A group 13 element 'X' reacts with chlorine gas to produce a compound XCl$$_3$$. XCl$$_3$$ is electron deficient and easily reacts with NH$$_3$$ to form Cl$$_3$$X $$\leftarrow$$ NH$$_3$$ adduct; however, XCl$$_3$$ does not dimerize. X is:

We have element $$X$$ belonging to group 13, so the possible elements are $$\text{B, Al, Ga, In, Tl}$$. The question tells us that $$X$$ reacts with chlorine gas to give the trichloride $$\text{XCl}_3$$.

Now, the given properties of $$\text{XCl}_3$$ are:

(i) $$\text{XCl}_3$$ is electron-deficient. (ii) It readily accepts a lone pair from $$\text{NH}_3$$ to form the Lewis-acid-base adduct $$\text{Cl}_3X \leftarrow \text{NH}_3$$. (iii) It does not dimerize.

First recall the concept of electron deficiency. A molecule is electron deficient when the central atom has fewer than eight electrons in its valence shell. All group-13 trihalides $$\text{MX}_3$$ (where $$\text{M = B, Al, Ga, In, Tl}$$) possess only $$6$$ valence electrons around the metal, so they are potential Lewis acids. Thus condition (i) is satisfied by every $$\text{MCl}_3$$ in the list.

Next, we use the experimentally observed tendency toward dimerization. A larger central atom can expand its coordination number by employing vacant $$p$$ orbitals to accept electron density from chloride bridges, leading to the dimer:

$$2\,\text{MCl}_3 \;\; \longrightarrow \;\; \text{Cl}_3\text{M}\!-\!\!\!\;\:\,\text{Cl}\;\!-\!\!\!\;\:\,\text{MCl}_3$$

The accepted textbook facts are:

• $$\text{BCl}_3$$ remains monomeric in the vapour and in non-coordinating solvents because the boron atom is too small to accommodate two extra chloride bridges. • $$\text{AlCl}_3, \text{GaCl}_3,$$ and $$\text{InCl}_3$$ dimerize to $$\text{M}_2\text{Cl}_6$$ in the gaseous state or in solution unless a stronger base is present.

Therefore the information that $$\text{XCl}_3$$ does not dimerize singles out boron trichloride:

$$\text{BCl}_3 \; ( \text{monomeric} ) , \qquad \text{AlCl}_3 \to \text{(Al}_2\text{Cl}_6), \; \text{GaCl}_3 \to \text{(Ga}_2\text{Cl}_6), \; \text{InCl}_3 \to \text{(In}_2\text{Cl}_6)$$

Finally, statement (ii) about forming the adduct $$\text{Cl}_3X \leftarrow \text{NH}_3$$ is perfectly consistent with $$\text{BCl}_3$$, which is the classic Lewis acid that picks up the nitrogen lone pair:

$$\text{BCl}_3 + \text{NH}_3 \;\; \longrightarrow \;\; \text{Cl}_3\text{B} \leftarrow \text{NH}_3$$

So every given property matches $$X = \text{Boron (B)}$$ and none of the heavier group-13 elements satisfy the “no dimerization” condition.

Hence, the correct answer is Option A.