Which of the following sequences of hybridisation, geometry and magnetic nature are correct for the given coordination compounds ?
A. $[\text{NiCl}_4]^{2-}$ $-$ sp$^3$, tetrahedral, paramagnetic
B. $[\text{Ni(NH}_3)_6]^{2+}$ $-$ sp$^3$d$^2$, octahedral, paramagnetic
C. $[\text{Ni(CO)}_4]$ $-$ sp$^3$, tetrahedral, paramagnetic
D. $[\text{Ni(CN)}_4]^{2-}$ $-$ dsp$^2$, square planar, diamagnetic
Choose the correct answer from the options given below :

In $[\text{NiCl}_4]^{2-}$, Ni is in $+2$ oxidation state with a $3d^8$ configuration. Since $\text{Cl}^-$ is a weak field ligand, no pairing of electrons takes place. The hybridisation is $sp^3$, the geometry is tetrahedral, and the presence of two unpaired electrons makes it paramagnetic. Thus, statement A is correct.

In $[\text{Ni(NH}_3)_6]^{2+}$, Ni is in $+2$ oxidation state with a $3d^8$ configuration. In an octahedral field, a $d^8$ ion cannot form inner orbital complexes. It undergoes $sp^3d^2$ hybridisation, forming an octahedral geometry. It has two unpaired electrons, making it paramagnetic. Thus, statement B is correct.

In $[\text{Ni(CO)}_4]$, Ni is in $0$ oxidation state with a $3d^8 4s^2$ configuration. Since $\text{CO}$ is a strong field ligand, the $4s$ electrons are forced to pair up in the $3d$ orbitals, resulting in a $3d^{10}$ configuration. The hybridisation is $sp^3$, the geometry is tetrahedral, and since there are no unpaired electrons, it is diamagnetic. Statement C incorrectly lists it as paramagnetic.

In $[\text{Ni(CN)}_4]^{2-}$, Ni is in $+2$ oxidation state with a $3d^8$ configuration. Since $\text{CN}^-$ is a strong field ligand, it causes the pairing of $3d$ electrons, leaving one $3d$ orbital empty. The hybridisation is $dsp^2$, the geometry is square planar, and the absence of unpaired electrons makes it diamagnetic. Thus, statement D is correct.

Therefore, only statements A, B, and D are correct.

Answer: A, B and D only