Animated Solution for Chemistry - Ionic Equilibrium: Which of the following are Lewis acids?

To tackle this problem, we must first anchor ourselves to the fundamental definition of a Lewis acid. A Lewis acid is defined as any chemical species that is electron-deficient and possesses the ability to accept at least one lone pair of electrons. This is the lens through which we must evaluate every molecule presented in the options.

Let's begin our investigation with ${\text{PH}}_3$ (Phosphine). In this molecule, the central phosphorus atom has formed three bonds with hydrogen and retains one lone pair. Its octet is completely satisfied. While it is true that phosphorus, being a third-period element, has vacant $3d$-orbitals, it practically shows no tendency to accept an incoming lone pair into these orbitals. Instead, it prefers to donate its own lone pair, making it a classic Lewis base, not an acid.

Next, we turn our attention to ${\text{BCl}}_3$ (Boron trichloride). Boron, the central atom, forms three single bonds with chlorine atoms. This gives it a total of only six valence electrons. Its octet is glaringly incomplete! If we look at its orbital configuration, boron is $s{p}^2$ hybridized, leaving one pure, unhybridized $2p$-orbital completely vacant. This empty orbital is perfectly primed to accept a lone pair of electrons. Because of this inherent electron deficiency, ${\text{BCl}}_3$ is a textbook example of a proper Lewis acid.

Following the same logic, let's examine ${\text{AlCl}}_3$ (Aluminium trichloride). Aluminium sits directly below boron in Group 13. In ${\text{AlCl}}_3$, aluminium also forms three bonds, resulting in an incomplete octet of six electrons. It possesses a vacant $3p$-orbital ready to accept electron density. Therefore, just like ${\text{BCl}}_3$, ${\text{AlCl}}_3$ is highly electron-deficient and acts as a strong Lewis acid.

Now we encounter a fascinating case: ${\text{SiCl}}_4$ (Silicon tetrachloride). Silicon has four valence electrons and forms four covalent bonds with chlorine. Its octet is completely full. At first glance, it shouldn't be a Lewis acid. However, silicon has vacant $3d$-orbitals. During specific reactions, such as hydrolysis with water, silicon can expand its octet by accepting a lone pair from the oxygen atom of ${\text{H}}_2\text{O}$ into these empty $d$-orbitals.

Because of this ability, ${\text{SiCl}}_4$ does exhibit Lewis acidic character. But here is the critical distinction: it is not 'electron-deficient' in its ground state. Its octet is complete.

We are faced with a choice where both options (b) and (d) seem plausible because ${\text{SiCl}}_4$ can act as a Lewis acid. However, in competitive examinations, we must select the most appropriate answer. The core spirit of a Lewis acid is its electron deficiency. ${\text{BCl}}_3$ and ${\text{AlCl}}_3$ perfectly embody this strict definition due to their incomplete octets. They are 'proper' Lewis acids. Therefore, the most suitable pair is ${\text{BCl}}_3$ and ${\text{AlCl}}_3$, making option (d) the correct choice.