A) Tetrahedral, Paramagnetic Tetrahedral, Diamagnetic
B) Tetrahedral, Diamagnetic Square planar, Diamagnetic
C) Square planar, Diamagnetic Square planar, Paramagnetic
D) Trigonal pyramidal Square pyramidal
Correct Answer: B
Solution :
(I) In \[[Ni{{(CO)}_{4}}]\], Ni is present as Ni atom. Since, CO is a strong field ligand, therefore, it pairs up the unpaired electrons. \[_{28}Ni=[Ar]\,3{{d}^{8}}\,\,4{{s}^{2}}\]You need to login to perform this action.
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