Question

$SiC{l}_4\stackrel{H_{2}O}{\to }X\stackrel{Heat}{\to }Y\stackrel{NaOH}{\to }Z$
X,Y and Z in the above reaction are :

• A. $X-Si{O}_2,Y-Si,Z-NaSi$
• B. $X-Si(OH{)}_4,Y-Si{O}_2,Z-N{a}_{2}Si{O}_3$
• C. $X-Si(OH{)}_4,Y-Si,Z-Si{O}_2$
• D. $X-Si{O}_2,Y-SiC{l}_4,Z-N{a}_{2}Si{O}_3$

Answer 1

The correct option is B $X-Si(OH{)}_4,Y-Si{O}_2,Z-N{a}_{2}Si{O}_3$

Silicon tetrachloride reacts with and water to produce orthosilicic acid and hydrogen chloride.

$Si{Cl}_4+4H_{2}O⟶Si{\left(OH\right)}_4+4HCl$

Thus, X is $Si{\left(OH\right)}_4$.

$Si{\left(OH\right)}_4$, on further heating breaks down into silica and water.

$Si{\left(OH\right)}_4\stackrel{△}{\to }Si{O}_2+H_{2}O$

Thus, Y is $Si{O}_2$.

Since, $Si{O}_2$ is acidic in nature, thus, on reaction with $NaOH\left(base\right)$, it forms salt.

$Si{O}_2\stackrel{2NaOH}{\to }\underset{Salt}{{Na}_{2}Si{O}_3}+H_{2}O$

Thus Z is ${Na}_{2}Si{O}_3$.

Hence, the correct answer is option $\text{B}$.