The wrong statement regarding sulphur oxides is

The S in ${\mathrm{SO}}_3$ has 2 electrons in s-shell and four electrons in p- shell, Thus in second excited state we get 1 electron in  s , 3 electron in p and 2 in d orbitals.

Therefore ${\mathrm{SO}}_3$ is formed in $2^{\text{nd }}$ excited state of ' $\mathrm{S}$ '

Molecule of ${\mathrm{SO}}_3$ contains both ${\mathrm{P}}_{\pi }={\mathrm{P}}_{\pi }$ bonds and ${\mathrm{P}}_{\pi }-{\mathrm{d}}_{\pi }$ bonds as there are 1 electron in  s , 3 electron in p and 2 in d orbitals.

As we know that ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ works as an oxidant.

The reactions are as follows:

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(\mathrm{aq}\right)+{\mathrm{SO}}_3^{2-}\left(\mathrm{aq}\right)⟶{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}$

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+3{\mathrm{SO}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{H}}_2\mathrm{O}$

$\mathrm{\gamma }-{\mathrm{SO}}_3$ is less stable than $\mathrm{\alpha }-{\mathrm{SO}}_3$