Question

Reaction of LiAlH\(_4\) with one equivalent of Me\(_3\)N·HCl gives a tetrahedral compound, which reacts with another equivalent of Me\(_3\)N·HCl to give compound N. The compound N and its geometry, respectively, are:

• A. LiAlH\(_4\)NMe\(_3\) and trigonal bipyramidal
• B. Li\(_2\)AlH\(_4\)Cl and square pyramidal
• C. AlH\(_3\)(NMe\(_3\))\(_2\) and trigonal bipyramidal
• D. AlH\(_3\)(NMe\(_3\))\(_2\) and pentagonal

Hint:

When analyzing coordination compounds, consider the steric and electronic requirements of the ligands to determine the geometry.

Answer 1

The Correct Option is C

Step 1: Understand the Reaction.
LiAlH\(_4\) reacts with Me\(_3\)N·HCl to form a tetrahedral complex. Upon further reaction with another equivalent of Me\(_3\)N·HCl, a new compound is formed, which contains two NMe\(_3\) groups. The geometry of this compound is trigonal bipyramidal.

Step 2: Analyze the Geometry.
The presence of two NMe\(_3\) groups coordinated to the Al center results in a trigonal bipyramidal geometry, as the two bulky NMe\(_3\) groups need to be placed in axial positions to minimize steric repulsion.

Step 3: Conclusion.
The correct answer is \( \text{AlH}_3(\text{NMe}_3)_2 \) with trigonal bipyramidal geometry.

Final Answer: \[ \boxed{\text{AlH}_3(\text{NMe}_3)_2 \text{ and trigonal bipyramidal}} \]