Question

Boron trifluoride on reaction with lithium aluminium hydride in ether gives LiF, AlF\(_3\) and X. X on reaction with NH\(_3\) gives Y. Y on further heating gives a compound Z. The number of \(\sigma\)-bonds and \(\pi\)-bonds in Z are x and y respectively. (x+y) is equal to

• A. 15
• B. 12
• C. 14
• D. 18

Hint:

Borazine (\(B_3N_3H_6\)), known as "inorganic benzene," has a structure very similar to benzene. It has 12 sigma bonds (6 in the ring, 6 to hydrogens) and 3 pi bonds (from dative bonding from N to B). The total bond count is 15.

Answer 1

The Correct Option is A

Step 1: Identify compound X.
The first reaction is the reduction of Boron trifluoride (BF\(_3\)) by Lithium aluminium hydride (LiAlH\(_4\)) in ether. This is a standard preparation method for Diborane (B\(_2\)H\(_6\)).
The balanced reaction is: \( 4\text{BF}_3 + 3\text{LiAlH}_4 \rightarrow 2\text{B}_2\text{H}_6 + 3\text{LiF} + 3\text{AlF}_3 \).
So, compound X is Diborane, B\(_2\)H\(_6\).
Step 2: Identify compound Y.
Diborane (X) reacts with ammonia (NH\(_3\)). The product depends on the conditions. Typically, at low temperatures, it forms an adduct, which is an ionic compound.
\( \text{B}_2\text{H}_6 + 2\text{NH}_3 \rightarrow [\text{BH}_2(\text{NH}_3)_2]^+[\text{BH}_4]^- \). This is compound Y. It's often written as B\(_2\)H\(_6\)\(\cdot\)2NH\(_3\).
Step 3: Identify compound Z.
When compound Y is heated, it undergoes further reaction to form Borazine (also known as inorganic benzene).
\( 3[\text{BH}_2(\text{NH}_3)_2]^+[\text{BH}_4]^- \xrightarrow{\text{heat}} 2\text{B}_3\text{N}_3\text{H}_6 + 12\text{H}_2 \).
So, compound Z is Borazine, B\(_3\)N\(_3\)H\(_6\).
Step 4: Determine the number of \(\sigma\) and \(\pi\) bonds in Z.
Borazine has a cyclic structure analogous to benzene, with alternating Boron and Nitrogen atoms in a six-membered ring. Each Boron and Nitrogen atom is bonded to one Hydrogen atom.
The structure consists of: - 3 B-N single bonds in the ring. - 3 N-B single bonds in the ring. - 3 B-H single bonds. - 3 N-H single bonds. All these are sigma bonds. Total number of \(\sigma\)-bonds = \(3+3+3+3 = 12\). So, x=12.
Additionally, there is a delocalized \(\pi\)-system involving the lone pair of electrons from each Nitrogen atom donating into the empty p-orbital of the adjacent Boron atoms. This results in 3 \(\pi\)-bonds (delocalized). So, y=3.
The total number of bonds is (x+y).
\( x+y = 12 + 3 = 15 \).