d and f Block Elements Part 3

Concept:This problem involves the chromyl chloride test and the formation of chromium pentoxide ($\mathrm{CrO}_5$).The key is to identify products from each reaction and then determine the number of peroxide units, total oxygen atoms, and oxidation state of chromium in $\mathrm{CrO}_5$.Explanation:Step 1: Reaction of $\mathrm{NaCl}$ with $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ and $\mathrm{H}_2\mathrm{SO}_4$ produces deep red vapors of chromyl chloride ($\mathrm{CrO}_2\mathrm{Cl}_2$).$4\mathrm{NaCl} + \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 6\mathrm{H}_2\mathrm{SO}_4 \rightarrow 2\mathrm{CrO}_2\mathrm{Cl}_2 + 2\mathrm{KHSO}_4 + 4\mathrm{NaHSO}_4 + 3\mathrm{H}_2\mathrm{O}$Product A is $\mathrm{CrO}_2\mathrm{Cl}_2$.Step 2: A reacts with $\mathrm{NaOH}$ to give sodium chromate ($\mathrm{Na}_2\mathrm{CrO}_4$).$\mathrm{CrO}_2\mathrm{Cl}_2 + 4\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{CrO}_4 + 2\mathrm{NaCl} + 2\mathrm{H}_2\mathrm{O}$Product B is $\mathrm{Na}_2\mathrm{CrO}_4$.Step 3: B reacts with $\mathrm{H}_2\mathrm{SO}_4$ and $\mathrm{H}_2\mathrm{O}_2$ to form deep blue chromium pentoxide ($\mathrm{CrO}_5$).$\mathrm{Na}_2\mathrm{CrO}_4 + \mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{CrO}_5 + \mathrm{Na}_2\mathrm{SO}_4 + 3\mathrm{H}_2\mathrm{O}$Product C is $\mathrm{CrO}_5$.Step 4: In $\mathrm{CrO}_5$, the structure has a chromium atom bonded to one oxygen via a double bond and two peroxide ($\mathrm{O}-\mathrm{O}$) linkages.Each peroxide linkage is an $\mathrm{O}_2^{2-}$ unit. Thus, number of $\mathrm{O}_2^{2-}$ units, $X = 2$.Total oxygen atoms in $\mathrm{CrO}_5$ are 5, so $Y = 5$.Step 5: To find oxidation state of Cr ($Z$): In $\mathrm{CrO}_5$, one oxygen is normal oxide (oxidation state $-2$), and four oxygens are in peroxide form (each $-1$).Let $Z$ be oxidation state of Cr. Then:$Z + 1(-2) + 4(-1) = 0$$Z - 2 - 4 = 0$$Z = +6$So $Z = 6$.Step 6: Calculate $X + Y + Z = 2 + 5 + 6 = 13$.Answer:13.