Number of compounds with one lone pair of electrons on central atom amongst following is _______
$$O_3, H_2O, SF_4, ClF_3, NH_3, BrF_5, XeF_4$$

The number of lone pairs on the central atom can be determined using

$$\text{Lone Pairs}=\frac{\text{Valence Electrons of Central Atom}-\text{Electrons Used in Bonding}}{2}$$

For $$O_3$$, the central oxygen has 6 valence electrons and uses 4 electrons in bonding. The remaining 2 electrons constitute 1 lone pair.

For $$H_2O$$, the central oxygen has 6 valence electrons and uses 2 electrons in bonding. The remaining 4 electrons constitute 2 lone pairs.

For $$SF_4$$, sulfur has 6 valence electrons and uses 4 electrons in bonding. The remaining 2 electrons constitute 1 lone pair.

For $$ClF_3$$, chlorine has 7 valence electrons and uses 3 electrons in bonding. The remaining 4 electrons constitute 2 lone pairs.

For $$NH_3$$, nitrogen has 5 valence electrons and uses 3 electrons in bonding. The remaining 2 electrons constitute 1 lone pair.

For $$BrF_5$$, bromine has 7 valence electrons and uses 5 electrons in bonding. The remaining 2 electrons constitute 1 lone pair.

For $$XeF_4$$, xenon has 8 valence electrons and uses 4 electrons in bonding. The remaining 4 electrons constitute 2 lone pairs.

Thus, the compounds containing exactly one lone pair on the central atom are $$O_3$$, $$SF_4$$, $$NH_3$$, and $$BrF_5$$.

Therefore, the total number of such compounds is, $$4$$