In acidic medium, $${H_2}{O_2}$$ changes $$C{r_2}{O_7}^{2 - }$$ to $$Cr{O_5}$$ which has two (-O-O-) bonds. Oxidation state of Cr in $$Cr{O_5}$$ is:(A) +5(B) +3(C) +6(D) -10

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Vedantu Content Team · Accepted Answer

Hint: In peroxide compounds, oxidation number of oxygen atoms is taken as -1 . We can calculate the overall charge of a compound by just adding respective oxidation numbers of all atoms present in the compound. From this expression, we can find the oxidation number of Cr.Complete answer: Let’s see the structure of $$Cr{O_5}$$ to get ahead.You need to remember that $$Cr{O_5}$$ contains two -O-O- linkages. So, in short $$Cr{O_5}$$ has four oxygen atoms that are in peroxide linkage.Now that is a rule that if oxygen atom is involved in a peroxide linkage, then both the oxygen have oxidation number +1 which is different from the normal oxidation state of oxygen atom.Let’s calculate the oxidation state of Cr in $$Cr{O_5}$$.Overall charge on compound= Oxidation state of Cr + 4(Oxidation state of oxygen in peroxide form) + Oxidation state of other oxygen atom0 = Oxidation state of Cr + 4(-1) + (-2)0 = Oxidation state of Cr - 4 - 2 Oxidation state of Cr = 4 + 2Oxidation state of Cr = +6Hence we can say that the correct answer of this question is (C) +6.Additional Information:- Remember that the normal oxidation state of an oxygen atom is -2 .- In case that compound contains -O-O- (peroxide) linkage, then the numbers of oxygen involved in peroxide linkage have an oxidation number equal to -1 .- When oxygen atom is present in superoxide from, those atoms have an oxidation number of -0.5 .Note:  Do not consider oxidation number of oxygen atoms in peroxide linkage equal to (-2) as this mistake often occurs. In case there is overall charge on the compound, do not forget to include it in the calculation of oxidation number.

In acidic medium, \[{H_2}{O_2}\] changes \[C{r_2}{O_7}^{2 - }\] to \[Cr{O_5}\] which has two (-O-O-) bonds. Oxidation state of Cr in \[Cr{O_5}\] is:
(A) +5
(B) +3
(C) +6
(D) -10

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In acidic medium, \[{H_2}{O_2}\] changes \[C{r_2}{O_7}^{2 - }\] to \[Cr{O_5}\] which has two (-O-O-) bonds. Oxidation state of Cr in \[Cr{O_5}\] is:(A) +5(B) +3(C) +6(D) -10

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