Question

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 1
B. both statement 1 and statement 2 are correct and statement 2 is not the correct explanation of statement 1
C. statement 1 is correct but statement 2 is not correct
D. statement 1 is not correct but statement 2 is correct
E. both the statement 1 and 2 are not correct.

Answer 1

. 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.

Complete step by step answer:
- 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:

& 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{aligned}$$ \- 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. \- It is found that in a neutral compound, the sum of oxidation numbers of all the atoms must equal zero. This is because if it is not zero then it cannot be neutral. This rule helps to calculate the oxidation number of an atom that may have multiple oxidation states. And hence the statement 2 is correct. \- Hence, we can conclude that the correct option is (D), that is statement 1 is not correct but statement 2 is correct. **So, the correct answer is “Option D”.** **Note:** \- Oxidation state is not constant for all elements, or we can say that it is not compulsory that each element will show only one oxidation state in every compound. It is found that there are many elements that show variable valency due to the electrons of the outermost shell. \- It is found that Transition metals have the ability to show variable valency states.