(i)`P+C("carbon")+Cl_(2)toQ+CO uarr` , (ii)`Q+H_(2)OtoR+HCl`
(iii)`BN+H_(2)O to R+NH_(3)` , (iv)`Q+LiAIH_(4) to S+LiCl+ AlCl_(3)`
(v)`S+H_(2)toR+H_(2) uarr` , (vi)`S+NaH to T`
(`P,Q,R,S` and `T` do not represent their chemical symbols)
Compound S is

Options

A. Sodium borohydride
B. Boron trichloride
C. Orthoboric acid
D. diborane

Explanation

To solve the problem step by step, we will analyze each reaction and identify the compounds involved.

### Step 1: Identify Compound P
The first reaction is:
\[ P + C + Cl_2 \rightarrow Q + CO \]

From the context, we can deduce that compound P is likely **Boric Oxide (B2O3)**. This is because boron compounds typically react with carbon and chlorine to form boron trichloride (Q) and carbon monoxide.

### Step 2: Identify Compound Q
Using the information from Step 1, we can conclude that:
\[ Q = BCl_3 \] (Boron Trichloride)

### Step 3: Identify Compound R
The second reaction is:
\[ Q + H_2O \rightarrow R + HCl \]

When boron trichloride (BCl3) reacts with water, it undergoes hydrolysis to form **Ortho Boric Acid (H3BO3)**. Therefore:
\[ R = H3BO3 \]

### Step 4: Analyze the Third Reaction
The third reaction is:
\[ BN + H_2O \rightarrow R + NH_3 \]

Boron nitride (BN) reacts with water to produce ortho boric acid (R) and ammonia (NH3). This confirms that R is indeed **H3BO3**.

### Step 5: Identify Compound S
The fourth reaction is:
\[ Q + LiAlH_4 \rightarrow S + LiCl + AlCl_3 \]

When boron trichloride (BCl3) reacts with lithium aluminum hydride (LiAlH4), it forms **Diborane (B2H6)**. Thus:
\[ S = B2H6 \]

### Step 6: Analyze the Fifth Reaction
The fifth reaction is:
\[ S + H_2 \rightarrow R + H_2 \]

When diborane (B2H6) reacts with hydrogen, it produces ortho boric acid (R) and releases hydrogen gas. This confirms that R is **H3BO3**.

### Step 7: Analyze the Sixth Reaction
The sixth reaction is:
\[ S + NaH \rightarrow T \]

When diborane (B2H6) reacts with sodium hydride (NaH), it forms **Sodium Borohydride (NaBH4)**, which is a common reducing agent. Therefore:
\[ T = NaBH4 \]

### Conclusion
After analyzing all the reactions, we conclude that the compound S is:
\[ \text{Compound S is Diborane (B2H6)} \]

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