Question

The correct polarity order among the following is?

• A. Boroxine > Borazine > Benzene
• B. Benzene > Boroxine > Borazine
• C. Benzene > Borazine > Boroxine
• D. Borazine > Boroxine > Benzene

Hint:

For isoelectronic aromatic rings, the reactivity and polar nature increase as the electronegativity difference between the ring atoms increases.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Polarity in these isoelectronic cyclic molecules depends on the electronegativity difference (\(\Delta EN\)) between the alternating atoms in the ring.
Step 2: Detailed Explanation:
1. Benzene (\(C_6H_6\)): It is a non-polar hydrocarbon. The \(C-C\) and \(C-H\) bonds have negligible \(\Delta EN\), and its high symmetry (\(D_{6h}\)) ensures a net dipole moment of zero.
2. Borazine (\(B_3N_3H_6\)): Known as "inorganic benzene". The \(B-N\) bond is polar because \(EN(N) = 3.0\) and \(EN(B) = 2.0\), giving \(\Delta EN = 1.0\). While the symmetric ring has \(\mu = 0\), it has significant "local" bond polarity.
3. Boroxine (\(B_3O_3H_3\)): Contains alternating B and O atoms. Since Oxygen is more electronegative (\(EN=3.5\)) than Nitrogen (\(EN=3.0\)), the \(B-O\) bond (\(\Delta EN = 1.5\)) is much more polar than the \(B-N\) bond.
Higher \(\Delta EN\) leads to higher ionic character and higher polarity in the molecule's chemical behavior.
Step 3: Final Answer:
The polarity order is Boroxine > Borazine > Benzene.