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INORGANIC CHEMISTRY REASONING QUESTIONS

INORGANIC CHEMISTRY REASONING QUESTIONS. 1. There is a considerable increase in covalent radius from N to P. However, from As to Bi only small increase in covalent radius is observed. Ans. This is due to the presence of completely filled  d  and/or  f  orbital in

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INORGANIC CHEMISTRY REASONING QUESTIONS

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  1. INORGANIC CHEMISTRY REASONING QUESTIONS

  2. 1. There is a considerable increase in covalent radius from N to P. However, from As to Bi only small increase in covalent radius is observed.

  3. Ans. This is due to the presence of completely filled d and/or f orbital in heavier members.

  4. 2. The ionization enthalpy of the group 15 elements is much greater than that of group 14 elements in the corresponding periods.

  5. Ans. Because of the extra stable half-filled p orbital electronic configuration and smaller size.

  6. 3. R3P=O exist but R3N=O does not.

  7. Ans. Due to the absence of d orbitals in  valence shell of nitrogen and because of inability of Nitrogen to expand its covalency beyond  four, nitrogen cannot form d π–pπ bond.

  8. 4. In solid state PCl5 exists as Ionic compound.

  9. Ans. Since solid phosphorous pentachloride exists as [PCl4]+ [PCl6]-and  hence exhibit some ionic character. [PCl4]+ is tetrahedral and the anion [PCl6]– is octahedral.

  10. 5. Tendency to show –2 oxidation state diminishes from Sulphur to polonium in group 16.

  11. Ans. The outer electronic configuration of group 16 elements is ns2 np4. These elements therefore have the tendency to gain two electrons to complete the octet. Since elctronegativity and I.E. decrease on going down the group, tendency to show –2 oxidation state diminishes.

  12. 6. Ozone (O3) act as a powerful oxidising agent.

  13. Ans. Due to the ease with which it liberates atoms of nascent oxygen (O3 → O2 + O), it acts as a powerful oxidizing agent.

  14. 7. Halogens are coloured.

  15. Ans. Halogens are coloured. This is due to absorption of radiations in visible region which results in the excitation of outer electrons to higher energy level. By absorbing different quanta of radiation, they display different colours. For example, F2 has yellow, Cl2 greenish yellow, Br2 red and I2 violet colour.

  16. 8. Cl2 bleaches a substance permanently  but SO2 does it temporarily.

  17. Ans. Cl2 bleaches a substance permanently because it is due to oxidation but SO2 does it temporarily because it is due to reduction.

  18. 9. Xe does not forms compounds such as XeF3 and XeF5.

  19. Ans. By the promotion of one, two or three electrons from filled p-orbital to the vacant d-orbital in the valence shell, 2,4 or 6 half filled orbitals are formed. Thus Xe can combine only with even number of fluorine atoms and not odd. 

  20. 10. Helium is used for inflating aeroplane tyres & filling balloons for meteorogical observations.

  21. Ans. Helium is a non-inflammable and light gas.

  22. Why does nitrogen does not form pentahalides?

  23. Nitrogen with n=2 has s and p orbitals only.It does not have d orbitals to expand its coelent beyond four.Thats why it does not form pentahalides.

  24. Why does PH3 has lowrer boiling point than NH3?

  25. Unlike NH3 PH3 molecules are not associated through hydrogen bonding in liquid state.That’s why the Boling point of PH3 is lower than NH3.

  26. Why are Pentahalides more covalent than Trihalides ?

  27. Higher the positive oxidation state of central atom ,more will be it’s polarising power which in turn increases the covalent character of bond formed between the central atom and the other atom.

  28. Why is BIH3 the strongest reducing agent amongst the hydrides of group 15 elements?

  29. BIH3 is the strongest reducing agent amongst all the hydrides of group 15 element.Because BIH3 is the least stable amongst the hydrides of group 15.

  30. Why does NH3 acts as a Lewis Base?

  31. Nitrogen atom in NH3 has one lone pair of electron which is available for donation.Therefore it acts as a Lewis base.

  32. Why is the highest oxidation state of metal exhibited in its oxides or fluorides only?

  33. The highest oxidation state of a metal exhibited in its oxides or fluorides because of small size and higher electronegatively oxygen or fluorine can oxidise the metal to its higher oxidation state.

  34. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?

  35. Mn2+ compounds are more stable than Fe2+ towards oxidation to their +3 state because Mn2+ has 3d configuration which has extra stabilty.

  36. Why is Actinoids contraction is greater from elements to elements than lanthanoids contraction?

  37. The 5F electrons are more effectively shielded from nuclear charge.In other words the 5F electrons themseles provide poor shielding from element to element in the series.

  38. Why Cu+ ions is not stable in aqueous solution?

  39. Cu+ ions in aqueous solution undergoes disproportionation ie 2Cu+ -> Cu2+ (aq) + Cu(s)

  40. Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration?

  41. Cr2+ is reducing as its configuration changes from d4 to d3 later having half filled t2g level.On the other hand ,the change from Mn2+ to Mn3+ results in the half filled d5 configuration which has extra stability.

  42. Q.Why is first ionisation energies of boron and aluminium are lower than that of beryllium

  43. Ans. In B and Al,the first electron to be removed lies in p-orbitals while in beryllium and magnesium the valency electrons of an orbital are strongly attracted by nucleus and hence, difficult to remove tan the p-orbitals.

  44. Q. Boric acid can be titrated against sodium hydroxide using phenolphthalein indicator only in presence of polyhydroxy compounds like catechal. Explain .

  45. Ans. Boric acid,B(OH)3 is a very weak lewis acid. It forms a stable complex with polyhydroxy compounds. Due to formation of this stable complex, boric acid acts as a strong acid and hence can be titrated against NaOH

  46. Q. Why (SiH3)3N is a weaker base than (CH3)3N.

  47. Ans. In (SiH3)3N , lone pair of electrons on nitrogen is involved in pπ-dπ back bonding, while in (CH3)3N pπ-dπ back bonding is not possible because of absence of d-orbitals. Hence, (CH3)3N is more basic than (SiH3)3N

  48. Q. The experiment determined N-F bond length in NF3 is greater than the sum of the single bond covalent radii of N ans F. why

  49. Ans. Nitrogen and flourine both are small and have high electron density and hence they repel the bonded pair of electrons leading to larger bond length than expected.

  50. Q. Why Nitrogen is a gas while other members of this group are solids.

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