Geometry of Ammonia Complexes of Nickel Platinum and Zinc

To determine the geometries of the ammine complexes formed by $\text{Ni}^{2+}$, $\text{Pt}^{2+}$, and $\text{Zn}^{2+}$ ions, we analyze their electron configurations, preferred coordination numbers, and hybridizations:

1. Ammine complex of $\text{Ni}^{2+}$:

   • The electronic configuration of $\text{Ni}^{2+}$ is $[\text{Ar}]\,3d^8$.
   • Nickel(II) forms an ammine complex with a coordination number of 6: $\left[\text{Ni}(\text{NH}_3)_6\right]^{2+}$.
   • Since $\text{NH}_3$ cannot cause pairing in the $3d$ orbitals of $\text{Ni}^{2+}$, outer $4d$ orbitals are used for hybridization ($sp^3d^2$).
   • Therefore, the geometry of $\left[\text{Ni}(\text{NH}_3)_6\right]^{2+}$ is octahedral.

2. Ammine complex of $\text{Pt}^{2+}$:

   • The electronic configuration of $\text{Pt}^{2+}$ is $[\text{Xe}]\,4f^{14}\,5d^8$.
   • Platinum is a $5d$ transition metal, where the crystal field splitting energy ($\Delta_o$) is very high. Thus, all ligands (including $\text{NH}_3$) behave as strong field ligands and force electron pairing in the $5d$ orbitals.
   • It forms a 4-coordinate complex: $\left[\text{Pt}(\text{NH}_3)_4\right]^{2+}$.
   • The hybridization involved is $dsp^2$, giving a square planar geometry.

3. Ammine complex of $\text{Zn}^{2+}$:

   • The electronic configuration of $\text{Zn}^{2+}$ is $[\text{Ar}]\,3d^{10}$.
   • Since the $3d$ subshell is completely filled ($d^{10}$ configuration), inner $d$-orbitals are not available for hybridization.
   • It forms a 4-coordinate complex: $\left[\text{Zn}(\text{NH}_3)_4\right]^{2+}$.
   • The metal uses one $4s$ and three $4p$ orbitals to undergo $sp^3$ hybridization, resulting in a tetrahedral geometry.

Thus, the respective geometries are octahedral, square planar, and tetrahedral.

Correct Option: A