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oxidation of state chemistry v2074-11th class​

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CBSE

Chemistry

Grade 11

Oxidation State or Oxidation N...

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Find the oxidation number of elements in each case.

$Cr$in$Cr{O_2}C{l_2}$, $N{a_2}C{r_3}{O_{10}}$, $C{r_2}{\left( {S{O_4}} \right)_3}$, and ${\left[ {Cr{O_8}} \right]^{3 - }}$

A) $ + 6$, $ + 6$, $ + 3$, $ + 5$.

B) $ + 5$, $ + 6$, $ + 3$, $ + 4$.

C) $ + 4$, $ + 6$, $ + 2$, $ + 5$.

D) None of the above.

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Hint: We can define oxidation state as the degree of loss of an electron in a chemical compound. The oxidation state by an element in a compound can be calculated using the rules of oxidation numbers.

Complete step by step answer:

Let us see few rules for oxidation numbers,

1.A free element will be zero as its oxidation number.

2.Monatomic ions will have an oxidation number equal to charge of the ion.

3.In hydrogen, the oxidation number is + 1, when combined with elements having less electronegativity, the oxidation number of hydrogen is -1.

4.In compounds of oxygen, the oxidation number of oxygen will be -2 and in peroxides it will be -1.

5.Group 1 elements will have +1 oxidation number.

6.Group 2 elements will have +2 oxidation numbers.

7.Group 17 elements will have -1 oxidation number.

8.Sum of oxidation numbers of all atoms in neutral compounds is zero.