`NaBH_(4)+I_(2)rarr X+Y+Z`
`BF_(3)+LiAlH_(4) overset("450 K")rarr X+P`
`BF_(3)+LiAlH_(4)rarrX+Q+R`
X, Y, Z, P, Q and R in the reactions are

Options

A. `{:("X","Y","Z","P","Q","R"),(Na_(4)B_(4)O_(7),Nal, HI,HF,AlF_(3),LiF):}`
B. `{:("X","Y","Z","P","Q","R"),(B_(2)H_(6),Nal,H_(2),NaF,LiF,AlF_(3)):}`
C. `{:("X","Y","Z","P","Q","R"),(B_(2)H_(6),BH_(3),Nal,B_(3)N_(3)H_(6),Al_(2)F_(6),AlF_(3)):}`
D. `{:("X","Y","Z","P","Q","R"),(BH_(3),B_(2)H_(6),H_(2),B_(3)N_(3)H_(6),LiF,AlF_(3)):}`

Answer

To solve the question step by step, we will analyze each reaction provided and identify the products X, Y, Z, P, Q, and R.

### Step 1: Analyze the first reaction
**Reaction:**
\[ \text{NaBH}_4 + \text{I}_2 \rightarrow X + Y + Z \]

**Explanation:**
- Sodium borohydride (NaBH₄) reacts with iodine (I₂).
- This reaction typically leads to the formation of diborane (B₂H₆), hydrogen gas (H₂), and sodium iodide (NaI).
- Therefore, we can assign:
- \( X = \text{B}_2\text{H}_6 \)
- \( Y = \text{NaI} \)
- \( Z = \text{H}_2 \)

### Step 2: Analyze the second reaction
**Reaction:**
\[ \text{BF}_3 + \text{LiAlH}_4 \overset{450 \, K}{\rightarrow} X + P \]

**Explanation:**
- Boron trifluoride (BF₃) reacts with lithium aluminum hydride (LiAlH₄) at elevated temperatures (450 K).
- This reaction produces diborane (B₂H₆) and sodium fluoride (NaF).
- Thus, we can assign:
- \( X = \text{B}_2\text{H}_6 \) (same as in the first reaction)
- \( P = \text{NaF} \)

### Step 3: Analyze the third reaction
**Reaction:**
\[ \text{BF}_3 + \text{LiAlH}_4 \rightarrow X + Q + R \]

**Explanation:**
- In the absence of temperature, the reaction of BF₃ with LiAlH₄ produces multiple products.
- The products typically include diborane (B₂H₆), lithium fluoride (LiF), and aluminum trifluoride (AlF₃).
- Therefore, we can assign:
- \( X = \text{B}_2\text{H}_6 \) (same as before)
- \( Q = \text{LiF} \)
- \( R = \text{AlF}_3 \)

### Summary of Products
- \( X = \text{B}_2\text{H}_6 \)
- \( Y = \text{NaI} \)
- \( Z = \text{H}_2 \)
- \( P = \text{NaF} \)
- \( Q = \text{LiF} \)
- \( R = \text{AlF}_3 \)

### Final Answer
The products are:
- \( X = \text{B}_2\text{H}_6 \)
- \( Y = \text{NaI} \)
- \( Z = \text{H}_2 \)
- \( P = \text{NaF} \)
- \( Q = \text{LiF} \)
- \( R = \text{AlF}_3 \)