Molecule AB has a bond length of 1.617 mathringA and a dipole moment of 0.38 D. The fractional charge on each atom (absolute magnitude) is: (e0 = 4.802 × 10-10esu)

Question

Molecule AB has a bond length of 1.617 mathringA and a dipole moment of 0.38 D. The fractional charge on each atom (absolute magnitude) is: (e0 = 4.802 × 10-10esu) (A) 0 (B) 0.05 (C) 0.5 (D) 1.0

Tardigrade Admin · Accepted Answer

μ=q × l ⇒ 0.38 × 3.33 × 10-30( cm ) =q × 1.617 × 10-10 ⇒ q=(0.38 × 3.00/1.67) × 10-20 =q=0.759 × 10-20 C with respect to electron, q=(0.759 × 10-20/1.6 × 10-19)=0.05

Q. Molecule AB has a bond length of 1.617A˚ and a dipole moment of 0.38D. The fractional charge on each atom (absolute magnitude) is : (e0​=4.802×10−10esu)

0

0.05

0.5

1.0

μ=q×l⇒0.38×3.33×10−30(cm) =q×1.617×10−10 ⇒q=1.670.38×3.00​×10−20 =q=0.759×10−20C with respect to electron, q=1.6×10−190.759×10−20​=0.05

Q. Molecule AB has a bond length of $1.617 \overset{˚}{A}$ and a dipole moment of $0.38 D$ . The fractional charge on each atom (absolute magnitude) is : ( $e_{0} = 4.802 \times 1 0^{- 10}$ esu)