Answers Assignment No – 2 d & f – Block Elements | Class 12th

Answers

Assignment No – 2

d & f – Block Elements

---

Ans 1. In aqueous solution Cu+ undergoes disproportionation to form a more stable Cu2+ion.

2Cu⁺ (aq) → Cu²⁺ (aq) + Cu(s)

The higher stability of Cu²⁺ in aqueous solution may be attributed to its greater negative ∆_hydH

 than that of Cu⁺. It compensates the second ionisation enthalpy of Cu involved in the formation of Cu²⁺ ions.

Ans 2. Cr²⁺ is a stronger reducing agent than Fe²⁺ because after the loss of one electron Cr²⁺ becomes Cr³⁺ which has more stable t_2g³ (half filled) configuration in medium like water.

Ans 3. CrO <Cr₂O₃<CrO₃. Higher the oxidation state, more will be acidic character.

Ans 4. 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

Ans 5. Zr and Hf have similar ionic size due to its lanthanoid contraction. So, they exhibit similar properties.

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

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.

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

Ans 9.   Lanthanoids exhibit oxidation states of +2, +3 and +4. This is because of 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 same.

Ans 10.  Lanthanoid contraction: Steady decrease in the size of the lanthanoids with 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 same radii as that of 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 separation of lanthanides. 

Ans 11. Check last Part of the chapter

Ans 12. a.  Mn³⁺ is less stable and changes to Mn²⁺ which is more stable due to 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).

Ans 13. i) Because of 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 ionisation 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.

Ans 14. i) Because of 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.

Ans 15: a) This is because transition elements have strong metallic bonds as they have 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 lower activation energy for the reaction.

ii) In some cases, the transitions 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) orbital's of suitable energy, small size of cations and higher nuclear charge.

e) Due to presence of unpaired electrons in   d‐orbitals which undergoes d‐d transition.