Answer:
(i) In\[Ga\], there are ten d-electrons
present in the penultimate shell which do not screen the nuclear charge
effectively. On account of this outermost electron is held firmly. As a result,
the ionisation enthalpy of \[Ga\]is slightly higher than aluminium.
(ii) Since, the sum of first,
second and third successive ionisation energies of the boron is very high (6887
kJ\[mo{{l}^{-1}}\]), it does not form \[{{B}^{3+}}\] ion.
(iii) Boron has no vacant d-orbitals
in valence shell hence excitation of electrons is not possible and hence \[{{[Al{{F}_{6}}]}^{3-}}\]
is possible but \[{{[B{{F}_{6}}]}^{3-}}\] is not possible.
(iv) Since, inert pair effect is
maximum in lead, the lower valency of lead, i.e., \[Pb{{X}_{2}}\] is more
stable than tetravalency, i.e.,\[Pb{{X}_{4}}\].
(v) In lead, \[P{{b}^{2+}}\]
state is more stable than \[P{{b}^{4+}}\] state due to inert pair effect. Thus,
\[P{{b}^{4+}}\] tries to attain the more stable state, i.e., \[P{{b}^{2+}}\] by
acting as an oxidising agent.
\[P{{b}^{4+}}+2e\xrightarrow{{}}P{{b}^{2+}}\]
In tin, \[S{{n}^{2+}}\] state is
less stable than\[S{{n}^{4+}}\]. Thus,\[S{{n}^{2+}}\] tries to attain the more
stable state, i.e., \[S{{n}^{4+}}\], by acting as a reducing agent.
\[S{{n}^{2+}}\xrightarrow{{}}S{{n}^{4+}}+2e\]
(vi) Atomic size of fluorine is
very small as compared to that of chlorine. Thus, electron density of fluorine
is very high. Repulsion takes place between electrons of fluorine already
present and electron to be added. This is the reason why electron gain enthalpy
of fluorine is less negative as compared to that of chlorine.
(vii) In thallium +1 oxidation state is
more stable than +3 oxidation state. Thus, \[Tl{{(N{{O}_{3}})}_{2}}\] acts as
an oxidising
\[T{{l}^{3+}}+2e\xrightarrow{{}}T{{l}^{+}}\]
(viii) C-C bond energy is high while \[PbPb\]bond
energy is negligible. Thus, carbon shows catenation while lead does not.
(ix) \[B{{F}_{3}}\]does not
hydrolyse but forms addition product with water as B-F bond is very strong.
\[B{{F}_{3}}+{{H}_{2}}O\xrightarrow{{}}{{H}^{+}}[B{{F}_{3}}OH]\]
(x) Silicon does not form \[p\pi
-p\pi \] multiple bonds due to its large size. Due to this reason silicon does
not form a graphite like structure.
You need to login to perform this action.
You will be redirected in
3 sec