Lithium aluminium hydride reacts with silicon tetrachloride to form:

We have the reducing agent lithium aluminium hydride whose formula is written as $$\text{LiAlH}_4$$. In this salt the aluminium-centred anion $$\text{AlH}_4^-$$ supplies hydride ions $$\text{H}^-$$ very readily.

Silicon tetrachloride is $$\text{SiCl}_4$$. Whenever a covalent chloride such as $$\text{SiCl}_4$$ is treated with a strong hydride donor, each $$\text{Cl}$$ atom can in principle be replaced by a hydride ion, because the hydride has a very high affinity for the electropositive silicon centre.

First we state the general hydride-transfer idea:

$$\text{SiCl}_4 + 4\,\text{H}^- \longrightarrow \text{SiH}_4 + 4\,\text{Cl}^-$$

Those four chloride ions set free in the process cannot remain free in the reaction mixture; they immediately combine with the available metal cations. There are two different cations present after the hydride has been delivered:

• $$\text{Li}^+$$ coming from $$\text{LiAlH}_4$$
• the trivalent aluminium ion $$\text{Al}^{3+}$$ left behind after it has lost the four hydrides

The chloride ions distribute themselves between these two cations according to their charges. One chloride ion pairs with each lithium ion to give lithium chloride:

$$\text{Li}^+ + \text{Cl}^- \longrightarrow \text{LiCl}$$

and three chloride ions bind to each newly formed $$\text{Al}^{3+}$$ to give aluminium trichloride:

$$\text{Al}^{3+} + 3\,\text{Cl}^- \longrightarrow \text{AlCl}_3$$

Putting all the pieces together, a convenient schematic equation is

$$4\,\text{LiAlH}_4 + \text{SiCl}_4 \;\longrightarrow\; \text{SiH}_4 + 4\,\text{LiCl} + 4\,\text{AlCl}_3$$

(The exact stoichiometric coefficients are not asked in the problem; the essential point is simply the identity of the products.) We see clearly that the reaction gives three distinct substances: $$\text{LiCl}$$, $$\text{AlCl}_3$$ and $$\text{SiH}_4$$.

Comparing this product set with the choices provided, only Option B contains the trio $$\text{LiCl}$$, $$\text{AlCl}_3$$ and $$\text{SiH}_4$$.

Hence, the correct answer is Option B.