Read and download the CBSE Class 12 Chemistry The d and f Block Elements Assignment Set A for the 2025-26 academic session. We have provided comprehensive Class 12 Chemistry school assignments that have important solved questions and answers for Unit 4 The d- and f-Block Elements. These resources have been carefuly prepared by expert teachers as per the latest NCERT, CBSE, and KVS syllabus guidelines.
Solved Assignment for Class 12 Chemistry Unit 4 The d- and f-Block Elements
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Unit 4 The d- and f-Block Elements Class 12 Solved Questions and Answers
Q.1-Explain briefly how +2 oxidation states becomes more and more stable in the first half of the first row transition elements with increasing atomic number.
Ans.1-In M2+ ions, 3d-orbitals get occupied gradually as the atomic number increases. Since, the number of empty d-orbitals decreases, the stability of cations increases from Sc2+ to Mn2+ .Mn2+ is most stable as all d-orbitals are singly occupied.
Q.2- Explain why transition elements have many irregularities in their electronic configurations?
Ans.2-In the transition elements, there is a little difference in the energy of (n-1) d-orbitals and nsorbitals. Thus, incoming electron can occupy either of shell. Hence, transition elements exhibit many irregularities in their electronic configurations.
Q.3-What are different oxidation states exhibited by Lanthanides?
Ans.3-The common stable oxidation state of lanthanides is +3.However some members also show oxidation states of +2 & +4.
Q.4-How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Ans.4-The transition elements use its (n-1)d, ns and np orbital and the successive oxidation states differ by unity. For example, Mn shows all the oxidation states from +2 to +7. On other hand non transition elements use its ns, np and nd orbitals and the successive oxidation states differ by two units e.g. Sn2+, Sn4+ etc.
Q.5- Why do transition elements show variable oxidation states?
Ans.5- The transition elements show variable oxidation state due to small energy difference between (n-1) d &ns orbital as a result both (n-1)d &ns electrons take part in bond formation.
Q.6-Why are Mn2+ compounds more stable than Fe2+ compounds towards oxidation to +3 state?
Ans.6-The electronic configuration of Mn2+ is [Ar] 3d5, i.e. all five d-orbitals are singly occupied. Thus this is stable electronic configuration and further loss of electron requires high energy .on other hand side the electronic configuration of Fe2+is [Ar] 3d6, i.e. Loss of one electron requires low energy.
Q.7-To what extend do the electronic configuration decide the stability of oxidation state in the first series of the transition elements? Illustrate your answer with an example.
Ans.7-In a transition series, the oxidation state which lead to exactly half filled or completely filled orbitals are more stable.e.g. the electronic configuration of Feis [Ar] 3d6 ,4s2. It shows various oxidation state but Fe(III) is more stable than Fe(II).
Q.8- Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Ans.8- Copper with configuration [Ar] 3d10 4s1 exhibits +1 oxidation state. Copper loses 4s1 electron easily and achieved a stable configuration 3d10 by forming Cu+.
Q.9- What are inner transition elements?
Ans.9- The f-block elements in which the last electron accommodated on (n-2) f-subshell are called inner transition elements. These include atomic numbers 58 to 71 and from 90 to 103.
Q.10- The paramagnetic character in 3d-transition series elements increases upto Mn and then decreases. Explain why?
Ans.10- In the 3d-transition series as we move from Sc (21) to Mn (25) the number of unpaired electrons increases and hence paramagnetic character increases. After Mn, the pairing of electrons in the d-orbital starts and the number of unpaired electrons decreases and hence, paramagnetic character decreases.
Q.11- Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition metal
Ans.11-The following points justify that the given statement is true:-
Question. Which of the following oxidation states is the most common among the lanthanoids?
(a) 4
(b) 2
(c) 5
(d) 3
Answer D
Question. Identify the incorrect statement among the following :
(a) Lanthanoid contraction is the accumulation of successive shrinkages.
(b) As a result of lanthanoid contraction, the properties of 4d series of the transition elements have no similarities with the 5d series of elements.
(c) Shielding power of 4f electrons is quite weak.
(d) There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu.
Answer B
Question. Lanthanoids are
(a) 14 elements in the sixth period (atomic no. 90 to 103) that are filling 4f sublevel
(b) 14 elements in the seventh period (atomic number = 90 to 103) that are filling 5f sublevel
(c) 14 elements in the sixth period (atomic number = 58 to 71) that are filling the 4f sublevel
(d) 14 elements in the seventh period (atomic number = 58 to 71) that are filling 4f sublevel.
Answer C
Question. The correct order of ionic radii of Y3+, La3+, Eu3+ and
Lu3+ is (At. nos. Y = 39, La = 57, Eu = 63, Lu = 71)
(a) Y3+ < La3+ < Eu3+ < Lu3+
(b) Y3+ < Lu3+ < Eu3+ < La3+
(c) Lu3+ < Eu3+ < La3+ < Y3+
(d) La3+ < Eu3+ < Lu3+ < Y3+
Answer B
Question. Bell metal is an alloy of
(a) Cu + Zn
(b) Cu + Sn
(c) Cu + Pb
(d) Cu + Ni
Question. In which of the following compounds transition metal has zero oxidation state?
(a) NOClO4
(b) NH2NH2
(c) CrO5
(d) [Fe(CO)5]
Question. Which one of the following ionic species will impart colour to an aqueous solution?
(a) Zn2+
(b) Cu+
(c) Ti4+
(d) Cr3+
Question. A transition element X has a configuration [Ar]3d4 in its +3 oxidation state. Its atomic number is
(a) 22
(b) 19
(c) 25
(d) 26
Question. The mercury is the only metal which is liquid at 0°C.
This is due to its
(a) high vapour pressure
(b) weak metallic bond
(c) high ionization energy
(d) both (b) and (c).
Question. The manganate and permanganate ions are tetrahedral, due to
(a) the p-bonding involves overlap of d-orbitals of oxygen with d-orbitals of manganese
(b) the p-bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese
(c) there is no p-bonding
(d) the p-bonding involves overlap of p-orbitals of oxygen with p-orbitals of manganese.
Question. When neutral or faintly alkaline KMnO4 is treated with potassium iodide, iodide ion is converted into ‘X’. ‘X’ is
(a) I2
(b) IO4–
(c) IO3–
(d) IO–
Question. Which one of the following ions exhibits d-d transition and paramagnetism as well?
(a) CrO42–
(b) Cr2O72–
(c) MnO4–
(d) MnO42–
Question. Name the gas that can readily decolourise acidified KMnO4 solution.
(a) SO2
(b) NO2
(c) P2O5
(d) CO2
Question. Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 solution?
(a) SO2 is reduced.
(b) Green Cr2(SO4)3 is formed.
(c) The solution turns blue.
(d) The solution is decolourised.
Question. Assuming complete ionisation, same moles of which of the following compounds will require the least amount of acidified KMnO4 for complete oxidation?
(a) FeSO3
(b) FeC2O4
(c) Fe(NO2)2
(d) FeSO4
Question. The reaction of aqueous KMnO4 with H2O2 in acidic conditions gives
(a) Mn4+ and O2
(b) Mn2+ and O2
(c) Mn2+ and O3
(d) Mn4+ and MnO2.
Question. Acidified K2Cr2O7 solution turns green when Na2SO3 is added to it. This is due to the formation of
(a) Cr2(SO4)3
(b) CrO42–
(c) Cr2(SO3)3
(d) CrSO4
Question. The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is
(a) one
(b) two
(c) five
(d) one fifth.
Question. The oxidation state of Cr in K2Cr2O7 is
(a) +5
(b) +3
(c) +6
(d) +7
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CBSE Class 12 Chemistry Unit 4 The d- and f-Block Elements Assignment
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