Dipole moment arises from the difference in electronegativity between bonded atoms and the molecular geometry. It is a vector quantity, and the net dipole moment is the vector sum of individual bond dipoles.

1. CH\u2084: Methane ($C{H}_4$) has a tetrahedral geometry. All C-H bonds are polar, but due to the symmetrical arrangement, the individual bond dipoles cancel each other out. Hence, its net dipole moment is zero.

2. NF\u2083: Nitrogen trifluoride ($N{F}_3$) has a trigonal pyramidal geometry with a lone pair on nitrogen. The N-F bonds are polar, with fluorine being more electronegative than nitrogen. The bond dipoles point towards fluorine atoms. The lone pair also contributes to the dipole moment, but its contribution is in the opposite direction to the resultant of the N-F bond dipoles. This leads to a relatively small net dipole moment (0.23 D).

3. NH\u2083: Ammonia ($N{H}_3$) also has a trigonal pyramidal geometry with a lone pair on nitrogen. The N-H bonds are polar, with nitrogen being more electronegative than hydrogen. The bond dipoles point towards the nitrogen atom. The lone pair's dipole moment also points in the same direction as the resultant of the N-H bond dipoles, reinforcing each other. This results in a significant net dipole moment (1.47 D).

4. H\u2082O: Water ($H_{2}O$) has a bent (V-shaped) geometry with two lone pairs on oxygen. The O-H bonds are highly polar, with oxygen being much more electronegative than hydrogen. Both the O-H bond dipoles and the lone pair dipoles add up constructively, resulting in a very high net dipole moment (1.85 D).

Comparing the dipole moments: $C{H}_4$ (0 D) < $N{F}_3$ (0.23 D) < $N{H}_3$ (1.47 D) < $H_{2}O$ (1.85 D).

Therefore, the correct order is $C{H}_4<N{F}_3<N{H}_3<H_{2}O$.

Option Analysis:

• A) $C{H}_4<N{F}_3<N{H}_3<H_{2}O$: This order correctly reflects the increasing dipole moments based on molecular geometry and electronegativity differences.
• B) $N{F}_3<C{H}_4<N{H}_3<H_{2}O$: Incorrect, as $C{H}_4$ has a zero dipole moment, which is less than $N{F}_3$.
• C) $N{H}_3<N{F}_3<C{H}_4<H_{2}O$: Incorrect, the order is completely scrambled.
• D) $H_{2}O<N{H}_3<N{F}_3<C{H}_4$: Incorrect, this is the reverse order of the correct one.

Correct Answer: (A)