${\mathrm{O}}_2$ undergoes photochemical dissociation into $1$ a normal oxygen atom $\left(\mathrm{O}\right)$ and a more energetic oxygen atom $\mathrm{O}*$. If $\left(\mathrm{O}\right)$ has $1.967 \mathrm{eV}$ more energy than $\left(\mathrm{O}\right)$ and normal dissociation energy of ${\mathrm{O}}_2$ is $498 \mathrm{kJ} {\mathrm{mol}}^{-1}$, what is the maximum wavelength effective for the photochemical dissociation of ${\mathrm{O}}_2$?

Suppose that a hypothetical atom gives a red, green, blue and violet line. Which electron jump would give off the red spectral line according to figure ?

Consider the following statements:

$\left(\mathrm{a}\right)$ Electron density in the $\mathrm{XY}$ plane in $3{\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$ orbital is zero

$\left(\mathrm{b}\right)$ Electron density in the $\mathrm{XY}$ plane in $3{\mathrm{d}}_{{\mathrm{z}}^2}$ orbital is zero.

$\left(\mathrm{c}\right)$ $2\mathrm{s}$ orbital has one nodal surface.

$\left(\mathrm{d}\right)$ $2{\mathrm{p}}_z$ orbital, $\mathrm{XY}$ is the nodal plane.

From the above statements, which are incorrect statements?