CBSE Class 12th Chemistry Examination 2020 will be held on 7th March 2020. For the ease of students, we have provided important questions and answers for chapter d & f block elements. The students appearing in the CBSE Class 12th Examination 2020 can go through the below- mentioned questions.
Key Points to be mentioned while writing the answers to the below mentioned important questions:
Question 1- Explain:
(i) Generally, there is an increase in the density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements.
(ii) Transition elements and their compounds are generally found to be good catalysts in chemical reactions. Also, they generally form coloured compounds.
(iii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series.
Answer: (i) From titanium (Z = 22) to copper (Z = 29) in the first series of transition elements, the atomic size of elements decreases and mass increases. Therefore, there is an increase in the density of these elements.
(ii) Transition elements and their compounds are generally found to be good catalysts in chemical reactions because of the presence of unpaired electrons in their incomplete d- orbitals and variable oxidation states. Transition elements generally form coloured compounds as they undergo d-d transition by absorption of energy from visible regions and then the emitted light shows complementary colours.
(iii) It is because they have their electrons of the outermost shell at a greater distance from the nucleus, in comparison to atoms of 3d transition metals.
Question 2- Account for the following:
(i) Copper (I) ion is not known in aqueous solution.
(ii) Actinoids exhibit a greater range of oxidation states than lanthanoids.
(iii) Manganese exhibits the highest oxidation state of +7 among the 3d series of transition elements.
(iv) The enthalpies of atomization of transition metals are quite high.
Answer: i) Cu2+(aq) is much more stable than Cu+(aq) because second ionization enthalpy of copper is large but Δhyd (hydration enthalpy) for Cu2+(aq) is much more negative than that for Cu2+(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds undergo disproportionation as follows :
2Cu+ → Cu2++ Cu
(ii) Actinoids exhibit a greater range of oxidation states than lanthanides as there is a very small energy gap between 5f, 6d and 7s subshells. Therefore, all the electrons can take part in bonding and shows variable oxidation states.
(iii) All the oxidation states are exhibited from +2 to +7 by Mn and no other element of this series shows this highest state of oxidation.
(iv) The enthalpies of atomization of transition metals are quite high due to a large number of unpaired electrons in their atoms. They have stronger interatomic interaction.
Question 3- Complete these chemical equations :
(i) MnO4 (aq) + S2O32- (aq) + H2O (l) →
(ii) Cr2O72- (aq) + Fe2+ (aq) + H+ (aq) →
Answer: (i) 8MnO4- (aq) + 3S2O32- (aq) + H2O (l) → 8MnO2 + 6S2O42- + 2OH-
(ii) Cr2O72- (aq) + 6Fe2+ (aq) + 14H+ (aq) → 2Cr3+ + 6Fe3+ + 7H2O
Question 4- (a) Which metal in the first transition series (3d series) exhibits +1 oxidation state most frequency 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 exhibits + 1 oxidation state more frequently because of its electronic configuration 3d104s1. It can easily lose 4s1 electrons to give a stable 3d10 configuration.
(b) SC3+ = 4S0 3d3+ = no unpaired electron
V3+ = 3d2 4s0 = 2 unpaired electron
Ti4+ = 3d0 4s0 = no unpaired electron
Mn2+ = 3d5 4s0 = 5 unpaired electron
Thus V3+ and Mn2+ are coloured in their aqueous solution due to the presence of unpaired electrons.
Question 5- What is Lanthanoid contraction? State its consequences.
Answer: The overall decrease in atomic and ionic radii with an increasing atomic number is known as lanthanoid contraction. From La+3 to Lu+3 in lanthanoid series, the size of ion decreases and this decrease is known as lanthanoid contraction. It arises due to imperfect shielding of one 4f electron by another present in the same subshell.
Consequences of lanthanoid contraction are :
(i) Similarity: Due to lanthanoid contraction, the size of elements which follow (Hf – Hg) is almost similar to the size of the elements of previous row (Zr – Cd) and are difficult to separate.
(ii) Basicity difference: Due to lanthanoid contraction, the size decreases from La+3 to Lu+3. Thus, the covalent character increases. Hence, the basic character of hydroxides also decreases. Thus, La(OH)3 is most basic while Lu(OH)3 is the least basic.
Question 6- Give reasons:
(i) The highest oxidation state of a transition metal is usually exhibited in its oxide.
(ii) Co2+ is easily oxidised in the presence of a strong ligand.
(iii) Actinoids show a wide range of oxidation states.
(iv) The silver atom has completely filled d-orbitals (4d10) in its ground state, yet it is regarded as a transition element.
(v) E° value for Mn3+ /Mn2+ couple is much more positive than Cr3+/Cr2+.
Answer: (i) The highest oxidation state of a transition metal is usually exhibited in its oxide because of high electronegativity, smaller size and lower ionization energy.
(ii) Co2+ is easily oxidised in the presence of a strong ligand as it has higher crystal field energy. This energy pairs the electrons to give inner orbital complexes (d2sp3).
(iii) Actinoids show a wide range of oxidation states as they have comparable energies of 5f, 6d and 7s orbitals.
(iv) The silver atom has completely filled d-orbitals (4d10) in its ground state, yet it is regarded as a transition element as it has incomplete d-orbital (4d9) in its +2 oxidation state.
(v) The large positive E° value for Mn3+ /Mn2+ indicates that Mn2+ is much more stable than Mn3+ due to stable half-filled configuration (3d5). Therefore, the 3rd ionisation energy of Mn will be very high and Mn3+ is unstable and can be easily reduced to Mn2+. E° value for Cr3+ / Cr2+ is positive but small. It means that Cr3+ can also be reduced to Cr2+ but less easily.
Question 7- What is disproportionation? Give an example of a disproportionation reaction in aqueous solution.
Answer: In disproportionation reactions, elements undergo self-oxidation and self-reduction states forming two different compounds. In other words, when a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation.
Example: 3MnO42- + 4H+ → 2MnO4- + MnO2 + 2H2O
Question 8- How is potassium permanganate made? How does the acidified permanganate solution react with oxalic acid? Write the equations for the reactions.
Answer: Potassium Permanganate (KMnO4) is prepared from pyrolusite ore (MnO2) which is fused with an alkali metal hydroxide like KOH in the presence of air or an oxidising agent like KNO3 to give dark green potassium manganate (K2MnO4). K2MnO4 disproportionates in a neutral or acidic solution to give potassium permanganate.
Question 9- Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with
(i) iodine (ii) H2S
Answer: Potassium dichromate (K2Cr2O7) acts as a strong oxidising agent in acidic medium using H2SO4.
K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O + 3[O]
Ionic reactions are as follows:
(i) It oxidises iodide ion (I-) to iodine (I2)
(ii) It oxidises H2S to S
Question 10- (a) 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 KC1 forms an orange coloured crystalline compound (C).
(i) Write the formulae of the compounds (A), (B) and (C).
(ii) Write one use of the compound (C).
(b) In the 3d series (Sc = 21 to Zn = 30) :
(i) Which element shows the maximum number of oxidation states?
(ii) Which element shows only +3 oxidation state?
(iii) Which element has the lowest enthalpy of atomization?
Answer: The chromite ore FeCr2O4 on fusion with NaOH in presence of air, forms a yellow coloured compound (A) that is Sodium Chromate.
Sodium dichromate (B) on reaction with KCl forms orange coloured compound Potassium dichromate (C).
Na2Cr2O7 + 2KCl → 2NaCl + K2Cr2O7 (C)
(i) Thus (A) → Sodium chromate Na2CrO4
(A) → Sodium dichromate Na2Cr2O7
(B) → Potassium dichromate K2Cr2O7
(ii) (C) is used as a strong oxidising agent in acidic medium in volumetric analysis.
(b) (i) Mn
These were some of the important questions and answers from chapter d & f block elements. The students can go through important questions of other chapters with the help of links mentioned below: