Explain the following :

Question:

 Explain the following :

(i) Gallium has a higher ionisation enthalpy than aluminium.

(ii) Boron does not exist as B3+ ion.

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

(iv) PbX2 is more stable than PbX4.

(v) Pb4+ acts as an oxidising agent but Sn2+ acts as a reducing agent.

(vi) Electron gain enthalpy of chlorine is more negative as compared to fluorine.

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

(viii) Carbon shows catenation property but lead does not.

(ix) BF3 does not hydrolyse.

(x) Why does the element silicon, not form graphite-like structure whereas carbon does.

Solution:

(i); Due to the ineffective shielding of valence electrons by the intervening 3d electrons, the effective nuclear charge on Ga is slightly higher than that on Al.

(ii); Boron has three electrons in the valence shell. Because of its small size and a high sum of the first three ionization enthalpies, it doesn’t form B3+ ion

(iii); Aluminum has empty d-orbital to accommodate the electrons from the fluorine atom but boron has no empty d-orbital to accommodate the electrons.

(iv); Pb is the member of the group 14 of the periodic table (carbon family). The valence shell electronic configuration of this element is ns2 np2 type. Pb can show variable oxidation states of +2 and +4.

On moving top to bottom in the group, the lower oxidation state i.e +2 is more stable than the higher one due to the inert pair effect.

Thus, Pbdue to the inert pair effect (poor shielding of the inner electronic orbitals) shows +2 as a stable oxidation state rather than +4.

(v); On moving top to bottom in the group, the lower oxidation state i.e +2 is more stable than the higher one due to the inert pair effect.

Thus, Pb due to the inert pair effect shows +2 as stable oxidation state rather than +4. So, by accepting two electrons Pb4+ will get converted into Pb2+. In the case of Sn, the inert pair effect is very less, the +4 oxidation state is more stable than +2.

(vi); Fluorine has a very small size thus the electrons of the 2p orbital experiences the interelectronic repulsion. Thus, the electron coming out of the orbital does not experience much of the attraction from the nucleus hence the negative electron gain enthalpy of fluorine is less than that of chlorine.

(vii); Due to the poor shielding effect of the inner electronic orbitals, the +3 oxidation state of Tl is less stable than its +1 oxidation state.

Since in Tl(NO3)3, the oxidation state of Tl is +3, therefore, it can easily gain two electrons to form TlNO3 in which the oxidation state of Tl is +1.

(viii); The catenation property depends on atom size and the M-M bond energy. Smaller the atomic radii and the greater the M-M bond energy, the greater is the tendency to show catenation. On moving down a group the atomic size increases and the M-M bond energy also reduces.

(ix); BF3 does not hydrolyze completely. Instead, it hydrolyzes incompletely to form boric acid and fluoroboric acid. This is because the HF first formed reacts with H3BO3

Thus, BF3 does not hydrolyze completely.

(x) Graphite is a macromolecule that is made up of hexagonal carbon rings.

This property is due to the catenation of the carbon atoms. Smaller the atomic radii and the greater the M-M bond energy, the greater is the tendency to show catenation.

On moving down a group the atomic size increases and the M-M bond energy also reduces.

Thus, Carbon shows the catenation whereas silica does not.

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