Question

The oxidation number of Cr in \[{K_2}C{r_2}{O_7}\] is:
A. +3
B. -3
C. +6
D. -6

Answer 1

Hint: In this question we will use the method of finding the oxidation number of an element in a compound. We have to remember that the algebraic sum of the oxidation numbers of all the atoms present in a compound must be equal to zero.

Complete answer:

We know that Oxidation number is defined as the charge that an atom of an element has in its ion or appears to have when present in the combined state with other atoms.
 The compound given is \[{K_2}C{r_2}{O_7}\].
\[{K_2}C{r_2}{O_7}\], Potassium dichromate, contains the dichromate ion, in which chromium is in its hexavalent form, i.e. chromium(VI).
We know that, In \[{K_2}C{r_2}{O_7}\], potassium has a +1 charge, and oxygen has a −2 charge.
Let the oxidation number of each Cr atom be $x$.
We know that the algebraic sum of the oxidation numbers of all the atoms present in a compound must be equal to zero.
So,
$
   \Rightarrow 2\left( { + 1} \right) + 2(x) + 7( - 2) = 0 \\
   \Rightarrow + 2 + 2x - 14 = 0 \\
   \Rightarrow 2x - 12 = 0 \\
   \Rightarrow 2x = 12 \\
   \Rightarrow x = \dfrac{{12}}{2} = 6 \\
$
Hence, we have obtained the value of x = +6.
Hence, we can say that the oxidation number of Cr in \[{K_2}C{r_2}{O_7}\] is +6.
Therefore, the correct answer is option (C).

Note: Whenever we are asked such types of questions we have to follow some basic steps. First we have to find out the charge on the other elements and the oxidation number of the asked element be x. After that we will multiply the charge of each element with the number of atoms present in the compound. Then we will equate the algebraic sum of the oxidation number of all atoms equal to zero. Then we will find out the oxidation state of the required element.