Question

Show that the Oxidation number of $Cr$ in ${K_3}Cr{O_8}$ is $ + 5$ .

Answer 1

Hint: Oxidation number generally determines the degree of oxidation of that atom. Oxidation number is also called the oxidation state of that atom. This is simply defined as the total number of electrons an atom loses or gains to make a bond with another atom. The substance when oxidized loses electrons and hence the overall reaction is called oxidation reaction.

Complete step by step answer:
According to the recorded data, the oxidation state may be positive, negative or even zero. It is the hypothetical charge which will be observed when all bonds to atoms of different elements are $100\% $ ionic that means no covalent component. Oxidation states are typically represented by integers which may be positive, zero, or negative. In some cases the average oxidation state of an element is a fraction too.
In case of ${K_3}Cr{O_8}$ ,
Let the oxidation state of chromium be $x$
As we know that oxidation state of oxygen in this case is $ - 1$ because it has peroxide linkages.
And also the oxidation state of potassium is $ + 1$
So, according to the theory, the sum of all the oxidation states in a compound is always equal to zero if there is no charge on the compound.
Now,
$\
   \Rightarrow 3 \times 1 + x + 8 \times ( - 1) = 0 \\
   \Rightarrow 3 + x - 8 = 0 \\
   \Rightarrow x - 5 = 0 \\
   \Rightarrow x = 5
\ $
So, the oxidation state of chromium in ${K_3}Cr{O_8}$ is $ + 5$ .

Note: There is a shielding effect that is observed between sublevels within the same principal energy level. An electron in the s-sublevels is capable of shielding electrons in the p-sublevels of the same principal energy level. This is because of the spherical shape of the s-orbital .
The oxidation state of an atom does not represent the real charge on that atom nor any other property of atom. This is particularly right for high oxidation states, because the ionization energy required to produce a multiply positive ion is far greater than the energies available in chemical reactions.