question_answer

  Explain the following observations. (i) Nitrogen is much less reactive than phosphorus. (ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water. (iii) Sulphur has greater tendency for catenation than oxygen in the same group. (b) Draw the structures of the following molecules. (i)                                                             (ii)                                                                                            [3 + 2] Or (a) Explain the following observations. (i)is more acidic than (ii) Fluorine does not exhibit any positive oxidation state. (iii) Helium forms no real chemical compound. (b) Draw the structures of the following molecules. (i)                                                                (ii)                                                                                                    [3 + 2 ]

Answer:

  (a) (i) Phosphorus exists as ?4 with structure as shown Nitrogen exists as N; with structure N =N BE of (P?P) bonds in  is 215 kJ while BE of (N = N) bond in N;, is 946 kJ . Greater the BE greater the stability and smaller the reactivity. Thus, nitrogen is very less reactive than phosphorus.                                                      [1] (ii) H?F molecules are joined by intermolecular H-bond such that each molecule is joined at two points by other H?F molecules. In case of , each molecule is joined to four water molecules. Hence, larger heat is required to break H-bonding in molecules to convert them into vapours as compared to that in H?F molecules. Hence, H?F boils at lower temperature than .                              [1] (iii) Oxygen has no d-orbital, hence unpaired electrons cannot be excited to expand its valency beyond two. Thus, oxygen exists as . Sulphur has vacant d-orbital, thus it can expand its valency beyond two. Also (BE) of (S?S) bond in S8 is smaller than that of (O=O) in . Thus, catenation in sulphur is larger than that in oxygen             [1] (b) (i)                                           [1] (ii) [1] Or (a) (i)                     As we go down the group, electro negativity decreases BE of (H?S) bond in  is smaller than that of (O?H) bond in . Thus, ionization of . Is larger than that of. Thus,  is more acidic than[1] (ii) Fluorine is the most electronegative element of the Periodic Table. It can gain electron to form stable Ne configuration . There is no d-orbital in which electron can be excited to form positive oxidation state.               [1] Thus, F shows only negative (-1) oxidation state. (ii) He (l52) Most stable and with maximum ionisation potential. Thus, it does not form or             [1] (b) (i) (ii)