Answers Assignment No – 3 d & f Block Elements | Class 12th

Answers

Assignment No – 3

D & f Block Elements

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Ans 1. Cr²⁺ is strong reducing agent than Fe²⁺.

Reason: d⁴ → d⁵ occurs in case of Cr²⁺ to Cr³⁺.

But d⁶ →d⁵ occurs in case of Fe²⁺ to Fe³⁺. In a medium (like water) d³ is more stable as compared to d⁵.

 

Ans 2. 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, +2 state becomes more an 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 starts 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).

 

Ans 3. 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 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 actinoides 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.

 

 

Ans 4. (i) 6Fe²⁺ + Cr₂O₇^2– + 14 H⁺ →6Fe³⁺ + 2Cr³⁺ + 7H₂O

(ii) 2MnO₄^– + 3S₂O₃^2– + H₂O → 2MnO₂ + 3SO₄^2– + 3S + 2OH^–.

 

Ans 5. (i) Most of the complexes of transition metals are coloured. This is because of the absorption of radiation from visible light region to promote an electron from one of the d-orbitals to another. In the presence of ligands, the d-orbitals split up into two sets of orbitals having different energies. Therefore, the transition of electrons can take place from one set to another. The energy required for these transitions is quite small and falls in the visible region of radiation. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.

(ii) This is due to electronic configuration of Mn is 3d⁵, 4s² due to which it excites it's multiple electrons to upper Orbit like 3d and these transition provide large no. of oxidation state.

 

 

Ans 6. On the basis of the given data, the black mineral (A) is silver glance, Ag₂S. It is confirmed by the following:

            (i)              It dissolves in sodium cyanide solution in presence of air.

 

Ans 7. The electronic configuration of Sc, cu and Eu are 

₂₁Sc= [Ar] 3d¹4s²

₅₈Ce= [Xe] 4f²5d¹6s¹

₆₀Eu= [Xe] 4f⁷5d⁰6s²

Due to stable configuration of Sc³⁺and Eu²⁺they will not gain electron where as Ce4+can take one electron and become stable thus Ce⁴⁺is good oxidizing agent.