Back to Directory
NEET CHEMISTRYMedium

The dipole moment of compound AB is 10.92 D and that of compound CD is 12.45 D. If the bond length of AB is 2.72 A˚\text{\AA} and that of CD is 2.56 A˚\text{\AA}, then for these compounds the correct statement is:

A

More ionic nature in AB

B

More ionic nature in CD

C

Equal in both

D

Cannot be predicted

Step-by-Step Solution

The ionic nature of a compound can be assessed by comparing its observed dipole moment to the theoretical dipole moment it would have if the bond were 100% ionic. According to the sources, the dipole moment (μ\mu) is defined as the product of the magnitude of charge (QQ) and the distance of separation (rr), so μ=Q×r\mu = Q \times r . The percentage ionic character is proportional to the ratio of the dipole moment to the bond length (μ/r\mu/r).

  1. For compound AB: The ratio is 10.922.724.015\frac{10.92}{2.72} \approx 4.015 D/A˚\text{\AA}.
  2. For compound CD: The ratio is 12.452.564.863\frac{12.45}{2.56} \approx 4.863 D/A˚\text{\AA}.

Comparing these values, the ratio for compound CD is significantly higher than that for AB. This indicates that for a given bond length, CD has a greater magnitude of charge separation, reflecting a higher percentage of ionic character. Therefore, compound CD exhibits more ionic nature.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started