Numerical/Formula

1. TITLE

How to solve Dipole Moment Calculations in NEET — Step-by-step

2. PROBLEM TYPE INTRO 

This is a Bond Polarity numerical from the Chemical Bonding chapter. To solve this, you need the formula $\mu = Q \times r$ and must master the unit conversion between Coulomb-meters (Cm) and Debye (D). The most common mistake students make is calculating the value in Coulomb-meters but forgetting to divide by the conversion factor to get the answer in Debye, or failing to convert bond length from Angstroms/picometers to meters.

3. MCQ 

Problem:
The bond length of a heteronuclear diatomic molecule A—B is $1.5$ Å. The partial charge on each atom is experimentally determined to be $1.112 \times 10^{-20}$ C. Calculate the dipole moment of the molecule in Debye (D).
(Given: $1 \text{ D} = 3.336 \times 10^{-30} \text{ C m}$)

Options:
A) $1.67 \text{ D}$
B) $0.5 \text{ D}$
C) $5.0 \text{ D}$
D) $1.67 \times 10^{-30} \text{ D}$

4. STEP-BY-STEP SOLUTION

SOLUTION

Given:
* Partial Charge ($Q$) = $1.112 \times 10^{-20}$ C
* Bond Length ($r$) = $1.5$ Å

To Find:
* Dipole Moment ($\mu$) in Debye (D)

Formula:

$$\mu = Q \times r$$


Use this formula to find the dipole moment in Coulomb-meters (Cm) first, then convert to Debye.

 

Step 1: Convert Bond Length to standard SI units (meters)
The bond length is given in Angstroms (Å). We must convert it to meters.

$$1 \text{ Å} = 10^{-10} \text{ m}$$

 

$$r = 1.5 \times 10^{-10} \text{ m}$$

 

Step 2: Calculate Dipole Moment in Coulomb-meters (Cm)
Multiply the charge ($Q$) by the distance ($r$).

$$\mu = (1.112 \times 10^{-20} \text{ C}) \times (1.5 \times 10^{-10} \text{ m})$$

 

$$\mu = 1.668 \times 10^{-30} \text{ C m}$$

 

Step 3: Convert Coulomb-meters to Debye (D)
The question asks for the answer in Debye. We use the given conversion factor:

$$1 \text{ D} = 3.336 \times 10^{-30} \text{ C m}$$

 

$$\mu (\text{in D}) = \frac{\text{Value in C m}}{\text{Conversion Factor}}$$

 

$$\mu = \frac{1.668 \times 10^{-30} \text{ C m}}{3.336 \times 10^{-30} \text{ C m/D}}$$

 

$$\mu = 0.5 \text{ D}$$

 

Answer: 0.5 D (Option B)

5. WHERE STUDENTS GO WRONG

  • ❌ Mistake 1: Forgetting the unit conversion. Students often calculate $1.668 \times 10^{-30}$ and look for that number in the options (choosing D), forgetting that the result is in C-m, not Debye.
  • ❌ Mistake 2: Distance unit errors. Students may use the value 1.5 directly without converting Angstroms ($10^{-10}$ m) or picometers ($10^{-12}$ m) to meters, leading to incorrect powers of 10.
  • ❌ Mistake 3: Using total electronic charge. Sometimes students use $e = 1.6 \times 10^{-19}$ C by default, even when the problem specifies a "partial charge" or "fractional charge," which is much smaller.

6. QUICK FORMULA REFERENCE

Formulas Used:

  • Dipole Moment Calculation:
    $\mu = Q \times r$
    Use when calculating the magnitude of polarity given charge and distance.

  • Unit Conversion:
    $1 \text{ D} = 3.33564 \times 10^{-30} \text{ C m}$
    Use to convert SI units (Coulomb-meter) to the CGS unit (Debye) commonly used in chemistry.

7. SIMILAR PROBLEM HINT

"If NEET changes the values, the method stays the same. Practice with: Calculating the percentage ionic character. In those problems, you calculate $\mu_{calculated}$ (assuming 100% ionic charge $1.6 \times 10^{-19}$ C) and compare it to the $\mu_{observed}$ given in the question using: $\text{\% Ionic Character} = (\frac{\mu_{observed}}{\mu_{calculated}}) \times 100$."

8. NCERT REFERENCE

Based on: NCERT Class 11, Chapter 4 (Chemical Bonding), Section 4.3.6 (Polarity of Bonds). See the discussion on dipole moment units and values for HF and $H_2O$.

9. CTA

More solved numericals → Chemical Bonding Complete Guide
Practice problems → Mock exams