chemistry

EAMCET QUESTIONS FROM CHAPTER D AND F-BLOCK ELEMENTS PART-2

Problem 9: In the melting point curves of transition metals, one observes a dip in the curves at the end i.e. Cu, Ag & Au and Zn, Cd & Hg have lower melting points when compared to other transition metals. Explain.

Solution: In the last two groups of transition elements i.e. Cu, Ag, Au, Zn, Cd and Hg all the electrons

are paired which can not take part in metallic bonding. As a result, metallic bond in these elements is weak resulting in the lower melting points of these metals.

Problem 10: Enthalpies of atomization of transition elements are higher than those of alkali and alkaline earth metals. Explain.

Solution: The number of unpaired electrons in transition elements are more when compared to those in alkali and alkaline earth metals. As a result, the metallic bonds in transition metals are stronger and enthalpies of atomization are higher than those of alkali and alkaline earth metals.

Problem 11:  Explain the following:

(a) Scandium forms no coloured ions, yet it is regarded as a transition element.

(b) Transition elements have many irregularities in electronic configurations.

Solution:

(a) Scandium in the ground state has one d electron. Hence it is regarded as transition element.

(b) In the transition elements, the (n – 1)d subshell and ns subshell have very small difference in energy. The incoming electron may enter into either ns or (n-1)d subshell. Hence they show irregularities in their electronic configurations.

Problem 12: Explain the following

(a) Chromium is a typical metal while mercury is a liquid metal.

(b) Cobalt (II) is stable in aqueous solution but in the presence of strong ligands, it is a easily oxidised to cobalt (III).

Solution:

(a) chromium has 5 unpaired electrons in its d – orbitals which make its metallic bond very stronger. Whereas in mercury there are no unpaired d electrons, so its metallic bond is very weak.

(b) CO(III) has greater tendency to form complex than CO(II) hence in the presence of ligands CO(II) changes to CO(III).

Problem 13: Write down the products of the following reactions.

(a) CuSO4 solution is treated with KI solution.

(b) AgNO3 solution is added to Na2S2O3 solution.

Solution:

(a) Free iodine is liberated along with the formation of a white precipitate of cupric iodide.

            CuSO4 + 2Kl ————→ Cul2 + K2SO4

             2Cul2 ————→ 2Cul + l2 

(b) A white precipitate of Ag2S2O3 is obtained which turns yellow, brown and finally black on keeping.

            2AgNO3 + Na2S2O3 ————→ Ag2S2O3 + 2NaNO3

            Ag2S2O3 + H2O ————→ Ag2S + H2SO4

                                                         black ppt.

Problem 14: Explain the following

(a) Zinc readily liberates H2 form cold dil.H2SO4 but not from cold conc. H2SO4.

(b) Blue colour of the CuSO4 solution is discharged slowly when an iron rod is dipped into it.

Solution:

(a) Conc. H2SO4 is a covalent compound. Hence does not contain H+ ions. Dilute H2SO4 contains H3O+ which reacts with Zn and liberates H2.

         H2SO4 + H2O ————→ 2H3O+ + SO42–

         Zn + 2H3O+ ————→ Zn2+ + 2H2O + H2O ↑

(b) Fe is more electropositive than Cu, hence it displaces copper form CuSO4solution.

         Fe(s) + CuSO(aq) ————→ FeSO4(aq) + Cu(s) 

Problem 15: An aqueous solution containing one mole of HgI2 and two moles of NaI is orange in colour. On addition of excess NaI the solution becomes colourless. The orange colour reappears on subsequent addition of NaOCl. Explain with equations.

Solution:

         Hgl2 + 2Nal ————→           Na2[Hgl4]

         (orange)                           coloured due to residual Hgl2)

          Hgl2 + Nal(excess) ————→                   Na2[Hgl4]

          (orange)                                        (colourless because there is no residual Hgl2)

         2Na2[Hgl4] + 2NaOCl + H2O ————→ 2Hgl2 + NaCl + 4NaOH + 2Nal3

                                                                          (orange)