Q.19 The dipole moment (μ) of BrF is 1.42 D and the bond length is 176 pm.
The atomic charge distribution (q) in the molecule is _____ (rounded off to two decimal places).
(Given: 1 D = 3.34 × 10-30 C m; the factor e (electronic charge) = 1.60 × 10-19 C)
Final Answer
The atomic charge distribution q in BrF is 0.17 (in units of electronic charge e).
Calculation Method
Dipole moment relates charge separation to bond distance via μ = q × d, rearranged to q = μ/d. Convert μ = 1.42 D to C·m using 1 D = 3.34 × 10-30 C·m, yielding μ = 4.74 × 10-30 C·m. Convert bond length d = 176 pm to meters as 1.76 × 10-10 m.
Step-by-Step Solution
- Convert units: μ = 1.42 × 3.34 × 10-30 = 4.74 × 10-30 C·m and d = 176 × 10-12 = 1.76 × 10-10 m.
- Compute charge: q = 4.74 × 10-30 / 1.76 × 10-10 = 2.69 × 10-20 C.
- Normalize by e: e = 1.60 × 10-19 C, so q/e = 0.168, rounded to 0.17.
CSIR NET Context
This calculation assesses partial charge in polar diatomic molecules like BrF, common in CSIR NET physical chemistry for electronegativity and bonding analysis. Fluorine bears negative charge (higher electronegativity), making Br partially positive. No options provided; direct numerical answer expected.
BrF dipole moment charge distribution problems test core concepts in molecular polarity for CSIR NET aspirants. This BrF dipole moment 1.42 D bond length 176 pm atomic charge calculation yields q = 0.17, using standard unit conversions and the formula μ = q × d.
Understanding Dipole Moment in Diatomics
Dipole moment quantifies charge asymmetry: μ = q × d, where q is partial charge and d is internuclear distance. For BrF dipole moment calculation, convert Debye to SI units first. BrF’s polarity arises from F’s higher electronegativity (4.0 vs Br’s 2.8), creating δ⁻ on F and δ⁺ on Br.
Detailed Numerical Breakdown
- Unit Conversion: μ = 1.42 D × 3.34 × 10-30 C m/D = 4.74 × 10-30 C m
- Bond Length: 176 pm = 1.76 × 10-10 m
- Charge in Coulombs: q = 4.74 × 10-30 / 1.76 × 10-10 = 2.69 × 10-20 C
- In Electron Units: Divide by e = 1.60 × 10-19 C: 0.168 ≈ 0.17 (two decimals)
This matches GATE/CSIR NET expectations, where ionic character ≈ 17% (q/e × 100).
CSIR NET Preparation Tips
- Practice similar: HCl (μ = 1.03 D, d = 127 pm) gives q/e ≈ 0.20
- Common pitfalls: Forgetting pm-to-m conversion or Debye factor
- Related topics: Percent ionic character = (observed μ / theoretical μ) × 100, with theoretical μ = e × d
Master atomic charge distribution in BrF for scoring in physical/chemical bonding sections.
