For Group 13

(a) Atomic Size On moving down the group for each successive member, one extra shell of electrons is added and therefore, atomic radius is expected to increase. However, a deviation can be seen.

Atomic radius of Ga is less than that of Al due to presence of additional 10 d - electrons, which offer poor screening effect to the outer electron.

(b) Ionisation Enthalpy The ionisation enthalpy values as expected from general trends do not decrease smoothly down the group. The decrease from B to Al is associated with increase in size.

The observed discontinued between Al and Ga and between In and TI due to low screening effect of $d$ and $f$-electrons which compensates increased nuclear charge.

(c) Metallic or Electropositive Character Boron is a semi-metal (metalloid) due to very high ionisation enthalpy. All others are metals and metallic character first increases from B to Al as size increases. From Al to TI decrease due to poor shielding of $d$ - and $f$-electrons.

(d) Oxidation States As we move down the group, the stability of +3 oxidation state decreases while that of +1 oxidation state progressively increases. In other words, the order of stability of +1 oxidation state increase in the order. $\mathrm{Al}<\mathrm{Ga}<\ln <\mathrm{TI}$. Infact, in Ga , In and TI , both +1 and +3 oxidation states are observed.

(e) Nature of Halides These elements react with halogens to form trihalids (except $\mathrm{TIl}_3$ )

$$2 E(s)+3 X_2(g) \longrightarrow 2 \mathrm{EX}_3(s) \quad[X=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}]$$

Boron in halides are electron deficient molecules and behave as Lewis acids. The Lewis character decreases in the order: $\mathrm{BI}_3>\mathrm{BBr}_3>\mathrm{BCl}_3>\mathrm{BF}_3$

For Group 14

(a) Atomic Size There is considerable increase in covalent radius from C to Si thereafter from Si to Pb as small increase in radius is observed. This is due to the presence of completely filled $d$ and $f$-orbitals in heavier member.

(b) Ionisation Enthalpy The first ionisation enthalpy of group 14 members is higher than the corresponding members of group 13. The influence of inner core electrons is visible here. In general the ionisation enthalpy decreases down the group. Small decrease in $\Delta$; H from Si to Ge to Sn and slight increase in $\Delta ; \mathrm{H}$ from Sn to Pb is the consequence of poor shielding effect of intervening $d$ and $f$-orbitals and increase in size of the atom.

(c) Metallic Character Metallic character increases down the group C (non-metal) $\mathrm{Si}, \mathrm{Ge}$ (metalloid) $\mathrm{Sn}, \mathrm{Pb}$ (metals).

(d) Oxidation States The group 14 elements have four electrons in the outermost shell. The common oxidation states exhibited by these elements are +4 and +2 . Carbon also exhibits negative oxidation states. Since, the sum of the first four ionisation enthalpies is very high, compounds in +4 oxidation states are generally covalent in nature. In heavier members the tendency to show +2 oxidation state increases in the $\mathrm{Ge}<\mathrm{Sn}<\mathrm{Pd}$ due to inert pair effect.

(e) Nature of Halides These elements can form halides of formula $M X_2$ and $M X_4$ (where, $X=F, C l, B r, I)$. Except carbon, all other members react directly with halogen under suitable condition to make halides. Most of $M X_4$ are covalent with $s p^3$ hybridisation and tetrahedral in structure. Exceptions are $\mathrm{SnF}_4$ and $\mathrm{PbF}_4$ which are ionic in nature. Heavier members Ge to Pb are able to make halides of formula $M X_2$. Stability of halides increases down the group.

(a) In $\mathrm{AlCl}_3, \mathrm{Al}$ has only six electrons in its valence shell. It is an electron deficient species. Therefore, it acts as a Lewis acid (electron acceptor).

(b) In $\mathrm{BF}_3$ boron has a vacant $2 p$-orbital and fluorine has one $2 p$-completely filled unutilised orbital. Both of these orbitals belong to same energy level therefore, they can overlap effectively and form $p \pi-p \pi$ bond. This type of bond formation is known as back bonding.

While back bonding is not possible in $\mathrm{BCl}_3$, because there is no effective overlapping between the $2 p$-orbital of boron and $3 p$-orbital of chlorine. Therefore, electron deficiency of B is higher in $\mathrm{BCl}_3$ than that of $\mathrm{BF}_3$. That's why $\mathrm{BF}_3$ is a weaker Lewis acid than $\mathrm{BCl}_3$.

(c) In $\mathrm{PbO}_2$ and $\mathrm{SnO}_2$, both lead and tin are present in +4 oxidation state. But due to stronger inert pair effect, $\mathrm{Pb}^{2+}$ ion is more stable than $\mathrm{Sn}^{2+}$ ion. In other words, $\mathrm{Pb}^{4+}$ ions i.e., $\mathrm{PbO}_4$ is more easily reduced to $\mathrm{Pb}^{2+}$ ions than $\mathrm{Sn}^{4+}$ ions reduced to $\mathrm{Sn}^{2+}$ ions. Thus, $\mathrm{PbO}_2$ acts as a stronger oxidising agent than $\mathrm{SnO}_2$.

(d) $\mathrm{TI}^{+}$is more stable than $\mathrm{TI}^{3+}$ because of inert pair effect.

When an aqueous solution of borax is acidified with HCl boric acid is formed.

$$\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+2 \mathrm{HCl}+5 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaCl}+\underset{\text { Boric acid }}{4 \mathrm{H}_3 \mathrm{BO}_3}$$

Boric acid is a white crystalline solid. It is soapy to touch because of its planar layered structure. Boric acid is a weak monobasic acid. It is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion.

$$\mathrm{B}(\mathrm{OH})_3+2 \mathrm{HOH} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}_3 \mathrm{O}^{+}$$

(a) TICl more stable than $\mathrm{TICl}_3$ due to inert pair effect. $\mathrm{TICl}_3$ is less stable and covalent in nature but TICl is more stable and ionic in nature.

(b) Due to absence of $d$-orbitals, Al does not show inert pair effect. Hence, its most stable oxidation state is +3 . Thus, $\mathrm{AlCl}_3$ is much more stable than AlCl . Further, in the solid or the vapour state, $\mathrm{AlCl}_3$ covalent in nature but in aqueous solutions, it ionises to form A $\mathrm{Al}^{3+}(a q)$ and $\mathrm{Cl}^{-}(a q)$ ions.

(c) Due to inert pair effect, indium exists in both +1 and +3 oxidation states out of which + 3 oxidation state is more stable than +1 oxidation state. In other words, $\mathrm{InCl}_3$ is more stable than InCl . Being unstable, In Cl undergoes disproportionation reaction.

$$3 \mathrm{InCl}(a q) \longrightarrow 2 \ln (s)+\ln ^{3+}(a q)+3 \mathrm{Cl}^{-}(a q)$$

Boron halides do not exist as dimer due to small size of boron atom which makes it unable to accommodate four large sized halide ions. $\mathrm{AlCl}_3$ exists as dimer. Al makes use of vacant $3 p$-orbital by coordinate bond i.e., Al atoms complete their octet by forming dimers.