The bond length of HCl bond is 2.29 × 10-10 m. The percentage ionic character of HCl, if measured dipole moment is 6.226 × 10-30 C m is

Question

The bond length of HCl bond is 2.29 × 10-10 m. The percentage ionic character of HCl, if measured dipole moment is 6.226 × 10-30 C m is (A) 8 % (B) 20 % (C) 17 % (D) 50 %

Tardigrade Admin · Accepted Answer

μcal = q × d = 1.6 × 10-19 × 2.29 × 10-10. = 3.664 × 10-29 C m % ionic character = (μexp/μcal) × 100 = (6.226 × 10-30/3.664 × 10-29) × 100 = 17 %

Q. The bond length of $H Cl$ bond is $2.29 \times 1 0^{- 10} m$ . The percentage ionic character of $H Cl$ , if measured dipole moment is $6.226 \times 1 0^{- 30} C m$ is

8 %

20 %

17 %

50 %

$μ_{c a l} = q \times d = 1.6 \times 1 0^{- 19} \times 2.29 \times 1 0^{- 10}$ .
$= 3.664 \times 1 0^{- 29} C m$
% ionic character = $\frac{μ _{e x p}}{μ _{c a l}} \times 100$
$= \frac{6.226 \times 1 0 ^{- 30}}{3.664 \times 1 0 ^{- 29}} \times 100 = 17%$

Q. The bond length of HCl bond is 2.29×10−10m. The percentage ionic character of HCl, if measured dipole moment is 6.226×10−30Cm is

8 %

20 %

17 %

50 %

μcal​=q×d=1.6×10−19×2.29×10−10. =3.664×10−29Cm % ionic character = μcal​μexp​​×100 =3.664×10−296.226×10−30​×100=17%

Q. The bond length of $H Cl$ bond is $2.29 \times 1 0^{- 10} m$ . The percentage ionic character of $H Cl$ , if measured dipole moment is $6.226 \times 1 0^{- 30} C m$ is

$μ_{c a l} = q \times d = 1.6 \times 1 0^{- 19} \times 2.29 \times 1 0^{- 10}$ .
$= 3.664 \times 1 0^{- 29} C m$
% ionic character = $\frac{μ _{e x p}}{μ _{c a l}} \times 100$
$= \frac{6.226 \times 1 0 ^{- 30}}{3.664 \times 1 0 ^{- 29}} \times 100 = 17%$