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The oxidation state of Cr in $[Cr(CO)_6]$ is (Z(Cr) = 24):
A+3
B+2
C+6
D0
Answer & Solution
Correct answer: D. 0
CO is a neutral ligand (zero charge). The complex is neutral, so Cr must have oxidation state **0**. EAN = 24 − 0 + 12 = 36 (Kr config) — rule obeyed.
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