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The oxidation state of Ni in $Ni(CO)_4$ is

$(a)\;II \\ (b)\;0 \\ (c)\;III \\ (d)\;VIII $

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Answer: The oxidation state of Ni in $Ni(CO)_4$ is 0.
Metal carbonyl s have symmetrical structures and are charge-neutral, resulting in their high volatility. In $Ni(CO)_4$, the nickel atom has a formal oxidation number of zero.
The sum o fthe oxidaton states = the charge on the complex $Ni(CO)_4$ being neutral has a zero charge. CO is a neutral ligand having zero charge as well. Therefore, Ni mush have an oxidation state of 0.
answered Jan 29, 2014 by sreemathi.v
edited Jul 29, 2014 by balaji.thirumalai
 

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