Which of the following sequences of hybridisation, geometry and magnetic nature are correct for the given coordination compounds ? A. [ NiCl _4]^ 2- - sp^3, tetrahedral, paramagnetic B. [ Ni(NH _3)_6]^ 2+ - sp^3d^2, octahedral, paramagnetic C. [ Ni(CO) _4] - sp^3, tetrahedral, paramagnetic D. [ Ni(CN) _4]^ 2- - dsp^2, square planar, diamagnetic Choose the correct answer from the options given below :
D. A, B and D only
In [ NiCl _4]^ 2- , Ni is in +2 oxidation state with a 3d^8 configuration. Since 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 [ 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 [ Ni(CO) _4], Ni is in 0 oxidation state with a 3d^8 4s^2 configuration. Since 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 [ Ni(CN) _4]^ 2- , Ni is in +2 oxidation state with a 3d^8 configuration. Since 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
Related: Chemistry — Coordination Compounds · All PYQ Banks