Question:
Silicon (IV) chloride on hydrolysis gives an acid \(X\). The number of \(-OH\) groups present in \(X\) is
Silicon (IV) chloride on hydrolysis gives an acid \(X\). The number of \(-OH\) groups present in \(X\) is
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Hydrolysis of \(SiCl_4\) produces orthosilicic acid,
\[
Si(OH)_4 \; (H_4SiO_4)
\]
which contains four hydroxyl groups.
Updated On: Jun 18, 2026
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The Correct Option is D
Solution and Explanation
Step 1: Write the hydrolysis reaction of silicon tetrachloride.
Silicon tetrachloride undergoes hydrolysis as \[ SiCl_4 + 4H_2O \rightarrow Si(OH)_4 + 4HCl \]
Step 2: Identify the acid formed.
The product \[ Si(OH)_4 \] is orthosilicic acid.
It may also be written as \[ H_4SiO_4 \]
Step 3: Count the hydroxyl groups.
The structure of orthosilicic acid contains \[ 4 \] hydroxyl (\(-OH\)) groups attached to silicon.
\[ Si(OH)_4 \] Therefore, the number of \(-OH\) groups present is \[ 4 \]
Step 4: Final conclusion.
Hence, \[ \boxed{4} \] Therefore, the correct option is (4).
Silicon tetrachloride undergoes hydrolysis as \[ SiCl_4 + 4H_2O \rightarrow Si(OH)_4 + 4HCl \]
Step 2: Identify the acid formed.
The product \[ Si(OH)_4 \] is orthosilicic acid.
It may also be written as \[ H_4SiO_4 \]
Step 3: Count the hydroxyl groups.
The structure of orthosilicic acid contains \[ 4 \] hydroxyl (\(-OH\)) groups attached to silicon.
\[ Si(OH)_4 \] Therefore, the number of \(-OH\) groups present is \[ 4 \]
Step 4: Final conclusion.
Hence, \[ \boxed{4} \] Therefore, the correct option is (4).
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