You visited us 1 times! Enjoying our articles? Unlock Full Access!

Dipole Moment

The percentag...

Question

The percentage ionic character of a bond having $1.257 A^{0}$ its length and $1.03$ D its dipole moment is:
( $q = 4.8 \times 10^{- 10}$ esu)

10 %

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

15 %

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

16.83 %

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

18.88 %

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

Open in App

Solution

The correct option is B $16.83 %$
$μ_{i o n i c} (i o n i c d i p o l e) = 4.8 \times 10^{- 10} \times 1.275 \times 10^{- 8}$
$μ_{i o n i c} = 6.12 \times 10^{- 18} e s u c m$ = $6.12 D$
Observed dipole moment $μ_{o b s e r v e d} = 1.032 D$
% ionic character = $\frac{μ_{o b s e r v e d}}{μ_{i o n i c}} \times 100$
= $\frac{1.032 D}{6.12 D} \times 100$
= $16.38 %$

Suggest Corrections

Similar questions

H C l

1.275 ˚ A

1.03 D

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

H C l

1.3 D

H C l

1.3 ˚ A

H - C l

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

H C l

1.275 A^{o}

1.03 D

2.6 \times 10^{- 30}

1.3 ˚ A

H B r

0.78 \times 10^{- 18}

H B r

1.41 A^{\circ}

H B r

Join BYJU'S Learning Program