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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 B.IncreasesC.DecreasesD.None Verified
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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.

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 ${K_2}Cr{O_4}$
$2 \times K + Cr + O \times 4 = 0 \\ \Rightarrow 2 \times 1 + Cr + ( - 2) \times 4 = 0 \\ \Rightarrow Cr = + 6 \\$