Boron Family

Introduction

The electronic configuration for elements from Boron family have outer-shell electronic configuration as $\mathrm{ns^2np^1}$. The existence of nd electrons, however, causes differences in the characteristics of the first two elements (B and Al) and the remainder of the elements from Ga onwards.

The electronic configuration of Group 13 elements is shown in the table below. Thus, boron $\mathrm{(1s^2)}$ has 2 electrons in the outermost shell, aluminium $\mathrm{(2s^22p^6)}$ has 8 electrons, and the remaining elements have 18 electrons $\mathrm{(ns^2np^6nd^{10})}$.

Elements	Electronic Configuration
Boron (B)	$\mathrm{[He] 2s^2 \:2p^1}$
Aluminium (Al)	$\mathrm{[Ne] 3s^2\:3p^1}$
Gallium (Ga)	$\mathrm{[Ar] 3d^{10}\: 4s^2\: 4p^1}$
Indium (In)	$\mathrm{[Kr] 4d^{10}\: 5s^2 \:5p^1}$
Thallium (Tl)	$\mathrm{[Xe] 4f^{14}\: 5d^{10}\: 6s^2 \:6p^1}$

Physical Properties of Boron Family

Atomic Size

Because of the enhanced screening effect, the atomic size jumps from B to Al. Due to the low shielding effect of d-electrons, the size of Al and Ga elements is the same. The size grows again from Ga to In, but due to poor-shielding f-electrons, the size of Tl is identical to that of In.

Ionisation Energy

Because of its smallest size and smallest screening effect, B has the greatest IE. The larger atomic size and efficient shielding of the nuclear charge leads in a decreased IE of Al as one moves down from B to Al. The extra ten d-electrons in Ga, on the other hand, demonstrate inadequate shielding, resulting in a tighter grip of the nucleus on the outer-shell electrons.

As a result, the impact of increasing atomic size is nearly balanced, and IE is nearly unchanged. Increased atomic size overcomes the inefficient shielding of d-electrons in the case of In, resulting in a lower IE. In addition, the shielding effect of 14 f-electrons is significantly worse than that of d electrons. As a result, in the case of Tl, the outer-shell electrons are more strongly retained, resulting in an unusually high IE.

Oxidation State

The stability of (+III) states reduces as one moves down the group, whereas (+I) state stability improves. This is owing to the s-electrons' non-participation in bonding, i.e. their inertness. The inert-pair effect is what it's called. The reason for this is that 14felectrons have a weak shielding effect, which strengthens the nuclear grip on ns electrons, causing them to remain coupled and inactive.

As a result, Tl (+I) compounds are more stable than Tl (+III) compounds and resemble alkali metal compounds in appearance. TlOH is a strong base that is also quite soluble.

Metallic Character

Because of its tiny size and high IE, boron is a notable nonmetal. The electropositive, or metallic, character rises from B to Al (with increasing atomic size and decreasing IE), whereas the electropositive character of subsequent elements diminishes due to the weak shielding effect of d- and f-electrons. The usual electrode potentials for $\mathrm{M^{3+}/M}$ are less negative, but they are more negative for $\mathrm{M^+/M}$.

Chemical Properties of Boron Family

Reaction with Acids/Bases

Hot concentrated sulphuric and nitric acids are reduced by amorphous boron. While other elements of the group react with dilute mineral acid, Al gets passivated due to formation of a coating of oxide layer on the surface.

$$\mathrm{B + 3HNO_3\:\rightarrow\:H_3BO_3 + 3NO}$$

$$\mathrm{2B + 3H_2SO_4 \:\rightarrow\:2H_3BO_3 + 3SO_2}$$

$$\mathrm{2M + 6HCl\:\rightarrow\:2MCl_3 + 3H_2}$$

Boron liberates $\mathrm{H_2}$ from fused caustic alkalis, resulting in the formation of borates. In aqueous $\mathrm{NaOH}$, Al and Ga dissolve and release $\mathrm{H_2}$.

$$\mathrm{2B + 6NaOH\:\rightarrow\:2Na_3BO_3 + 3H_2}$$

$$\mathrm{M + NaOH + H_2O\:\rightarrow\:NaMO_2 + H_2O}$$

Reaction with Moisture

Due to the formation of a thin protective oxide layer on its surface, Al, on the other hand, is relatively stable in air and water. When it is burned in dinitrogen at a very high temperature, it produces AlN. Other elements in group 13 are unaffected by interactions with air and water.

$$\mathrm{4B + 3O_2\:\rightarrow\:2B_2O_3}$$

$$\mathrm{2B + N_2\:\rightarrow\:2BN}$$

$$\mathrm{2Al + N_2\:\rightarrow\:2AlN}$$

Reaction with Halogen

Except for Tl, which generates monohalides, all metals react with halogens to form trihalides.

Boron only reacts at high temperatures, whereas others do so even at low temperatures. Because of their more stable (+III) oxidation state, monohalides of Al, Ga, and In are reductants.

Conclusion

The last electron of group 13 elements is located in the p-orbital, making them p-block elements. In the outermost orbit, all of these elements contain three electrons wherein two are present in the s-orbital and one is present in the p-orbital.

As a result, the electronic arrangement of these atoms' outermost energy levels might be represented as $\mathrm{ns^2np^1}$.

FAQs

Q1. Explain the nature of $\mathrm{B(OH)_3, \:Al(OH)_3}$, and $\mathrm{Tl(OH)_3}$.

Ans: Please note that $\mathrm{B(OH)_3, \:Al(OH)_3}$, and $\mathrm{Tl(OH)_3}$ are acidic, amphoteric and basic in nature respectively. From B through Tl, the electropositive or metallic property of group 13 elements rises. Because B is a nonmetal, it produces an acidic hydroxide. Tl, the most metallic of basic hydroxides, is formed by Al and amphoteric hydroxide.

Q2. Why does pπ – pπ backbonding occurs in boron halides but does not occur for aluminium halides?

Ans: The tendency to display pπ – pπ back bonding is greatest in boron halides and rapidly declines as the size of the central atom and halogen atom increases. Back-bonding does not occur since Al is greater than B.

Q3. Why does boron have higher melting and boiling point when compared to other congeners of Group 13 elements?

Ans: Boron has an extremely high melting point and boiling point because it exists as a massive covalent, polymeric structure in both solid and liquid form.

Q4. What do you mean by Borides?

Ans: Boron reacts with numerous metals to generate binary compounds termed borides, such as $\mathrm{MgB_2,\: VB}$, and $\mathrm{Fe_2B}$, when heated. Metal borides are refractory materials that are exceedingly hard, chemically inert, and non-volatile. Their melting points are high, and their thermal and electrical conductivities are high. $\mathrm{Ti, Zr, Hf, Nb}$, and Ta diborides all have melting temperatures above $\mathrm{3200 \:K. \:TiB_2}$ and ZrB have thermal and electrical conductivities that are 10 times higher than Ti and Zr metals.

Q5. State the difference in behavior of Boron and Aluminium, when hydrolysed.

Ans: The enthalpy of $\mathrm{AlCl_3}$ solution is quite negative. As a result, $\mathrm{AlCl_3}$ occurs in aqueous solution as $\mathrm{Al^{3+}\: (aq)}$ and $\mathrm{Cl^-\:(aq)}$. The similar argument could be made for $\mathrm{BCl_3}$, but in order for the enthalpy of solution of $\mathrm{BCl_3}$ to be negative, the enthalpy of hydration of $\mathrm{B^{3+}}$ would have to be -6009 kJ, which is improbable for a tiny $\mathrm{B^{3+}}$ cation. $\mathrm{BCl_3}$ hydrolyzes as

$$\mathrm{BCl_3 + 3H_2O\:\rightarrow\:H_3BO_3 + 3HCl}$$