To find the oxidation state of Chromium (Cr) in the compound CrO_6, we must consider the oxidation states of each element within the molecule.
In most compounds, the oxidation state of oxygen (O) is -2. This is a general rule for determining oxidation states in oxides.
In CrO_6, there are six oxygen atoms. Therefore, the total oxidation state contributed by oxygen atoms is calculated as:
6 \times (-2) = -12.
Let the oxidation state of chromium (Cr) be x. For the compound to be electrically neutral, the sum of oxidation states of chromium and oxygen should be zero.
Therefore, the equation is:
x + (-12) = 0.
Solving for x gives:
x = +12.
Thus, the oxidation state of chromium (Cr) in CrO_6 is +12.
However, in practice, such high oxidation states are often not observed for chromium, indicating potentially incorrect interpretation or formulation in the given compound. For the provided options, the closest matching and feasible answer based on usual constraints for chromium containing oxides should be reconsidered or verified contextually.