Co-Ordination Compounds Question 287
Question: Which of the following statements is correct? (Atomic number of Ni = 28)
Options:
A) $ Ni{{( CO )}_4} $ is diamagnetic and $ {{[NiCl_4]}^{2-}} $ and $ {{[Ni{{(CN)}_4}]}^{2-}} $ are paramagnetic
B) $ Ni{{(CO)}_4} $ and $ {{[Ni{{(CN)}_4}]}^{2-}} $ are diamagnetic and $ {{[NiCl_4]}^{2-}} $ is paramagnetic
C) $ Ni{{(CO)}_4} $ and $ {{[NiCl_4]}^{2-}} $ are diamagnetic and $ {{[Ni{{(CN)}_4}]}^{2-}} $ is paramagnetic
D) $ {{[NiCl_4]}^{2-}} $ and $ {{[Ni{{(CN)}_4}]}^{2-}} $ are diamagnetic and $ Ni{{( CO )}_4} $ is paramagnetic
Show Answer
Answer:
Correct Answer: B
Solution:
Though both $[NiCl_4]^{2−}$ and $[Ni(CO)_4]$ are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl− is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, $[NiCl_4]^{2−}$ is paramagnetic.
In $[Ni(CO)_4]$, Ni is in the zero oxidation state i.e., it has a configuration of $3d^8 4s^2$.
But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since no unpaired electrons are present in this case, $[Ni(CO)_4]$ is diamagnetic.
