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)
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