Co-Ordination Compounds Question 162
Question: Amongst $ Ni{{(CO)}_4},{{[Ni{{(CN)}_4}]}^{2-}} $ and $ NiCl_4^{2-} $
Options:
A) $ Ni{{(CO)}_4} $ and $ NiCl_4^{2-} $ are diamagnetic and $ {{[Ni{{(CN)}_4}]}^{2-}} $ is paramagnetic
B) $ {{[Ni{{(CN)}_4}]}^{2-}} $ and $ {{[Ni{{(CN)}_4}]}^{2-}} $ are diamagnetic and $ Ni{{(CO)}_4} $ is paramagnetic
C) $ Ni{{(CO)}_4} $ and $ {{[Ni{{(CN)}_4}]}^{2-}} $ are diamagnetic and $ NiCl_4^{2-} $ is paramagnetic
D) $ Ni{{(CO)}_4} $ is diamagne and $ NiCl_4^{2-} $ and $ {{[Ni{{(CN)}_4}]}^{2-}} $ are paramagnetic
Show Answer
Answer:
Correct Answer: C
Solution:
$ [Ni{{(CO)}_4}]: $ Oxidation state of $ Ni $ is 0 $ Ni=(Z=28)\Rightarrow 3d^{8}4s^{2} $ Co is a strong field ligand, it causes pairing. $ sp^{3} $ -Hybridisation (tetrahedral) There are no unpaired electrons, so the complex is diamagnetic.
Spin magnetic moment = Zero. $ {{[NiCl_4]}^{2-}}: $ $ N{i^{2+}}=3d^{8} $ Cyano is a strong field ligand, it causes pairing. $ dsp^{2} $ -hybridsation (square planar) There are no unpaired electrons so, the complex is diamagnetic.
Spin magnetic moment = zero. $ {{[NiCl_4]}^{2-}}: $ $ N{i^{2+}}=3d^{8} $ . Chloride is a weak field ligand, no pairing. $ sp^{3} $ -hybridisation (tetrahedral) There are two unpaired electrons, so the complex is paramagnetic. Spin magnetic moment $ \mu =\sqrt{n(n+2)}BM=\sqrt{2(2+2)}BM=\sqrt{8}BM $
