The number of nodal planes present in s × s antibonding orbitals is

The correct option is 1.

To understand the number of nodal planes in an s-s antibonding molecular orbital, we must look at how atomic orbitals combine to form molecular orbitals.
When two atomic s orbitals (which are spherically symmetrical) overlap along the internuclear axis, they can combine in two different ways depending on the phase relationship of their wave functions: bonding and antibonding.

Bonding Molecular Orbital: This is formed by the constructive combination (addition) of the wave functions of the two s orbitals. The electron density increases in the region between the nuclei, and there is no node between them.

Antibonding Molecular Orbital: This is formed by the destructive combination (subtraction) of the wave functions of the two s orbitals. This orbital is represented as:

${\sigma }^{*}$

Because the wave functions of the combining s orbitals have opposite phases, they cancel each other out in the region exactly halfway between the two nuclei.
This region where the probability of finding an electron is zero forms a plane perpendicular to the internuclear axis, known as a nodal plane.

Since there is exactly one such plane separating the two lobes of the antibonding orbital, the number of nodal planes present in an s × s antibonding molecular orbital is 1.