Question

which has square planar geometry?
a.) $Pt{{(N{{H}_{3}})}_{2}}C{{l}_{2}}$
b.) ${{[Ni{{(CN)}_{4}}]}^{2-}}$
c.) Both (a) and (b)
d.) None of these

Answer 1

Hybridization of $Pt{{(N{{H}_{3}})}_{2}}C{{l}_{2}}$ is $ds{{p}^{2}}$. Hybridization of ${{[Ni{{(CN)}_{4}}]}^{2-}}$ is $ds{{p}^{2}}$. Square planar complexes have $ds{{p}^{2}}$ hybridization. Nickel and platinum has oxidation number as +2 and electronic configuration $3{{d}^{8}}$.

Complete step by step answer:
In ${{[Ni{{(CN)}_{4}}]}^{2-}}$ ,
Oxidation state of nickel can be calculated as follows:
Consider x as oxidation state of Nickel, oxidation state of CN is -1, so
x + 4(-1) =-2
x= +2
Electronic configuration of Ni is $[Ar]3{{d}^{8}}4{{s}^{2}}$.
Electronic configuration of $N{{i}^{2+}}$is $[Ar]3{{d}^{8}}$.
As $C{{N}^{-}}$ strong field ligand, electrons are paired, four pair of electrons from $C{{N}^{-}}$ occupies 3d, 4s and two 4p hybrid orbitals, so hybridization is $ds{{p}^{2}}$. So, ${{[Ni{{(CN)}_{4}}]}^{2-}}$ has square planar geometry.
In $Pt{{(N{{H}_{3}})}_{2}}C{{l}_{2}}$,
Oxidation state of platinum is +2 as the oxidation state of ammonia molecules is zero and for Cl oxidation state is -1.
Electronic configuration of Pt is $[Xe]5{{d}^{8}}6{{s}^{2}}$.
Electronic configuration of $P{{t}^{2+}}$ is $[Xe]5{{d}^{8}}$.
Eight electrons from 5d occupy for 5d orbitals, two electron pairs from ammonia molecule and two electron pairs from chloro ligand that is total four electron pairs occupy 5d, 6s and two 6p orbitals leading to hybridization $ds{{p}^{2}}$.
All complexes of Pt have geometry as square planar.
So, the geometry of $Pt{{(N{{H}_{3}})}_{2}}C{{l}_{2}}$ is square planar.
So, the correct answer is “Option C”.

Note: Complexes having hybridization as $ds{{p}^{2}}$ have square planar geometry. Strong field ligands like ammonia, $C{{N}^{-}}$ cause pairing of electrons. Weak field ligand does not cause pairing. Complexes with electronic configuration as ${{d}^{8}}$ usually have square planar geometry.