 Statement I: The oxidation state of Cr in $A{{l}_{2}}{{(C{{r}_{2}}{{O}_{7}})}_{3}}$ is +3.Statement II: as a neutral compound, the sum of oxidation numbers of all the atoms must equal zero.A. statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1B. both statement 1 and statement 2 are correct and statement 2 is not the correct explanation of statement 1C. statement 1 is correct but statement 2 is not correctD. statement 1 is not correct but statement 2 is correctE. both the statement 1 and 2 are not correct. Verified
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Hint:. As we know that oxidation number is basically the charge that an element has in its ions or we can say appears to have when present in the combined state with other atoms. It is found that in a polyatomic ion the sum of all the oxidation numbers is equal to the charge on the ion.

- As we know that the oxidation number of O is -2 and oxidation state of Al is +3. Therefore, we can calculate the oxidation state of Cr in $A{{l}_{2}}{{(C{{r}_{2}}{{O}_{7}})}_{3}}$ by considering it as X.
- Thus, oxidation state of Cr in $A{{l}_{2}}{{(C{{r}_{2}}{{O}_{7}})}_{3}}$ is:
\begin{align} & A{{l}_{2}}{{(C{{r}_{2}}{{O}_{7}})}_{3}} \\ & \Rightarrow (2\times 3)+6X+\left[ 21\times \left( -2 \right) \right]=0 \\ & \Rightarrow 6+6X+\left[ -42 \right]=0 \\ & X\Rightarrow +6 \\ \end{align}
- Here, we get the value of X as +6. Therefore, we can say that the oxidation state of Cr in $A{{l}_{2}}{{(C{{r}_{2}}{{O}_{7}})}_{3}}$ is +6. And hence the statement 1 is incorrect.