Solved Important Questions d & f Block Elements 

Solved Important Questions d & f Block Elements:

Que 1. Cu+ is not stable in an aqueous solution. Why?
Ans 1. In an aqueous solution, Cu+ undergoes disproportionation to form a more stable Cu2+ ion. 2Cu+ (aq) → Cu2+ (aq) + Cu(s)
The higher stability of Cu2+ in an aqueous solution may be attributed to its greater negative ∆hydH
than that of Cu+. It compensates for the second ionization enthalpy of Cu involved in the formation of Cu2+ ions.

Que 2. Which is a stronger reducing agent Cr2+ or Fe2+ and why?
Ans 2. Cr2+ is a stronger reducing agent than Fe2+ because after the loss of one electron Cr2+ becomes Cr3+ which has a more stable t2g3 (half-filled) configuration in a medium like water.

Que 3. Arrange the following increasing order of acidic character: CrO3, CrO, Cr2O3
Ans 3. CrO <Cr2O3<CrO3. The higher the oxidation state the more will be the acidic character.

Que 4. Calculate the ‘spin only’ magnetic moment of M2+(aq) ion. (Z=27)
Ans 4. The electronic configuration of the M2+ ion (Z=27) would be M2+(aq): (Ar) 3d7 It would contain three unpaired electrons. The ‘spin only’ magnetic moment is given by the relation:
µ = n (n + 2) BM = 3 (3 + 2) BM = 3.87 BM

Que 5. Why do Zr and Hf exhibit almost similar properties?
Ans 5. Zr and Hf have similar ionic sizes due to their lanthanoid contraction. So, they exhibit similar properties.

Que 6. Why are Zn, Cd, and Hg not regarded as transition elements?
Ans 6. Zn, Cd, Hg neither in their ground state nor in the oxidized state have partially filled d‐orbital. Thus, they are not regarded as transition elements.

Que 7. What are alloys? Name one important alloy which contains some of the lanthanoid metals. Mention its use.
Ans 7. Alloys are homogeneous mixtures of metals with metals or non‐metals. Misch metal (pyrophoric alloy) consists of lanthanoid metal Ce = 40.5%, neodymium 44%, iron 4‐5%, and traces of S, C, Ca, and Al. Misch metal is used to make bullets, shells, and light flints.

Que 8. La(OH)3 is a stronger base than Lu(OH)3. Why?
Ans 8.  Lu3+ is smaller in size than La3+ due to lanthanoid contraction. Due to the smaller size of Lu3+, the Lu‐O bond is stronger than the La‐O bond in the respective hydroxides. Due to the weaker La‐O bond, La(OH)3 behaves as a stronger base.

Que 9. Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical.
Ans 9.   Lanthanoids exhibit oxidation states of +2, +3, and +4. This is because of the large energy gap between 4f, 5d, and 6s subshells. Actinoids show +3, +4, +5, +6, and +7 oxidation states because 5f, 6d, and 7s energy levels are nearly the same.

Que 10. What is meant by the term lanthanoid contraction? What is it due to and what consequences does it have on the chemistry of elements following lanthanoids in the periodic table?
Ans 10.  Lanthanoid contraction: Steady decrease in the size of the lanthanoids with an increase in the atomic number across the period. The electrons of 4f orbitals offer imperfect/poor shielding effect in the same subshell.
Consequence:
i) Due to this 5d series elements have nearly the same radii as that of the 4d series.
ii) Decrease in the basic strength from La(OH)3 to Lu(OH)3.
iii) Due to similar atomic size there is difficulty in the separation of lanthanides.

Solved Important Questions d & f Block Elements

Que 11. Compare lanthanoids and actinoids with reference to their:
a. Electronic configuration of atoms.
b. Atomic and ionic sizes.
c. Oxidation states of elements.
d. The general chemical reactivity of elements.
Ans 11. (a) Electronic configuration: lanthanoids have general electronic configuration [Xe]4f1−145d0−16s2. Actinoids have general electronic  configuration [Rn]5f1−146d0−17s2.
In lanthanoids, 4f orbital is filled and in actinoids, 5f orbital is filled.
(b) Atomic and ionic sizes: The atomic and ionic sizes of lanthanoids and actinoids decrease with an increase in the atomic number due to lanthanoid and actinoid contraction.
(c) Oxidation state: Apart from the +3 oxidation state, lanthanoids show +2 and +4 oxidation states due to a large energy gap between 4f and 5d subshells. Whereas actinoids show a large number of oxidation states due to a small energy gap between 5f and 6d subshells.
(d) Chemical reactivity: Highly electropositive lanthanoids have almost similar chemical reactivity. Actinoids (electropositive and highly reactive) are more reactive (especially in the finely divided state) than lanthanoids.

Que 12. How would you account for the following?
a). Mn(III) undergoes disproportionation reaction easily.
b). Co(II) is easily oxidized in the presence of strong ligands.
Ans 12. a).  Mn3+ is less stable and changes to Mn2+ which is more stable due to the half‐filled d‐ orbital configuration. That is why Mn3+ undergoes disproportionation reaction.
b). Co(II) has electronic configuration 3d74s0, i.e., it has three unpaired electrons. In the presence of strong ligands, two unpaired electrons in 3d‐subshell pair‐up and third unpaired electron shifts to higher energy subshell from where it can be easily lost and hence oxidized to Co(III).

Que. 13 Give reasons for the following:
i) First ionization energies of 5d elements are higher than those of 3d and 4d elements.
ii) Actinoid contraction is greater from element to element than lanthanoid contraction.
Ans 13. i) Because of the weak shielding (or screening) effect of 4f electrons, the effective nuclear charge acting on the valence electrons in 5d elements is quite high. Hence, the first ionization energies of 5d elements are higher than those of 3d and 4d elements.
ii) This is because the 5f electrons themselves provide poor shielding from element to element in the series.

Que 14. Explain the following observations about the transition/inner transition elements:
i) There is, in general, an increase in the density of elements from titanium (Z=22) to copper (Z=29).
ii) There occurs much more frequent metal‐metal bonding in compounds of heavy transition elements (3rd series).
iii) The greatest number of oxidation states are exhibited by the members in the middle of a transition series.
Ans 14. i) Because of the decrease in atomic size from titanium to copper.
ii) Because of high enthalpies of atomization of heavy transition elements.
iii) Because of the involvement of both (n‐1)d and ns electrons in bonding.

Q 15: Assign reasons for the following:
a) The enthalpies of atomization of transition elements are high.
b) The transition metals and many of their compounds act as good catalysts.
c) There is a gradual decrease in the atomic sizes of transition elements in a series with increasing atomic numbers.
d) The transition elements have a great tendency for complex formation.
e) Transition metals generally form coloured compounds.
Ans 15: a) This is because transition elements have strong metallic bonds as they have a large number of unpaired electrons, therefore they have greater interatomic overlap.
b) The catalytic activity of transition metals is attributed to the following reasons‐
i) Because of their variable oxidation state, transition metals form unstable intermediate compounds and provide a new path with the lower activation energy for the reaction.
ii) In some cases, the transition metal provides a suitable large surface area with free valencies on which reactants are adsorbed.
c) There is a gradual decrease in the atomic sizes of transition elements in a series with increasing atomic numbers due to poor shielding effects of the d‐electrons, the net electrostatic attraction between the nucleus and the outermost electrons increases
d) orbitals of suitable energy, the small size of cations, and higher nuclear charge.
e) Due to the presence of unpaired electrons in d‐orbitals which undergoes a d‐d transition.

Solved Important Questions d & f Block Elements

Que 16. Explain briefly how the +2 state becomes more and more stable in the first half of the first-row transition elements with increasing atomic number?
Ans 16. In transition elements, there are greater horizontal similarities in the properties in contrast to the main group elements because of similar ns2 common configuration of the outermost shell. An examination of common oxidation states reveals that excepts scandium, the most common oxidation state of first-row transition elements is +2 which arises from the loss of two 4s electrons. This means that after scandium, d-orbitals become more stable than the s-orbital. Further, the +2 state becomes more and more stable in the first half of first-row transition elements with increasing atomic number because 3d orbitals acquire only one electron in each of five 3d orbitals (i.e. remains half-filled) and electronic repulsion is the least, and nuclear charge increases. In 2nd half of first-row transition elements, electrons start pairing up in 3d orbitals. (Ti2+ to Mn2+ electronic configuration changes from 3d2 to 3d5 but in 2nd half i.e. Fe2+ to Zn2+ it changes from d6 to d10).

Que 17. The Chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
Ans 17. Among the actinoids, there is a greater range of oxidation states as compared to lanthanoids. This is in part due to the fact that 5f, 6d, and 7s levels are of very much comparable energies and the frequent electronic transition among these three levels is possible. This 6d-5f transition and a larger number of oxidation states among actinoids make their Chemistry more complicated particularly among the 3rd to 7th elements. following examples of oxidation states of actinoids. Justify the complex nature of their Chemistry.
(i) Uranium exhibits oxidation states of +3, +4, +5, +6 in its compounds. However, the dominant oxidation state in actinoids is +3.
(ii) Nobelium, No is stable in +2 state because of completely filled f14 orbitals in this state.
(iii) Berkelium, Bk in +4 oxidation state is more stable due to f7 (exactly half-filled) configuration.

Que 18. Write complete chemical equations for:
(i) Oxidation of Fe2+ by Cr2O72– in acid medium.
(ii) Oxidation of S2O32– by MnO4 in neutral aqueous solution.
Ans 18. (i) 6Fe2+ + Cr2O72– + 14 H+ →6Fe3+ + 2Cr3+ + 7H2O
(ii) 2MnO4 + 3S2O32– + H2O → 2MnO2 + 3SO42– + 3S + 2OH.

Que 19. (i) A black mineral (A) on treatment with dilute sodium cyanide solution in presence of air gives a clear solution of (B) and (C).
(ii) The solution (B) on reaction with zinc gives a precipitate of a metal (D).
(iii) (D) is dissolved in dilute HNO3 and the resulting solution gives a white precipitate (E) with dilute HCl.
(iv) (E) on fusion with sodium carbonate gives (D).
(v) (E) dissolves in ammonia solution giving a colourless solution of
(F). Identify (A) to (F) and give chemical equations for reactions at steps (i) to (v).
Ans 19. On the basis of the given data, the black mineral (A) is a silver glance, Ag2S. It is confirmed by the following:
(i) It dissolves in sodium cyanide solution in presence of air.

Que 20. Explain which one of the following is a good oxidizing agent: Sc3+, Ce4+or Eu2+ (atomic of Sc=21, Ce=58, Eu=63)
Ans 20. The electronic configuration of Sc, Ce and Eu are
21Sc= [Ar] 3d14s2
58Ce= [Xe] 4f25d16s1
60Eu= [Xe] 4f75d06s2
Due to the stable configuration of Sc3+and Eu2+they will not gain electron whereas Ce4+can take one electron and become stable thus Ce4+is good oxidizing agent.

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