The d- and f-Block Elements - MCQs with Explanations

The d- and f-Block Elements - MCQs with Explanations

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The d- and f-Block Elements - MCQs with Explanations

1. The radius of \(La^{3+}\) (atomic number of \(La=57\)) is 1.06 Å. Which one of the following given values will be closest to the radius of \(Lu^{3+}\) (atomic number of \(Lu=71\))?

  • (a) 0.85 Å
  • (b) 1.60 Å
  • (c) 1.40 Å
  • (d) 1.06 Å

Correct Answer: (a)

Explanation: Due to the phenomenon of lanthanoid contraction, the ionic radius of the elements decreases steadily as the atomic number increases across the series. Because of this, \(Lu^{3+}\) is expected to have a smaller ionic radius than \(La^{3+}\).

2. \([Cr(H_{2}O)_{6}]Cl_{3}\) (at. no. of \(Cr=24\)) has a magnetic moment of 3.83 B.M. The correct distribution of 3d electrons in the chromium of the complex is:

  • (a) \(3d_{xy}^{1}, 3d_{yz}^{1}, 3d_{xz}^{1}\)
  • (b) \(3d_{xy}^{1}, 3d_{yz}^{1}, 3d_{z^{2}}^{1}\)
  • (c) \(3d_{x^{2}-y^{2}}^{1}, 3d_{z^{2}}^{1}, 3d_{xz}^{1}\)
  • (d) \(3d_{xy}^{1}, 3d_{x^{2}-y^{2}}^{1}, 3d_{yz}^{1}\)

Correct Answer: (a)

Explanation: Chromium in this complex is in the +3 oxidation state with an electronic configuration of \([Ar] 3d^3\). In an octahedral field, these three electrons occupy the lower energy \(t_{2g}\) orbitals (\(3d_{xy}, 3d_{xz}, 3d_{yz}\)) to provide stability.

3. In the dichromate dianion:

  • (a) 4 Cr-O bonds are equivalent
  • (b) 6 Cr-O bonds are equivalent
  • (c) all Cr-O bonds are equivalent
  • (d) all Cr-O bonds are non-equivalent.

Correct Answer: (b)

Explanation: The structure of the \(Cr_{2}O_{7}^{2-}\) ion consists of two tetrahedra sharing one oxygen atom. It contains 6 equivalent terminal Cr-O bonds (three on each chromium), while the two Cr-O bonds involved in the bridging Cr-O-Cr linkage are different.

4. Zinc gives \(H_{2}\) gas with dil. \(H_{2}SO_{4}\) and dil. HCl but not with dil. \(HNO_{3}\) because:

  • (a) \(NO_{3}^{-}\) ion is reduced in preference to hydronium ion
  • (b) dil. \(HNO_{3}\) is a weaker acid than dil. \(H_{2}SO_{4}\) and dil. HCl
  • (c) dil. \(HNO_{3}\) acts as a reducing agent
  • (d) zinc is more reactive than \(H_{2}\)

Correct Answer: (a)

Explanation: When reacting with dilute \(HNO_3\), the \(NO_3^-\) ion is reduced to \(N_2O\) or other nitrogen oxides rather than the hydronium ion being reduced to hydrogen gas. This occurs because nitric acid is a stronger oxidising agent than sulphuric or hydrochloric acid.

5. Amphoteric oxides of Mn and Cr are:

  • (a) MnO and CrO
  • (b) \(Mn_{2}O_{3}\) and \(Cr_{2}O_{3}\)
  • (c) \(MnO_{2}\) and \(Cr_{2}O_{3}\)
  • (d) \(Mn_{2}O_{7}\) and \(CrO_{3}\)

Correct Answer: (c)

Explanation: In transition metals, oxides in intermediate oxidation states often show amphoteric behaviour. \(MnO_2\) and \(Cr_2O_3\) are amphoteric, while oxides in lower states like MnO are basic and those in very high states like \(Mn_2O_7\) are acidic.

6. Which of the following is most basic?

  • (a) \(Ce(OH)_{3}\)
  • (b) \(Lu(OH)_{3}\)
  • (c) \(Yb(OH)_{3}\)
  • (d) \(Tb(OH)_{3}\)

Correct Answer: (a)

Explanation: As the size of the lanthanoid ions decreases from \(Ce^{3+}\) to \(Lu^{3+}\), the covalent character of the M-OH bond increases and the basic strength decreases. Therefore, \(Ce(OH)_3\) is the most basic hydroxide in this group.

7. \(FeCr_{2}O_{4} \xrightarrow{I} Na_{2}CrO_{4} \xrightarrow{II} Cr_{2}O_{3} \xrightarrow{III} Cr\). I, II and III are:

  • (a) \(Na_{2}CO_{3}/air, NH_{4}Cl, \Delta, Al\)
  • (b) \(NaOH/air, C, heat, C, heat\)
  • (c) \(Na_{2}CO_{3}/air, C, heat, C, heat\)
  • (d) \(NaOH/air, heat, Al, heat, C, heat\)

Correct Answer: (a)

Explanation: Chromite ore is first fused with sodium carbonate in air to produce sodium chromate. This is then converted to chromium trioxide using ammonium chloride with heating, and finally reduced to metallic chromium using aluminium (aluminothermic process).

8. The dissolution of \(V_{2}O_{5}\) in 30% \(H_{2}O_{2}\) gives:

  • (a) \(V(O_{2})_{4}^{3-}\)
  • (b) \(VO(O_{2})_{2}^{-}\)
  • (c) \(V_{2}O_{3}\)
  • (d) \(VO_{2}^{-}\)

Correct Answer: (b)

Explanation: Vanadium pentoxide reacts with hydrogen peroxide to form the peroxovanadate ion, \(VO(O_2)_2^-\).

9. Acidified solution of chromic acid on treatment with hydrogen peroxide yields:

  • (a) \(CrO_{5} + H_{2}O\)
  • (b) \(H_{2}Cr_{2}O_{7} + H_{2}O + O_{2}\)
  • (c) \(Cr_{2}O_{3} + H_{2}O + O_{2}\)
  • (d) \(CrO_{3} + H_{2}O + O_{2}\)

Correct Answer: (a)

Explanation: The reaction of an acidified dichromate or chromic acid solution with \(H_2O_2\) produces \(CrO_5\) (chromium pentoxide), which forms a deep blue solution.

10. \((NH_{4})_{2}Cr_{2}O_{7}\) on heating liberates a gas. The same gas will be obtained by:

  • (a) heating \(NH_{4}NO_{2}\)
  • (b) heating \(NH_{4}NO_{3}\)
  • (c) treating \(Mg_{3}N_{2}\) with \(H_{2}O\)
  • (d) heating \(H_{2}O_{2}\) on \(NaNO_{2}\)

Correct Answer: (a)

Explanation: Heating ammonium dichromate liberates nitrogen (\(N_2\)) gas. The same gas is produced by the thermal decomposition of ammonium nitrite (\(NH_4NO_2\)).

11. The stability of complexes of \(Cu^{2+}, Ni^{2+}, Co^{2+}\) and \(Fe^{2+}\) varies in the order:

  • (a) \(Cu^{2+} > Ni^{2+} > Co^{2+} > Fe^{2+}\)
  • (b) \(Cu^{2+} > Fe^{2+} > Ni^{2+} > Co^{2+}\)
  • (c) \(Ni^{2+} > Co^{2+} > Fe^{2+} > Cu^{2+}\)
  • (d) \(Cu^{2+} < Ni^{2+} < Co^{2+} < Fe^{2+}\)

Correct Answer: (a)

Explanation: Generally, complex stability increases as the ionic radius of the metal cation decreases across the series. Following the Irving-Williams series, the stability order for these divalent ions is \(Cu^{2+} > Ni^{2+} > Co^{2+} > Fe^{2+}\).

12. Which of the following nitrates on strong heating leaves the metal as the residue?

  • (a) \(AgNO_{3}\)
  • (b) \(Pb(NO_{3})_{2}\)
  • (c) \(Cu(NO_{3})_{2}\)
  • (d) \(Al(NO_{3})_{3}\)

Correct Answer: (a)

Explanation: Silver is low in the electrochemical series and its oxide is thermally unstable. Consequently, strong heating of silver nitrate leads to its decomposition directly into metallic silver, nitrogen dioxide, and oxygen.

13. Lanthanoids are:

  • (a) 14 elements in the sixth period (atomic no. 90 to 103) filling 4f sublevel
  • (b) 14 elements in the seventh period (atomic number = 90 to 103) filling 5f sublevel
  • (c) 14 elements in the sixth period (atomic number = 58 to 71) filling the 4f sublevel
  • (d) 14 elements in the seventh period (atomic number = 50 to 71) filling 4f sublevel.

Correct Answer: (c)

Explanation: Lanthanoids are a series of 14 elements following lanthanum (atomic numbers 58 to 71) in the 6th period, where the 4f orbitals are being filled.

14. Calomel (\(Hg_{2}Cl_{2}\)) on reaction with \(NH_{4}OH\) gives:

  • (a) \(HgNH_{2}Cl\)
  • (b) \(NH_{2}-Hg-Hg-Cl\)
  • (c) \(Hg_{2}O\)
  • (d) HgO

Correct Answer: (a)

Explanation: Calomel reacts with ammonium hydroxide to produce a black mixture of \(HgNH_2Cl\) and metallic mercury.

15. \(KMnO_{4}\) reacts with ferrous sulphate according to the equation \(MnO_{4}^{-} + 5Fe^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_{2}O\). Here 10 mL of 0.1 M \(KMnO_{4}\) is equivalent to:

  • (a) 30 mL of 0.1 M \(FeSO_{4}\)
  • (b) 40 mL of 0.1 M \(FeSO_{4}\)
  • (c) 20 mL of 0.1 M \(FeSO_{4}\)
  • (d) 50 mL of 0.1 M \(FeSO_{4}\)

Correct Answer: (d)

Explanation: Based on the stoichiometry of the reaction, 1 mole of permanganate reacts with 5 moles of ferrous ions. Therefore, 10 mL of 0.1 M \(KMnO_4\) will react with 50 mL of 0.1 M \(FeSO_4\).

16. In nitroprusside ion, the iron and NO exist as \(Fe^{2+}\) and \(NO^{+}\) rather than \(Fe^{3+}\) and NO. These forms can be differentiated by:

  • (a) estimating the concentration of iron
  • (b) measuring the solid state magnetic moment
  • (c) thermally decomposing the compound
  • (d) measuring the concentration of \(CN^{-}\).

Correct Answer: (b)

Explanation: The oxidation state and electronic configuration of the metal can be established by measuring the magnetic moment, which reflects the number of unpaired electrons (iron as \(Fe^{2+}\) would have zero unpaired electrons in this low-spin complex).

17. The colourless species is:

  • (a) \(VCl_{3}\)
  • (b) \(VOSO_{4}\)
  • (c) \(Na_{3}VO_{4}\)
  • (d) \([V(H_{2}O)_{6}SO_{4}]\cdot H_{2}O\)

Correct Answer: (c)

Explanation: In \(Na_3VO_4\), vanadium is in the +5 oxidation state, which means it has a \(d^0\) configuration. Without any d-electrons, d-d transitions are not possible, making the ion colourless.

18. An aqueous solution of \(FeSO_{4}\), \(Al_{2}(SO_{4})_{3}\) and chrome alum is heated with excess of \(Na_{2}O_{2}\) and filtered. The material obtained are:

  • (a) a colourless filtrate and a green residue
  • (b) a yellow filtrate and a green residue
  • (c) a yellow filtrate and a brown residue
  • (d) a green filtrate and a brown residue.

Correct Answer: (c)

Explanation: \(Fe^{2+}\) is oxidised to \(Fe^{3+}\) and precipitates as brown \(Fe(OH)_3\) residue. Aluminium and chromium form soluble aluminate and yellow chromate, which pass into the filtrate.

19. Which of the following pair of compounds is expected to exhibit same colour in aqueous solution?

  • (a) \(FeCl_{2}, CuCl_{2}\)
  • (b) \(VOCl_{2}, CuCl_{2}\)
  • (c) \(VOCl_{2}, FeCl_{2}\)
  • (d) \(FeCl_{2}, MnCl_{2}\)

Correct Answer: (b)

Explanation: Colour often depends on the number of unpaired d-electrons. Both \(V^{4+}\) (in \(VOCl_2\)) and \(Cu^{2+}\) have one unpaired electron (\(d^1\) and \(d^9\) configurations respectively), which leads to them exhibiting similar light absorption and colour.

20. What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

  • (a) \(Cr^{3+}\) and \(Cr_{2}O_{7}^{2-}\) are formed.
  • (b) \(Cr_{2}O_{7}^{2-}\) and \(H_{2}O\) are formed.
  • (c) \(CrO_{4}^{2-}\) is reduced to + 3 oxidation state of Cr.
  • (d) None of the above.

Correct Answer: (b)

Explanation: Chromate ions are stable in alkaline solutions but are converted to orange dichromate (\(Cr_2O_7^{2-}\)) and water upon acidification with an acid like \(HNO_3\).

21. The reason for the stability of \(Gd^{3+}\) ion is:

  • (a) half-filled 4f-subshell
  • (b) completely filled 4f-subshell
  • (c) possesses noble gas configuration
  • (d) empty 4f-subshell.

Correct Answer: (a)

Explanation: Gadolinium (\(Gd\)) has an atomic number of 64. The \(Gd^{3+}\) ion has a half-filled \(4f^7\) subshell, which provides extra electronic stability.

22. Which of the following is expected to have the highest second ionization enthalpy?

  • (a) V
  • (b) Cr
  • (c) Mn
  • (d) Fe

Correct Answer: (b)

Explanation: The first electron removal from chromium leaves it with a stable half-filled \(3d^5\) configuration. Removing a second electron requires breaking this stable configuration, resulting in a very high second ionization enthalpy.

23. The oxidation state of Ag in Tollens' reagent is:

  • (a) 0
  • (b) +1
  • (c) +2
  • (d) +1.5

Correct Answer: (b)

Explanation: Tollens' reagent is an ammoniacal silver nitrate solution containing the complex \([Ag(NH_3)_2]^+\). In this complex, silver is in the +1 oxidation state.

24. Which of the following nitrates will leave behind a metal on strong heating?

  • (a) Ferric nitrate
  • (b) Copper nitrate
  • (c) Manganese nitrate
  • (d) Silver nitrate

Correct Answer: (d)

Explanation: Nitrates of metals low in the activity series, such as silver nitrate, decompose upon heating to form the metal residue directly rather than a metal oxide.

25. The colour of \(CuCr_{2}O_{7}\) solution in water is green because:

  • (a) \(Cr_{2}O_{7}^{2-}\) ions are green
  • (b) \(Cu^{2+}\) ions are green
  • (c) both ions are green
  • (d) \(Cu^{2+}\) ions are blue and \(Cr_{2}O_{7}^{2-}\) ions are yellow.

Correct Answer: (d)

Explanation: In aqueous solution, the blue colour of \(Cu^{2+}\) ions and the yellow colour of chromate/dichromate ions mix to produce a characteristic green appearance.

26. For a transition metal ion having seven electrons in its d-orbital, the effective magnetic moment will be:

  • (a) 7.98 B.M.
  • (b) 4.90 B.M.
  • (c) 3.87 B.M.
  • (d) 2.83 B.M.

Correct Answer: (c)

Explanation: A \(d^7\) configuration has 3 unpaired electrons. Using the spin-only formula \(\mu = \sqrt{n(n+2)}\), where \(n=3\), the magnetic moment is \(\sqrt{15} \approx\) 3.87 B.M..

27. Most transition metals: (I) form sets of compounds with different oxidation states, (II) form coloured ions, (III) burn vigorously in oxygen, (IV) replace \(H_{2}\) from dilute acids.

  • (a) I, II, III are correct
  • (b) II, III, IV are correct
  • (c) I, II are correct
  • (d) All are correct.

Correct Answer: (c)

Explanation: The characteristic properties of transition metals include variable oxidation states and the formation of coloured ions. However, vigorous burning in oxygen and universal displacement of hydrogen from dilute acids are not properties shared by most members of the block.

28. Which one of the following first row transition elements is expected to have the highest third ionization enthalpy?

  • (a) Vanadium (Z = 23)
  • (b) Manganese (Z = 25)
  • (c) Chromium (Z = 24)
  • (d) Iron (Z = 26)

Correct Answer: (b)

Explanation: Manganese in its +2 state has a stable half-filled \(3d^5\) configuration. Removing a third electron requires breaking this stable arrangement, making its third ionization enthalpy the highest among the choices.

29. Acidified \(KMnO_{4}\) dropped over sodium peroxide produces:

  • (a) hydrogen peroxide
  • (b) mixture of hydrogen and oxygen
  • (c) colourless gas hydrogen
  • (d) colourless gas dioxygen.

Correct Answer: (d)

Explanation: The reaction between acidified permanganate and sodium peroxide yields dioxygen (\(O_2\)) gas through a vigorous redox process.

30. Which of the following metals do not give a metal nitrate on treatment with concentrated \(HNO_{3}\)?

  • (a) Fe and Zn
  • (b) Fe and Pt
  • (c) Pb, Ag and Pt
  • (d) Fe, Ag and Pt

Correct Answer: (b)

Explanation: Iron becomes passive in concentrated nitric acid due to the formation of a protective oxide layer, and platinum is a noble metal that is unreactive towards it.

31. Which of the following is not a characteristic 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) They are chemically very reactive.

Correct Answer: (d)

Explanation: Interstitial compounds are typically chemically inert rather than reactive. They are also known for being very hard and having high melting points while retaining metallic conductivity.

32. Zn reacts with (A) conc. \(HNO_{3}\) and (B) very dil. \(HNO_{3}\) to give X and Y. X and Y are:

  • (a) \(NO_{2}\) and NO
  • (b) \(NO_{2}\) and \(NO_{2}\)
  • (c) NO and \(NO_{2}\)
  • (d) \(NO_{2}\) and \(NH_{4}NO_{3}\)

Correct Answer: (d)

Explanation: Zinc reduces concentrated nitric acid to \(NO_2\) and very dilute nitric acid to \(NH_4NO_3\).

33. The brown ring test for \(NO_{3}^{-}\) is due to the formation of:

  • (a) \([Fe(H_{2}O)_{5}NO]^{2+}\)
  • (b) \([Fe(H_{2}O)_{5}NO]^{+}\)
  • (c) \(Fe(OH)_{2}\)
  • (d) \([Fe(H_{2}O)_{5}NO_{3}]^{-}\)

Correct Answer: (a)

Explanation: The characteristic brown ring in this test is caused by the formation of the complex \([Fe(H_2O)_5NO]^{2+}\).

34. The titanium compound that does not exist is:

  • (a) TiO
  • (b) \(K_{2}TiO_{4}\)
  • (c) \(K_{2}TiF_{6}\)
  • (d) \(TiCl_{3}\)

Correct Answer: (b)

Explanation: Titanium commonly shows oxidation states of +2, +3, and +4. A compound like \(K_2TiO_4\), which would require titanium to be in a +6 oxidation state, does not exist.

35. Heating \(K_{2}Cr_{2}O_{7}\), NaCl and conc. \(H_{2}SO_{4}\) forms:

  • (a) \(CrO_{2}Cl_{2}\)
  • (b) \(CrCl_{2}\)
  • (c) \(Cr_{2}(SO_{4})_{3}\)
  • (d) \(Na_{2}CrO_{4}\)

Correct Answer: (a)

Explanation: This procedure is known as the chromyl chloride test and results in the formation of \(CrO_2Cl_2\) (chromyl chloride) as orange-red vapours.

36. In \(K_{2}Cr_{2}O_{7}\), every Cr is linked to:

  • (a) two O-atoms
  • (b) three O-atoms
  • (c) four O-atoms
  • (d) five O-atoms.

Correct Answer: (c)

Explanation: In the dichromate ion structure, each chromium atom is tetrahedrally coordinated to four oxygen atoms (three terminal and one shared bridging oxygen).

37. Colour of transition metal ions is due to:

  • (a) d-s transition
  • (b) d-d transition
  • (c) f-f transition
  • (d) d-f transition.

Correct Answer: (b)

Explanation: The colour observed in transition metal ions is primarily due to electronic transitions between split d-orbitals (d-d transitions) within the visible light spectrum.

38. Match the following ions with their appearance: P. Aquated \(Mn^{2+}\) ion; Q. \(FeSO_{4}\cdot 7H_{2}O\); R. Aquated \(V^{4+}\) ion; S. Anhydrous \(CuSO_{4}\).

  • (a) 4 2 3 1
  • (b) 3 4 2 1
  • (c) 4 1 3 2
  • (d) 1 2 4 3

Correct Answer: (b)

Explanation: Standard appearances are: \(Mn^{2+}\) is pink (3), \(FeSO_4\cdot 7H_2O\) is green (4), \(V^{4+}\) is blue (2), and anhydrous \(CuSO_4\) is white (1).

39. Given standard potentials \(E^{\circ}\) for \(Cr^{3+}/Cr\) (-0.74V), \(MnO_{4}^{-}/Mn^{2+}\) (1.51V), \(Cr_{2}O_{7}^{2-}/Cr^{3+}\) (1.33V), and \(Cl/Cl^{-}\) (1.36V). The strongest oxidising agent is:

  • (a) \(MnO_{4}^{-}\)
  • (b) Cl
  • (c) \(Cr^{3+}\)
  • (d) \(Mn^{2+}\)

Correct Answer: (a)

Explanation: A substance with a higher standard reduction potential is a more powerful oxidising agent. Here, \(MnO_4^-\) has the highest value (1.51 V), making it the strongest oxidising agent.

40. Which of the following does not represent the correct order of the property stated?

  • (a) \(Sc < Ti < Cr < Mn\): number of oxidation states
  • (b) \(V^{2+} < Cr^{2+} < Mn^{2+} < Fe^{2+}\): paramagnetic behaviour
  • (c) \(Ni^{2+} < Co^{2+} < Fe^{2+} < Mn^{2+}\): ionic size
  • (d) \(Co^{3+} < Fe^{3+} < Cr^{3+} < Sc^{3+}\): stability in aqueous solution

Correct Answer: (b)

Explanation: Paramagnetic behaviour depends on the number of unpaired electrons. \(Mn^{2+}\) (\(d^5\)) has the most unpaired electrons (5) and should be the most paramagnetic, but the order in (b) incorrectly places \(Fe^{2+}\) (\(d^6\), 4 unpaired) as higher.

41. Which of the following is an amphoteric oxide?

  • (a) \(CrO_{3}\)
  • (b) \(Cr_{2}O_{3}\)
  • (c) \(V_{2}O_{3}\)
  • (d) TiO

Correct Answer: (b)

Explanation: As noted previously, transition metal oxides in intermediate oxidation states, such as \(Cr_2O_3\), typically exhibit amphoteric properties.

42. \(X + K_{2}CO_{3} + air \rightarrow Y; Y + Cl_{2} \rightarrow Z\) (Pink). Which of the following is correct?

  • (a) X=Black \(MnO_{2}\), Y=Blue \(K_{2}CrO_{4}\), Z=\(KMnO_{4}\)
  • (b) X=Green \(Cr_{2}O_{3}\), Y=Yellow \(K_{2}CrO_{4}\), Z=\(K_{2}Cr_{2}O_{7}\)
  • (c) X=Black \(MnO_{2}\), Y=Green \(K_{2}MnO_{4}\), Z=\(KMnO_{4}\)
  • (d) X=Black \(Bi_{2}O_{3}\), Y=Colourless \(KBiO_{2}\), Z=\(KBiO_{3}\)

Correct Answer: (c)

Explanation: Fusing black \(MnO_2\) with potassium carbonate in air yields green \(K_2MnO_4\), which then reacts with chlorine gas to form pink \(KMnO_4\).

43. Arrange \(Ce^{3+}, La^{3+}, Pm^{3+}\) and \(Yb^{3+}\) in increasing order of their ionic radii:

  • (a) \(Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}\)
  • (b) \(Ce^{3+} < Yb^{3+} < Pm^{3+} < La^{3+}\)
  • (c) \(Yb^{3+} < Pm^{3+} < La^{3+} < Ce^{3+}\)
  • (d) \(Pm^{3+} < La^{3+} < Ce^{3+} < Yb^{3+}\)

Correct Answer: (a)

Explanation: Due to lanthanoid contraction, ionic radii decrease as the atomic number increases. The correct increasing order based on atomic numbers (57 to 70) is \(Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}\).

44. In which of the following series do all metal ions have a \(3d^{2}\) electronic configuration?

  • (a) \(Ti^{3+}, V^{2+}, Cr^{3+}, Mn^{4+}\)
  • (b) \(Ti^{2+}, V^{4+}, Cr^{6+}, Mn^{7+}\)
  • (c) \(Ti^{4+}, V^{3+}, Cr^{2+}, Mn^{3+}\)
  • (d) \(Ti^{2+}, V^{3+}, Cr^{4+}, Mn^{5+}\)

Correct Answer: (d)

Explanation: Metal ions like \(Ti^{2+}, V^{3+}, Cr^{4+}\), and \(Mn^{5+}\) all have two electrons remaining in their 3d sublevels after losing their respective numbers of valence electrons.

45. The ionic radii of Group 12 metals Zn, Cd and Hg are smaller than those of Group 2 metals because they have:

  • (a) 10 d-electrons which shield the nuclear charge poorly
  • (b) 10 d-electrons which shield the nuclear charge strongly
  • (c) 10 d-electrons which have a large radius ratio
  • (d) 10 d-electrons which have a large exchange energy.

Correct Answer: (a)

Explanation: Group 12 metals have filled d-subshells. Because d-electrons provide poor shielding, the effective nuclear charge increases, pulling the outer electrons closer and resulting in smaller ionic radii compared to Group 2 metals.

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