How many non-axial $d$ -orbitals are involved in the hybridization of ${\mathrm{CrO}}_2{\mathrm{Cl}}_2$?

In ${\mathrm{CrO}}_2{\mathrm{Cl}}_2, \mathrm{Cr}$ is in +6 oxidation state which can be easily understood as follows:

$$\begin{gathered} x+2(-2)+2(-1)=0 \\ x-4-2=0 \\ x-6=0 \\ x=+6 \end{gathered}$$

Hence, the electronic configuration of this ion can be written as follows:

${\mathrm{Cr}}^{+6}=\left[\mathrm{Ar}\right]4{\mathrm{s}}^{0}3{\mathrm{d}}^0$

The chromium atom in $Cr{O}_{2}C{l}_2$ has $$d^{3}s$$ hybridization. In $$d^{3}s$$ hybridization of chromium, the atomic orbitals involved are 4s, 3d_xy, 3d_yz and 3d_xz. The set of orbitals whose lobes lie in between the axis and are called non-axial orbitals. There are three non axial d-orbitals involved in ${\mathrm{CrO}}_2{\mathrm{Cl}}_2.$