Which of the following has $ds{p}^2$ hybridization and is diamagnetic in nature?

(P) ${\mathrm{Na}}_4\left[\mathrm{Cr}(\mathrm{CO}{)}_4\right]$              (Q) $\left[\mathrm{Ni}(\mathrm{DMG}{)}_2\right]$

(R) $\left[\mathrm{PtHBr}{\left({\mathrm{PEt}}_3\right)}_2\right]$              (S) ${\left[\mathrm{Ag}(\mathrm{SCN}{)}_4\right]}^{3-}$

(T) ${\left[{\mathrm{AuBr}}_4\right]}^{-}$

To determine which compounds exhibit dsp2 hybridization and are diamagnetic in nature, let us analyze the given options:

1. Na₄Cr(CO)₄: This compound does not exhibit dsp2 hybridization because of its electronic configuration and bonding nature.
2. Ni(DMG)₂: The nickel ion in this complex undergoes dsp2 hybridization as its d-orbital contains 10 electrons, and s-orbital contains 2 electrons. This allows the ligand to form a square planar geometry. Additionally, the electronic configuration results in a diamagnetic nature.
3. PtHBr₂(PEt₃)₂: The platinum ion in this compound also exhibits dsp2 hybridization, leading to a square planar structure. Its electronic configuration ensures that it is diamagnetic.
4. Ag(SCN)₄³⁻: This complex does not undergo dsp2 hybridization due to its geometry and bonding characteristics.
5. AuBr₄⁻: The gold ion in this compound exhibits dsp2 hybridization, resulting in a square planar shape, and its filled d-orbital configuration makes it diamagnetic.

Hence, the correct examples of dsp2 hybridization that are also diamagnetic include:

• Ni(DMG)₂
• PtHBr₂(PEt₃)₂
• AuBr₄⁻

These examples of dsp2 hybridization demonstrate how specific complexes form square planar geometries and exhibit diamagnetic properties. This combination arises due to the full filling of the d-orbitals and the hybridization involving one d, one s, and two p orbitals.

Correct Option: C

For more dsp2 hybridization examples in coordination chemistry, understanding the electronic configuration and ligand effects is essential. This helps predict the geometry and magnetic behavior of the compounds.