Answer
Bond angle in a square planar structure is $90^{\circ}$ and bond angle in a tetrahedral structure is $109.5^{\circ}$. Therefore, repulsions in a square planar structure is higher than repulsions in a tetrahedral structure. Hence, the tetrahedral $CH_{4}$ molecule is more stable than the square planar $CH_{4}$ molecule.
Work Step by Step
Bond angle in a square planar structure is $90^{\circ}$ and bond angle in a tetrahedral structure is $109.5^{\circ}$. Therefore, repulsions in a square planar structure is higher than repulsions in a tetrahedral structure. Hence, the tetrahedral $CH_{4}$ molecule is more stable than the square planar $CH_{4}$ molecule.