Among $$H_{2}S$$, $$H_{2}O$$, $$NF_{3}$$, $$NH_{3}$$ and $$CHC1_{3}$$, identify the molecule (X) with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule (X) is :

We need to identify the molecule with the lowest dipole moment among $$H_2S$$, $$H_2O$$, $$NF_3$$, $$NH_3$$, and $$CHCl_3$$, and then find the number of lone pairs on its central atom.

The approximate dipole moments (in Debye) are:

- $$H_2O$$: 1.85 D (bent geometry, two lone pairs on O)

- $$H_2S$$: 0.97 D (bent geometry, bond angle ~92 degrees)

- $$NH_3$$: 1.47 D (trigonal pyramidal, one lone pair on N)

- $$NF_3$$: 0.23 D (trigonal pyramidal, one lone pair on N)

- $$CHCl_3$$: 1.04 D (tetrahedral-like geometry)

$$NF_3$$ has the lowest dipole moment (0.23 D). This is because in $$NF_3$$, the lone pair on nitrogen points in one direction while the three highly electronegative fluorine atoms pull electron density in the opposite direction. The bond dipoles of the three N-F bonds nearly cancel the lone pair contribution, resulting in a very small net dipole moment.

This contrasts with $$NH_3$$ where both the lone pair and the N-H bond dipoles point in the same direction (toward nitrogen), reinforcing each other.

Nitrogen in $$NF_3$$ has the electronic configuration: 5 valence electrons, 3 used in bonds with F, leaving 1 lone pair on nitrogen.

The correct answer is Option 4: 1.