The dipole moment of H - Br molecule is $2.6 \times 10^{- 30}$ coulomb.metre. Its bond length is $1.3 ˚ A$ Then, the percentage ionic character of H - Br bond is:

20%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

21.5%

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12.5%

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
12.5%

% of ionic character = dipole moment(observed)/ calculated value $\times$ 100

Calculated dipole moment = charge $\times$ distance, charge as you know is $1.6 \times 10^{- 19}$ C

Suggest Corrections

Similar questions

2.6 \times 10^{- 30}

1.3 ˚ A

2.6 \times 10^{- 30}

1.3 ˚ A

The dipole moment of H-Br molecule is 2.6×10^-30. It's bond length is 1.3A°. Then the percentage ionic character of H-Br ?

H C l

1.3 D

H C l

1.3 ˚ A

H - C l

= 4.8 \times 10^{- 10} e s u

H X

2.59 \times 10^{- 30}

H X

1.39 ˚ A

Join BYJU'S Learning Program