NCERT Solutions for Class 11 Chemistry Chapter 11 The p-Block element

Ans : (i) B to Tl
The electric configuration of group 13 elements is \( \mathrm{ns}^{2} \mathrm{np}^{1}\). Therefore, the most common oxidation state exhibited by them should be +3. However, it is only boron and aluminium which practically show the +3 oxidation state. The remaining elements, i.e., Ga, In, Tl, show both the +1 and +3 oxidation states. On moving down the group, the +1 state becomes more stable. For example, Tl (+1) is more stable than Tl (+3). This is because of the inert pair effect. The two electrons present in the s-shell are strongly attracted by the nucleus and do not participate in bonding. This inert pair effect becomes more and more prominent on moving down the group. Hence, Ga (+1) is unstable, In (+1) is fairly stable, and Tl (+1) is very stable.

The stability of the +3 oxidation state decreases on moving down the group.


(ii) C to Pb
The electronic configuration of group 14 elements is \( n s^{2} n p^{2}\) . Therefore, the most common oxidation state exhibited by them should be +4. However, the +2 oxidation state becomes more and more common on moving down the group. C and Si mostly show the +4 state. On moving down the group, the higher oxidation state becomes less stable. This is because of the inert pair effect. Thus, although Ge, Sn, and Pb show both the +2 and + 4 states, the stability of the lower oxidation state increases and that of the higher oxidation state decreases on moving down the group.
 

Ans : Boron and thallium belong to group 13 of the periodic table. In this group, the +1 oxidation state becomes more stable on moving down the group. 
 \( \mathrm{BCl}_{3}\) is more stable than \( \mathrm{TICl}_{3}\)   because the +3 oxidation state of B is more stable than the +3 oxidation state of Tl. In Tl, the +3 state is highly oxidising and it reverts back to the more stable +1 state. 

Ans : The electric configuration of boron is \( n s^{2} n p^{1}\). It has three electrons in its valence shell. Thus, it can form only three covalent bonds. This means that there are only six electrons around boron and its octet remains incomplete. When one atom of boron combines with three fluorine atoms, its octet remains incomplete. Hence, boron trifluoride remains electron-deficient and acts as a Lewis acid.


 

Ans : Being a Lewis acid, BCl3 readily undergoes hydrolysis. Boric acid is formed as a result
\( \mathrm{BCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow 3 \mathrm{HCl}+\mathrm{B}(\mathrm{OH})_{3}\)
\( \mathrm{CCl}_{4}\) completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When 
\( \mathrm{CCl}_{4}\) and water are mixed, they form separate layers.
\( \mathrm{CCl}_{4}+\mathrm{H}_{2} \mathrm{O} \) -> NO reaction 

Ans : Boric acid is not a protic acid. It is a weak monobasic acid, behaving as a Lewis acid.


\( \mathrm{B}(\mathrm{OH})_{3}+2 \mathrm{HOH} \longrightarrow\left[\mathrm{B}
(\mathrm{OH})_{4}\right]^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\)
It behaves as an acid by accepting a pair of electrons from –OH ion.