A complex with the composition $\left[ {MA3B} \right]n \pm $ is found to have no geometrical isomers. The possible structure(s) of the complex is (Where A and B are monodentate ligands).
A) Tetrahedral
B) Square planar
C) Both (a) and (b)
D) Cannot be predicted

Hint: We need to remember that in the field of inorganic coordination buildings it is the mathematical example framed by the atoms in the ligands that are attached to the center particle in an atom or a coordination complex. The mathematical course of action will change as per the number and kind of ligands attached to the metal community, and to the coordination inclination of the center molecule, commonly a metal in a coordination complex. The quantity of particles fortified as named the coordination number.

Complete step by step answer:
We must know that the square planar sub-atomic calculation in science portrays the stereochemistry (spatial course of action of particles) that is received by certain substance mixes. As the name recommends, particles of this math have their molecules situated at the edges of a square on a similar plane about a center atom.
As we know that in a tetrahedral sub-atomic math, a center atom is situated at the middle with four substituents that are situated at the sides of a tetrahedron. The bond angle is $109.5^\circ $ when each of the four substituents is the equivalent, as in methane. Methane and other totally even tetrahedral atoms have a place with tetrahedral point groups, yet most tetrahedral particles have lower evenness. Tetrahedral atoms can be chiral.
In both square planar and tetrahedral geometries, just a single course of action is conceivable. Subsequently, it can't have geometrical isomers.

So, the correct answer is Option C.

Note: We need to remember that the square planar coordination is normal when two ligands on the z-pivot of an octahedron are taken out from the mind boggling, leaving just the ligands in the x-y plane. As the z-ligands move away, the ligands in the square plane draw a little nearer to the metal. The orbital parting graph for square planar coordination would thus be able to be gotten from the octahedral outline. As ligands move away along the z-hub, d-orbitals with a z-segment will fall in energy. The energy of \[{{\text{d}}_{{{\text{z}}^{\text{2}}}}}\] orbital falls the most, as its electrons are amassed in projections along the z-axis. The \[{{\text{d}}_{{\text{xz}}}}\] and \[{{\text{d}}_{{\text{yz}}}}\] orbitals drop in energy, but not as much. On the other hand, the ${{\text{d}}_{{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}}}$ orbital and the \[{{\text{d}}_{{\text{xy}}}}\] orbitals increase in energy.