Which of the statement is incorrect for XeO4 ?

The correct option is "It has a square planar shape".

Let us analyze the structure, hybridization, and bonding in Xenon tetroxide (${\mathrm{XeO}}_4$) step-by-step to understand why this statement is incorrect:
1. Valence Electrons and Hybridization of Xenon ($\mathrm{Xe}$):
Xenon is a noble gas belonging to Group 18, so it has 8 valence electrons in its outermost shell. In ${\mathrm{XeO}}_4$, xenon forms four double bonds with four oxygen atoms (each oxygen atom requires two electrons to complete its octet). Therefore, all 8 valence electrons of xenon are involved in bonding, leaving zero lone pairs on the central xenon atom.
The steric number (SN) is calculated as:
$\mathrm{Steric\; Number}=\left(\mathrm{Number\; of\; sigma\; bonds}\right)+\left(\mathrm{Number\; of\; lone\; pairs}\right)$
$\mathrm{Steric\; Number}=4+0=4$
A steric number of 4 corresponds to ${\mathrm{sp}}^3$ hybridization of the central xenon atom.

2. Molecular Geometry and Shape:
With ${\mathrm{sp}}^3$ hybridization and no lone pairs on the central atom, the molecular geometry of ${\mathrm{XeO}}_4$ is perfect tetrahedral. Thus, the statement "It has a tetrahedral shape" is correct, which makes the statement "It has a square planar shape" incorrect.

3. Sigma ($\sigma$) and Pi ($\pi$) Bonding:
Each of the four $\mathrm{Xe}=O$ double bonds consists of one $\sigma$ bond and one $\pi$ bond:
- The four $\sigma$ bonds are formed by the overlap of the ${\mathrm{sp}}^3$ hybridized orbitals of Xenon with the half-filled $p$ orbitals of the oxygen atoms. Therefore, four ${\mathrm{sp}}^3-p$ $\sigma$ bonds are present (which makes this option a correct statement).
- Xenon uses its $d$ orbitals to form the $\pi$ bonds because its valence $s$ and $p$ orbitals are already used in ${\mathrm{sp}}^3$ hybridization. Oxygen, being in the second period, uses its $\mathrm{2p}$ orbitals for $\pi$ bonding. Thus, all four $\pi$ bonds are formed by $d\left(\mathrm{Xe}\right)-p\left(O\right)$ overlap. Therefore, four $\mathrm{p\pi }-\mathrm{d\pi }$ bonds are present (which also makes this option a correct statement).

Since ${\mathrm{XeO}}_4$ is tetrahedral in shape and not square planar, the statement "It has a square planar shape" is incorrect.