Question:
NaBH$_4$ reacts with I$_2$ and gives a salt and two gases Y, Z. The gas Y is toxic in nature. Z is a combustible gas. The correct statements regarding Y, Z are:
• I. Y with NaH forms a compound which acts as a good reducing agent.
• II. Y on hydrolysis gives a monobasic acid.
• III. Z is used in Haber’s process.
NaBH$_4$ reacts with I$_2$ and gives a salt and two gases Y, Z. The gas Y is toxic in nature. Z is a combustible gas. The correct statements regarding Y, Z are:
• I. Y with NaH forms a compound which acts as a good reducing agent.
• II. Y on hydrolysis gives a monobasic acid.
• III. Z is used in Haber’s process.
• I. Y with NaH forms a compound which acts as a good reducing agent.
• II. Y on hydrolysis gives a monobasic acid.
• III. Z is used in Haber’s process.
Show Hint
Diborane contains 3-center-2-electron (banana) bonds due to electron deficiency.
Updated On: Jun 10, 2026
- I, II only
- I, III only
- I, II, III
- II, III only
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
NaBH$_4$ reacts with iodine producing diborane (B$_2$H$_6$) and hydrogen gas.
Step 1: Reaction \[ 2NaBH_4 + I_2 \rightarrow 2NaI + B_2H_6 + H_2 \] Thus,
• Y = B$_2$H$_6$ (diborane)
• Z = H$_2$ (hydrogen)
Step 2: Verify statements
• I Correct: B$_2$H$_6$ reacts with NaH to form NaBH$_4$ (reducing agent system).
• II Correct: Hydrolysis of B$_2$H$_6$ gives boric acid (weak monobasic acid behavior).
• III Correct: H$_2$ is used in Haber process for NH$_3$ production.
All statements are correct.
Step 1: Reaction \[ 2NaBH_4 + I_2 \rightarrow 2NaI + B_2H_6 + H_2 \] Thus,
• Y = B$_2$H$_6$ (diborane)
• Z = H$_2$ (hydrogen)
Step 2: Verify statements
• I Correct: B$_2$H$_6$ reacts with NaH to form NaBH$_4$ (reducing agent system).
• II Correct: Hydrolysis of B$_2$H$_6$ gives boric acid (weak monobasic acid behavior).
• III Correct: H$_2$ is used in Haber process for NH$_3$ production.
All statements are correct.
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