Question

The correct order of dipole moment is:

• A. $C{H}_4<N{F}_3<N{H}_3<H_{2}O$
• B. $N{F}_3<C{H}_4<N{H}_3<H_{2}O$
• C. $N{H}_3<N{F}_3<C{H}_4<H_{2}O$
• D. $H_{2}O<N{H}_3<N{F}_3<C{H}_4$

Answer 1

The correct option is A $C{H}_4<N{F}_3<N{H}_3<H_{2}O$

$C{H}_4$ is regular tetrahedral and the electronegativity difference (EN) between $C$ and $H$ are negligible, which cancel out.

$N{F}_3$ has less EN between $N$ and $F$ as compared to $N$ and $H$ difference, therefore will have a lower dipole.

$H_{2}O$ has maximum EN between $H$ and $O$, therefore will have a maximum dipole.

$C{H}_4\left(0\right)<N{F}_3\left(0.24\right)<N{H}_3\left(1.46\right)<H_{2}O\left(1.85\right)$.