Question

A compound \(X\) reacts with \(\mathrm{LiAlH_4}\) in diethyl ether and gives a colourless, highly toxic gas. This gas when heated with \(\mathrm{NH_3}\) gives inorganic benzene. \(X\) is \[ \begin{aligned} \text{I. } & \text{Electron deficient molecule} \text{II. } & \text{Trigonal planar molecule} \text{III. } & \text{On hydrolysis forms octahedral ion} \end{aligned} \] Correct answer is

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

Hint:

Remember: \[ \boxed{ \mathrm{BF_3} \text{ is electron deficient and trigonal planar.} } \] Also, \[ \boxed{ \mathrm{B_2H_6 + NH_3 \xrightarrow{\Delta} B_3N_3H_6} } \] (Borazine = inorganic benzene).

Answer 1

The Correct Option is A

Step 1: Identify compound \(X\). \(\mathrm{LiAlH_4}\) reduces \[ \mathrm{BF_3} \] to \[ \mathrm{B_2H_6}, \] a colourless and highly toxic gas. Further, \[ \mathrm{B_2H_6} \] reacts with ammonia on heating to form borazine, \[ \boxed{\mathrm{B_3N_3H_6}} \] called inorganic benzene. Hence, \[ \boxed{X=\mathrm{BF_3}.} \]

Step 2: Check the statements. For \(\mathrm{BF_3}\): \[ \boxed{\text{I. Electron deficient molecule -- Correct}} \] \[ \boxed{\text{II. Trigonal planar molecule -- Correct}} \] On hydrolysis, \[ \mathrm{BF_3} \rightarrow \mathrm{H_3BO_3} \] and does not form an octahedral ion. Hence, \[ \boxed{\text{III is incorrect}.} \] Therefore, \[ \boxed{\text{I and II only}} \] Hence, \[ \boxed{(A)} \] is the correct answer.