Question

Lithium aluminium hydride reacts with silicon tetrachloride to form?
(A) \[LiCl,Al{{H}_{3}}\text{ and }Si{{H}_{4}}\]
(B) $LiCl,AlC{{l}_{3}}\text{ and }Si{{H}_{4}}$
(C) $LiH,AlC{{l}_{3}}\text{ and }SiC{{l}_{2}}$
(D) $LiH,Al{{H}_{3}}\text{ and }Si{{H}_{4}}$

Answer 1

Hint: Understanding the nature of both the reactants in this case is important. It is a redox reaction.

Complete step by step solution:
Lithium aluminium hydride ($LiAl{{H}_{4}}$) is a reducing agent and therefore in this case it will reduce silicon tetrachloride ($SiC{{l}_{4}}$) into silicon hydride ($Si{{H}_{4}}$), aluminium chloride ($AlC{{l}_{3}}$) and lithium chloride ($LiCl$). The reaction is as follows:

\[SiC{{l}_{4}}+LiAl{{H}_{4}}\to Si{{H}_{4}}+LiCl+AlC{{l}_{3}}\]

The reaction is basically a shuffle of atoms. The oxidation state of silicon is $+4$ in case of $SiC{{l}_{4}}$and that of lithium and aluminium in case of $LiAl{{H}_{4}}$are $-1$ and $-3$ respectively. Let’s not forget this is the scene before the reaction takes place. In the product, silicon has an oxidation state of $-4$ in$Si{{H}_{4}}$while lithium and aluminium do not have a change in their states. It is clear that silicon has undergone a reduction.

As per the discussion the answer to this question is option (B)$LiCl,AlC{{l}_{3}}\text{and }Si{{H}_{4}}$.

Additional information:
- The above reaction is a redox reaction. Reduction is always accompanied by oxidation and vice-versa. As we saw above silicon tetrachloride reduced to silicon hydride and simultaneously hydrogen oxidised from $-1$ to $+1$ with the loss of two electrons.
- Oxidation state of any atom in a molecule is the total charge it will have when it separates from that particular molecule by the means of heterolytic cleavage. The electronegativity of an atom is a major influencer in deciding its oxidation state.
-Reduction is basically a gain of electrons. In the above case silicon gains a total of $8$ electrons to reduce.
- Lithium aluminium hydride ($LiAl{{H}_{4}}$) is a reducing agent. In many organic reactions performed at laboratory, this reagent is extensively used to reduce carboxylic acid compounds- as other reagents such as sodium borohydride ($NaB{{H}_{4}}$) are not powerful enough- into their corresponding carbonyl compounds or alcohols. For example ethanoic acid when reacts with $LiAl{{H}_{4}}$reduces to ethanal, an aldehyde. This reaction is as follows:

\[C{{H}_{3}}COOH\xrightarrow{LiAl{{H}_{4}}}C{{H}_{3}}CHO\]


Note: Even if a student knows that $LiAl{{H}_{4}}$will reduce $SiC{{l}_{4}}$he/she may make mistakes in forming the appropriate products. Let’s break it down here:
-\[LiCl,Al{{H}_{3}}\text{ and }Si{{H}_{4}}\](A)
This is not possible because the equation of chemical reaction is not balanced due to the extra number of hydrogen atoms.
-$LiH,AlC{{l}_{3}}\text{ and }SiC{{l}_{2}}$(C)
This is not possible because the valency of silicon is not satisfied.
- $LiH,Al{{H}_{3}}\text{ and }Si{{H}_{4}}$(D)
This is not possible as chlorine atoms are missing and the equation of chemical reaction will be unbalanced due to extra hydrogen atoms.