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Question: In the reaction the stoichiometric coefficients of \(C{r_2}{O_7}^{2 - }\), \(N{O_2}^ - \) and \({H^ ...

In the reaction the stoichiometric coefficients of Cr2O72−C{r_2}{O_7}^{2 - }, NO2−N{O_2}^ - and H+{H^ + } respectively are:
Cr2O72−+NO2−+H+→Cr3++NO3−+H2OC{r_2}{O_7}^{2 - } + N{O_2}^ - + {H^ + } \to C{r^{3 + }} + N{O_3}^ - + {H_2}O
A.1,31,3 and 88
B.1,41,4 and 88
C.1,31,3 and 1212
D.1,51,5 and 1212

Explanation

Solution

Stoichiometric coefficients: They are defined as the coefficients of the compound in the reaction after balancing the reaction. For balancing the ionic reactions we have to first write the separate reactions for each reactant. And we have to balance them.

Complete step by step solution:
For balancing the ionic reactions we have to first write the separate reactions for each reactant. And we have to balance them. Here we are given with the reaction Cr2O72−+NO2−+H+→Cr3++NO3−+H2OC{r_2}{O_7}^{2 - } + N{O_2}^ - + {H^ + } \to C{r^{3 + }} + N{O_3}^ - + {H_2}O
Oxidation number of an element: It is defined as the total number of electrons that an atom accepts or loses in order to make a chemical bond which results in the formation of a chemical compound.
First compound here in the reaction is Cr2O72−C{r_2}{O_7}^{2 - } (oxidation number of chromium is 66) and the corresponding product is Cr3+C{r^{3 + }} (oxidation number of chromium is 33) . So the reaction will be: Cr6++3e−→Cr3+C{r^{6 + }} + 3{e^ - } \to C{r^{3 + }} and the second compound is NO2−N{O_2}^ - (oxidation number of nitrogen is as 33) and the corresponding product is NO3−N{O_3}^ - (oxidation number of nitrogen is 55) so the reaction is as: N3+→N5++2e−{N^{3 + }} \to {N^{5 + }} + 2{e^ - }. Now multiply the first equation with 22 and second equation by 33 and then add these we will get the reaction as: Cr2O72−+3NO2−→2Cr3++3NO3−C{r_2}{O_7}^{2 - } + 3N{O_2}^ - \to 2C{r^{3 + }} + 3N{O_3}^ - But in the question we are given with hydrogen ion and water as well. So now balance them according to the charge i.e. Cr2O72−+3NO2−+8H+→2Cr3++3NO3−+4H2OC{r_2}{O_7}^{2 - } + 3N{O_2}^ - + 8{H^ + } \to 2C{r^{3 + }} + 3N{O_3}^ - + 4{H_2}O because in the reaction charge on both the sides is balanced and number of atoms is also balanced. Hence it is a balanced chemical reaction and the stoichiometry coefficients of Cr2O72−C{r_2}{O_7}^{2 - }, NO2−N{O_2}^ - and H+{H^ + } respectively are 1,31,3 and 88 respectively.

Hence,option A is correct.

Note:
If a compound has net charge either positive or negative then the sum of all charges in the compound will be equal to that net charge on the compound. By using this concept we can calculate the oxidation number of elements in charged species.