Question

A square planar complex is formed by hybridization of which atomic orbitals:
(A) s, ${p_x}$, ${p_y}$, ${d_{yz}}$
(B) s, ${p_x}$, ${p_y}$,${d_{{x^2} - {y^2}}}$
(C) s, ${p_x}$, ${p_y}$,${d_{{z^2}}}$
(D) s, ${p_x}$, ${p_y}$,${d_{xy}}$

Answer 1

For a compound to have square planar geometry it will have a coordination number of 4 and the involved orbitals will lie in the same plane at right angles to each other.

Complete step by step answer:
-Square planar geometry is a type of molecular geometry where the atoms are positioned at the corners of a square on the same plane about the central atom.
Here the central atom bonds with 4 other atoms. So for this type of geometry the coordination number will be 4 and the bond angle between the orbitals involved is 90$^ \circ C$. So, the orbitals involved should be oriented at right angles to each other and thus lie in the same plane.
-Now we will see all the options to check which one of them has all the orbitals lying in the same plane.
For (A) s, ${p_x}$, ${p_y}$, ${d_{yz}}$: s-orbital is spherical at the centre and so lies in all planes. The ${p_x}$ and ${p_y}$ orbitals lie on the x and y axis but the ${d_{yz}}$ orbital will lie between y and z axis. These orbitals do not lie in the same plane and hence they cannot form a square planar complex.
For (B) s, ${p_x}$, ${p_y}$,${d_{{x^2} - {y^2}}}$: The ${p_x}$ and ${p_y}$ orbitals lie on the x and y axis and the ${d_{{x^2} - {y^2}}}$ orbital also lies on the x and y axis. These orbitals will lie in the same plane which is the x-y plane and hence they can form a square planar complex.
For (C) s, ${p_x}$, ${p_y}$,${d_{{z^2}}}$: The ${p_x}$ and ${p_y}$ orbitals lie on the x and y axis but the ${d_{{z^2}}}$ orbital will lie on the z axis. These orbitals do not lie in the same plane and hence they cannot form a square planar complex.
For (D) s, ${p_x}$, ${p_y}$,${d_{xy}}$: The ${p_x}$ and ${p_y}$ orbitals lie on the x and y axis but the ${d_{xy}}$ orbital will lie between the x and y axis (not on the x and y axis). These orbitals do not lie in the same plane and hence they cannot form a square planar complex.

Hence the correct option will be: (B) s, ${p_x}$, ${p_y}$,${d_{{x^2} - {y^2}}}$

Note: Square planar geometry is stabilized by ligands like porphyrins and is generally shown by transition metal complexes with ${d^8}$ configuration like Rh(l), Pd(ll), Au(lll). It is shown by $Xe{F_4}$, $PtCl_4^{ - 2}$, anticancer drugs like cisplatin $\left[ {PtC{l_2}{{(N{H_3})}_2}} \right]$ and carboplatin, etc.