Question

Arrange the following in order of increasing dipole moment:
$H_{2}O,H_{2}S,B{F}_3$

• A. $B{F}_3<H_{2}S<H_{2}O$
• B. $H_{2}S<B{F}_3<H_{2}O$
• C. $H_{2}O<H_{2}S<B{F}_3$
• D. $B{F}_3<H_{2}O<H_{2}S$

Answer 1

The correct option is A $B{F}_3<H_{2}S<H_{2}O$

In $B{F}_3$, dipole moment is zero due to its symmetrical structure. Summations of all dipoles are zero.

In $H_{2}S$ and $H_{2}O$ due to unsymmetrical structure net +ve dipole is there. $H_{2}O$ has a higher dipole due to the higher electronegativity of oxygen than sulphur.

Hence, the correct order is: $B{F}_3<H_{2}S<H_{2}O$.