2]+: IUPAC name is Dibromodobis(ethane-1,2-diamine)cobalt(III) ion.
(c) In aqueous solution of the complex, 3 ions are produced: [Co(NH3)6]Cl2 -> [Co(NH3)6]2+ + 2Cl-.
2023 · 3 marks
(a) On the basis of crystal field theory write the electronic configuration for d5 ion with a strong field ligand for which Delta_0 > P.
(b) [Ni(CO)4] has tetrahedral geometry while [Ni(CN)4]2- has square planar yet both exhibit diamagnetism. Explain. [Atomic number: Ni = 28]
Answer(a) The magnitude difference in energy between the two sets of d-orbital i.e. t2g and eg. Electronic configuration of d5 if Delta_0 > P is t2g5 eg0. Because in a strong field ligand pairing of electrons takes place, e.g. [Ni(CN)4]2-. CN is a strong field ligand.
(b) Ni = 28, 3d8 4s2. Ni2+ = 3d8 4s0.
[Ni(CN)4] is a strong field ligand: dsp2 hybridization showing square planar geometry. All electrons are paired so it is diamagnetic in nature. [Ni(CO)4]: Ni = 28, 3d8 4s2. Ni valency is zero. sp3 hybridization showing tetrahedral geometry. All electrons are paired so it is diamagnetic in nature.
2023 · 3 marks
(a) On the basis of crystal field theory write the electronic configuration for d5 ion with a weak ligand for which delta 0 < P.
(b) Explain [Fe(CN)6]3- is an inner orbital complex whereas [FeF6]3- is an outer orbital complex. [Atomic number: Fe = 26]
Answer(a) For d5 ion with weak ligand where delta_0 < P, the configuration is t2g3 eg2. Because it follows weak field ligand phenomena - when weak field ligands are present in a structure, pairing of electrons does not take place.
(b) [Fe(CN)6]3-: CN is a strong field ligand. Fe = 26, [Ar] 3d6 4s2. Fe3+ = 3d5. Hybridization = d2sp3. Inner d-complex. If strong field ligand is available then delta_0 > P = t2g5 eg0.
[FeF6]3-: In [Fe(F6)]3-, Fe = [Ar] 3d6 4s2. Fe3+ = 3d5. Hybridization = sp3d2. Outer d-complex. With a weak field ligand delta_0 < P, so there is no pairing of electrons in 3d orbitals.
2023 · 3 marks
(b) A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CO)4] is colourless. Explain. [Atomic number: Ni = 28]
AnswerIn [Ni(H2O)6]2+ Ni has configuration of 3d8. The two unpaired electrons do not pair up in presence of weak H2O ligand, but in [Ni(CO)4] due to strong CO ligand, it pairs up. As there is unpaired electron in [Ni(H2O)6]2+ it is coloured and [Ni(CO)4] is colourless, due to absence of unpaired electrons.
2023 · 3 marks
(a) Draw the geometrical isomers of [Co(en)2Cl2]2+. Which geometrical isomer of [Co(en)2Cl2]2+ is not optically active and why?
(b) Write the hybridisation and magnetic behaviour of [CoF6]3-.
[Given: Atomic number of Co = 27]
Answer(a) Geometrical isomers of [Co(en)2Cl2]2+: Trans-isomer: Cl atoms on opposite sides. Optically inactive due to superimposable mirror images. Cis-isomer: Cl atoms on same side. Optically active due to non-superimposable mirror images.
The trans isomer is not optically active because it has a plane of symmetry (superimposable mirror images).
(b) [CoF6]3-: Co (Z=27), ground state: [Ar] 3d7 4s2. Co3+ = 3d6. F- is a weak field ligand, so no pairing occurs. Hybridization = sp3d2. It is paramagnetic as it contains unpaired electrons.
2025 · 3 marks
Using valence bond theory, explain the hybridisation and magnetic character of the following:
(a) [Co(NH3)6]3+
(b) [Ni(CO)4]. [At. no.: Co = 27, Ni = 28]
Answer(a) [Co(NH3)6]3+: Co3+ has [Ar]3d6 configuration. In the presence of ammonia (strong field ligand), electrons in 3d level get paired up leaving two of the 3d orbitals for hybridisation. Hence, d2sp3 hybridisation. The complex is diamagnetic as there are no unpaired electrons.
(b) [Ni(CO)4]: The oxidation state of Ni is 0. The valence shell electronic configuration is 3d8 4s2, but all 10 electrons are pushed into 3d orbital due to the strong field CO ligands, forming sp3 hybridisation. The complex is tetrahedral and diamagnetic.
2025 · 3 marks
Write IUPAC names of the following coordination entities:
(a) Using valence bond theory, explain the hybridisation and magnetic behaviour of the following:
(i) [Co(NH3)6]Cl3
(ii) K2[NiCl4]. (At. no.: Co = 27, Ni = 28)
(b) Write the electronic configuration of d5 ion when delta_o > P.
Answer(a)
(i) [Co(NH3)6]Cl3: Co3+ has [Ar]3d6 configuration. NH3 is strong field ligand. Electrons in 3d get paired up. d2sp3 hybridisation (inner orbital octahedral complex). Diamagnetic due to absence of unpaired electrons.
(ii) K2[NiCl4]: Ni2+ has [Ar]3d8 4s0 configuration. Cl- is a weak field ligand. sp3 hybridisation with tetrahedral geometry. Paramagnetic due to two unpaired electrons.
(b) When CFSE (delta_o) is more than pairing energy, it indicates a low spin complex. The electron will pair in lower energy level before filling in higher energy level. Electronic configuration will be t2g5 eg0.
2024 · 3 marks
Write the IUPAC names of the following coordination compounds (any three):
(a) Draw the geometrical isomers of the given complex: [Pt(en)2Cl2]2+
(b) Write the electronic configuration for d4 ion if Delta_0 < P on the basis of crystal field theory.
(c) What is meant by a unidentate ligand? Give an example.
Answer(a) [Pt(en)2Cl2]2+ has two geometrical isomers: cis (both Cl on same side) and trans (Cl on opposite sides).
(b) If Delta_0 < P, the electrons will occupy the higher energy orbitals first (high spin). The electron configuration becomes t2g3 eg1.
(c) An unidentate ligand is a type of ligand that can form only one coordinate bond with a central metal ion. In other words, unidentate ligands have a single donor atom. Examples: H2O, NH3, Cl-.
2024 · 3 marks
(a) Draw the geometrical isomers of the given complex: [Co(en)2Cl2]+
(b) Write the electronic configuration of d4 ion if Delta_0 > P.
(c) What is a didentate ligand? Give one example.
Answer(a) [Co(en)2Cl2]+ has two geometrical isomers: cis and trans.
(b) If Delta_0 > P, electrons occupy the lower energy t2g orbitals first. The electron configuration becomes t2g4 eg0.
(c) A didentate ligand is a ligand that can form coordination bonds to a metal ion using two donor atoms. These donor atoms are typically separated by a specific number of atoms in the ligand's structure, forming a chelate complex. A common example is ethylenediamine (en).
2024 · 3 marks
(a) Draw the geometrical isomers of the given complex: [Pt(NH3)2Cl2]2+
(b) Write the electronic configuration of d5 ion if Delta_0 < P.
(c) What is an ambidentate ligand?
Answer(a) [Pt(NH3)2Cl2]2+ has two geometrical isomers: cis (both Cl and NH3 on same side) and trans (Cl on opposite sides).
(b) Electronic configuration of d5 ion if Delta_0 < P: t2g3 eg2 (high spin, all electrons unpaired).
(c) The ligand which has two different donor atom is called ambidentate ligand. E.g., NO2- (can bind through N or O).
2015 · 3 marks
(i) Indicate the principle behind the method used for the refining of zinc.
(ii) What is the role of silica in the extraction of copper?
(iii) Which form of the iron is the purest form of commercial iron?
Answer(i) Distillation is used for refining of zinc. It is based on the difference in boiling points of metal and impurities.
(ii) Silica acts as a flux and combines with FeO (impurity) to form FeSiO3 (slag): FeO + SiO2 → FeSiO3
(iii) Wrought iron is the purest form of commercial iron.
2017 · 3 marks
(a) Write the principle of method used for the refining of germanium.
(b) Out of PbS and PbCO3 (ores of lead), which one is concentrated by froth flotation process?
(c) What is significance of leaching in the extraction of aluminium?
Answer(a) Zone refining: Based on the principle that impurities are more soluble in the melt than in the solid state.
(b) PbS (sulphide ore is concentrated by froth flotation).
(c) Leaching is used to concentrate bauxite ore (Al2O3) by removing impurities like SiO2, Fe2O3, TiO2 using Bayer's process with NaOH.
2017 · 3 marks
Write the principles of the following methods:
(i) Vapour phase refining
(ii) Zone refining
(iii) Chromatography
Answer(i) Vapour phase refining: Metal is converted into volatile compound which is decomposed to give pure metal.
(ii) Zone refining: Impurities are more soluble in melt than in solid state of the metal.
(iii) Chromatography: Based on differential adsorption of different components of a mixture on an adsorbent.
2017 · 3 marks
Write the principle of the following:
(a) Zone refining
(b) Froth flotation process
(c) Chromatography
Answer(a) Zone refining: Impurities are more soluble in melt than in solid state of metal.
(b) Froth flotation: Based on preferential wetting of ore particles by oil and gangue by water.
(c) Chromatography: Based on differential adsorption of different components on an adsorbent.
2017 · 3 marks
(a) Write the principle of vapour phase refining.
(b) Write the role of dilute NaCN in the extraction of silver.
(c) What is the role of collectors in the froth flotation process? Give an example of a collector.
Answer(a) Metal is converted into its volatile compound which is then decomposed to give pure metal. E.g., Mond process for Ni.
(b) NaCN dissolves silver from ore: Ag2S + 4NaCN → 2Na[Ag(CN)2] + Na2S. Silver is recovered by adding Zn.
(c) Collectors enhance non-wettability of ore particles. E.g., Sodium ethyl xanthate, pine oil.
2018 · 3 marks
Write the chemical reactions involved in the process of extraction of Gold. Explain the role of dilute NaCN and Zn in this process.
Answer4Au(s) + 8NaCN(aq) + 2H2O(l) + O2(g) → 4Na[Au(CN)2](aq) + 4NaOH(aq) Role of NaCN: Acts as leaching agent, dissolves gold as soluble sodium aurocyanide complex. 2Na[Au(CN)2](aq) + Zn(s) → Na2[Zn(CN)4](aq) + 2Au(s) Role of Zn: Acts as reducing agent to displace gold from the complex (cementation).
2013 · 3 marks
How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) Zr(Z = 40) and Hf(Z = 72) have almost identical radii.
(iii) Transition metals and their compounds act as catalyst.
Answer(i) Transition metals show variable oxidation state because they have tendency to loose electron from their penultimate d-subshell and exhibit various oxidation state.
(ii) Zr and Hf have almost same radii because of lanthanoid contraction. Filling up of electrons in 4f series shows poor shielding effect.
(iii) 4d series and other transition metals act as catalyst because they participate in various type of chemical reactions. E.g., N2 + 3H2 ⟶[Fe/Mo, High temp, High pressure] 2NH3
(i) Actinoid contraction is greater than lanthanoid contraction.
(ii) Transition metals form coloured compounds.
(b) Complete the following equation:
2MnO4− + 6H+ + 5NO2− →
Answer(a)
(i) 5f orbital electrons have poor shielding effect than 4f orbital electrons, so actinoid contraction is greater.
(ii) Due to d-d transition or the energy of excitation of an electron from lower d orbital to higher d-orbital lies in the visible region, and presence of unpaired electrons in the d-orbital.
(b) 2MnO4− + 6H+ + 5NO2− → 2Mn2+ + 3H2O + 5NO3−
2016 · 3 marks
(i) Name the method of refining of metals such as germanium.
(ii) In the extraction of Al, impure Al2O3 is dissolved in conc. NaOH to form sodium aluminate and leaving impurities behind. What is the name of this process?
(iii) What is the role of coke in the extraction of iron from its oxides?
Answer(i) Zone refining is used for refining of germanium.
(ii) The process is called Baeyer's process (or leaching).
(iii) Coke acts as a reducing agent: Fe2O3 + 3CO -> 2Fe + 3CO2. Coke first converts to CO which then reduces iron oxide.
2016 · 3 marks
Give reasons:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Transition metals show variable oxidation states.
(iii) Actinoids show irregularities in their electronic configurations.
Answer(i) Oxygen can form multiple bonds (p-pi - d-pi) with Mn enabling Mn2O7. Fluorine has no d-orbitals and cannot form multiple bonds, so maximum oxidation state with F is +4 (MnF4).
(ii) Due to comparable energies of ns and (n-1)d orbitals, electrons of both can participate in bonding, giving variable oxidation states.
(iii) Due to comparable energies of 5f, 6d, and 7s orbitals, electrons can rearrange for more stable configurations (f0, f7, f14 stability).
2016 · 3 marks
(i) Name the method of refining of nickel.
(ii) What is the role of cryolite in the extraction of aluminium?
(iii) What is the role of limestone in the extraction of iron from its oxides?
Answer(i) Mond process (vapour phase refining): Ni + 4CO -> Ni(CO)4 -> Ni + 4CO.
(ii) Cryolite (Na3AlF6) lowers the melting point of alumina from ~2050°C to ~950°C and increases conductivity of electrolyte.
(iii) Limestone (CaCO3) acts as a flux; it decomposes to CaO which combines with SiO2 (gangue) to form slag CaSiO3.
2016 · 3 marks
Give reasons:
(i) SO2 is reducing while TeO2 is an oxidizing agent.
(ii) Nitrogen does not form pentahalide.
(iii) ICl is more reactive than I2.
Answer(i) SO2 is reducing because S(+4) can be oxidized to +6. TeO2 is oxidizing because Te(+4) tends to get reduced due to inert pair effect.
(ii) Nitrogen has no d-orbitals to expand its octet, maximum covalency is 4.
(iii) ICl is more reactive because the I-Cl bond is weaker and more polar than I-I bond, making heterolytic fission easier.
2018 · 3 marks
Give reasons:
(a) E° value for Mn3+/Mn2+ couple is much more positive than that for Fe3+/Fe2+.
(b) Iron has higher enthalpy of atomization than that of copper.
(c) Sc3+ is colourless in aqueous solution whereas Ti3+ is coloured.
Answer(a) Because Mn2+ is more stable than Mn3+ due to half-filled d5 configuration, whereas Fe2+ becomes unstable after losing an electron from half-filled orbital. Much larger third ionization energy of Mn (d5 to d4).
(b) Due to higher number of unpaired electrons present in iron. Fe (3d64s2) has more unpaired electrons than Cu (3d104s1). More unpaired electrons leads to stronger metallic bonding and higher enthalpy of atomization.
(c) Sc3+ (3d0) is colourless due to absence of unpaired electron and no d-d transition. Ti3+ (3d1) is coloured due to presence of one unpaired electron and d-d transition absorbing light in visible region.
2019 · 3 marks
(a) Name the method of refining which is
(i) used to obtain semiconductor of high purity,
(ii) used to obtain low boiling metal.
(b) Write chemical reactions taking place in the extraction of copper from Cu2S.
Write the name and principle of the method used for refining of
(a) Zinc,
(b) Germanium,
(c) Titanium.
Answer(a) Zinc: Distillation - based on difference in boiling points.
(b) Germanium: Zone refining - impurities more soluble in melt than solid.
(c) Titanium: Van Arkel method - forming volatile compound with I2 followed by thermal decomposition.
2019 · 3 marks
Give reasons:
(a) Transition metals form complex compounds.
(b) E° values for (Zn2+/Zn) and (Mn2+/Mn) are more negative than expected.
(c) Actinoids show wide range of oxidation states.
Answer(a) Due to small size, high ionic charge, availability of vacant d-orbital.
(b) Due to stable 3d10 configuration in Zn2+ and 3d5 in Mn2+.
(c) Due to comparable energies of 5f, 6d and 7s orbitals.
2019 · 3 marks
Write the name and principle of the method used for refining of
(a) Tin,
(b) Copper,
(c) Nickel.
Answer(a) Tin: Liquation - based on difference in melting points.
(b) Copper: Electrolytic refining - impure Cu as anode, pure Cu as cathode, CuSO4 electrolyte.
(c) Nickel: Mond process - formation and decomposition of volatile Ni(CO)4.
2019 · 3 marks
Give reasons:
(a) Transition metals show variable oxidation states.
(b) E° value for (Zn2+/Zn) is negative while that of (Cu2+/Cu) is positive.
(c) Higher oxidation state of Mn with fluorine is +4 whereas with oxygen is +7.
Answer(a) Due to partially filled orbitals and comparable energies of ns and (n-1)d orbitals.
(b) Zn2+/Zn is negative due to stable 3d10 configuration; Cu2+/Cu is positive due to low hydration enthalpy and high atomization enthalpy.
(c) Mn forms multiple bonds with O using 2p-3d overlap showing +7; F cannot form multiple bonds, so max oxidation state is +4.
2019 · 3 marks
Write the principle of:
(a) Hydraulic washing
(b) Chromatography
(c) Froth floatation process
Answer(a) Based on density difference; lighter gangue washed away, heavier ore settles.
(b) Based on differential adsorption of components on adsorbent.
(c) Based on difference in wetting properties; ore wetted by oil, gangue by water.
2019 · 3 marks
Give reasons:
(a) Transition metals have high enthalpies of atomization.
(b) Manganese has lower melting point even though it has higher number of unpaired electrons.
(c) Ce4+ is a strong oxidizing agent.
Answer(a) Due to strong interatomic interactions/strong metallic bonding.
(b) Due to stable 3d5 configuration, interatomic interaction is poor between unpaired electrons.
(c) Ce is more stable in +3 oxidation state; Ce4+ readily gains electron to achieve +3.
2020 · 3 marks
Following ions are given: Cr2+, Cu2+, Cu+, Fe2+, Fe3+, Mn3+. Identify the ion which is
(i) a strong reducing agent
(ii) unstable in aqueous solution
(iii) a strong oxidising agent. Give suitable reason in each.
Answer(i) Cr2+, because its configuration changes from d4 to d3 and having a half-filled t2g level.
(ii) Cu+ in aqueous medium energy is required to remove one electron from Cu+ to Cu2+; high hydration energy of Cu2+ compensates. Therefore Cu+ is unstable: 2Cu+ -> Cu2+(aq) + Cu(s).
(iii) Mn3+, because its configuration changes from Mn3+ to Mn2+ results in the half-filled d5 configuration, which has extra stability.
2022-II · 3 marks
(a) Write any two consequences of Lanthanoid Contraction.
(b) Name the element of 3d series which exhibits the largest number of oxidation states. Give reason.
Answer(a) Two consequences of lanthanide contraction: (i) Due to close similarity in electronic configuration and ionic radii, the lanthanides have identical chemical properties which makes their separation difficult. (ii) Due to lanthanide contraction, the size of Ln3+ ions decrease regularly with increase in atomic number. According to Fajan's rule, decrease in size of Ln3+ ions increase the covalent character and decrease the basic character between Ln3+ and OH- ion in Ln(OH)3. Order of size: La3+ > Ce3+ > ... > Lu3+.
(b) The element which shows largest number of oxidation states in 3d series is Manganese (Mn). It shows oxidation states +2, +3, +4, +5, +6 and +7. It is because it contains the maximum number of unpaired electrons in the outermost shell, i.e., 3d5 4s2.
2022-II · 3 marks
Give reasons for the following statements:
(a) Copper does not displace hydrogen from acids.
(b) Transition metals and most of their compounds show paramagnetic behaviour.
(c) Zn, Cd and Hg are soft metals.
Answer(a) Copper cannot displace hydrogen from an acid because copper is less reactive element than hydrogen and it is present below hydrogen in the activity series of metals.
(b) Transition metals and many other metals show paramagnetic behaviour because they possess number of unpaired electrons in d-orbital.
(c) Metal atoms like Zn, Cd, Hg have completely filled d-orbitals (d10 configuration). Thus, d-electrons are not available for metallic bond formation. As a result, these metals are quite soft and also have low melting points.
2022-II · 3 marks
Give reason for the following statements:
(a) Scandium (Z = 21) is a transition element but Zn (Z = 30) is not.
(b) [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.
(c) Physical and chemical properties of the 4d and 5d series of the transition elements are quite similar to expected.
Answer(a) On the basis of incompletely filled 3d-orbitals in case of Scandium atom (3d1) having one electron in its ground state, it is regarded as a transition element. On the other hand, zinc atom having completely filled d-orbitals (3d10) in its ground state as well as in its oxidized state, it is not considered as a transition element.
(b) [Ti(H2O)6]3+ is coloured because Ti3+ (3d1) has one unpaired electron which undergoes d-d transition so it is coloured. While [Sc(H2O)6]3+ (3d0) has no unpaired electrons so it will not undergo d-d transition so it is colourless.
(c) Physical and chemical properties of 4d and 5d series elements are similar because they have same atomic and ionic radii due to lanthanide contraction. Due to equal atomic radii the size of Zr and Hf, Nb and Ta, Mo and W, etc., the two elements of each pair have the same properties.
2022-II · 3 marks
Account for the following:
(a) Cu2+ salts are coloured while Zn2+ salts are white.
(b) E° value of the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+.
(c) Transition metals form alloys.
Answer(a) Cu2+ salts are coloured whereas Zn2+ salts are white because Cu2+ ([Ar]3d9) has one unpaired electron in d-orbital which allows electron transition in visible region which imparts colour whereas Zn2+ ([Ar]3d10) does not possess any unpaired electron hence no electron transition takes place thus shows no colour.
(b) E° values for Mn3+/Mn2+ couple is much more positive than for Cr3+/Cr2+ because the conversion of Mn2+ from Mn3+ possesses extra stability due to half filled valence electronic configuration (3d5) whereas Cr ion changes from Cr3+ to Cr2+ undergoes change in outer configuration from 3d3 to 3d4 making it less stable comparatively.
(c) Transition metals have very similar atomic sizes as a result one metal can easily replace the other metal from its lattice to form solid solution (alloy). Transition metals are also miscible with one another in the molten state. Thus, the molten state solution of two or more transition metals on cooling forms alloy.
2022-II · 3 marks
Define transition metals. Why Zn, Cd and Hg are not called transition metals? How is the variability in oxidation states of transition metals different from that of p-block elements?
AnswerTransition metals: Elements having partially filled d-orbitals in ground state or in excited state, are known as transition elements. The orbitals of the elements like Zn, Cd, Hg are completely filled when they are in their ground state as well as in their general oxidation state. Therefore, these elements are not considered as transition elements. In transition elements, the variability in the oxidation state is due to the participation of (n-1) d orbitals and ns orbitals. Thus, oxidation states differ by unity. For example, Fe+3 to Fe+2, Cr+3 to Cr+2, etc. On the other hand, the variable oxidation states shown by some p block elements differ by two units. For example, Sn4+ to Sn2+, Pb4+ to Pb2+.
2025 · 3 marks
The elements of 3d transition series are given as: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn. Answer the following:
(a) Copper has exceptionally positive E°(M2+/M) value, why?
(b) Which element is a strong reducing agent in +2 oxidation state and why?
(c) Zn2+ salts are colourless. Why?
Answer(a) The sum of enthalpy of atomization and ionisation enthalpy for copper is higher than hydration enthalpy due to this copper has positive value of E°(Cu2+/Cu).
(b) Cr, because Cr2+ which has [Ar]3d4 4s0 configuration, tends to lose one more electron forming Cr3+ to attain the more stable half-filled t2g(d3) configuration. This is proved by its negative reduction potential value.
(c) Zn in its +2 state has the configuration [Ar]3d10 4s0, indicating all electrons are paired in d orbitals. This does not allow d-d transition to occur and its salts are thus colourless.
2025 · 3 marks
Complete and balance the following chemical equations:
(a) Of the d4 species, Cr2+ is strongly reducing while Mn3+ is strongly oxidising. Why?
(b) Write two consequences of lanthanoid contraction.
(c) Which element of 3d series has lowest enthalpy of atomisation and why?
Answer(a) Cr2+ is a strong reducing agent, it has 3d4 electronic configuration. Hence, it loses electron to form 3d3 configuration, which has half filled t2g and is more stable. But in case of Mn3+, the electronic configuration 3d4 gains electron to become 3d5 (Mn2+). It attains extra stability due to 3d5 electronic configuration. Hence it is a strong oxidising agent. (b)(1) It causes the radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. (2) The basic strength decreases from La(OH)3 to Lu(OH)3.
(c) In 3d elements, d-orbitals are involved in metallic bonding but not in case of zinc. Hence it shows lowest enthalpy of atomisation.
2025 · 3 marks
(a) E°(Mn2+/Mn) is -1.18 V. Why is this value highly negative in comparison to neighbouring d-block elements?
(b) What is lanthanoid contraction?
(c) Zn, Cd and Hg are soft metals. Why?
Answer(a) When Mn loses electron and forms Mn2+, it attains stable half-filled 3d5 electronic configuration, so it exhibits more negative reduction potential as compared to neighbouring d block elements.
(b) The gradual decrease in the size of lanthanide atoms and ions as their atomic number increases from La-57 to Lu-71 is called lanthanoid contraction. This is due to poor shielding effect of 4f orbital.
(c) Zn, Cd and Hg have completely filled d orbitals, they do not have unpaired electrons in d subshell, so it is difficult to make strong metallic bonds that's why these are soft metals.
2014 · 3 marks
(a) Draw the structures of the following molecules:
(i) XeOF4
(ii) H3SO4 (should be H2SO4)
(b) Write the structural difference between white phosphorus and red phosphorus.
Answer(a)
(i) XeOF4: Square pyramidal structure with one O at apex.
(ii) H2SO4: Tetrahedral structure around S with 2 OH groups and 2 =O.
(b) White phosphorus exists as discrete P4 tetrahedral molecules. Red phosphorus has a polymeric chain structure where P4 units are linked together.
2014 · 3 marks
Account for the following:
(i) PCl5 is more covalent than PCl3.
(ii) Iron on reaction with HCl forms FeCl2 and not FeCl3.
(iii) The two O–O bond lengths in the ozone molecule are equal.
Answer(i) PCl5 is more covalent than PCl3 because P in PCl5 has higher positive charge (+5), which polarizes the Cl− more according to Fajan's rule.
(ii) Fe + 2HCl → FeCl2 + H2. HCl is not a strong enough oxidizing agent to oxidize Fe to +3 state.
(iii) Ozone is a resonance hybrid of two structures; both O–O bonds are equivalent due to resonance, having equal bond length.
2014 · 3 marks
Give reasons for the following:
(i) (CH3)3P = O exists but (CH3)3N = O does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) H3PO2 is a stronger reducing agent than H3PO3.
Answer(i) P has vacant d-orbitals and can expand its octet to form (CH3)3P=O using dπ−pπ bonding. N does not have d-orbitals so cannot expand its octet.
(ii) Due to small size of O, high electron density causes inter-electronic repulsion when an extra electron is added, so less electron gain enthalpy than S.
(iii) H3PO2 has two P–H bonds and is a stronger reducing agent, while H3PO3 has one P–H bond.
2017 · 3 marks
Give reasons:
(i) Thermal stability decreases from H2O to H2Te.
(ii) Fluoride ion has higher hydration enthalpy than chloride ion.
(iii) Nitrogen does not form pentahalide.
Answer(i) As size of central atom increases from O to Te, bond length increases and bond strength decreases, hence thermal stability decreases.
(ii) Due to small size and high charge density of fluoride ion compared to chloride ion.
(iii) Nitrogen does not have d-orbitals to expand its octet, so maximum covalency is 4.
2024 · 4 marks
The oxidation number of the central atom in a complex is defined as the charge it would carry if all the ligands are removed along with the electron pairs that are shared with the central atom. Similarly the charge on the complex is the sum of the charges of the constituent parts. Based on this information, answer:
(a) Define ambidentate ligand with an example.
(b) What type of isomerism is shown by [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl?
(c) Define chelate effect. How it affects the stability of complex?
OR
(c) Find the coordination number and oxidation state of chromium in Na3[Cr(C2O4)3].
Answer(a) A ligand which has two different donor atoms and either of the two sites in the ligand can form coordinate bonds with the central atom is called an ambidentate ligand. Ex: -O-N=O (NO2- can bind through N or O)
(b) Ionisation isomerism
(c) When a ligand attaches to the metal ion in a manner that forms a ring, then a chelate is formed. A bidentate or polydentate ligand positioned such that when they coordinate with the central metal ion, a 5 or 6 membered ring is formed, the effect is known as the chelate effect. The complex resulting from chelation is much more thermodynamically stable.
OR
(c) Coordination no. = No. of ligands x denticity of ligand = 3 x 2 = 6. Since overall charge of the ligand + metal ion is +3, the oxidation no. of Cr is +3.
2024 · 4 marks
The nature of bonding, structure of the coordination explained to some extent by valence bond theory. The central metal atom/ion makes available a number of vacant orbitals equal to its coordination number. Based on this:
(a) Predict whether [CoF6]3- is diamagnetic or paramagnetic and why [Atomic number: Co = 27]
(b) What is the coordination number of Co in [Co(en)2Cl2]+?
(c)
(i) Write the IUPAC name of the given complex: [Pt(NH3)2Cl2]2+
(ii) Explain (Co(NH3)6)3+ is an inner orbital or outer orbital complex.
OR
(c) Using valence bond theory, deduce the shape and hybridisation of [Ni(NH3)6]2+ (Atomic number of Ni = 28).
Answer(a) [CoF6]3-: Co = 27, Co3+ = 1s2 2s2 2p6 3s2 3p6 3d6 4s0 4p0. Fluorido is a weak ligand so it will undergo high spin, sp3d2 hybridisation. It will be paramagnetic due to 4 unpaired electrons.
(b) [Co(en)2Cl2]+: en = ethylenediamine, a bidentate ligand providing two lone pairs. Coordination number = 2 x 2 + 2 = 6.
(c)
(i) Diamminedichloridoplatinum(IV) ion.
(ii) [Co(NH3)6]3+: Since NH3 is a borderline ligand it will form an outer orbital complex with sp3d2 hybridisation and is diamagnetic in nature.
OR
(c) [Ni(NH3)6]2+: Ni = 28, Ni2+ = 1s2 2s2 2p6 3s2 3p6 3d8. The coordination number is six, that means the compound is octahedral in shape. Since [Ni(NH3)6]2+ is an outer orbital complex it undergoes sp3d2 hybridisation. It is paramagnetic with 2 unpaired electrons.
2024 · 4 marks
The oxidation number of the central atom in a complex is defined as the charge it would carry if all the ligands are removed. Based on this:
(a) What is the secondary valence of Co in [Co(NH3)4Cl2]+?
(b) What type of isomerism is shown by the complex [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2.H2O?
(c) Write the electronic configuration of d4 ion on the basis of crystal field theory when (i) Delta_0 < P (ii) Delta_0 > P
OR
(c) Find the oxidation state and coordination number of the central metal ion in [Co(H2O)(CN)(en)2]2+
Answer(a) The secondary valency of Co in [Co(NH3)4Cl2]+ is 6, meaning it forms 6 bonds with surrounding ligands (4 NH3 and 2 Cl-).
(b) Solvate isomerism (Hydrate isomerism). The two complexes differ by water/solvent molecules that are directly bonded to metal ion or present as free solvent molecules.
(c)
(i) If Delta_0 < P: electrons occupy higher energy orbitals first. Configuration becomes t2g3 eg1 (high spin).
(ii) If Delta_0 > P: electrons occupy lower energy t2g orbitals first. Configuration becomes t2g4 eg0 (low spin).
OR
(c) The coordination number of cobalt is 6 and the oxidation state of cobalt (Co) in the complex is +3.
2025 · 5 marks
(A) Answer the following:
(a) Low spin tetrahedral complexes are not known.
(b) Co2+ is easily oxidised to Co3+ in the presence of a strong ligand [At. No. of Co = 27].
(c) What type of isomerism is shown by the complex [Co(NH3)6][Cr(CN)6]?
(d) Why a solution of [Ni(H2O)6]2+ is green while a solutions of [Ni(CN)4]2- is colourless. (At. No. of Ni = 28)
(e) Write the IUPAC name of the following complex: [Co(NH3)5(CO3)]Cl.
OR (B)(a) What is meant by 'Chelate effect'? Give an example.
(b) Write the hybridisation and magnetic behaviour of [Fe(CN)6]4-. (Atomic number: Fe = 26)
(c) If PtCl2.2NH3 does not react with AgNO3, what will be its formula?
Answer(A)(a) A tetrahedral complex has sp3 hybridisation. During crystal field splitting, the splitting energy of d orbitals is much lower than that of the pairing energy, and thus the electrons jump to the higher energy level d-orbitals rather than pairing, resulting in high-spin complexes being formed.
(b) Co2+ can be oxidised to Co3+ in the presence of strong field ligands. Strong field ligands create a strong ligand field that stabilizes the higher oxidation state of the metal ion, making the +3 oxidation state more favourable.
(c) Coordination isomerism - both positive and negative ions of a salt are complex ions and the two isomers differ in the distribution of ligands between the cation and anion.
(d) In both complexes, Ni is in +2 oxidation state with 3d8 configuration. In the presence of weak field water ligands, two unpaired electrons do not pair up. The [Ni(H2O)6]2+ has two unpaired electrons which result in green colour due to d-d transition. In presence of strong field cyanide ligand, the unpaired electrons in 3d orbital pair up. Due to absence of unpaired electrons, no d-d transitions are possible and [Ni(CN)4]2- is colourless.
(e) Pentaamminecarbonatocobalt(III)chloride.
OR (B)(a) When a ligand uses its two or more donor atoms simultaneously a ring-like structure is formed. The formation of such ring-like structures is called a chelate effect. Compounds having chelate rings are more stable.
(b) The hybridisation is d2sp3 and it forms an inner orbital complex. Since there are no unpaired electrons, the complex is diamagnetic.
(c) [Pt(NH3)2Cl2] - since it does not react with AgNO3, the chloride ions are inside the coordination sphere.
2025 · 5 marks
(A) For the complex [Fe(en)2Cl2] Cl, identify:
(a) the oxidation number of iron.
(b) the hybridisation and the shape of the complex.
(c) the magnetic behaviour of the complex
(d) whether there is an optical isomer of the complex? If so draw its structure.
(e) IUPAC name of the complex. (At. no. of Fe = 26) OR (B)(a) Using IUPAC norms write the names of the following:
(i) [Co(NH3)4 Cl(NO2)]Cl
(ii) K3[Fe(CN)6]
(iii) [Cr(C2O4)3]3-.
(b) What is crystal field splitting energy? Why low spin tetrahedral complexes are not formed?
Answer(A)(a) Oxidation number of iron: x + 0 + 2(-1) + (-1) = 0, so x = +3.
(b) The complex has two bidentate ligands and two monodentate ligands. Hybridisation is d2sp3 and shape is octahedral.
(c) Fe3+ has d5 configuration with 1 unpaired electron in d-orbital, so the complex is paramagnetic.
(d) The cis-isomer of [Fe(en)2Cl2]Cl shows optical isomerism because it has non-superimposable mirror images. The trans-isomer does not show optical isomerism as it lacks chirality.
(b) Crystal field splitting energy refers to the energy difference between the split d-orbitals of a transition metal ion when it is surrounded by ligands in a coordination complex. Low spin tetrahedral complexes are not formed because in a tetrahedral field, the splitting energy (delta_t) is smaller than in an octahedral field. The energy difference between the higher and lower sets of orbitals is too small to favour pairing of electrons.
2025 · 5 marks
(A)(a) In the following complex ions, explain the type of hybridisation, shape and magnetic property:
(a) [Fe(H2O)6]2+
(b) [NiCl4]2-. (At. Nos.: Fe = 26, Ni = 28) OR (B)(a) Write IUPAC names of the following:
(i) [Co(H2O)(CN)(en)2]2+
(ii) [PtCl4]2-
(iii) [Cr(NH3)4Cl(ONO)]+.
(b) What is spectrochemical series? Write the difference between a strong field ligand and a weak field ligand.
Answer(A)(a) [Fe(H2O)6]2+: Fe2+ has [Ar]3d6 configuration. H2O is a weak field ligand, so no pairing occurs. Hybridisation is sp3d2, shape is octahedral, and the complex is paramagnetic with 4 unpaired electrons.
(b) [NiCl4]2-: Ni2+ has [Ar]3d8 configuration. Cl- is a weak field ligand. Hybridisation is sp3, shape is tetrahedral, and the complex is paramagnetic with 2 unpaired electrons.
(b) The spectrochemical series is the arrangement of ligands in increasing order of their crystal field splitting energy. Weak field ligands: I- < Br- < SCN- < Cl- < S2- < F- < OH- < C2O42- < H2O. Strong field ligands: NCS- < edta4- < NH3 < en < CN- < CO. Strong field ligands cause high crystal field splitting and often lead to low-spin complexes. Weak field ligands cause low crystal field splitting and often lead to high-spin complexes.
2014 · 5 marks
(a) How do you prepare:
(i) KMnO4 from MnO2
(ii) Na2Cr2O7 from Na2CrO4
(b) Account for the following:
(i) Mn2+ is more stable than Fe2+ towards oxidation to +3 state.
(ii) The enthalpy of atomization is lowest for Zn in 3d series of the transition elements.
(iii) Actinoid elements show wide range of oxidation states.
(ii) Na2Cr2O7 from Na2CrO4: 2Na2CrO4 + 2H+ ⟶[Δ] Na2Cr2O7 (orange) + 2Na+ + H2O
(b)
(i) Mn in +2 state has stable d5 (half-filled) configuration, shows resistance to oxidation to Mn3+. Fe2+ has 3d6 and easily loses one electron to get more stable 3d5 configuration of Fe3+.
(ii) Zn has lowest enthalpy of atomization because it lacks unpaired electrons, making its inter-atomic metallic bonding the weakest.
(iii) Actinoids show wide range of oxidation states because of very small energy gap between 5f, 6d and 7s sub-shells. All can be involved in bonding.
2014 · 5 marks
(i) Name the element of 3d transition series which shows maximum number of oxidation states. Why does it show so?
(ii) Which transition metal of 3d series has positive E°(M2+/M) value and why?
(iii) Out of Cr3+ and Mn3+, which is a stronger oxidizing agent and why?
(iv) Name a member of the Lanthanoid series which is well known to exhibit +2 oxidation state.
(v) Complete the following equation: MnO4− + 8H+ + 5e− →
Answer(i) Manganese ([Ar] 3d5 4s2) shows maximum number of oxidation states as its atoms have five unpaired electrons in 3d orbitals. It shows all the oxidation states from +2 to +7.
(ii) Cu has positive E°(M2+/M) value because the sum of enthalpies of sublimation and ionization is not balanced by hydration enthalpy.
(iii) Mn3+ is stronger oxidizing agent as the charge from Mn3+ to Mn2+ results in half filled d5 configuration which has extra stability.
(iv) Europium (Eu2+) is formed by losing the two 5s electrons and its electronic configuration becomes [Xe]4f7 which is quite stable configuration.
(v) MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
2014 · 5 marks
(a) Complete the following equations:
(i) Cr2O72− + 2OH− →
(ii) MnO4− + 4H+ + 3e− →
(b) Account for the following:
(i) Zn is not considered as a transition element.
(ii) Transition metals form a larger number of complexes.
(iii) The E value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple.
Answer(a)
(i) Cr2O72− + 2OH− → 2CrO42− + H2O
(ii) MnO4− + 4H+ + 3e− → MnO2 + 2H2O
(b)
(i) Zn has completely filled d-orbitals (3d10) in both ground state and its common +2 oxidation state. A transition metal must have incomplete d subshell.
(ii) Transition metals form many complexes because of variable oxidation states, small highly charged ions, and availability of vacant d-orbitals for bonding.
(iii) Mn2+ has stable half-filled d5 configuration, so Mn3+ → Mn2+ is highly favourable. Cr3+ (d3, t2g3) is more stable than Cr2+.
2014 · 5 marks
(i) With reference to structural variability and chemical reactivity, write the difference between lanthanoids and actinoids.
(ii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
(iii) Complete the following equation: MnO4− + 8H+ + 5e− →
(iv) Out of Mn3+ and Cr3+, which is more paramagnetic and why? (Mn = 25, Cr = 24)
Answer(i)(a) Actinoids are radioactive; lanthanoids are not.
(b) Ionisation enthalpies of early actinoids are lower. Actinoids show oxidation states from +3 to +7; lanthanoids show +3 (most common) up to +4.
(c) Actinoids are more reactive and have more complex magnetic properties.
(ii) Cerium (Ce) is known to exhibit +4 oxidation state.
(iii) MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
(iv) Mn2+ is more paramagnetic because it has more number of unpaired electrons (5) than Cr3+ (3).
2016 · 5 marks
(a) Account for the following:
(i) Ozone is thermodynamically unstable.
(ii) Solid PCl5 is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
(b) Draw the structure of
(i) BrF3
(ii) XeF4.
OR
(i) Compare the oxidising action of F2 and Cl2 by considering bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
(ii) Write conditions to maximize yield of H2SO4 by contact process.
(iii) Arrange in increasing order:
(a) H3PO3, H3PO4, H3PO2 (Reducing character).
(b) NH3, PH3, AsH3, SbH3, BiH3 (Base strength).
Answer(a)
(i) Ozone decomposes to oxygen as delta-G is negative for decomposition.
(ii) Solid PCl5 exists as [PCl4]+[PCl6]- (tetrahedral cation and octahedral anion).
(iii) Due to high electronegativity and small size of fluorine. (b)
(i) BrF3: T-shaped (sp3d, 2 lone pairs).
(ii) XeF4: Square planar (sp3d2, 2 lone pairs).
OR
(i) F2 is stronger oxidizing agent due to low bond dissociation enthalpy and very high hydration enthalpy of F-.
(ii) Low temperature (~720 K), high pressure, V2O5 catalyst, excess O2. (iii)(a) H3PO4 < H3PO3 < H3PO2.
(b) BiH3 < SbH3 < AsH3 < PH3 < NH3.
2016 · 5 marks
(a) Account for:
(i) Mn shows highest oxidation state +7 with oxygen but +4 with fluorine.
(ii) Cr2+ is a strong reducing agent.
(iii) Cu2+ salts are coloured while Zn2+ salts are white.
(b) Complete:
(i) 2MnO2 + 4KOH + O2 -> ?
(ii) Cr2O72- + 14H+ + 6I- -> ?
OR The elements of 3d-transition series: Sc Ti V Cr Mn Fe Co Ni Cu Zn. Answer:
(i) Which shows maximum oxidation states? Why?
(ii) Which has highest m.p.?
(iii) Which shows only +3 oxidation state?
(iv) Which is strong oxidizing agent in +3 state and why?
Answer(a)
(i) Oxygen forms multiple bonds (p-pi-d-pi) with Mn; fluorine cannot.
(ii) Cr2+ -> Cr3+ gives stable d3/t2g3 configuration; more negative electrode potential.
(iii) Cu2+ (3d9) has unpaired electrons for d-d transitions giving colour; Zn2+ (3d10) has no unpaired electrons, hence white. (b)
(i) 2K2MnO4 + 2H2O.
(ii) 2Cr3+ + 3I2 + 7H2O.
OR
(i) Mn - maximum unpaired electrons, can achieve half-filled/completely-filled configurations.
(ii) Cr - maximum unpaired electrons (6), strongest metallic bonds.
(iii) Sc - too few electrons, loses one 3d + two 4s electrons.
(iv) Co - highest E value for M3+/M2+ redox couple.
2017 · 5 marks
(a) Account for the following:
(i) Transition metals form large number of complex compounds.
(ii) The lowest oxide of transition metal is basic whereas the highest oxide is amphoteric or acidic.
(iii) E° value for the Mn3+/Mn2+ couple is highly positive (+1.57 V) as compared to Cr3+/Cr2+.
(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements.
OR (a)
(i) How is the variability in oxidation states of transition metals different from that of the p-block elements?
(ii) Out of Cu+ and Cu2+, which ion is unstable in aqueous solution and why?
(iii) Orange colour of Cr2O72- ion changes to yellow when treated with an alkali. Why?
(b) Chemistry of actinoids is complicated as compared to lanthanoids. Give two reasons.
Answer(a)
(i) Due to small size, high ionic charge and availability of d-orbitals.
(ii) In lowest oxides, metal has low oxidation state with electrons available for donation (basic). In highest oxides, high effective nuclear charge makes them acidic.
(iii) Mn2+ has d5 (half-filled, stable) configuration. Mn3+ is d4 which readily converts to d5. Cr3+ is d3 (half-filled t2g) and stable.
(b) Similarity: Both show +3 oxidation state. Difference: Actinoids are radioactive, show wider range of oxidation states; lanthanoids are not radioactive.
OR (a)
(i) Transition elements differ by one unit in oxidation state; p-block elements differ by two units (inert pair effect).
(ii) Cu+ is unstable. Cu2+ is more stable due to higher hydration energy compensating for ionization.
(iii) Cr2O72- + 2OH- → 2CrO42- + H2O. Chromate ion (CrO42-) is yellow.
(b) Actinoids show many oxidation states; actinoids are radioactive.
2017 · 5 marks
(a) Account for the following:
(i) Transition metals show variable oxidation states.
(ii) Zn, Cd and Hg are soft metals.
(iii) E° value for the Mn3+/Mn2+ couple is highly positive (+1.57 V) as compared to Cr3+/Cr2+.
(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements.
OR
(a) Following are the transition metal ions of 3d series: Ti4+, V2+, Mn3+, Cr3+ (Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24). Answer:
(i) Which ion is most stable in an aqueous solution and why?
(ii) Which ion is a strong oxidising agent and why?
(iii) Which ion is colourless and why?
(b) Complete the following equations:
(i) 2MnO4- + 16H+ + 5S2- →
(ii) KMnO4 --heat--→
Answer(a)
(i) Due to small energy difference between ns and (n-1)d orbitals.
(ii) Completely filled d-orbitals / absence of unpaired d electrons cause weak metallic bonding.
(iii) Mn2+ has d5 stable configuration while Cr3+ is d3 (half-filled t2g), stable. Conversion from d4 to d5 has high E°.
(b) Similarity: Both show +3 oxidation state. Difference: Actinoids are radioactive; lanthanoids are not.
OR (a)
(i) Cr3+ (half-filled t2g3, stable).
(ii) Mn3+ (d4 to d5 has highest E°(M3+/M2+)).
(iii) Ti4+ (no d-electrons, no d-d transition).
(b)
(i) 2MnO4- + 16H+ + 5S2- → 5S + 2Mn2+ + 8H2O
(ii) 2KMnO4 → K2MnO4 + MnO2 + O2
2019 · 5 marks
(a) Give reasons:
(i) Sulphur in vapour state shows paramagnetic behaviour.
(ii) N-N bond is weaker than P-P bond.
(iii) Ozone is thermodynamically less stable than oxygen.
(b) Write the name of the gas released when Cu is added to
(i) Dilute HNO3
(ii) Conc. HNO3.
Answer(a)
(i) S2 molecules in vapour state have two unpaired electrons in antibonding molecular orbitals.
(ii) N-N single bond is weaker due to high interelectronic repulsion of lone pairs on small nitrogen atoms.
(iii) Ozone has higher energy and tendency to decompose to more stable O2. (b)
(i) Dilute HNO3: NO (nitric oxide).
(ii) Conc. HNO3: NO2 (nitrogen dioxide).
2019 · 5 marks
(a)
(i) Write the disproportionation reaction of H3PO3.
(ii) Draw the structure of XeF4.
(b) Account for:
(i) Although fluorine has less negative electron gain enthalpy yet F2 is a strong oxidizing agent.
(ii) Acidic character decreases from N2O3 to Bi2O3 in group 15.
(c) Write a chemical reaction to test sulphur dioxide gas.
Answer(a)
(i) 4H3PO3 -> 3H3PO4 + PH3.
(ii) XeF4 has square planar geometry. (b)
(i) F2 is strong oxidizing agent due to low F-F bond dissociation energy, high hydration enthalpy of F-.
(ii) Metallic character increases down group 15; N2O3 is acidic (non-metallic) while Bi2O3 is basic (metallic).
(c) SO2 turns acidified K2Cr2O7 from orange to green: K2Cr2O7 + 3SO2 + H2SO4 -> K2SO4 + Cr2(SO4)3 + H2O.
2019 · 5 marks
(a) Account for:
(i) Tendency to show -3 oxidation state decreases from N to Bi in group 15.
(ii) Acidic character increases from H2O to H2Te.
(iii) F2 is more reactive than ClF3, whereas ClF3 is more reactive than Cl2.
(b) Draw structure of
(i) XeF2,
(ii) H4P2O7.
Answer(a)
(i) Size increases and ionization enthalpy decreases down group, reducing tendency to gain 3 electrons.
(ii) As size of central atom increases, H-X bond weakens, easier to break.
(iii) F2 has weak F-F bond; ClF3 has more reactive Cl-F bonds than Cl-Cl. (b)
(i) XeF2: Linear geometry (sp3d, 3 lone pairs).
(ii) H4P2O7: Two PO4 tetrahedra sharing one oxygen.
2019 · 5 marks
(a) Give one example of anomalous reaction of fluorine.
(b) Structural difference between white and red phosphorus?
(c) What happens when XeF6 reacts with NaF?
(d) Why is H2S a better reducing agent than H2O?
(e) Arrange in increasing acidic character: HBr, HI, HCl, HF.
(b) White P has discrete P4 tetrahedral molecules; red P has polymeric chain.
(c) XeF6 + NaF -> Na[XeF7].
(d) H-S bond is weaker than H-O bond, easier to donate electrons.
(e) HF < HCl < HBr < HI.
2020 · 5 marks
(a) Give reasons:
(i) Transition metals and their compounds show catalytic activities.
(ii) Separation of a mixture of Lanthanoid elements is difficult.
(iii) Zn, Cd and Hg are soft and have low melting point.
(b) Write the preparation of the following:
(i) Na2Cr2O7 from Na2CrO4
(ii) K2MnO4 from MnO2.
OR
(a) Account for the following:
(i) Ti3+ is coloured whereas Sc3+ is colourless in aqueous solution.
(ii) Cr2+ is a strong reducing agent.
(b) Write two similarities between chemistry of lanthanoids and actinoids.
(c) Complete the ionic equation: 3MnO4(2-) + 4H+ ->
Answer(a)
(i) Catalytic activities of transition metals is due to ability to adopt variable oxidation states and to form complexes, providing large surface area for reactants.
(ii) Separation of lanthanoid elements is difficult because all have almost similar physical and chemical properties due to lanthanoid contraction.
(iii) Zn, Cd and Mg are soft and have low melting point because no d-orbitals are available for metallic bond formation and bonds are very weak. (b)
(i) 2Na2CrO4 + H2SO4 -> Na2Cr2O7 + Na2SO4 + H2O.
(ii) 2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O.
OR (a)
(i) Ti3+ has incomplete d (3d1) orbital whereas Sc3+ has empty (3d0) d-orbital.
(ii) Cr2+ ion can lose electron to form Cr3+, so acts as strong reducing agent.
(b) Both show +3 oxidation state; Both are strong reducing agents.
(c) 3MnO4(2-) + 4H+ -> 2MnO4- + MnO2 + 2H2O.
2023 · 5 marks
(a) Why is chemistry of actinoids complicated as compared to lanthanoids?
(b) Complete the following reaction and justify that it is a disproportionation reaction:
3 MnO42- + 4H+ -> _____ + _____ + 2 H2O.
(c) The given graph shows the trends in melting points of transition metals (Ti, Cr, Mn, Mo, W shown on graph with atomic number on x-axis and M.P./103 K on y-axis). Explain the reason why Cr has highest melting point and manganese (Mn) a lower melting point.
Answer(a) Actinoids are present below the series of lanthanoid. They are radioactive in nature. To study them is more complicated as compared to lanthanoid. Second, they have variable oxidation state.
(b) 3MnO42- + 4H+ -> 2MnO4- + MnO2 + 2H2O. It is disproportionation reaction because oxidation take place in acidic medium and oxidation numbers are: MnO42- = 6, MnO4- = 7, MnO2 = 4.
(c) Cr = 24 [Ar] 3d5 4s1. Mn = 25 [Ar] 3d5 4s2. Cr has maximum number of unpaired electrons in d-orbital. So it can have maximum pairing and form bond. Number of unpaired electrons are six in Cr. But Mn has five unpaired electrons in d-orbital which is less than Cr. That's why boiling point of Cr is maximum in transition metal. Mn is half filled which have extra stability and less tendency to form bonds.
2023 · 5 marks
(a) A transition element X has electronic configuration |Ar| 4s2 3d3. Predict its likely oxidation states. (1+1+3=5)
(b) Complete the reaction mentioning all the products formed: 2KMnO4 --heat →
(c) Account for the following:
(i) In the 3d transition series, zinc has the lowest enthalpy of atomisation.
(ii) Cu+ ion is unstable in aqueous solution.
(iii) Actinoids show more number of oxidation states than lanthanoids.
Answer(a) The X-element is vanadium (V). Its electronic configuration is [Ar] 3d3 4s2. Likely oxidation states: +2, +3, +4, +5.
(b) 2KMnO4 --heat → K2MnO4 + MnO2 + O2
(c)
(i) Zn = 30 [Ar] 4s2 3d10. The d-orbital is fully filled in zinc (Zn). There is no unpaired electrons are available. Inter atomic bonding is weak because of which metallic bonding is also weak.
(ii) Cu = [Ar] 3d9 4s2 = 3d10 4s1. In aqueous medium it exist as Cu+ loosing one e- but because of high hydration enthalpy it readily converted into Cu2+. It also shows disproportionation reaction: 2Cu2+ -> Cu2+ + Cu
(iii) Actinoids are radioactive in nature. They have almost similar enthalpy of actinides. Therefore, they show more actinoids contraction than lanthanoids contraction. They have more poor shielding of 5f orbitals than 4f orbitals.
2023 · 5 marks
(a) Write the number of unpaired electrons in Cr3+. (Atomic number of Cr = 24) (1+2+2=5)
(b) Complete the reaction mentioning all the products formed: Cr2O72- + 3H2S + 8H+ ->
(c) Account for the following:
(i) Mn2+ is more stable than Fe2+ towards oxidation to +3 state.
(ii) Copper has exceptionally positive E(M2+/M) value.
(iii) Eu2+ with electronic configuration [Xe] 4f7 6s2 is a strong reducing agent.
Answer(a) Cr = 25, [Ar] 3d4 4s2. Cr3+ = [Ar] 3d4 (Note: Cr = 24, so Cr3+ = [Ar] 3d3). There are three unpaired electrons present in Cr3+. (Note: The book says four but with Cr atomic number 24, Cr3+ has 3 unpaired electrons.)
(i) Electronic configuration: Mn = 25, [Ar] 3d5 4s2. Mn2+ = [Ar] 3d5 (half-filled, stable). Fe = 26, [Ar] 3d6 4s2. Fe2+ = [Ar] 3d6 -> Fe3+ = [Ar] 3d5 (donating 3e-). Mn = after losing 2 electrons, 3d orbital is half filled and Mn2+ is stable. Fe = after losing 2 electrons, Fe3+ is stable at +3 oxidation state and Mn2+ is stable in +2 oxidation state.
(ii) E0 value of copper is +ve i.e. 0.34 V. This is due to presence of high enthalpy of atomization and low enthalpy of hydrogen which make it exceptionally positive.
(iii) Electronic configuration of Eu2+: [Xe] 4f7 6s2. It has the tendency to lose two electrons and attain a stable half filled configuration. So, it is oxidized by losing 2 electrons and reduces other species.
2023 · 5 marks
(a) (I) Account for the following:
(i) E° value of Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+.
(ii) Sc3+ is colourless whereas Ti3+ is coloured in an aqueous solution.
(iii) Actinoids show wide range of oxidation states.
(II) Write the chemical equations for the preparation of KMnO4 from MnO2.
Answer(a) (I) (i) Because Mn3+ has the outer electronic configuration of 3d4 & Mn2+ has the outer electron configuration of 3d5. Thus the conversion is favourable. However in case of Cr3+/Cr2+ undergoes a change in outer configuration from 3d3 to 3d4 which is not stable.
(ii) Due to absence of unpaired electron Sc3+ is colourless and due to presence of one unpaired electron d-d transition takes place making Ti3+ coloured in nature.
(iii) It is due to comparable energies of 5f, 6d and 7s orbitals.
(II) Write one similarity and one difference between chemistry of Lanthanoids and Actinoids. (III) Complete the following ionic equation: Cr2O72- + 2OH- -> (OR)
Answer(b) (I) (i) The atomic radii of the transition elements in any series are not much different from each other. As a result they can very easily replace each other in the lattice and form alloys.
(ii) Ce4+ has the tendency to accept one electron to get the +3 oxidation state, hence Ce4+ is a good oxidising agent.
(II) In case of lanthanoids, differentiating electron enters in 4f orbital whereas in case of Actinoids it enters in 5f orbital. They both have (+3) as their most common oxidation state. (III) Cr2O72- + 2OH- -> 2CrO42- + H2O
2024 · 5 marks
Attempt any five of the following:
(a) Ce(III) is easily oxidised to Ce(IV). Comment.
(b) E°(Mn2+/Mn) is -1.18 V. Why is this value highly negative in comparison to neighbouring d block elements?
(c) Which element of 3d series has lowest enthalpy of atomisation and why?
(d) What happens when sodium chromate is acidified?
(e) Zn, Cd and Hg are soft metals. Why?
(f) Why is permanganate titration not carried out in the presence of HCl? (g) The lower oxides of transition metals are basic whereas the highest are amphoteric/acidic. Give reason.
Answer(a) Ce has configuration 4f1 5d1 6s2. Ce(III) has 4f1 5d0 6s0, it can easily lose an electron to acquire the stable configuration of 4f0 5d0 6s0 (noble gas Xe configuration), forming Ce(IV).
(b) Mn has half-filled 3d5 configuration. Negative E° values of Mn2+/Mn is because of the stabilities of half-filled (3d5: Mn2+) configuration.
(c) Zn has no unpaired electrons. The greater the number of unpaired electrons, the stronger the resultant bonding and hence higher the enthalpy of atomisation. Zn exhibits weak interactions and thus possesses least enthalpy of atomisation.
(d) At acidic pH, sodium chromate converts to sodium dichromate: 2Na2CrO4 + 2HCl -> Na2Cr2O7 + 2NaCl + H2O
(e) Zinc, cadmium and mercury have a fully filled d-subshell, due to which metallic bonds are weak. That's why they are soft and have low melting point.
(f) Permanganate titration is not carried out in presence of hydrochloric acid because some of the hydrochloric acid gets oxidised to chlorine gas. Hence we do not get the correct endpoint. (g) In low oxidation state, valence electrons of the metal atom are not involved in bonding. Hence it can donate electrons and behave as a base. In higher oxidation state, valence electrons are involved in bonding and are not available; effective nuclear charge is high and hence it can accept electrons and behave as an acid.
2024 · 5 marks
Attempt any five of the following:
(a) Why zinc is not regarded as a transition element?
(b) What is Lanthanoid contraction?
(c) Why is first ionization enthalpy of chromium lower than that of Zn?
(d) Why are transition elements good catalysts?
(e) Compounds of transition metals are generally coloured. Give reason.
(f) Out of KMnO4 and K2MnO4, which one is paramagnetic and why? (g) Complete the following ionic equation: Cr2O72- + 14H+ + 6e- ->
Answer(a) Zinc is not considered a transition element because it does not have partially filled d or f orbitals in its electronic configuration. Its configuration is [Ar]3d10 4s2, where the d orbitals are completely filled.
(b) The lanthanoid contraction refers to the gradual decrease in the size of the atoms as you move across the lanthanoid series. This contraction occurs because of poor shielding of the added electrons in the 4f subshell.
(c) Chromium has electron configuration [Ar] 3d5 4s1. Removing an electron from the 4s orbital is easier than removing an electron from the half-filled 3d orbital due to increased stability associated with half-filled orbitals. Therefore, first ionization energy of chromium is lower than that of zinc.
(d) Transition elements have high surface area, variable oxidation states, formation of complex compounds, adsorption of reactant molecules onto their surface, formation of intermediates makes them good catalysts.
(e) Due to d-d transition of electrons. When transition metal ions absorb light, electrons in d-orbitals can be excited to higher energy levels. The energy absorbed corresponds to specific wavelengths of light, causing the compound to appear colored.
(f) Potassium manganate (K2MnO4) is paramagnetic because it has unpaired electrons in the d orbitals of the manganese (Mn) atom. In K2MnO4, manganese is in its +6 oxidation state, and it has one unpaired electron in its d-orbitals. (g) Cr2O72- + 14H+ + 6e- -> 2Cr3+ + 7H2O
2013 · 5 marks
(a) Give reasons for the following:
(i) Bond enthalpy of F2 is lower than that of Cl2.
(ii) PH3 has lower boiling point than NH3.
(b) Draw the structures of the following molecules:
(i) BrF3
(ii) (HPO3)3
(iii) XeF4
Answer(a)
(i) F2 has lower bond enthalpy because of small size of F atoms, there is high inter-electronic repulsion between lone pairs.
(ii) NH3 has hydrogen bonding due to high electronegativity of N, so higher boiling point. PH3 does not have H-bonding.
(ii) Fluorine does not exhibit positive oxidation state.
(iii) Oxygen shows catenation behaviour less than sulphur.
(b) Draw the structures of the following molecules:
(i) XeF2
(ii) H2S2O8
Answer(a)
(i) Helium is used in diving apparatus because it is very slightly soluble in blood, preventing nitrogen narcosis (bends).
(ii) Fluorine does not exhibit positive oxidation state because it is the most electronegative element and has no d-orbitals available.
(iii) Oxygen shows less catenation than sulphur because O–O bond is weaker than S–S bond due to high inter-electronic repulsion in small oxygen atoms.
(b)
(i) XeF2: Linear structure (sp3d hybridization)
(ii) H2S2O8: Peroxodisulphuric acid structure with O–O peroxide linkage
2015 · 5 marks
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(b) Draw the structures of the following:
(i) ClF3
(ii) XeF4
Answer(a)
(i) Bond dissociation enthalpy decreases from HF to HI (as size of halogen increases), making it easier to release H+, so acidic character increases from HF to HI.
(ii) Oxygen exists as diatomic O2 molecules with weak van der Waals' forces (low mp/bp). Sulphur exists as S8 rings with much stronger intermolecular forces (high mp/bp).
(iii) Nitrogen does not have d-orbitals to expand its octet beyond 4 bonds, so it cannot form pentahalide.
(ii) XeF4: Square planar (sp3d2 hybridization, 2 lone pairs)
2015 · 5 marks
(i) Which allotrope of phosphorus is more reactive and why?
(ii) How the supersonic jet aeroplanes are responsible for the depletion of ozone layers?
(iii) F2 has lower bond dissociation enthalpy than Cl2. Why?
Answer(i) White phosphorus is more reactive because it is a discrete P4 tetrahedron with angular strain (60° bond angle). It is unstable and highly reactive.
(ii) Supersonic jet aeroplanes release NO (nitric oxide) in the stratosphere which acts as a catalyst for ozone decomposition: NO + O3 → NO2 + O2; NO2 + O → NO + O2
(iii) F2 has lower bond dissociation enthalpy than Cl2 because F atoms are very small, leading to high lone pair-lone pair repulsion between the two F atoms, weakening the F–F bond.
2018 · 5 marks
(a) Give reasons:
(i) H3PO3 undergoes disproportionation reaction but H3PO4 does not.
(ii) When Cl2 reacts with excess of F2, ClF3 is formed and not FCl3.
(iii) Dioxygen is a gas while sulphur is a solid at room temperature.
(b) Draw the structures of the following:
(i) XeF4
(ii) HClO3
OR
(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube, a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).
(i) Identify (A) and (B).
(ii) Write the structures of (A) and (B).
(iii) Why does gas (A) change to solid on cooling?
(b) Arrange the following in the decreasing order of their reducing character: HF, H, HBr, HI
(c) Complete the following reaction: XeF4 + SbF5 →
Answer(a)
(i) H3PO3 has P-H bond which can be oxidized; H3PO4 has no P-H bond.
(ii) Fluorine is more electronegative and smaller; it cannot be central atom. Cl has vacant d-orbitals to expand octet.
(iii) O2 is diatomic with weak van der Waals forces; S8 has larger molecular mass with stronger forces.
(b)
(i) XeF4: Square planar. (ii) HClO3: Cl bonded to two =O, one -OH and one lone pair.
OR (a)
(i) A = NO2 (brown gas), B = N2O4 (colourless solid)
(ii) NO2 is bent molecule; N2O4 is two NO2 linked.
(iii) 2NO2 ⇌ N2O4. On cooling, equilibrium shifts forward (dimerization is exothermic).