In the structure of ClF 3 , the number of lone pairs of electrons on the central atom Cl is

Number of hybrid orbitals

$=\frac{1}{2}\left[\mathrm{V}.\mathrm{E} + \mathrm{M}-\mathrm{C}+\mathrm{A}\right]$

$$\begin{gathered} =\frac{1}{2}\left[7+3-0+0\right] \\ =\frac{10}{2} = 5 \end{gathered}$$

There are three side atoms, so, $\text{bond pair} =3 \text{and lone pair} = 2$

${\mathrm{ClF}}_3=3 \mathrm{bp}+2 \mathrm{lp}$

$2$ lone pair at equatorial position.