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.

• A. I, II only
• B. I, III only
• C. I, II, III
• D. II, III only

Hint:

Diborane contains 3-center-2-electron (banana) bonds due to electron deficiency.

Answer 1

The Correct Option is C

Step 1: Write the reaction of $NaBH_4$ with iodine.
Sodium borohydride reacts with iodine to give a salt and two gases: \[ 2NaBH_4 + I_2 \rightarrow 2NaI + B_2H_6 + H_2 \] So the salt is $NaI$, and the two gases are diborane and hydrogen.

Step 2: Name the two gases Y and Z.
Diborane $B_2H_6$ is toxic, so it is the toxic gas Y. Hydrogen $H_2$ is combustible, so it is the burnable gas Z.

Step 3: Check statement I about Y with NaH.
Diborane reacts with sodium hydride to form sodium borohydride again, which is a well-known good reducing agent. So statement I is correct.

Step 4: Check statement II about hydrolysis of Y.
When diborane is hydrolysed, it gives boric acid. Boric acid behaves as a weak monobasic acid in water. So statement II is correct.

Step 5: Check statement III about Z.
Hydrogen gas is used in the Haber process to make ammonia from nitrogen. So statement III is correct.

Step 6: Combine the verdicts.
Since statements I, II, and III are all correct, the right option lists all three.
\[ \boxed{\text{I, II, III}} \]