Question:
The reaction between \(X\) and \(Y\) produces a solid substance and a mixture of two gases, one of which is a colourless toxic gas. \(X\) and \(Y\) respectively are
The reaction between \(X\) and \(Y\) produces a solid substance and a mixture of two gases, one of which is a colourless toxic gas. \(X\) and \(Y\) respectively are
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Important reaction:
\[
2NaBH_4 + I_2
\rightarrow
B_2H_6 + 2NaI + H_2
\]
\[
B_2H_6
\]
(diborane) is a colourless, highly toxic gas and is an important boron hydride.
Updated On: Jul 29, 2026
- \(LiAlH_4,\ BF_3\)
- \(NaBH_4,\ I_2\)
- \(NaH,\ BF_3\)
- \(LiBH_4,\ BCl_3\)
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The Correct Option is B
Solution and Explanation
Concept:
Sodium borohydride reacts with iodine to produce diborane. Diborane further decomposes to give hydrogen and boron hydrides. The reaction ultimately yields a solid and a mixture of gases including a colourless toxic gas.
Step 1: Consider the reaction of \(NaBH_4\) with \(I_2\). \[ 2NaBH_4 + I_2 \rightarrow B_2H_6 + 2NaI + H_2 \] Here, \[ NaI \] is a solid.
Step 2: Identify the gaseous products. The gases formed are \[ B_2H_6 \] and \[ H_2. \] Diborane, \[ B_2H_6, \] is a colourless toxic gas.
Step 3: Match with the given condition. The reaction produces:
• a solid (\(NaI\))
• a mixture of two gases (\(B_2H_6\) and \(H_2\))
• one gas is colourless and toxic (\(B_2H_6\))
Therefore, \[ \boxed{X=NaBH_4,\qquad Y=I_2} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Consider the reaction of \(NaBH_4\) with \(I_2\). \[ 2NaBH_4 + I_2 \rightarrow B_2H_6 + 2NaI + H_2 \] Here, \[ NaI \] is a solid.
Step 2: Identify the gaseous products. The gases formed are \[ B_2H_6 \] and \[ H_2. \] Diborane, \[ B_2H_6, \] is a colourless toxic gas.
Step 3: Match with the given condition. The reaction produces:
• a solid (\(NaI\))
• a mixture of two gases (\(B_2H_6\) and \(H_2\))
• one gas is colourless and toxic (\(B_2H_6\))
Therefore, \[ \boxed{X=NaBH_4,\qquad Y=I_2} \] \[ \boxed{\text{Answer = (B)}} \]
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