(i) `P+C("carbon")+Cl_(2) to Q+CO (ii) `Q+H_(2)O to R+HCl`
(iii) `BN+H_(2)O to R+NH_(3) uarr`
(iv) `Q+LiAlH_(4) to S+LiCl+AlCl_(3)`
(v) `S+H_(2) to R+H_(2) uarr`
(vi) `S+NaH to T`
(P,Q,R,S annd T do not represent their chemical symbols)
Q. Compound S is:
(I) an odd `e^(-)` compound
(II) `(2c-3e^(-))` compound
(III) a electron deficient compound
(IV) a `sp^(2)` hybridized compound
Choose the correct code:

Options

A. III
B. I,III
C. II,III,IV
D. I,II,IV

Explanation

To solve the problem step by step, we will analyze each reaction and determine the compounds involved (P, Q, R, S, and T) and their characteristics.

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

To identify P, we consider that it reacts with carbon and chlorine to produce Q (which we will identify later) and carbon monoxide. A common compound that fits this description is boron oxide, specifically \( B_2O_3 \).

Thus, we can assume:
- **P = \( B_2O_3 \)**

### Step 2: Identify Compound Q
The second reaction is:
\[ Q + H_2O \rightarrow R + HCl \]

From the previous step, we hypothesized that Q is produced from the reaction of \( B_2O_3 \) with carbon and chlorine. A likely candidate for Q is \( BCl_3 \) (boron trichloride), which can hydrolyze to produce orthoboric acid and hydrochloric acid.

Thus, we can assume:
- **Q = \( BCl_3 \)**

### Step 3: Identify Compound R
The third reaction is:
\[ BN + H_2O \rightarrow R + NH_3 \]

Boron nitride (BN) reacts with water to produce orthoboric acid and ammonia. Therefore, R can be identified as:
- **R = \( H_3BO_3 \)** (orthoboric acid)

### Step 4: Identify Compound S
The fourth reaction is:
\[ Q + LiAlH_4 \rightarrow S + LiCl + AlCl_3 \]

Substituting Q with \( BCl_3 \):
\[ BCl_3 + LiAlH_4 \rightarrow S + LiCl + AlCl_3 \]

The product S from this reaction is diborane, \( B_2H_6 \).

Thus, we can assume:
- **S = \( B_2H_6 \)**

### Step 5: Identify Compound T
The fifth reaction is:
\[ S + H_2 \rightarrow R + H_2 \uparrow \]

When diborane (\( B_2H_6 \)) reacts with hydrogen, it can produce orthoboric acid and hydrogen gas. However, this reaction is not typically straightforward. Instead, we can look at the next reaction to identify T.

The sixth reaction is:
\[ S + NaH \rightarrow T \]

When \( B_2H_6 \) reacts with sodium hydride, it forms sodium borohydride (\( NaBH_4 \)).

Thus, we can assume:
- **T = \( NaBH_4 \)**

### Summary of Compounds
- **P = \( B_2O_3 \)**
- **Q = \( BCl_3 \)**
- **R = \( H_3BO_3 \)**
- **S = \( B_2H_6 \)**
- **T = \( NaBH_4 \)**

### Step 6: Analyze Compound S
Now we need to analyze the properties of compound S (\( B_2H_6 \)):
1. **Odd Electron Compound**: \( B_2H_6 \) has 12 valence electrons (not odd).
2. **(2c-3e) Compound**: This is incorrect; \( B_2H_6 \) has 3-center 2-electron bonds.
3. **Electron Deficient Compound**: Correct, as it requires 14 electrons for bonding but has only 12.
4. **sp² Hybridized Compound**: Incorrect; \( B_2H_6 \) is sp³ hybridized due to the tetrahedral geometry around boron.

### Conclusion
The only correct statement about compound S is that it is an electron deficient compound.

### Final Answer
The correct code is:
- **(III) a electron deficient compound**