Explain the following

1.  Gallium has higher ionisation enthalpy than aluminium.

2.  Boron does not exist as B3+ ion.

3.  Aluminium forms [AlF6]3- ion but boron does not form [BF6]3- ion.

4.  PbX2 is more stable than PbX4.

5.  Pb4+ acts as an oxidising agent but Sn2+ acts as a reducing agent.

6.  Electron gain enthalpy of chlorine is more negative as compared to fluorine.

7.  Tl(NO3)3 acts as an oxidising agent.

8.  Carbon shows catenation property but lead does not.

9.  BF3 does not hydrolyse.

10.  Why does the element silicon, not form a graphite like structure whereas carbon does?


(a) In gallium, due to poor shielding of valence electrons by the intervening 3d electrons. The nuclear charge becomes effective, thus, atomic radius decreases and hence, the ionisation enthalpy of gallium is higher than that of aluminium.
(b) Due to small size of boron, the sum of its first three ionisation enthalpies is very high. This prevent it to form +3 ions and force it to form only covalent compound. That is why boron does not exist as B3+ ion.
(c) Aluminium forms [AlF6]3- ion because of the presence of vacant d-orbitals so it can expand its coordination number from 4 to 6. In this complex, Al undergoes sp3d2 hybridisation.
On the other hand, boron does not form [BF6]3- ion, because of the unavailability of d0orbitals as it cannot expand its coordination number beyond four. Hence, it can form [BF4]- ion (sp3 hybridisation).
(d) Due to inert pair effect, Pb in +2 oxidation state is more stable than in +4 oxidation state hence PbX2 is more stable than PbX4.
(e) Due to inert pair effect, tendency to form +2 ions increases down the group, hence Pb2+ is more stable than Pb4+. That is why Pb4+ acts as an oxidising agent while Sn2+ is less stable than Sn4+ and hence Sn2+ acts as a reducing agent.
Sn2+Reducing agentSn4++2e-Pb4+Oxidising agent+2e-Pb2+
(f) Electron gain enthalpy of Cl is more negative than electron gain enthalpy of fluorine because when an electron is added to F, the added electron goes to the smaller n=2. For n=3 quantum level (in Cl) the added electron occupies a larger region of space and the electron-electron repulsion is much less
(g) Due to inert pair effect, Tl in +1 oxidation state is more stable than that of +3 oxidation state. Therefore, Tl(NO3)3 acts as an oxidising agent.
(h) Property of catenation depends upon the atomic size of the element. Down the group size increases and electronegativity decreases, thus the tendency to show catenation decreases. As the size of C is much smaller than Pb, therefore, carbon show property of catenation but lead does not show catenation
(i) Unlike other boron halides, BF3 does not hydrolyse completely. However, it form boric acid and fluoroboric acid. This is because the HF first formed reacts with H3BO3.
BF3+3H2OH3BO3+3HF×4H2BO3+4HFH+[BF4]+3H2O×34BF3+3H2OH3BO3+3[BF4]+3H
(j) In graphite, C is sp2 hybridised. Carbon due to its smallest size and highest electronegativity among group 14 elements has strong tendency to form - multiple bonds while silicon due to its larger size and less electronegativity has poor ability to form - multiple bonds. That is why the element silicon does not form a graphite like structure.