The IUPAC name for $[N i (C O)_{4}]$ is :

Tetracarbonylnickel (II)

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Tetracarbonylnickel (0)

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Tetracarbonylnickelate (II)

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Tetracarbonylnickelate (0)

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Solution

The correct option is A Tetracarbonylnickel (0)
When we write the IUPAC name for a coordination complex, if the coordination entity is neutral, then the name of Ligands come first followed by the name of the metal and the oxidation state in ().

Thus, the IUPAC name for $[N i (C O)_{4}]$ is Tetracarbonylnickel (0).

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Similar questions

The IUPAC nomenclature for [Ni(CO)₄] is:

[N i (C O)_{4}]

K_{2} [H g I_{4}]

[C o (N H_{3})_{6}] C l_{3}

K_{3} [F e (C N)_{6}]

K_{2} [P d (C l)_{4}]

[N i (C o)_{4}]

N i (C O)_{4}

[C o {(N H_{3})}_{5} C l] S O_{4}

[N i {(C O)}_{4}]

{[N i {C l}_{4}]}^{2 -}

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