The P Block Elements

Chemistry Class 11


1   Discuss the pattern of variation in the oxidation states of$\\$ (i) $B$ to $Tl$ and$\\$ (ii)$ C$ to $Pb.$

Solution :

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

(ii) $C $ to $Pb$$\\$ The electronic configuration of group $14$ elements is $ns^2 np^ 2 $. Therefore, the most common oxidation state exhibited by them should be $+4$. However, the $+2$ oxidation state becomes more and more common on moving down the group. $C$ and $Si$ mostly show the $+4$ state. On moving down the group, the higher oxidation state becomes less stable. This is because of the inert pair effect. Thus, although $Ge, Sn,$ and $Pb$ show both the $+2$ and $+ 4$ states, the stability of the lower oxidation state increases and that of the higher oxidation state decreases on moving down the group.$\\$ $\begin{array}{|c|c|c|c|} \hline \text{Group 14 element}&\text{Oxidation state}\\ \hline C&+4 \\ \hline Si & +4 \\ \hline Ge,Sn,Pb & +2,+4 \\ \hline \end{array}$$\\$ $\overset{\text{stability of +2 state increases}}\rightarrow$$\\$ $ C \ Si\ Ge \ Sn \ Pb $$\\$ $\underset{\text{stability of +4 state decreases}}{\rightarrow}$$\\$

2   How can you explain higher stability of $BCl _3$ as compared to $TlCl _3$ ?

Solution :

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. $BCl_ 3$ is more stable than $TlCl _3$ 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.

3   Why does boron trifluoride behave as a Lewis acid?

Solution :

The electric configuration of boron is $ns ^2 np ^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 (sextet is available). When one atom of boron combines with three fluorine atoms, it form sextet. Hence, boron trifluoride remains electron-deficient and acts as a Lewis acid and always ready to accept electron pair.

4   Consider the compounds,$ BCl_ 3$ and $CCl _4$ . How will they behave with water? Justify.

Solution :

Being a Lewis acid, $BCl _3$ readily undergoes hydrolysis. Boric acid is formed as a result. $BCl _3 + 3 H _2 O \to 3 HCl + B (OH ) _3$$\\$ $CCl_4$ completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When $CCl _4$ and water are mixed, they form separate layers. $CCl _4 + H _2 O \to$ No reaction

5   Is boric acid a protic acid? Explain.

Solution :

Boric acid is not a protic acid. It is a weak monobasic acid, behaving as a Lewis acid.$\\$ $B ( OH )_3 +2 HOH \to[ B ( OH )_4 ]^-+ H _3 O^+$$\\$ It behaves as an acid by accepting a pair of electrons from $-OH$ ion.

6   Explain what happens when boric acid is heated.

Solution :

On heating orthoboric acid $(H _3 BO_ 3 )$ at $370 K$ or above, it changes to metaboric acid $(HBO _2 ).$ $\\$ On further heating, this yields boric oxide $B_ 2 O_ 3 .$$\\$ $H_3BO_3\quad \overset{\Delta}{\underset{370K}{\longrightarrow}}\quad \underset{\text{Metaboric acid}} {HBO_2}\quad \overset{\Delta}{\underset{red hot}{\longrightarrow}}\quad \underset{\text{Boric oxide}}{B_2O_3}$

7   Describe the shapes of $BF_ 3$ and $BH _4^{-}$ . Assign the hybridisation of boron in these species.

Solution :

(i) $BF_ 3$ As a result of its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of three $sp^ 2$ hybridised orbitals of boron with the$p $orbitals of three halogen atoms. Boron is sp^2 hybridised in $ BF_3$

(ii) $BH _ 4^{-}$ $\\$ Boron-hydride ion ( $BH ^-_ 4$ ) is formed by the sp 3 hybridisation of boron orbitals. Therefore, it is tetrahedral in structure.

8   Write reactions to justify amphoteric nature of aluminium.

Solution :

A substance is called amphoteric if it displays characteristics of both acids and bases.$\\$ Aluminium dissolves in both acids and bases, showing amphoteric behaviour.$\\$ (i) $ 2 Al ( s ) + 6 HCl ( aq ) $ $\longrightarrow $ $2 Al^{3+} ( aq )$ $+ 6 Cl^-( aq ) + 3 H_ 2( g )$ $\\$ (ii)$ 2 Al ( s ) + 2 NaOH ( aq ) + 6 H_ 2 O ( l ) \longrightarrow 2 Na ^+[ Al ( OH )4 ]^- ( aq ) +3 H _2( g )$ $\\$

9   What are electron deficient compounds? Are $BCl_ 3$ and $SiCl _4$ electron deficient species? Explain.

Solution :

In an electron-deficient compound, the octet of electrons is not complete, i.e., the central metal atom has an incomplete octet. Therefore, it needs electrons to complete its octet.$\\$ (i) $BCl_ 3$ $\\$ $BCl _3$ is an appropriate example of an electron-deficient compound. B has 3 valence electrons. After forming three covalent bonds with chlorine, the number of electrons around it increases to 6. However, it is still short of two electrons to complete its octet.$\\$ (ii) $SiCl_ 4$ $\\$ The electronic configuration of silicon is $ns ^2 np^ 2$ . This indicates that it has four valence electrons. After it forms four covalent bonds with four chlorine atoms, its electron count increases to eight. Thus, $SiCl _4 $ is not an electron-deficient compound.

10   Write the resonance structures of $CO _3 ^{2 -}$ and $HCO _3 ^- .$

Solution :

(a)$CO _3 ^{2 -}$

(b)$HCO _3 ^- .$

There are only two resonating structures for the bicarbonate ion.