Answer
a) In the compound $CrO$, oxidation state of $Cr = +2$.
b) In the compound $CrO_{3}$, oxidation state of $Cr = +6$.
c) In the compound $Cr_{2}O_{3}$, oxidation state of $Cr = +3$.
Work Step by Step
Algebraic sum of oxidation for any compound is equal to Zero.
a) In $CrO$, oxidation state of O = -2. Let's say oxidation state of Cr = $x$.
Hence, $x + (-2) = 0$, by solving this we get $x = +2$
So, oxidation state of $Cr = +2$ here.
b)In $CrO_{3}$, oxidation state of O = -2. Let's say oxidation state of Cr = $x$.
Hence, $x + (-2)\times3 = 0$ by solving this we get $x = +6$
So, oxidation state of $Cr = +6$ here.
c) In $Cr_{2}O_{3}$, oxidation state of O = -2. Let's say oxidation state of Cr = $x$.
Hence, $x\times2 + (-2)\times3 = 0$ by solving this we get $x = +3$
So, oxidation state of $Cr = +3$ here.
