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In the conversion of \[{K_2}C{r_2}{O_7}\] to ${K_2}Cr{O_4}$, the oxidation number of chromium:
A.Remains same

Last updated date: 29th May 2024
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Hint: Chromium is a d block element responsible for the formation of many compounds. It has twenty-four electrons in total which is also equal to the number of protons in the nucleus. It has six electrons in its last subshell.

Complete answer:
Here chromium is forming two compounds. As we know that the chromium has six electrons in its last subshell, here the twenty-four six electrons are to be distributed in the last shell such that the subshell is stable. Now the subshells are the most stable when either all the electrons present in it are paired or all the electrons present there are unpaired. These two configurations are more stable than any of the configurations when allocating electrons and giving spin to them.
Oxidation number of an element in a compound is calculated by taking the oxidation states of the other elements and then equating them to the total charge on the compound. Also, there is oxygen and potassium in both the compounds along with chromium. Potassium is a highly electropositive element and oxygen is an electronegative element. Potassium exhibits a $ + 1$ and oxygen exhibits an oxidation state of $ - 2$.
Calculate the OS of chromium in \[{K_2}C{r_2}{O_7}\]
  2 \times K + 2 \times Cr + O \times 7 = 0 \\
   \Rightarrow 2 \times 1 + 2Cr + ( - 2) \times 7 = 0 \\
   \Rightarrow Cr = + 6 \\
Calculating the OS of chromium in ${K_2}Cr{O_4}$
  2 \times K + Cr + O \times 4 = 0 \\
   \Rightarrow 2 \times 1 + Cr + ( - 2) \times 4 = 0 \\
   \Rightarrow Cr = + 6 \\
 As we can see that the oxidation amount of chromium in both the compounds are the same.

Chromium is a d block element which has the ability to lose electrons or gain six electrons in all. Also, chromium reacts differently in an acidic medium in which there is a drop of number five in the oxidation number while it acts differently in basic where the drop in oxidation number isn’t that much.
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