Question

What is the oxidation number of $C{r_2}{O_7}^{ - 2}$?

Answer 1

The concept of oxidation is very useful during the balancing of any chemical redox reaction. The oxidation number is defined as the magnitude of net charge which is possessed by an atom of a particular element in its ion form or appears when it is combined with another element.

Complete answer:
There are some basic rules which are followed for calculating the oxidation number of molecule-
$\to$ Write the chemical formula of the molecules whose oxidation number has to be calculated.
$\to$ Now write all the constituent elements at some distance.
$\to$ Write the oxidation number of known compounds below their chemical symbol.
$\to$ Mention the unknown oxidation number as $x$.
$\to$ Add all the oxidation numbers and equate the equation with zero in case of a neutral atom.
$\to$ If atoms possess the charge then equate the equation with magnitude of the charge.
$\to$ Now calculate the value of $x$.
$C{r_2}{O_7}^{ - 2}$ is known as a dichromate ion which contains two atoms of chromium metal and seven atoms of oxygen. The general oxidation number of oxygen atoms is $\left( { - 2} \right)$. Molecule of $C{r_2}{O_7}^{ - 2}$ contains a negative charge with magnitude of $2$. Oxidation number of chromium atom which is let $\left( x \right)$ is calculated by following the rules mentioned above: -
$2\left( x \right) + 7 \times \left( { - 2} \right) = - 2$
After simplifying the above equation, we get
$2x = - 2 + 14$
So, value of $\left( x \right)$ calculated from above equation is –
$x = \dfrac{{12}}{2}$
$x = 6$
$\Rightarrow$ Oxidation number of $C{r_2}{O_7}^{ - 2}$ is $6$.

Note:
Oxidation number also helps in calculating the net charge which an atom has when all the other atoms are separated from it as ions. Oxidation number in the free state of an element is always considered as zero. Chromium is a metal so its oxidation number will always be a positive value.