11th Class Chemistry The p-Block Elements-I

  • question_answer 79)   Explain the following : (i) Gallium has higher ionisation enthalpy than aluminium. (ii) Boron does not exist as \[{{B}^{3+}}\] ion. (iii) Aluminium forms \[{{[Al{{F}_{6}}]}^{3-}}\] ion but boron does not form\[{{[B{{F}_{6}}]}^{3-}}\] ion. (iv) \[Pb{{X}_{2}}\]is more stable than \[Pb{{X}_{4}}\]. (v) \[P{{b}^{4+}}\]acts as an oxidising agent but \[S{{n}^{2+}}\] acts as a reducing agent. (vi) Electron gain enthalpy of chlorine is more negative as compared to fluorine. (vii) acts as an oxidising agent. (viii) Carbon shows catenation property but lead does not. (ix) \[B{{F}_{3}}\]does not hydrolyse. (x) Why does the element silicon not form a graphite like structure whereas carbon does?

    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 spinner