The d and f-Block Elements

Multiple Choice Questions (MCQs)

1. Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d5. What is its atomic number?

(a) 25

(b) 26

(c) 27

(d) 24

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Answer

(b) Electronic configuration of X3+ is [Ar]3d5

It repersents the total number of es and oxidation state.

Therefore, atomic number of X=18+5+3=26

Hence, option (b) is correct.

  • Option (a) 25: If the atomic number of element X were 25, then in its +3 oxidation state, it would have lost 3 electrons. The electronic configuration of X would be [Ar]3d54s2. Removing 3 electrons would result in [Ar]3d4, not [Ar]3d5.

  • Option (c) 27: If the atomic number of element X were 27, then in its +3 oxidation state, it would have lost 3 electrons. The electronic configuration of X would be [Ar]3d74s2. Removing 3 electrons would result in [Ar]3d6, not [Ar]3d5.

  • Option (d) 24: If the atomic number of element X were 24, then in its +3 oxidation state, it would have lost 3 electrons. The electronic configuration of X would be [Ar]3d54s1. Removing 3 electrons would result in [Ar]3d3, not [Ar]3d5.

2. The electronic configuration of Cu(II) is 3d9 where as that of Cu(I) is 3d10. Which of the following is correct?

(a) Cu(II) is more stable

(b) Cu(II) is less stable

(c) Cu(I) and Cu(II) are equally stable

(d) Stability of Cu(I) and Cu(II) depends on nature of copper salts

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Answer

(a) Cu(II) is more stable than Cu(I). As it is known that, Cu(I) has 3d10 stable configuration while Cu(II) has 3d9 configuration. But Cu(II) is more stable due to greater effective nuclear charge of Cu(II) i.e., it hold 17 electrons instead of 18 in Cu(I).

  • (b) Cu(II) is less stable: This is incorrect because Cu(II) is actually more stable than Cu(I) due to the greater effective nuclear charge in Cu(II), which allows it to hold its electrons more tightly despite having one less electron than Cu(I).

  • (c) Cu(I) and Cu(II) are equally stable: This is incorrect because Cu(II) is more stable than Cu(I). The effective nuclear charge in Cu(II) makes it more stable despite the 3d9 configuration, compared to the 3d10 configuration of Cu(I).

  • (d) Stability of Cu(I) and Cu(II) depends on nature of copper salts: This is incorrect because the intrinsic stability of Cu(II) over Cu(I) is due to the effective nuclear charge and not dependent on the nature of the copper salts.

3. Metallic radii of some transition elements are given below. Which of these elements will have highest density?

Element Fe Co Ni Cu
Metallic radii/pm 126 125 125 128

(a) Fe

(b) Ni

(c) Co

(d) Cu

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Answer

(d) On moving left to right along period, metallic radius decreases while mass increases. Decrease in metallic radius coupled with increase in atomic mass results in increase in density of metal.

Hence, among the given four choices Cu belongs to right side of Periodic Table in transition metal, and it has highest density (89g/cm3).

  • (a) Fe: Iron (Fe) has a larger metallic radius (126 pm) compared to Co and Ni, and it is positioned to the left of Cu in the periodic table. This means it has a lower atomic mass and a larger atomic radius, resulting in a lower density compared to Cu.

  • (b) Ni: Nickel (Ni) has a smaller metallic radius (125 pm) compared to Cu, but it is positioned to the left of Cu in the periodic table. Although it has a smaller radius, its atomic mass is lower than that of Cu, leading to a lower density.

  • (c) Co: Cobalt (Co) has a metallic radius of 125 pm, similar to Ni, and is also positioned to the left of Cu in the periodic table. Like Ni, it has a lower atomic mass than Cu, resulting in a lower density.

4. Generally, transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?

(a) Ag2SO4

(b) CuF2

(c) ZnF2

(d) Cu2Cl2

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Answer

(b) Transition elements form coloured salt due to the presence of unpaired electrons. In CuF2,Cu(II) contain one unpaired electron hence, CuF2 is coloured in solid state.

  • (a) Ag2SO4: Silver (Ag) in Ag2SO4 is in the +1 oxidation state, which has a completely filled d^10 configuration. There are no unpaired electrons, so it does not exhibit color.

  • (c) ZnF2: Zinc (Zn) in ZnF2 is in the +2 oxidation state, which has a completely filled d^10 configuration. There are no unpaired electrons, so it does not exhibit color.

  • (d) Cu2Cl2: Copper (Cu) in Cu2Cl2 is in the +1 oxidation state, which has a completely filled d^10 configuration. There are no unpaired electrons, so it does not exhibit color.

5. On addition of small amount of KMnO4 to concentrated H2SO4, a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following.

(a) Mn2O7

(b) MnO2

(c) MnSO4

(d) Mn2O3

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Answer

(a) On addition of KMnO4 to concentrated H2SO4, a green oily compound Mn2O7 is obtained which is highly explosive in nature.

2KMnO4+2H2SO4 (Conc.) Mn2O7+2KHSO4+H2O

  • (b) MnO2: This compound is manganese dioxide, which is a solid and not a green oily liquid. It is also not highly explosive in nature.

  • (c) MnSO4: This compound is manganese(II) sulfate, which is a solid and typically forms a pink or pale red solution in water. It is not a green oily liquid and is not highly explosive.

  • (d) Mn2O3: This compound is manganese(III) oxide, which is a solid and not a green oily liquid. It is also not highly explosive in nature.

6. The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows highest magnetic moment.

(a) 3d7

(b) 3d5

(c) 3d8

(d) 3d2

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Thinking Process

This problem is based on calculation of magnetic moment can be done as Magnetic moment (μ)=n(n+2)BM.

Answer

(b) Greater the number of unpaired electron, higher will be its value of magnetic moment. Since, 3d5 has 5 unpaired electrons hence highest magnetic moment.

μ=5(5+2)=35=5.95BM

  • (a) 3d7: This configuration has 3 unpaired electrons. The magnetic moment is calculated as: μ=3(3+2)=153.87,BM Since it has fewer unpaired electrons than 3d5, its magnetic moment is lower.

  • (c) 3d8: This configuration has 2 unpaired electrons. The magnetic moment is calculated as: μ=2(2+2)=82.83,BM Since it has fewer unpaired electrons than 3d5, its magnetic moment is lower.

  • (d) 3d2: This configuration has 2 unpaired electrons. The magnetic moment is calculated as: μ=2(2+2)=82.83,BM Since it has fewer unpaired electrons than 3d5, its magnetic moment is lower.

7. Which of the following oxidation state is common for all lanthanoids?

(a) +2

(b) +3

(c) +4

(d) +5

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Answer

(b) Lanthanoids show common oxidation state of +3 . Some of which also show +2 and +4 stable oxidation state alongwith +3 oxidation state. These are shown by those elements which by losing 2 or 4 electrons acquire a stable configuration of f0,f7 or f14, e.g., Eu2+ is [Xe]4f7,Yb2+ is [Xe]4f14,Ce4+ is [Xe]4f0 and Tb4f is [Xe]4f7.

  • (a) +2: While some lanthanoids can exhibit a +2 oxidation state, it is not common to all lanthanoids. The +2 state is typically less stable and is observed in only a few specific lanthanoids such as Eu and Yb.

  • (c) +4: The +4 oxidation state is also not common to all lanthanoids. It is observed in a limited number of lanthanoids like Ce and Tb, where the elements achieve a stable electronic configuration by losing four electrons.

  • (d) +5: The +5 oxidation state is extremely rare and not observed in lanthanoids. Lanthanoids typically do not achieve this high oxidation state due to the high energy required to remove five electrons.

8. Which of the following reactions are disproportionation reactions?

(i) Cu+Cu2++Cu

(ii) 3MnO4+4H+2MnO4+MnO2+2H2O

(iii) 2KMnO4K2MnO4+MnO2+O2

(iv) 2MnO4+3Mn2++2H2O5MnO2+4H+

(a) (i)

(b) (i), (ii) and (iii)

(c) (ii), (iii) and (iv)

(d) (i) and (iv)

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Answer

(a) The reaction in which oxidation as well as reduction, occur upon same atom simultaneously is known as disproportionation reaction.

  • Option (b) (i), (ii) and (iii):

    • Reaction (ii) is not a disproportionation reaction because it involves the reduction of MnO4 to MnO2 and the oxidation of MnO4 to MnO42, but these changes do not occur on the same atom simultaneously.
    • Reaction (iii) is not a disproportionation reaction because it involves the decomposition of KMnO4 into K2MnO4, MnO2, and O2, but there is no simultaneous oxidation and reduction of the same species.
  • Option (c) (ii), (iii) and (iv):

    • Reaction (ii) is not a disproportionation reaction because it involves the reduction of MnO4 to MnO2 and the oxidation of MnO4 to MnO42, but these changes do not occur on the same atom simultaneously.
    • Reaction (iii) is not a disproportionation reaction because it involves the decomposition of KMnO4 into K2MnO4, MnO2, and O2, but there is no simultaneous oxidation and reduction of the same species.
    • Reaction (iv) is not a disproportionation reaction because it involves the reduction of MnO4 to MnO2 and the oxidation of Mn2+ to MnO2, but these changes do not occur on the same atom simultaneously.
  • Option (d) (i) and (iv):

    • Reaction (iv) is not a disproportionation reaction because it involves the reduction of MnO4 to MnO2 and the oxidation of Mn2+ to MnO2, but these changes do not occur on the same atom simultaneously.

9. When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because

(a) CO2 is formed as the product

(b) reaction is exothermic

(c) MnO4catalyses the reaction

(d) Mn2+ acts as autocatalyst

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Answer

(d) When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because Mn2+ acts as autocatalyst.

Reduction half .MnO4+8H++5eMn2++4H2O]×2

Oxidation half .C2O422CO2+2e]×5

Overall equation

2MnO4+16H++5C2O422Mn2++10CO2+8H2O

End point of this reaction Colourless to light pink

  • (a) CO2 is formed as the product: The formation of CO2 as a product does not explain the change in reaction rate. The decolourisation becoming instantaneous is due to the catalytic effect of Mn2+, not the production of CO2.

  • (b) reaction is exothermic: While the reaction may release heat, the exothermic nature of the reaction does not account for the initial slowness followed by a rapid decolourisation. The key factor is the autocatalytic role of Mn2+.

  • (c) MnO4 catalyses the reaction: MnO4 is actually a reactant in this reaction, not a catalyst. The catalyst in this reaction is Mn2+, which is produced during the reaction and speeds up the process.

10. There are 14 elements in actinoid series. Which of the following elements does not belong to this series?

(a) U

(b) Np

(c) Tm

(d) Fm

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Answer

(c) Tm(Z=69) do not belong to actinoid series. The actinoid series is with atomic numbers 90 to 103. Thulium (Tm) has atomic number 69 belongs to lanthanoids (4f series).

  • (a) U (Uranium) is incorrect because it belongs to the actinoid series with an atomic number of 92.
  • (b) Np (Neptunium) is incorrect because it belongs to the actinoid series with an atomic number of 93.
  • (d) Fm (Fermium) is incorrect because it belongs to the actinoid series with an atomic number of 100.

Q.11 KMnO4 acts as an oxidising agent in acidic medium. The number of moles of KMnO4 that will be needed to react with one mole of sulphide ions in acidic solution is

(a) 25

(b) 35

(c) 45

(d) 15

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Answer

The reaction of KMnO4 in which it acts as an oxidising agent in acidic medium is

2KMnO4+3H2SO4K2SO4+2MnSO4+3H2O+5[O]

H2S+[O]H2O+S]×52KMnO4+3H2SO4+5H2SK2SO4+2MnSO4+8H2O+5S

5 moles of S2 ions react with 2 moles of KMnO4. So, 1 mole of S2 ion will react with 25 moles of KMnO4

  • Option (b) 35 is incorrect because it suggests that 1 mole of S2 ions would react with 35 moles of KMnO4, which is not supported by the stoichiometry of the balanced chemical equation. The correct stoichiometry indicates that 1 mole of S2 ions reacts with 25 moles of KMnO4.

  • Option (c) 45 is incorrect because it overestimates the amount of KMnO4 needed to react with 1 mole of S2 ions. According to the balanced equation, only 25 moles of KMnO4 are required for 1 mole of S2 ions.

  • Option (d) 15 is incorrect because it underestimates the amount of KMnO4 needed to react with 1 mole of S2 ions. The balanced chemical equation shows that 25 moles of KMnO4 are necessary for 1 mole of S2 ions.

12. Which of the following is amphoteric oxide?

Mn2O7,CrO3,Cr2O3,CrO,V2O5,V2O4

(a) V2O5,Cr2O3

(b) Mn2O7,CrO3

(c) CrO,V2O5

(d) V2O5,V2O4

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Answer

(a) V2O5 and Cr2O3 are amphoteric oxide because both react with alkalies as well as acids.

Note In lower oxides, the basic character is predominant while in higher oxides, the acidic character is predominant.

  • Option (b) Mn2O7,CrO3: Both Mn2O7 and CrO3 are not amphoteric oxides. They are acidic oxides because they react with bases to form salts and water but do not react with acids.

  • Option (c) CrO,V2O5: CrO is a basic oxide, not amphoteric. It reacts with acids to form salts and water but does not react with bases. V2O5 is amphoteric, but the presence of CrO makes this option incorrect.

  • Option (d) V2O5,V2O4: V2O4 is a basic oxide, not amphoteric. It reacts with acids to form salts and water but does not react with bases. V2O5 is amphoteric, but the presence of V2O4 makes this option incorrect.

13. Gadolinium belongs to 4f series. Its atomic number is 64 . Which of the following is the correct electronic configuration of gadolinium?

(a) [Xe]4f75d16s2

(c) [Xe]4f86d2

(b) [Xe]4f65d26s2

(d) [Xe]4f95s1

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Answer

(a) Gadolinium belongs to 4f series and has atomic number 64 . The correct electronic configuration of gadolinium is

64Gd=54[Xe]4f75d16s2

It has extra stability due to half-filled 4f subshell.

  • Option (b): [Xe]4f65d26s2 is incorrect because it does not account for the extra stability provided by the half-filled 4f subshell. Gadolinium has 7 electrons in the 4f subshell, not 6.

  • Option (c): [Xe]4f86d2 is incorrect because it suggests that there are 8 electrons in the 4f subshell and 2 electrons in the 6d subshell, which does not match the actual electron count for gadolinium. Additionally, it omits the 6s electrons.

  • Option (d): [Xe]4f95s1 is incorrect because it suggests that there are 9 electrons in the 4f subshell and 1 electron in the 5s subshell, which is not the correct electron configuration for gadolinium. The 5s subshell should not be involved, and the 4f subshell should have 7 electrons.

14. Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following is not the characteristic property of interstitial compounds?

(a) They have high melting points in comparison to pure metals

(b) They are very hard

(c) They retain metallic conductivity

(d) The are chemically very reactive

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Answer

(d) Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Some of their important characteristics are as follows

(i) They are very hard and rigid.

(ii) They have high melting point which are higher than those of the pure metals.

(iii) They show conductivity like that of the pure metal.

(iv) They acquire chemical inertness.

  • (a) They have high melting points in comparison to pure metals: This is incorrect because interstitial compounds indeed have high melting points, often higher than those of the pure metals due to the strong bonding between the metal atoms and the small atoms trapped in the lattice.

  • (b) They are very hard: This is incorrect because interstitial compounds are known for their hardness and rigidity, which is a result of the small atoms fitting into the interstices of the metal lattice and strengthening the overall structure.

  • (c) They retain metallic conductivity: This is incorrect because interstitial compounds retain the metallic conductivity of the pure metals, as the presence of small atoms in the lattice does not significantly disrupt the flow of electrons.

15. The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr3+ ion is

(a) 2.87BM

(b) 3.87BM

(c) 3.47BM

(d) 3.57BM

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Answer

(b) The magnetic moment is associated with its spin angular momentum and orbital angular momentum.

Spin only magnetic moment value of Cr3+ ion is 3d3

Hence, magnetic moment

(μ)=n(n+2)BM=3(3+2)=15=3.87BM

  • Option (a) 2.87BM: This value is incorrect because it corresponds to a different number of unpaired electrons. For a Cr3+ ion, which has 3 unpaired electrons, the correct calculation yields 153.87BM, not 2.87BM.

  • Option (c) 3.47BM: This value is incorrect because it does not match the result of the spin-only magnetic moment formula for 3 unpaired electrons. The correct calculation is 153.87BM, not 3.47BM.

  • Option (d) 3.57BM: This value is incorrect because it also does not match the result of the spin-only magnetic moment formula for 3 unpaired electrons. The correct calculation is 153.87BM, not 3.57BM.

Q.16 KMnO4 acts as an oxidising agent in alkaline medium. When alkaline KMnO4 is treated with KI, iodide ion is oxidised to…… .

(a) I2

(b) IO

(c) IO3

(d) IO4

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Answer

(c) KMnO4 acts as an oxidising agent in alkaline medium. When alkaline KMnO4 is treated with KI, iodide ion is oxidised to IO3.

Reaction 2KMnO4+H2O+KI2MnO4+2KOH+KIO3

or, vI+6OHIO3+3H2O+6e

  • (a) I2: In an alkaline medium, KMnO4 is a strong oxidizing agent and it oxidizes iodide ions (I) to higher oxidation states. Molecular iodine (I2) is typically formed in acidic or neutral conditions, not in alkaline conditions.

  • (b) IO: The oxidation state of iodine in IO is +1. However, in the presence of a strong oxidizing agent like KMnO4 in an alkaline medium, iodide ions are oxidized to a higher oxidation state than +1, specifically to +5 in IO3.

  • (d) IO4: The oxidation state of iodine in IO4 is +7. While KMnO4 is a strong oxidizing agent, in an alkaline medium, it typically oxidizes iodide ions to the +5 oxidation state, forming IO3, rather than the +7 oxidation state in IO4.

17. Which of the following statements is not correct?

(a) Copper liberates hydrogen from acids

(b) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine

(c) Mn3+ and Co3+ are oxidising agents in aqueous solution

(d) Ti2+ and Cr2+ are reducing agents in aqueous solution

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Answer

(a) Copper lies below hydrogen in the electrochemical series and hence does not liberate H2 from acids. Therefore, option (a) is not correct.

Other three options ( b,c,d) are correct.

  • (b) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine: This statement is correct. Manganese in higher oxidation states (like +4, +6, and +7) forms stable compounds such as ( \text{MnO}_2 ), ( \text{MnO}_3 ), and ( \text{MnF}_4 ).

  • (c) ( \text{Mn}^{3+} ) and ( \text{Co}^{3+} ) are oxidising agents in aqueous solution: This statement is correct. Both ( \text{Mn}^{3+} ) and ( \text{Co}^{3+} ) have a tendency to gain electrons and get reduced, thus acting as oxidizing agents in aqueous solution.

  • (d) ( \text{Ti}^{2+} ) and ( \text{Cr}^{2+} ) are reducing agents in aqueous solution: This statement is correct. Both ( \text{Ti}^{2+} ) and ( \text{Cr}^{2+} ) have a tendency to lose electrons and get oxidized, thus acting as reducing agents in aqueous solution.

18. When acidified K2Cr2O7 solution is added to Sn2+ salt then Sn2+ changes to

(a) Sn

(b) Sn3+

(c) Sn4+

(d) Sn+

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Answer

(c) When acidified K2Cr2O7 solution is added to Sn2+ salts then Sn2+ changes to Sn4+. The reaction is given below

  • Option (a) Sn: This option is incorrect because Sn2+ is being oxidized in the reaction with acidified K2Cr2O7. Oxidation involves an increase in the oxidation state, not a reduction to elemental tin (Sn), which would be a decrease in the oxidation state.

  • Option (b) Sn3+: This option is incorrect because the oxidation state of tin in this reaction changes from +2 to +4. There is no intermediate oxidation state of +3 involved in this specific redox reaction.

  • Option (d) Sn+: This option is incorrect because Sn+ would represent a reduction in the oxidation state from +2 to +1. However, in the reaction with acidified K2Cr2O7, Sn2+ is oxidized, meaning its oxidation state increases, not decreases.

19. Highest oxidation state of manganese in fluoride is +4(MnF4) but highest oxidation state in oxides is +7(Mn2O7) because

(a) fluorine is more electronegative than oxygen

(b) fluorine does not possess d orbitals

(c) fluorine stabilises lower oxidation state

(d) in covalent compounds, fluorine can form single bond only while oxygen forms double bond

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Answer

(d) Highest oxidation state of manganese in fluoride is +4(MnF4) but highest oxidation state in oxides is +7(Mn2O7). The reason is that in covalent compounds fluorine can form single bond while oxygen forms double bond.

  • (a) Fluorine is more electronegative than oxygen, but this does not directly explain why manganese achieves a higher oxidation state in oxides compared to fluorides. The ability to form multiple bonds (as in the case of oxygen) is more relevant to the stabilization of higher oxidation states.

  • (b) The fact that fluorine does not possess d orbitals is not relevant to the oxidation state of manganese. The oxidation state is determined by the ability of the element to stabilize different charges, which is more influenced by bonding characteristics rather than the presence of d orbitals in fluorine.

  • (c) While fluorine can stabilize lower oxidation states due to its high electronegativity, this does not explain why manganese achieves a higher oxidation state in oxides. The key factor is the ability of oxygen to form multiple bonds, which stabilizes higher oxidation states of manganese.

20. Although zirconium belongs to 4d transition series and hafniun to 5d transition series even then they show similar physical and chemical properties because…… .

(a) both belong to d-block

(b) both have same number of electrons

(c) both have similar atomic radius

(d) both belong to the same group of the Periodic Table

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Answer

(c) Due to lanthanoide contraction, Zr and Hf possess nearly same atomic and ionic radii i.e., Zr=160pm and Hf=159pm,Zr4+=79pm and Hf4+=78pm. Therefore, these two elements show similar properties (physical and chemical properties).

  • (a) Both belong to d-block: While it is true that both zirconium and hafnium belong to the d-block, this is not the primary reason for their similar physical and chemical properties. Many elements in the d-block do not exhibit such close similarities in properties.

  • (b) Both have same number of electrons: Zirconium and hafnium do not have the same number of electrons. Zirconium has 40 electrons, while hafnium has 72 electrons. The number of electrons is not the reason for their similar properties.

  • (d) Both belong to the same group of the Periodic Table: Although zirconium and hafnium do belong to the same group (Group 4) of the Periodic Table, this alone does not account for their nearly identical physical and chemical properties. The primary reason is the lanthanoid contraction, which results in their nearly identical atomic and ionic radii.

21. Why is HCl not used to make the medium acidic in oxidation reactions of KMnO4 in acidic medium?

(a) Both HCl and KMnO4 act as oxidising agents

(b) KMnO4 oxidises HCl into Cl2 which is also an oxidising agent

(c) KMnO4 is a weaker oxidising agent than HCl

(d) KMnO4 acts as a reducing agent in the presence of HCl

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Answer

(b) HCl is not used to make the medium acidic in oxidation reactions of KMnO4 in acidic medium. The reason is that if HCl is used, the oxygen produced from KMnO4+HCl is partly utilised in oxidising HCl to Cl, which itself acts as an oxidising agent and partly oxidises the reducing agent.

  • (a) Both HCl and KMnO4 act as oxidising agents: This is incorrect because HCl is not an oxidizing agent; it is a strong acid that can be oxidized by KMnO4.

  • (c) KMnO4 is a weaker oxidising agent than HCl: This is incorrect because KMnO4 is a much stronger oxidizing agent compared to HCl.

  • (d) KMnO4 acts as a reducing agent in the presence of HCl: This is incorrect because KMnO4 is a strong oxidizing agent and does not act as a reducing agent in the presence of HCl.

Multiple Choice Questions (More Than One Options)

22. Generally transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured?

(a) KMnO4

(b) Ce(SO4)2

(c) TiCl4

(d) Cu2Cl2

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Answer

(a,b)

KMnO4 is coloured due to the charge transfer and not because of the presence of unpaired electrons. Similarly, oxidation state of Ce in Ce(SO4)2 is +4 with 4f electronic configuration. It is also coloured (yellow) due to charge transfer and not due to ff transition.

  • TiCl4 is not coloured because titanium in TiCl4 is in the +4 oxidation state, which has a 3d0 electronic configuration, meaning there are no unpaired electrons to cause colour.
  • Cu2Cl2 is not coloured because copper in Cu2Cl2 is in the +1 oxidation state, which has a 3d10 electronic configuration, meaning all electrons are paired and there are no unpaired electrons to cause colour.

23. Transition elements show magnetic moment due to spin and orbital motion of electrons. Which of the following metallic ions have almost same spin only magnetic moment?

(a) Co2+

(b) Cr2+

(c) Mn2+

(d) Cr3+

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Answer

(a,d)

Electronic configuration of Co2+=[Ar]3d7; Number of unpaired electrons =3

Electronic configuration of Cr2+=[Ar]3d4; Number of unpaired electrons =4

Electronic configuration of Mn2+=[Ar]3d5; Number of unpaired electrons =5

Electronic configuration of Cr3+=[Ar]3d3; Number of unpaired electrons =3

Hence, it is clearly seen that both Co2+ and Cr3+ have same number of unpaired electrons. i.e., 3 .

  • Cr2+: The electronic configuration of Cr2+ is [Ar]3d4, which means it has 4 unpaired electrons. This is different from the 3 unpaired electrons in Co2+ and Cr3+.

  • Mn2+: The electronic configuration of Mn2+ is [Ar]3d5, which means it has 5 unpaired electrons. This is different from the 3 unpaired electrons in Co2+ and Cr3+.

24. In the form of dichromate, Cr(VI) is a strong oxidising agent in acidic medium but Mo(VI) in MoO3 and W(VI) in WO3 are not because

(a) Cr(VI) is more stable than Mo(VI) and W(VI).

(b) Mo(VI) and W(VI) are more stable than Cr(VI).

(c) Higher oxidation states of heavier members of group-6 of transition series are more stable.

(d) Lower oxidation states of heavier members of group-6 of transition series are more stable.

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Answer

(b,c)

In d-block elements, for heavier elements, the higher oxidation states are more stable. Hence, Mo(VI) and W(VI) are more stable than Cr(VI). Thats why, Cr(VI) in the form of dichromate is a stronger oxidising agent in acidic medium whereas MoO3 and WO3 are not.

  • (a) Cr(VI) is more stable than Mo(VI) and W(VI): This is incorrect because, in reality, Mo(VI) and W(VI) are more stable than Cr(VI). The stability of higher oxidation states increases as we move down the group in the periodic table, making Mo(VI) and W(VI) more stable than Cr(VI).

  • (d) Lower oxidation states of heavier members of group-6 of transition series are more stable: This is incorrect because, for heavier elements in the d-block, the higher oxidation states are generally more stable. Therefore, the statement that lower oxidation states are more stable for heavier members of group-6 is not accurate.

25. Which of the following actinoids show oxidation states upto +7 ?

(a) Am

(b) Pu

(c) U

(d) Np

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Answer

(b,d)

The oxidation states of the following actinoids are

(a) Americium (Z=95); Electronic configuration =[Rn]5f76d07s2 Oxidation states shown by Am=+3,+4,+5,+6.

(b) Plutonium (Z=94); Electronic configuration =[Rn]5f66d07s2 Oxidation states shown by Pu=+3,+4,+5,+6,+7.

(c) Uranium ( Z=92 ); Electronic configuration =[Rn]5f36d17s2 Oxidation states shown by U=+3,+4,+5,+6.

(d) Neptunium ( Z=93 ); Electronic configuration =[Rn]5f46d17s2 Oxidation states shown by Np=+3,+4,+5,+6,+7.

  • Americium (Am): Americium does not show an oxidation state of +7. The highest oxidation state it exhibits is +6.

  • Uranium (U): Uranium does not show an oxidation state of +7. The highest oxidation state it exhibits is +6.

26. General electronic configuration of actinoids is (n2)f114(n1)d02ns2. Which of the following actinoids have one electron in 6d orbital?

(a) (Atomic number. 92)

(b) Np (Atomic number. 93)

(c) Pu (Atomic number. 94)

(d) Am (Atomic number. 95)

Show Answer

Answer

( a,b )

General electronic configuration of actinoids is (n1)f114(n1)d02ns2.U and Np each have one electron in 6d orbital. (Also, refer to Q. 25)

  • Pu (Atomic number 94): Plutonium typically has the electronic configuration [Rn] 5f^6 7s^2. It does not have an electron in the 6d orbital.

  • Am (Atomic number 95): Americium typically has the electronic configuration [Rn] 5f^7 7s^2. It does not have an electron in the 6d orbital.

27. Which of the following lanthanoids show +2 oxidation state besides the characteristic oxidation state +3 of lanthanoids?

(a) Ce

(b) Eu

(c) Yb

(d) Ho

Show Answer

Answer

(b,c)

(a) Cerium (Z=57) Electronic configuration =[Xe]4f55d06s2 Oxidation state of Ce=+3,+4

(b) Europium (Z=63) Electronic configuration =[Xe]4f75d06s2 Oxidation state of Eu=+2,+3

(c) Ytterbium (Z=70) Electronic configuration =[Xe]4f145d06s2 Oxidation state of Yb=+2,+3

(d) Holmium (Z=67) Electronic configuration =[Xe]4f115d06s2 Oxidation state of Ho=+3

  • Cerium (Ce): Cerium typically exhibits oxidation states of +3 and +4. It does not commonly show a +2 oxidation state.

  • Holmium (Ho): Holmium predominantly exhibits an oxidation state of +3 and does not commonly show a +2 oxidation state.

28. Which of the following ions show higher spin only magnetic moment value?

(a) Ti3+

(b) Mn2+

(c) Fe2+

(d) Co3+

Show Answer

Answer

(b,c)

As,

Ti3+=[Ar]3d1Mn2+=[Ar]3d5,(t2g3eg2)Fe2+=[Ar]3d6(t2g4e2g2)Co3+=[Ar]3d6(t2g6eg0)

Crystal field splitting energy (CFSE) is high in Co3+, thus electrons pair up in t2g. Hence, only Fe2+ and Mn2+ show higher spin magnetic moment value.

  • Ti3+: This ion has a single unpaired electron in the 3d orbital (3d1 configuration). The spin-only magnetic moment is directly related to the number of unpaired electrons. With only one unpaired electron, Ti3+ has a lower spin-only magnetic moment compared to ions with more unpaired electrons.

  • Co3+: This ion has a 3d6 configuration. Due to the high crystal field splitting energy (CFSE) in Co3+, the electrons pair up in the t2g orbitals, resulting in no unpaired electrons (t2g6eg0). Therefore, Co3+ has a very low or zero spin-only magnetic moment.

29. Transition elements form binary compounds with halogens. Which of the following elements will form MF3 type compounds?

(a) Cr

(b) Co

(c) Cu

(d) Ni

Show Answer

Answer

(a, b)

Transition elements such as Cr and Co form binary compounds with halogens, i.e., CrF3 and CoF3 whereas Cu and Ni do not form CuF3 and NiF3.

  • Cu: Copper typically forms compounds in the +1 and +2 oxidation states, such as CuF and CuF2. The +3 oxidation state is less stable for copper, making CuF3 unlikely to form.

  • Ni: Nickel commonly forms compounds in the +2 oxidation state, such as NiF2. The +3 oxidation state is not stable for nickel, hence NiF3 is not typically formed.

30. Which of the following will not act as oxidising agents?

(a) CrO3

(b) MoO3

(c) WO3

(d) CrO42

Show Answer

Answer

(b,c)

A species can act as oxidising agent only when metal is present in high oxidation state but lower oxidation state show stability. As higher oxidations states of W and Mo are more stable, therefore they will not act as oxidising agents.

  • (a) CrO3: Chromium trioxide (CrO3) contains chromium in the +6 oxidation state, which is a high oxidation state. Chromium in this state can readily accept electrons and be reduced to a lower oxidation state, making it a strong oxidizing agent.

  • (d) CrO42: Chromate ion (CrO42) also contains chromium in the +6 oxidation state. Similar to CrO3, chromium in this state can act as an oxidizing agent by accepting electrons and being reduced to a lower oxidation state.

31. Although +3 is the characteristic oxidation state for lanthanoids but cerium also shows +4 oxidation state because

(a) it has variable ionisation enthalpy

(b) it has a tendency to attain noble gas configuration

(c) it has a tendency to attain f0 configuration

(d) it resembles Pb4+

Show Answer

Answer

(b,c)

Electronic configuration of 58Ce=54[Xe]4f25d06s2.

Therefore, electronic configuration of Ce4+=54[Xe]4f0.

Thus, it has a tendency to attain noble gas configuration and attain f0 configuration.

  • (a) it has variable ionisation enthalpy: This option is incorrect because the primary reason for cerium showing a +4 oxidation state is not due to variable ionisation enthalpy but rather its ability to achieve a stable electronic configuration.

  • (d) it resembles Pb4+: This option is incorrect because the resemblance to Pb4+ is not a significant factor in cerium exhibiting a +4 oxidation state. The key reasons are related to its electronic configuration and stability.

Short Answer Type Questions

32. Why does copper not replace hydrogen from acids?

Show Answer

Answer

Copper not replace hydrogen from acids because Cu has positive E value, i.e., less reactive than hydrogen which has electrode potential 0.00V.

33. Why Evalues for Mn,Ni and Zn are more negative than expected?

Show Answer

Answer

Negative values of Mn2+ and Zn2+ are related to stabilities of half-filled and completely filled configuration respectively. But for Ni2+,E value is related to the highest negative enthalpy of hydration.

Hence, Es values for Mn,Ni and Zn are more negative than expected.

34. Why first ionisation enthalpy of Cr is lower than that of Zn ?

Show Answer

Answer

Ionisation enthalpy of Cr is less than that of Zn because Cr has stable d5 configuration. In case of zinc, electron comes out from completely filled 4s-orbital. So, removal of electron from zinc requires more energy as compared to the chromium.

35. Transition elements show high melting points. Why?

Show Answer

Answer

Transition elements show high melting point, due to involvement of greater number of electrons in the interatomic bonding from (n1)d-orbitals in addition to ns electrons in forming metallic bond. Thus, large number of electrons participate forming large number of metallic bond.

36. When Cu2+ ion is treated with KI, a white precipitate is formed. Explain the reaction with the help of chemical equation.

Show Answer

Answer

When Cu2+ ion is treated with KI, it produces Cu2I2 white precipitate in the final product.

2Cu2++4ICu2I2 (White ppt.) +I2

(In this reaction, CuI2 is formed which being unstable, dissociates into Cu2I2 and I2 ).

37. Out of Cu2Cl2 and CuCl2, which is more stable and why?

Show Answer

Answer

Among Cu2Cl2 and CuCl2,CuCl2 is more stable. Stability of complex can be explained on the basis of stability of oxidation state of copper. Stability of Cu2+(aq) rather than Cu+(aq) is due to much more negative value of [ΔhydHs. of .Cu2+(aq)] than Cu+which more than compensates for the second ionisation enthalpy of Cu.

38. When a brown compound of manganese (A) is treated with HCl it gives a gas (B). The gas taken in excess, reacts with NH3 to give an explosive compound (C). Identify compounds A, B and C.

Show Answer

Thinking Process

This problem is based on the properties of MnO2 and preparation of NCl3.

Answer

MnO2 is the brown compound of Mn which reacts with HCl to give Cl2 gas. This gas forms an explosive compound NCl3 when treated with NH3. Thus, A=MnO2;B=Cl2;C=NCl3 and reactions are as follows

(i) MnO2+4HCl[A]MnCl2+Cl2[B]+2H2O

(ii) NH3+3Cl2 (Excess) NCl3 [C] +3HCl

39. Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?

Show Answer

Answer

Oxygen can form multiple bonds with metals, while fluorine can’t form multiple bond with metals. Hence, oxygen has more ability to stabilize higher oxidation state rather than fluorine

40. Although Cr3+ and Co2+ ions have same number of unpaired electrons but the magnetic moment of Cr3+ is 3.87BM and that of Co2+ is 4.87BM, Why?

Show Answer

Answer

Magnetic moment of any metal ion can be decided on the basis of spin as well as orbital contribution of electron. Due to symmetrical electronic configuration, there is no orbital contribution in Cr3+ ion. However, appreciable orbital contribution takes place in Co2+ ion.

41. Ionisation enthalpies of Ce,Pr and Nd are higher than Th,Pa and U. Why?

Show Answer

Answer

Ce,Pr and Nd are lanthanoids and have incomplete 4f shell while Th,Pa and U are actinoids and have 5f shell incomplete.

In the beginning, when 5 -orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The 5f-electrons will therefore, be more effectively shielded from the nuclear charge than 4f electrons of the corresponding lanthanoids.

Therefore, outer electrons are less firmly held and they are available for bonding in the actinoids

42. Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them, Why?

Show Answer

Answer

Separation of Zr and Hf are quite difficult because of lanthanoid contraction. Due to lanthanoid contraction, they have almost same size (Zr=160pm and Hf=159pm) and thus, similar chemical properties. That’s why it is very difficult to separate them by chemical methods.

43. Although +3 oxidation states is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?

Show Answer

Answer

It is due to the fact that after losing one more electron Ce acquires stable 4f electronic configuration. So, Ce shows +4 oxidation state also alongwith +3 oxidation state.

44. Explain why does colour of KMnO4 disappear when oxalic acid is added to its solution in acidic medium?

Show Answer

Answer

When oxalic acid is added to acidic solution of KMnO4, its colour disappear due to reduction of MnO4ion to Mn2+. Chemical reaction occurring during this neutralisation reaction is as follows

5C2O42+2MnO4 (Coloured) +16H+2Mn2+ (Colourless) +8H2O+10CO2

45. When orange solution containing Cr2O72 ion is treated with an alkali, a yellow solution is formed and when H+ions are added to yellow solution, an orange solution is obtained. Explain why does this happen?

Show Answer

Answer

When orange solution containing Cr2O72 ion is treated with an alkali, a yellow solution of CrO42 is obtained. On the same way, Cr2O72 Dischromate  (orange) H+OHCrO42 Chromate  (yellow) 

when H+ions are added to yellow solution, an orange solution is obtained due to interconversion.

Q.46 A solution of KMnO4 on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

Show Answer

Answer

Oxidising behaviour of KMnO4 depends on pH of the solution.

In acidic medium ( pH<7 )

MnO4+8H++5eMn2+ Colourless +4H2O

In alkaline medium (pH>7)

MnO4+eMnO42( Green )

In neutral medium (pH=7)

MnO4+2H2O+3eMnO2 (Brown ppt) +4OH

47. The second and third rows of transition elements resemble each other much more than they resemble the first row. Explain, why?

Show Answer

Answer

Due to lanthanoid contraction, the atomic radii of the second and third row transition elements is almost same. So, they resemble each other much more as compared to first row elements and show similar character.

48. Es of Cu is +0.34V while that of Zn is 0.76V. Explain.

Show Answer

Answer

ES value of Cu is positive because of the fact that sum of sublimation enthalpy and ionisation enthalpy to convert Cu(s) to Cu2+(aq) is so high that it is not compensate by its hydration enthalpy. E value for Zn is negative because of the fact that after removal of electrons from 4s orbital, stable 3d10 configuration is obtained.

49. The halides of transition elements become more covalent with increasing oxidation state of the metal. Why?

Show Answer

Thinking Process

This problem is based on concept of Fajan’s rule and its application.

Answer

As the oxidation state increases, size of the ion of transition element decreases. As per Fajan’s rule, as the size of metal ion decreases, covalent character of the bond formed increases.

Therefore, the halide of transition elements become more covalent with increasing oxidation state of the metal.

50. While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but reverse happens during the ionisation of the atom. Explain why?

Show Answer

Answer

During filling up of electrons follow (n+l) rule. Here 4s has lower energy than 3d orbital. After the orbitals are filled 4s goes beyond 3d, i.e., 4s is farther from nucleus than 3d. So, electron from 4s is removed earlier than from 3d.

51. Reactivity of transition elements decreases almost regularly from Se to Cu. Explain.

Show Answer

Answer

Reactivity of transition elements depends mostly upon their ionisation enthalpies. As we move from left to right in the periodic table ( Se to Cu ), ionisation enthalpies increase almost regularly.

Hence, their reactivity decreases almost regularly from Se to Cu.

Matching The Columns

52. Match the catalysts given in Column I with the processes given in Column II.

Column I
(Catalyst)
Column II
(Process)
A. Ni in the presence of hydrogen 1. Ziegler-Natta catalyst
B. Cu2Cl2 2. Contact process
C. V2O5 3. Vegetable oil to ghee
D. Finely divided iron 4. Sandmeyer reaction
E. TiCl4+Al(CH3)3 5. Haber’s process
6. Decomposition of KClO3
Show Answer

Answer

A. (3)

B. (4)

C. (2)

E. (1)

Catalyst Process
A. Ni in the poresence of hydrogen Vegetable oil to ghee
B. Cu2Cl2 Sandmeyer reaction
C. V2O5 Contact process
SO2V2O5SO3
D. Finely divided iron Haber’s process
N2+3H2Fe2NH3
E. TiCl4+ I(CH3)3

53. Match the compounds/elements given in Column I with uses given in Column II.

Column I
(Compound/element)
Column II
(Use)
A. Lanthanoid oxide 1. Production of iron alloy
B. Lanthanoid 2. Television screen
C. Misch metall 3. Petroleum cracking
D. Magnesium based alloy is constituent of 4. Lanthanoid metal + iron
E. Mixed oxides of lanthanoids are employed 5. Bullets
Show Answer

Answer

A. (2)

B. (1)

C. (4)

D. (5)

E. (3)

Compound /Element Use
A. Lanthanoid oxide Television screen
B. Lanthanoid Production of iron alloy
C. Misch metall Lanthanoid metal + iron
D. Magnesium based alloy is constitute of Bullets
E Mixed oxides of lanthanoids are employed Petroleum cracking

54. Match the properties given in Column I with the metals given in Column II.

Column I
(Property)
Column II
(Metal)
A. An element which can show +8 oxidation state 1. Mn
B. 3d block element that can show upto +7
oxidation state
2. Cr
C. 3d block element with highest melting point 3. Os
4. Fe
Show Answer

Answer

A. (3)

B. (1)

C. (2)

A. Osmium is an element which can show +8 oxidation state.

B. 3d block element that can show upto +7 oxidation state is manganese.

C. 3d block element with highest melting point is chromium.

55. Match the statements given in Column I with the oxidation states given in Column II.

Column I Column II
A. Oxidation state of Mn in MnO2 is 1. +2
B. Most stable oxidation state of Mn is 2. +3
C. Most stable oxidation state of Mn in oxides is 3. +4
D. Characteristic oxidation state of lanthanoids is 4. +5
5. +7
Show Answer

Answer

A. (3)

B. (1)

C. (5)

D. (2)

A. Oxidation state of Mn in MnO2 is +4 .

B. Most stable oxidation state of Mn is +2 .

C. Most stable oxidation state of Mn in oxides is +7 .

D. Characteristic oxidation state of lanthanoids is +3 .

56. Match the solutions given in Column I and the colours given in Column II.

Column I
(Aqueous solution of salt)
Column II
(Colour)
A. FeSO47H2O 1. Green
B. NiCl24H2O 2. Light pink
C. MnCl24H2O 3. Blue
D. CoCl26H2O 4. Pale green
E. Cu2Cl2 5. Pink
6. Colourless
Show Answer

Answer

A. (4)

B. (1)

C. (2)

D. (5)

E. (6)

Aqueous solution of salt Colour
A. FeSO47H2O Pale green
B. NiCl24H2O Green
C. MnCl24H2O Light pink
D. CoCl26H2O Pink
E. Cu2Cl2 Colourless

57. Match the property given in Column I with the element given in Column II.

Column I
(Property)
Column II
(Element)
A. Lanthanoid which shows +4 oxidation state 1. Pm
B. Lanthanoid which can show +2 oxidation
state
2. Ce
C. Radioactive lanthanoid 3. Lu
D. Lanthanoid which has 4f7 electronic
configuration in +3 oxidation state
4. Eu
E. Lanthanoid which has 4f14 electronic
configuration in +3 oxidation state
5. Gd
Show Answer

Answer

A. (2)

B. (4)

C. (1)

D. (5)

E. (3)

A. Lanthanoid which shows +4 oxidation state is cerium.

58Ce=[Xe]4f25d06s2; Oxidation state =+3,+4

B. Lanthanoid which can show +2 oxidation state is europium.

63Eu=[Xe]4f75d06s2; Oxidation state =+2,+3

C. Radioactive lanthanoid is promethium. It is the only synthetic (man-made) radioactive lanthanoid.

D. Lanthnoid which has 4f7 electronic configuration in +3 oxidation state is gadolinium.

64Gd=[Xe]4f75d16s2; Oxidation state =+3

E. Lanthanoid which has 4f14 electronic configuration in +3 oxidation state is lutetium

71Lu=[Xe]4f145d16s2; Oxidation state =+3

58. Match the properties given in Column I with the metals given in Column II.

Column I
(Property)
Column II
(Metal)
A. Element with highest second ionisation enthalpy 1. Co
B. Element with highest third ionisation enthalpy 2. Cr
C. M in M(CO)6 is 3. Cu
D. Element with highest heat of atomisation 4. Zn
5. Ni
Show Answer

Answer

A. (3)

B. (4)

C. (2)

D. (1)

A. Cu+=3d10 which is very stable configuration due to full-filled orbitals. Hence, removal of second electron requires very high energy.

B. Zn2+=3d10 which is very stable configuration. Hence, removal of third electron requires very high energy.

C. Metal carbonyl with formula M(CO)6 is Cr(CO)6.

D. Nickel is the element with highest heat of atomisation.

Assertion and Reason

In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct answer out of the following choices.

(a) Both assertion and reason are true, and reason is the correct explanation of the assertion.

(b) Both assertion and reason are true but reason is not the correct explanation of assertion.

(c) Assertion is not true but reason is true.

(d) Both assertion and reason are false.

59. Assertion (A)Cu2+ iodide is not known. Reason ( R)Cu2+ oxidises Ito iodine.

Show Answer

Answer

(a) Assertion and reason both are correct and reason is the correct explanation of assertion.

Copper (II) iodide (CuI2) is not known because Cu2+ oxidises Ito lodine.

60. Assertion (A) Separation of Zr and Hf is difficult. Reason (R) Because Zr and Hf lie in the same group of the Periodic Table.

Show Answer

Answer

(b) Assertion and reason are true but reason is not correct explanation of assertion.

Separation of Zr and Hf is difficult; it is not because of they lie in the same group of Periodic Table. This is due to lanthanoid contraction which causes almost similar radii of both of them.

61. Assertion (A) Actinoids form relatively less stable complexes as compared to lanthanoids.

Reason (R) Actinoids can utilise their 5 f orbitals alongwith 6d orbitals in bonding but lanthanoids do not use their 4f orbital for bonding.

Show Answer

Answer

(c) Assertion is not true but reason is true.

Actinoids form relatively more stable complexes as compared to lanthanoids because of actinoids can utilise their 5f orbitals along with 6d orbitals in bonding but lanthanoids do not use their 4f orbitals for bonding.

62. Assertion (A) Cu cannot liberate hydrogen from acids.

Reason (R) Because it has positive electrode potential.

Show Answer

Answer

(a) Assertion and reason both are correct and reason is correct explanation of assertion. Cu can not liberate hydrogen from acids because it has positive electrode potential. Metals having negative value of electrode potential liberate H2 gas.

63. Assertion (A) The highest oxidation state of osmium is +8 .

Reason (R) 0smium is a 5d-block element.

Show Answer

Answer

(b) Assertion and reason both are correct but reason is not the correct explanation of assertion.

The highest oxidation state of osmium is +8 due to its ability to expand their octet by using its all 8 electrons ( 2 from 6 s and 6 from 5d ).

Long Answer Type Questions

64. Identify A to E and also explain the reaction involved.

Show Answer

Answer

The substances from A to E are

A=Cu;B=Cu(NO3)2;C=[Cu(NH3)4]2+;D=CO2;E=CaCO3

Reactions:

(i) CuCO3ΔCuO+CO2]×2

(ii) 2CuO+CuS3Cu[A]+SO2

(iii) Cu[A]+4HNO3(conc.)Cu(NO3)2[B]+2NO+2H2O

(iv) Cu2+[B]+NH3[Cu(NH3)4][C] (Blue Solution) 

(v) Ca(OH)2+CO2[D]CaCO3[E](Milky)+H2O

(vi) CaCO3+H2O+CO2Ca(HCO3)2

65. When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallised from the solution. When compound (C) is treated with KCl, orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.

Show Answer

Answer

K2Cr2O7 is an orange compound. It is formed when Na2Cr2O7 reacts with KCl. In acidic medium, yellow coloured CrO42 (chromate ion) changes into dichromate.

The given process is the preparation method of potassium dichromate from chromite ore.

A=FeCr2O4;B=Na2CrO4;C=Na2Cr2O7;D=K2Cr2O7.

(i) 4FeCr2O4[A]+8Na2CO3+7O28Na2CrO4[B]+2Fe2O3+8CO2

(ii) 2Na2CrO4+2H+Na2Cr2O7+2Na++H2O

(iii) Na2Cr2O7[C]+2KClK2Cr2O7[D]+2NaCl

66. When an oxide of manganese (A) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.

Show Answer

Thinking Process

This problem based on the concept of preparation and properties of potassium permanganate.

Answer

It is the method of preparation of potassium permanganate (purple). Thus, (A)=MnO2

(C)=KMnO4

(B)=K2MnO4

(D)=KIO3

2MnO2[A]+4KOH+O22K2MnO4[B]+2H2O3MnO42+4H+2MnO4[C]+MnO2+2H2O2MnO4+H2O+KI2MnO2[A]+2OH[D]+KIO3

67. On the basis of lanthanoid contraction, explain the following:

(i) Nature of bonding in La2O3 and Lu2O3.

(ii) Trends in the stability of oxo salts of lanthanoids from La to Lu.

(iii) Stability of the complexes of lanthanoids.

(iv) Radii of 4d and 5d block elements.

(v) Trends in acidic character of lanthanoid oxides.

Show Answer

Answer

(i) As the size decreases covalent character increases. Therefore, La2O3 is more ionic and Lu2O3 is more covalent.

(ii) As the size decreases from La to Lu, stability of oxo salts also decreases.

(iii) Stability of complexes increases as the size of lanthanoids decreases.

(iv) Radii of 4d and 5d-block elements will be almost same.

(v) Acidic character of oxides increases from La to Lu.

68. (a) Answer the following questions

(i) Which element of the first transition series has highest second ionisation enthalpy?

(ii) Which element of the first transition series has highest third ionisation enthalpy?

(iii) Which element of the first transition series has lowest enthalpy of atomisation?

(b) Identify the metal and justify your answer.

(i) Carbonyl M(CO)5

(ii) MO3F

Show Answer

Answer

(a) (i) Cu, because the electronic configuration of Cu is 3d104s1. So, second electron needs to be removed from completely filled d-orbital which is very difficult.

(ii) Zinc, because of electronic configuration of Zn=3d104s2 and Zn2+=3d10 which is fully filled and hence is very stable. Removal of third electron requires very high energy.

(iii) Zinc, because of it has completely filled 3d subshell and no unpaired electron is available for metallic bonding.

(b) (i) Carbonyl M(CO)5 is Fe(CO)5

According to EAN rule, the effective number of a metal in a metal carbonyl is equal to the atomic number of nearest inert gas EAN is calculated as

EAN = number of electrons in metal +2×(CO) = atomic number of nearest inert gas

lnM(CO)5=x+2×(5)=36 ( Kr is the nearest inert gas)

x=26 (atomic number of metal)

So, the metal is Fe (iron).

(ii) MO3F is MnO3F.

In MO3F

Let us assume that oxidation state of M is x

x+3×(2)+(1)=0

or, x=+7 i.e., M is in +7 oxidation state of +7 . Hence, the given compound is MnO3F.

69. Mention the type of compounds formed when small atoms like H,C and N get trapped inside the crystal lattice of transition metals. Also give physical and chemical characteristics of these compounds.

Show Answer

Answer

When small atoms like H,C and N get trapped inside the crystal lattice of transition metals.

(a) Such compounds are called interstitial compounds.

(b) Their characteristic properties are;

(i) They have high melting points, higher than those of pure metals.

(ii) They are very hard.

(iii) They retain metallic conductivity.

(iv) They are chemically inert.

70. (a) Transition metals can act as catalysts because these can change their oxidation state. How does Fe (III) catalyse the reaction between iodide and persulphate ions?

(b) Mention any three processes where transition metals act as catalysts.

Show Answer

Answer

(a) Reaction between iodide and persulphate ions is

2I+S2O82Fe(III)I2+2SO42 Role of Fe(III) ions Fe3++2I2Fe2++I22Fe2++S2O822Fe3++2SO42

(b) (i) Vanadium ( V ) oxide used in contact process for oxidation of SO2 to SO3.

(ii) Finely divided iron in Haber’s process in conversion of N2 and H2 to NH3.

(iii) MnO2 in preparation of oxygen from KClO3.

71. A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H2SO4 and NaCl, chlorine gas is liberated and a compound ( D ) of manganese alongwith other products is formed. Identify compounds A to D and also explain the reactions involved.

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Thinking Process

This problem is based on preparation and properties of KMnO4K2MnO4 and MnO2.

Answer

Since, compound (C) on treating with conc. H2SO4 and NaCl gives Cl2 gas, so it is manganese dioxide (MnO2). It is obtained alongwith MnO42 when KMnO4 (violet) is heated.

Thus,

(A)=KMnO4

(B)=K2MnO4

(C)=MnO2

(D)=MnCl2

KMnO4[A]ΔK2MnO4[B]+MnO2[C]+O22MnO2+4KOH+O22K2MnO4+2H2OMnO2+4NaCl+4H2SO4MnCl2+4NaHSO4+2H2O+Cl2