Identify Correct Sequences of Hybridisation Geometry and Magnetic Nature for Coordination Compounds

To determine the correct statements, let us analyze each of the given coordination compounds individually:

1. Compound A: $[\text{NiCl}_4]^{2-}$

   • Oxidation state of $\text{Ni}$: $\text{Ni}^{2+}$
   • Electronic configuration of $\text{Ni}^{2+}$: $[\text{Ar}] 3\text{d}^8 4\text{s}^0$
   • $\text{Cl}^-$ is a weak field ligand, so no pairing of $3\text{d}$ electrons takes place.
   • The complex utilizes one $4\text{s}$ and three $4\text{p}$ orbitals for bonding, resulting in $\text{sp}^3$ hybridisation and a tetrahedral geometry.
   • Due to the presence of 2 unpaired electrons in the $3\text{d}$ subshell, the complex is paramagnetic.
   • Statement A is correct.

2. Compound B: $[\text{Ni}(\text{NH}_3)_6]^{2+}$

   • Oxidation state of $\text{Ni}$: $\text{Ni}^{2+}$
   • Electronic configuration of $\text{Ni}^{2+}$: $[\text{Ar}] 3\text{d}^8 4\text{s}^0$
   • For a $3\text{d}^8$ ion in an octahedral field, the crystal field electronic distribution is $\text{t}_{2\text{g}}^6 \text{e}_\text{g}^2$. Two inner $3\text{d}$ orbitals are not available for $\text{d}^2\text{sp}^3$ hybridisation.
   • Thus, it uses outer orbitals ($4\text{s}$, $4\text{p}$, and $4\text{d}$) to undergo $\text{sp}^3\text{d}^2$ hybridisation, yielding an octahedral geometry.
   • With 2 unpaired electrons in the $\text{e}_\text{g}$ orbitals, the complex is paramagnetic.
   • Statement B is correct.

3. Compound C: $[\text{Ni}(\text{CO})_4]$

   • Oxidation state of $\text{Ni}$: $\text{Ni}^0$
   • Electronic configuration of $\text{Ni}^0$: $[\text{Ar}] 3\text{d}^8 4\text{s}^2$
   • $\text{CO}$ is a strong field ligand, causing complete pairing of electrons. The two $4\text{s}$ electrons shift to the $3\text{d}$ subshell, giving a $3\text{d}^{10} 4\text{s}^0$ configuration.
   • The empty $4\text{s}$ and three $4\text{p}$ orbitals hybridise to form $\text{sp}^3$ hybridisation with a tetrahedral geometry.
   • Since there are no unpaired electrons, the complex is diamagnetic (not paramagnetic).
   • Statement C is incorrect.

4. Compound D: $[\text{Ni}(\text{CN})_4]^{2-}$

   • Oxidation state of $\text{Ni}$: $\text{Ni}^{2+}$
   • Electronic configuration of $\text{Ni}^{2+}$: $[\text{Ar}] 3\text{d}^8 4\text{s}^0$
   • $\text{CN}^-$ is a strong field ligand, causing the pairing of $3\text{d}$ electrons. This leaves one $3\text{d}$ orbital vacant.
   • The vacant $3\text{d}$, $4\text{s}$, and two $4\text{p}$ orbitals undergo $\text{dsp}^2$ hybridisation, forming a square planar geometry.
   • Since all electrons are paired, the complex is diamagnetic.
   • Statement D is correct.

Thus, the correct statements are A, B and D only.

Correct Option: D