JEE Main 2026 — Chemistry Inorganic Chemistry

In $[\text{NiCl}_4]^{2-}$[NiCl4​]2−, Ni is in $+2$+2 oxidation state with a $3d^8$3d8 configuration. Since $\text{Cl}^-$Cl− is a weak field ligand, no pairing of electrons takes place. The hybridisation is $sp^3$sp3, 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(NH3​)6​]2+, Ni is in $+2$+2 oxidation state with a $3d^8$3d8 configuration. In an octahedral field, a $d^8$d8 ion cannot form inner orbital complexes. It undergoes $sp^3d^2$sp3d2 hybridisation, forming an octahedral geometry. It has two unpaired electrons, making it paramagnetic. Thus, statement B is correct.

In $[\text{Ni(CO)}_4]$[Ni(CO)4​], Ni is in $0$0 oxidation state with a $3d^8 4s^2$3d84s2 configuration. Since $\text{CO}$CO is a strong field ligand, the $4s$4s electrons are forced to pair up in the $3d$3d orbitals, resulting in a $3d^{10}$3d10 configuration. The hybridisation is $sp^3$sp3, 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(CN)4​]2−, Ni is in $+2$+2 oxidation state with a $3d^8$3d8 configuration. Since $\text{CN}^-$CN− is a strong field ligand, it causes the pairing of $3d$3d electrons, leaving one $3d$3d orbital empty. The hybridisation is $dsp^2$dsp2, 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