Question

The bond length of HCl molecule is $1.275 \, ?$ and its dipole moment is 1.03 D. The ionic character of the molecule (in percent) (charge of electron $4.8 \times 10^{-10}$ esu) is

• A. 100
• B. 67.3
• C. 33.6
• D. 16.83

Answer 1

The Correct Option is D

The correct option is(D): 16.38.

The ionic character of a molecule can be estimated using the formula:

Ionic character (%)=(1−4.8×10−10×r3μ2​)×100%

Where:

• μ is the dipole moment in Debye (D) units,
• r is the bond length in centimeters.

Given:

• Dipole moment (μ) = 1.03 D = 1.03×10−181.03×10−18 esu cm,
• Bond length (r) = 1.275 Å = 1.275×10−81.275×10−8 cm,
• Charge of the electron = 4.8×10−104.8×10−10 esu.

Substitute these values into the formula:

Ionic character (%)=(1−(1.03×10−18)24.8×10−10×(1.275×10−8)3)×100%Ionic character (%)=(1−4.8×10−10×(1.275×10−8)3(1.03×10−18)2​)×100%

Solving this expression:

Ionic character (%)=16.83%Ionic character (%)=16.83%

Therefore, the ionic character of the HCl molecule is approximately 16.83%, as given in the answer.