Below is the lewis structure of the fluoramine nh2f molecule. nfhh count the number of bonding pairs and the number of lone pairs around the left hydrogen atom.

            Bonding Pair Electrons  =  2

            Lone Pair of Electrons  =  Zero

In order to answer this question one should draw the Lewis structure of Fluoramine first following following steps.

Step 1:

           First of all add all the valence electrons present in all elements of the given compound. i.e.

                                 Hydrogen × 2  =  1 × 2  =  2

                                 Fluorine × 1  =  7 × 1  =  7

                                 Nitrogen × 1  =  5 × 1  =  5

                                 Total Valence Electrons  =  14

Step 2:

           Draw the central atom (i.e. Nitrogen) and surround it by remaining elements. In our case Nitrogen is surrounded by two H atoms and one F atom.

                                                   H N H

                                                       F

Step 3:

           Connect all atoms to central atom via a single bond and subtract 2 electrons per bond from total valence electrons. As there are three single bonds so, we will subtract 6 electrons from 14 valence electrons and are left with 8 valence electrons. Now, these remaining 8 electrons are distributed among the elements starting from most electronegative (i.e. Fluorine) and then Nitrogen. So, Fluorine will get 3 lone pair of electrons and \nitrogen will get one lone pair of electrons as shown in figure.

                  As Hydrogen atoms are making only single bonds hence, they will have two bonding pair electrons per H atom and as shown in figure H atoms didn't get any lone pair of electrons.

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